├── .gitignore ├── MATH-350 ├── Lectures │ ├── 1.tex │ ├── 10.tex │ ├── 11.tex │ ├── 12.tex │ ├── 13.tex │ ├── 14.tex │ ├── 15.tex │ ├── 16.tex │ ├── 17.tex │ ├── 18.tex │ ├── 19.tex │ ├── 2.tex │ ├── 20.tex │ ├── 21.tex │ ├── 22.tex │ ├── 23.tex │ ├── 24.tex │ ├── 25.tex │ ├── 26.tex │ ├── 27.tex │ ├── 28.tex │ ├── 29.tex │ ├── 3.tex │ ├── 30.tex │ ├── 31.tex │ ├── 32.tex │ ├── 33.tex │ ├── 34.tex │ ├── 35.tex │ ├── 36.tex │ ├── 4.tex │ ├── 5.tex │ ├── 6.tex │ ├── 7.tex │ ├── 8.tex │ └── 9.tex ├── notes.pdf ├── notes.tex └── references.bib ├── MATH-354 ├── Lectures │ ├── 1.tex │ ├── 10.tex │ ├── 11.tex │ ├── 12.tex │ ├── 15.tex │ ├── 16.tex │ ├── 2.tex │ ├── 3.tex │ ├── 4.tex │ ├── 5.tex │ ├── 6.tex │ ├── 7.tex │ ├── 8.tex │ └── 9.tex ├── notes.pdf └── notes.tex ├── MATH-370 ├── 15.tex ├── Lectures │ ├── 1.tex │ ├── 10.tex │ ├── 11.tex │ ├── 12.tex │ ├── 13.tex │ ├── 14.tex │ ├── 15.tex │ ├── 16.tex │ ├── 17.tex │ ├── 2.tex │ ├── 3.tex │ ├── 4.tex │ ├── 5.tex │ ├── 6.tex │ ├── 7.tex │ ├── 8.tex │ └── 9.tex ├── notes.pdf ├── notes.tex └── references.bib ├── PHIL-126 ├── Lectures │ ├── 1.tex │ ├── 2.tex │ ├── 3.tex │ ├── 4.tex │ ├── 5.tex │ ├── 6.tex │ └── 7.tex ├── notes.pdf └── notes.tex ├── PHIL-175 ├── Lectures │ ├── 1.tex │ ├── 2.tex │ ├── 3.tex │ ├── 4.tex │ ├── 5.tex │ ├── 6.tex │ ├── 7.tex │ └── 8.tex ├── notes.pdf └── notes.tex ├── PHIL-427 ├── Lectures │ ├── 1.tex │ ├── 2.tex │ └── 3.tex ├── notes.pdf └── notes.tex └── README.md /.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # cprotect 68 | *.cpt 69 | 70 | # elsarticle (documentclass of Elsevier journals) 71 | *.spl 72 | 73 | # endnotes 74 | *.ent 75 | 76 | # fixme 77 | *.lox 78 | 79 | # feynmf/feynmp 80 | *.mf 81 | *.mp 82 | *.t[1-9] 83 | *.t[1-9][0-9] 84 | *.tfm 85 | 86 | #(r)(e)ledmac/(r)(e)ledpar 87 | *.end 88 | *.?end 89 | *.[1-9] 90 | *.[1-9][0-9] 91 | *.[1-9][0-9][0-9] 92 | *.[1-9]R 93 | *.[1-9][0-9]R 94 | *.[1-9][0-9][0-9]R 95 | *.eledsec[1-9] 96 | *.eledsec[1-9]R 97 | *.eledsec[1-9][0-9] 98 | *.eledsec[1-9][0-9]R 99 | *.eledsec[1-9][0-9][0-9] 100 | *.eledsec[1-9][0-9][0-9]R 101 | 102 | # glossaries 103 | *.acn 104 | *.acr 105 | *.glg 106 | *.glo 107 | *.gls 108 | *.glsdefs 109 | 110 | # gnuplottex 111 | *-gnuplottex-* 112 | 113 | # gregoriotex 114 | *.gaux 115 | *.gtex 116 | 117 | # htlatex 118 | *.4ct 119 | *.4tc 120 | *.idv 121 | *.lg 122 | *.trc 123 | *.xref 124 | 125 | # hyperref 126 | *.brf 127 | 128 | # knitr 129 | *-concordance.tex 130 | # TODO Comment the next line if you want to keep your tikz graphics files 131 | *.tikz 132 | *-tikzDictionary 133 | 134 | # listings 135 | *.lol 136 | 137 | # makeidx 138 | *.idx 139 | *.ilg 140 | *.ind 141 | *.ist 142 | 143 | # minitoc 144 | *.maf 145 | *.mlf 146 | *.mlt 147 | *.mtc[0-9]* 148 | *.slf[0-9]* 149 | *.slt[0-9]* 150 | *.stc[0-9]* 151 | 152 | # minted 153 | _minted* 154 | *.pyg 155 | 156 | # morewrites 157 | *.mw 158 | 159 | # nomencl 160 | *.nlg 161 | *.nlo 162 | *.nls 163 | 164 | # pax 165 | *.pax 166 | 167 | # pdfpcnotes 168 | *.pdfpc 169 | 170 | # sagetex 171 | *.sagetex.sage 172 | *.sagetex.py 173 | *.sagetex.scmd 174 | 175 | # scrwfile 176 | *.wrt 177 | 178 | # sympy 179 | *.sout 180 | *.sympy 181 | sympy-plots-for-*.tex/ 182 | 183 | # pdfcomment 184 | *.upa 185 | *.upb 186 | 187 | # pythontex 188 | *.pytxcode 189 | pythontex-files-*/ 190 | 191 | # thmtools 192 | *.loe 193 | 194 | # TikZ & PGF 195 | *.dpth 196 | *.md5 197 | *.auxlock 198 | 199 | # todonotes 200 | *.tdo 201 | 202 | # easy-todo 203 | *.lod 204 | 205 | # xmpincl 206 | *.xmpi 207 | 208 | # xindy 209 | *.xdy 210 | 211 | # xypic precompiled matrices 212 | *.xyc 213 | 214 | # endfloat 215 | *.ttt 216 | *.fff 217 | 218 | # Latexian 219 | TSWLatexianTemp* 220 | 221 | ## Editors: 222 | # WinEdt 223 | *.bak 224 | *.sav 225 | 226 | # Texpad 227 | .texpadtmp 228 | 229 | # LyX 230 | *.lyx~ 231 | 232 | # Kile 233 | *.backup 234 | 235 | # KBibTeX 236 | *~[0-9]* 237 | 238 | # auto folder when using emacs and auctex 239 | ./auto/* 240 | *.el 241 | 242 | # expex forward references with \gathertags 243 | *-tags.tex 244 | 245 | # standalone packages 246 | *.sta -------------------------------------------------------------------------------- /MATH-350/Lectures/1.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 29 August 2018} 2 | 3 | Most of today's lecture was administrata covering how the course will be run. 4 | 5 | Towards the end of the period we began to play with the very basic concepts of group theory. First and foremost is the definition of a group. 6 | 7 | \begin{definition}[Group] 8 | A group is an ordered pair $(G, \star)$, where $G$ is a set and $\star$ is a binary operation on $G$, which obeys the following axioms. 9 | \begin{itemize} 10 | \item There exists an element $e \in G$ known as the \emph{identity} with the property that $e \star g = g \star e = g$ for all $g \in G$. 11 | \item For all $g \in G$ there exists a $g^{-1} \in G$ known as the \emph{inverse} of $g$ which has the property that $g \star g^{-1}= g^{-1} \star g = e$. 12 | \item The operation $\star$ is associative. 13 | \end{itemize} 14 | \end{definition} 15 | 16 | You will sometimes see a fourth axiom included in this list, namely that $G$ is closed under $\star$, but since $\star$ is a binary operation on $G$ which definitionally means it is a map $\star : G \times G \to G$, and so $G$ is always implicitly closed under $\star$. When \emph{checking} whether or not $(G, \star)$ is a group, though, it is a very good idea to check that $\star$ actually is a binary operation. 17 | 18 | You're probably already familiar with lots of groups already. Consdier the rational numbers $\Q$, the real numbers $\R$, the set of symmetries of the square, and the integers (under addition, which is important). -------------------------------------------------------------------------------- /MATH-350/Lectures/10.tex: -------------------------------------------------------------------------------- 1 | \section{September 21, 2018} 2 | 3 | \epigraph{``Oh I erased my smiley face again. How sad.'' (She did not sound sad.)}{Miki} 4 | 5 | \begin{problem}[Warm up] 6 | Draw the lattice diagram for $C_{12}$. 7 | \end{problem} 8 | 9 | \begin{figure}[h] 10 | \begin{center} 11 | \end{center} 12 | \caption{Lattice for $C_{12}$}\note{Finish this} 13 | \end{figure} -------------------------------------------------------------------------------- /MATH-350/Lectures/11.tex: -------------------------------------------------------------------------------- 1 | \section{September 24, 2018} 2 | 3 | \epigraph{``This is where the fun begins.'' (slightly paraphrased)}{Miki} 4 | 5 | \subsection{Quotient Groups} 6 | 7 | For the rest of this section, keep in mind the example of $\Z/n\Z$. This is kind of like the prototypical example for quotient groups. 8 | 9 | \begin{definition}[Coset] 10 | Let $H \leq G$. The left coset of $H$ in $G$ is a set of the form $aH = \{ah \mid h \in H\} \subset G$ for a fixed $a \in G$. The right coset of $H$ in $G$ is a set of the form $Hb = \{hb \mid h \in H\} \subset G$ for a fixed $b \in G$. 11 | \end{definition} 12 | 13 | We said previously that left multiplication permutes the elements of $H$ (this was called the left regular action), and in particular we know that $\abs{aH} = \abs{H}$. We can see this trivially by simply multiplying each $ah$ by $a^{-1}$. Note that this coset is usually \emph{not} a subgroup; if $a^{-1} \notin H$ then $e \notin aH$. 14 | 15 | \begin{example} 16 | Let $G = \Z$, and let $H = 2\Z$. Consider the cosets $0 + H$ and $1 + H$ (these are just the even integers and the odd integers). In particular, $0 \notin 1 + H$ and so $1 + H \not\leq G$.\note{The Right and Left cosets here are equal, which is always true of $G$ is abelian.} 17 | \end{example} 18 | 19 | Notice that in the above example, the cosets are disjoint and partition the group into equivalence classes. In general this is a true statement. 20 | 21 | \begin{lemma} 22 | The costs of $H$ partition $G$ into equivalence classes, with the relation $a \sim b$ if and only if $a = bh$ for some $h \in H$. In particular, $a \sim b$ if and only if $aH = bH$, and so the cosets defined by those elements are identical. 23 | \end{lemma} 24 | 25 | \begin{corollary} 26 | The order of the cosets divides the order of $G$. In particular, $\abs{G} = \abs{H} \cdot [G : H]$ where $[G : H]$ is the \emph{index} of $H$ in $G$ and is the number of (left OR right) cosets of $H$ in $G$. 27 | \end{corollary} 28 | 29 | In the example with $\Z$ and $2\Z$, lets try to make these cosets behave like groups. Consider that $(0 + H) + (1 + H) = (1 + H)$ (which just says that and even plus an odd equals an odd). We also have a homomorphism $\pi : \Z \to 2\Z : n \mapsto \bar{n} = n+H$. This maps integers to cosets. Note that $\pi$ respects the operations in each group! This is kind of what defines ``adding even and odd integers'' in the languages of sets. 30 | 31 | \begin{notation} 32 | Let $0+H = \pi^{-1}(\bar{0}) = \{n \in \Z \mid \pi(n) = \bar{0}\}$; this is the preimage of $\pi$ or the fiber of $\pi$ above $0$. Yes this is overloaded notaiton, and no $\pi$ does not have an inverse (it's pretty clearly \emph{not} injective.) 33 | \end{notation} 34 | 35 | Note that it doesn't really matter which elements we send into $\pi$ as long as they are both of the same coset, so $\pi(a) = \pi(b)$ if and only if $a \sim b$. Additionally, note that since $\pi$ is a homomorphism we can say that $\pi(\bar{1} + \bar{20}) = \pi(\bar{1}) + \pi(\bar{20})$. 36 | 37 | Now, in making these cosets into groups we want them to inherit their operation from the parent group (so we can't just make up multiplications to suit our needs). 38 | 39 | \begin{definition} 40 | Let $A,B \subset G$. Then $AB = \{ab \mid a \in A, b \in B\} \subset G$. In particular, note that $HH = H$ and $(1H \cdot 1H = 1H)$. 41 | \end{definition} 42 | 43 | \begin{example}[Things Go Wrong] 44 | Let $G = S_3$ and let $H = \langle (23) \rangle = \{1, (23)\}$. The cosets of $H$ are $1H = (23)H$, $(12)H = (123)H = \{(12), (123)\}$, and $(13)H = (132)H = \{(13), (132)\}$. Now consider $1H \cdot (12)H = \{(12), (123), (132), (13)\}$. In particular, note that this isn't a coset (it has too many elements!). We would have wanted that $1H \cdot (12)H = (12)H$ but this doesn't happen. Then there is not quotient group $G / H$. 45 | \end{example} 46 | 47 | What just happened? Why can't we create a group out of the cosets of $S_3$? We wanted that $aH \cdot bH = abH$ but this didn't happen; essentially, we want $b$ and $H$ to commute, so we want that the left and right cosets to be equal to one another. 48 | 49 | \begin{example} 50 | Let $G = S_3$ and let $H = \langle (123) \rangle = A_3$. This is the alternating group on three letters. As always, $1H = H1 = H$. Note that $(12)H$ contains \emph{the only other elements of $G$} which aren't in $1H$, and so $(12)H = H(12) = G \setminus 1H = G \setminus H1$. This happens when $[G : H] = 2$ even though $G$ is not abelian. Then $S_3 \setminus A_3 = G \setminus H$ and $\bar{a}\bar{b} = \bar{ab}$ so multiplication is well defined. 51 | \end{example} 52 | 53 | -------------------------------------------------------------------------------- /MATH-350/Lectures/12.tex: -------------------------------------------------------------------------------- 1 | \section{September 26, 2018} 2 | 3 | \epigraph{Let $N$ be a group...I'll call it $N$ suggestively}{Miki} 4 | 5 | Recall from last class that we found and example of a non-abelian group and a subgroup for which the left and right cosets of the group were the same; in this case, it was $G = S_3$ and $H = A_3$. 6 | 7 | \begin{definition}[Quotient Group] 8 | Let $H \leq G$. The \emph{quotient group} $G/H$ is a group whose elements are the left cosets of $H$. The set for this group is known as the quotient set, and the operation for the group is inherited from $G$ such that $gH \star kH = gk H$. Note that $(gH, \star)$ does not always form a group, so it isn't guaranteed that $G/H$ exists for any $G,H$. 9 | \end{definition} 10 | 11 | \subsection{Mapping from \texorpdfstring{$G$}{G} to \texorpdfstring{$G/H$}{G/H}} 12 | 13 | Given a group $G$ and a quotient group $G/H$ we can find a very natural mapping $\pi : G \to G/H$ where $g \mapsto gH$. This map sends elements to their coset, and $\pi(a) = \pi(b)$ if and only if $aH = bH$; thenthe fibers of $\pi$ are the left cosets of $H$, and $\ker \pi = H$. This is why we call it the quotient group --- it's like we're dividing out by $H$. Note that this homomorphism is always going to be surjective since there's no member of $G$ which isn't in some coset of $H$ as they partition $G$. 14 | 15 | \begin{definition}[Normal Subgroup] 16 | Let $N \leq G$. Then $N$ is normal if and only if the left and right cosets are the same, so $gN = Ng$. If $N$ is normal then $G/N$ forms a quotient group. Note that this does not mean that $gn = ng$ so $g$ and $n$ do not commute necessarily, but the cosets are preserved. This is equivalent to saying that $\mathrm{N}_N(G) = G$ but $\mathrm{C}_N(G)$ is not necessarily $G$. 17 | \end{definition} 18 | 19 | \begin{notation}[$\normal$] 20 | We write $N \normal G$ to mean that $N$ is a normal subgroup of $G$. 21 | \end{notation} 22 | 23 | \begin{theorem} 24 | The quotient group $G / N$ exists if and only if $N \normal G$. 25 | \end{theorem} 26 | 27 | \begin{proof}[Proof that $N \normal G$ is sufficient] 28 | Observe that $(aN)(bN) = abN$ if $N$ is normal. Then group multiplication is well defined. Observe also that $(aN)^{-1} = a^{-1}N$, so the group is closed under inversion, and by definition our multiplication is associative. Then $G/N$ forms a group if $N$ is normal in $G$. 29 | \end{proof} 30 | 31 | \begin{proof}[Proof that $N \normal G$ is necessary] 32 | Suppose $H \leq G$ is not normal. Then there is some $g \in G$ for which $gH \not= Hg$. Then we know that $1HgH \not= gH$, and our group operation $\star$ cannot hold. 33 | \end{proof} 34 | 35 | Not that $\abs{G/N} = \abs{G} / \abs{N} = [G : N]$ if $G$ is finite, which we already knew but it's worth remembering it. 36 | 37 | \subsection{Testing Normality} 38 | 39 | \begin{proposition} 40 | The following are equivalent: 41 | \begin{itemize} 42 | \item $N \normal G$; 43 | \item $gNg^{-1} \subset N$ for all $g \in G$ (note this implies they are equal since conjugation is injective); 44 | \item $N$ is the kernel of some homomorphism $\pi : G \to H$ for some $H \leq G$. 45 | \end{itemize} 46 | \end{proposition} 47 | 48 | \begin{proof}[Proof that 1 $\implies$ 2] 49 | Let $g \in G$ and $n \in N$. We know that $gN = Ng$, so there exists $n' \in N$ such that $ng = n'g$. Multiply on the right by $g^{-1}$ and we see that $gng^{-1} = n'$, and so $gng^{-1} \in N$ for all $g,n$. 50 | \end{proof} 51 | 52 | \begin{proof}[Proof that 2 $\implies$ 1] 53 | Literally just reverse the above procedure. 54 | \end{proof} 55 | 56 | \begin{proof}[Proof that 1 $\implies$ 3] 57 | Let $H = G / N$. Then we know that $\ker \pi = N$ where $\pi : G \to G/N : g \mapsto gN$. Then, rather trivially, we know $N$ is the kernel for some homomorhpism if $N \normal G$. 58 | \end{proof} 59 | 60 | \begin{proof}[Proof that 3 $\implies$ 2] 61 | We know that $N = \ker \pi$ for some $\pi : G \to H$. Then take any $g \in G$ and $n \in N$, and consider that $\pi(gng^{-1}) = \pi(g)\pi(n)\pi(g^{-1}) = \pi(g)\pi(g^{-1})$ since $n \in \ker \pi$, and then we conclude that $\pi(g)\pi(g^{-1}) = 1$ and so we know that $gng^{-1} \in \ker\pi$ so $gng^{-1} \in N$ for all $n \in N$ and for all $g \in G$. 62 | \end{proof} -------------------------------------------------------------------------------- /MATH-350/Lectures/14.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 1 October 2018} 2 | 3 | \epigraph{``I've got H's on the brain.''}{} 4 | 5 | \epigraph{``That's the third isomorphism theorem, I knew you wouldn't like it. It should take you anywhere from a day to seven years to become comfortable with it.''}{Miki} 6 | 7 | \epigraph{``It's math....it keeps doing things like that.''}{Miki} 8 | 9 | \subsection{Isomorphism Theorems Continued} 10 | 11 | Recall from last lecture we developed the first two isomorphism theorems. Today, we'll cover the last two (or one, depending on your perspective). 12 | 13 | \begin{theorem}[Third Isomorphism Theorem] 14 | Let $G$ be a group and let $H,N$ be normal subgroups of $G$ with $N \subseteq H$. THen $G/N \big/ H/N \cong G/H$. 15 | \end{theorem} 16 | 17 | \begin{proof} 18 | Consider a map $\phi : G/N \to G/H : gN \mapsto gH$. We need this map to be well-defined. Suppose that $g_1N = g_2N$. Then $g_1^{-1}g_2 \in N$, but $N \subseteq H$, and so $g_1H = g_2H$ and $\phi$ is well defined. We also need to know that this is a homomorphism. Consider $\phi(g_1N)\phi(g_2H) = g_1g_2H = \phi(g_1g_2N)$, and in fact we also know that $\phi$ is surjective. Consider $gH \in G/H$ and suppose that $gH = \phi(gN)$. Since $N \subset H$ this is well defined. Consider then that $\ker \phi : \phi(gN) = gH$. This happens if and only if $g \in H$ so $gN \subset H$ is a coset of $N$ in $H$ and $gN \in H/N$, so $\ker\phi = H/N$. Then by the First Isomorphism Theorem, we know that $G/N \big/ \ker\phi \cong G/H$. 19 | \end{proof} 20 | 21 | \begin{example} 22 | Let $G = \Z$ with $N = \langle 10 \rangle$ and $H = \langle 2 \rangle$. Then $G/N = \{0 + N,\dotsc,9+N\}$ and $G/H = \{0 + H, 1 + H\}$. Then $H / N = \{0 + N, 2 + N,\dotsc,8 + N\}$. The idea here is that if you take $\Z \pmod{10}$, and then modulo the result by $2$, then it didn't really matter than we modded out by $10$ to begin with. 23 | \end{example} 24 | 25 | \begin{theorem}{The Totally not fourth isomorphism theorem} 26 | Let $N \normal G$. THere is a correspondence (bijection) between subgroups of $G$ which contain $N$ and subgroups of $G/N$. That is, 27 | \[ \pi : H \mapsto \pi(H), \quad \bar{H} \mapsto \pi^{-1}(\bar{H}). \] 28 | Note that for any $H \leq G$ we know that $\pi(H) \leq G/N$. We require normality to ensure that $\pi$ is injective. 29 | \end{theorem} 30 | 31 | \begin{example} 32 | Consider $G = S_3$ with $N = A_3$. Then $\pi(S_3) = G/N$ and $\pi(A_3) = N$. What is $\pi(\langle (12) \rangle )$? It's all of $G/N$. 33 | \end{example} 34 | 35 | \subsection{Why do people care about normal groups?} 36 | 37 | \begin{definition}[Simple] 38 | A group $G$ is simple if $\abs{G} > 1$ and $G$ contains no proper normal subgroups. 39 | \end{definition} 40 | 41 | \begin{definition}[Composition Series] 42 | Consdier something like $1 = N_0 \normal N_1 \normal \cdots \normal N_r = G$ where $N_{i+1}/N_i$ is simple for all $0 \leq i \leq r-1$. As an example, $1 \normal A_3 \normal S_3$. Then $S_3/A_3 \cong Z_2$ and $A_3 / 1 \cong Z_3$. These series allow us to construct large groups whose multiplication is unknown, since normal subgroups multiply to form subgroups of something larger. For more information on this, see the \emph{Holder Program}, which was started in 1890 to classify simple groups and it took 103 years to actually classify them all. These series are \emph{almost} unique, where the quotient groups are unique up to a permutaiton, so the set of quotient groups are unique. 43 | \end{definition} 44 | 45 | \begin{definition}[Solvable groups] 46 | A group $G$ is solvalble is $1 = N_0 \normal \cdots N_r = G$ and $N_{i+1}/N$ is abelian. This kind of object shows up a lot in Galois Theory. As it turns out, $A_1$ through $A_4$ are solvalble but $A_5$ and higher is not solvable, which is why we can't solve arbitrary quintics. 47 | \end{definition} -------------------------------------------------------------------------------- /MATH-350/Lectures/15.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 3 October 2018} 2 | 3 | \epigraph{``I will not try to decide whether that was happy or sad.''}{Miki} 4 | 5 | \epigraph{``Try it if you don't believe me.''}{Miki} 6 | 7 | \epigraph{``If you don't have surjectivity, you have nothing.''}{Miki} 8 | 9 | Recall from last time that we defined a simple group to be a non-trivial group which has no proper normal subgroups. Observe that if $G$ is abelian and simple then it has no proper subgroups at all, since all subgroups would be normal. 10 | 11 | \subsection{Permutations} 12 | 13 | We'll take a shortcut throught linear algebra to talk about the signs of permutaitons; the book constructs the notion from scratch. Recall that we can switch the rows of a matrix using the permutation matrix $P_{mn}$, by which left multiplication swaps the rows $m$ and $n$. Now, we talk about this as the cycle $(mn)$, so for example 14 | \[ \begin{pmatrix} 15 | 0 & 1 & 0 \\ 16 | 1 & 0 & 0 \\ 17 | 0 & 0 & 1 18 | \end{pmatrix} \sim \sigma = (12) \in S_3. \] 19 | Essentially, we start with $I_n$ and permute the rows according to $\sigma$ to yield the corresponding permutation matrix $P_\sigma$. 20 | 21 | \begin{definition}[Sign of Permutation] 22 | Let $\varepsilon : S_n \to \{\pm 1\} \cong Z_2$ by $\varepsilon(\sigma) = \det P_\sigma$. Then $\varepsilon$ is the \emph{sign} of $\sigma$. 23 | \end{definition} 24 | 25 | Note that $\varepsilon$ is actually a group homomorphism since the determinant is multiplicative; that is $\varepsilon(\tau\sigma) = \det(P_{\tau\sigma}) = \det(P_\tau)\det(P_\sigma) = \varepsilon(\tau)\varepsilon(\sigma)$. Then we can quite naturally ask, what is the kernel of $\varepsilon$. We define the terms \emph{even} and \emph{odd} to mean permutations whose sign is $+1$ and $-1$ respectively. Then $\ker\varepsilon = A_n \leq S_n$ is the set of all even permutations. This gives us a rigorous definition of the alternating group. 26 | 27 | Let's note that a two-cycle in $S_n$ is a transposition, and we have already proven on homework that every element in $S_n$ can be written as the product of two-cycles. We can quite easily conclude that every transposition has a sign of $-1$. 28 | 29 | \begin{proposition} 30 | Let $\sigma \in S_n$ be a $k$-cycle. Then $\varepsilon(\sigma) = (-1)^{k-1}$. 31 | \end{proposition} 32 | 33 | \begin{problem} 34 | How large is $A_n$? 35 | \end{problem} 36 | 37 | Since $\varepsilon$ is surjective, we know by the First Isomorphism Theorem that $S_n / A_n \cong Z_2$ (since $\Im\varepsilon = Z_2$), so $\abs{A_n} = n!/2$. 38 | 39 | \begin{theorem} 40 | The alternating group on $n$ letters is simple if $n \geq 5$. This was proven by Galois in the 1830's and is the reason for quintic insolubility. 41 | \end{theorem} 42 | 43 | \subsection{Actions} 44 | 45 | Recall that an action is a map $\phi : G \times A \to A$ by $\phi(g,a) = g \cdot a$. This yields a homomorphism $G \to S_A$ by $g \mapsto \sigma_g$, where $\sigma_g$ is bijective for a fixed $g \in G$. Recall also for $a \in A$ the stabalizer $G_a$ is the set of $g$ for which $ga = a$, and the kernel of the action is the set of $g \in G$ for which $ga = a$ for all $a \in A$. We said that an action is \emph{faithful} if the kernel of the action is the identity; that is, different elements of $g$ give different permutaitons on $A$. Furthermore, the orbit of $a$ is the set of $ga$ for all $g \in G$. We proved on homework that the orbits partition $A$. 46 | 47 | \begin{definition}[Transitive] 48 | An action is transitive if all elements of $A$ are in a single orbit; i.e., $a \sim b$ for all $a,b \in A$. 49 | \end{definition} -------------------------------------------------------------------------------- /MATH-350/Lectures/16.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 5 October 2018} 2 | 3 | \epigraph{``I mean, $\infty!$ is a big number!''}{Miki} 4 | \epigraph{``Oh well, we'll cry later.''}{Miki} 5 | 6 | \subsection{Orbits and Stabalizers} 7 | 8 | Miki introduced a new proposition today which I think is just the Orbit-Stabalizer theorem. 9 | 10 | \begin{proposition} 11 | Given an action $G \times A \to A$ and an $a \in A$ we know that $\abs{O_a} = [G : G_a]$ which tells us that $\abs{G} = \abs{O_a}\abs{G_a}$. 12 | \end{proposition} 13 | 14 | \begin{proof} 15 | Define a map $\pi : \{\text{cosets of $G_a$ in $G$}\} \to O_a$. Note that this is just a map, not a homomorphism. Define $\pi$ to be $gG_a \mapsto g \cdot a$. We'll show it's well-defined and injective at the same time. Suppose we have that $gG_a = hG_a$ which happens if and only if $g^{-1}h \in G_a$ or $g^{-1}h \cdot a = a$, so $ha = ga$. Since anything in the orbit is $g \cdot a$ for some $g$, we also know that $\pi$ is surjective; then $\pi$ is a bijection and so $\abs{O_a} = [G : G_a]$. 16 | \end{proof} 17 | 18 | \subsection{Cycles in \texorpdfstring{$S_n$}{Sn}} 19 | 20 | Let $\sigma \in S_n$ be or order $k$. We want to write it as the product of disjoint cycles. 21 | Consider the set $A = \{1,\dotsc,n\}$ and let $G = \langle \sigma \rangle$. We construct the action $\langle \sigma \rangle \times A \to A$. Consider the orbit $O_a$ of $a \in A$ under $\langle \sigma \rangle$. We know by the orbit-stabalizer theorem that there is a bijection between the cosets $G_a$ and the orbit of $a$. Since $\langle \sigma \rangle$ is cyclic we know that $G_a = \langle \sigma^r \rangle$ is also cyclic. By the definition of our map $\pi$ from the Orbit-Stabalizer theorem, we know that $\pi(\sigma^iG_a) = \sigma^i a$. Then $O_a = \{a, \sigma a, \dotsc, \sigma^{r-1}a\}$ then on $O_a$ we can say that $\sigma$ acts as an $r$-cycle. Since the orbits collectively partition $A$ we know that they are disjoint, and so we know that we can write $\sigma$ as the product of disjoint cycles, which is unique up to the order of the cycles and up to cyclic permutation within each cycle. 22 | Note that since $\langle \sigma \rangle$ is cyclic (and therefore abelian) the cosets $G_a$ are simply $G/G_a$. 23 | 24 | \subsection{Actions of \texorpdfstring{$G$}{G} on itself} 25 | 26 | Previously we defined two cannonical actions of $G$ on itself, via \emph{left multiplication} where $G \times G \to G : (g,a) \mapsto g\cdot a$, and \emph{conjugation}, where $G \times G \to G : (g,a) \mapsto gag^{-1}$. In the first case, we know that the action of left multiplication is faithful, and gave us an injective homomorphism from $G$ to $S_G$ (i.e., finite $G$ always is isomorphism to a subgroup of $S_n$). 27 | 28 | \begin{example} 29 | Let $G = \Z$ and our action is $(i,j) \mapsto i+j$, so $\sigma_i(j) = i+j$. Let's consider the orbits of $0$ and $1$ under $\sigma_2$. Observe that $\cdots -4 \to -2 \to 0 \to 2 \to 4 \to \cdots = O_0$ while $O_1$ is just the odd integers. Now consider $H = 4\Z \subset G$, and let's consider how $\sigma_2$ acts of $G/H$, or on the cosets of $H$ in $G$. Well, we know that $H = \{\bar{0}, \bar{1},\bar{2},\bar{3}\}$. Note that $\sigma_2$ becomes $(\bar{0}\bar{2})(\bar{1}\bar{3})$. 30 | \end{example} 31 | 32 | \begin{theorem} 33 | Consider $G \times G \to G$ via left multiplication, and consider how this action acts on $H \leq G$. We know it acts like $g \cdot (aH) = gaH$. We can show this is well defined by noting that if $aH = bH$ then we know that $ga = gbh$ and so our action is well defined. Let $A$ be set set of cosets of $H$ in $G$, and we get a map $\pi : G \to S_A$. We know the following things. 34 | \begin{enumerate} 35 | \item $G$ acts transitively on $A$; 36 | \item $G_{1H} = H$; 37 | \item The kernel of $\pi$ is the intersection of all $gHg^{-1}$ for all $g \in G$. This is actually the largest normal subgroup of $G$ contained in $H$. 38 | \end{enumerate} 39 | \end{theorem} 40 | 41 | \begin{proof} 42 | Left as an exercize to the reader (me). 43 | \end{proof} -------------------------------------------------------------------------------- /MATH-350/Lectures/17.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 8 October 2018} 2 | 3 | \epigraph{``Remember that $1+1 = 2$.''}{Miki} 4 | \epigraph{``This proof should make you feel better after your exam.''}{Miki} 5 | 6 | Recall from last lecture we described how $G$ can act on itself by either left multiplication or conjufgation. Today we'll cover in detail conjugation. 7 | Remember that conjugation is an action defined as 8 | \[ G \times G \to G : ga \mapsto gag^{-1}. \] 9 | We define the orbit $O_a$ of $a$ under conjugation to be the \emph{conjugacy class} of $a$. This is an equivalence relation (since we already know this holds for orbits). Then consider $S_1, S_2 \subset G$. There are conjugate if there exists a $g \in G$ such that $gS_1g^{-1} = S_2$. Note that these subsets better have the same cardinality. 10 | 11 | For any $x \in G$, we know that $\abs{O_x} = [G : G_x]$ where $G_x = G_G(x)$ is the centralizer of $x$ in $G$, as it turns out. 12 | 13 | \begin{example}[Conjugacy Classes] 14 | \begin{enumerate} 15 | \part Consider $G = C_6$. Since $G$ is abelian, we know that $gxg^{-1} = x$ for all $x,g \in G$ so the orbit of $x$ is simply $\{x\}$. 16 | \part Consider $D_8 = \langle r,s \mid \cdots \rangle$. What is the center of $G$? It's $Z(G) = \{1,r^2\}$. Let's consider that orbit of $1$ is $1$ and the orbit of $r^2$ is just $r^2$. This tells us that if $x \in Z(G)$ then $O_x = \{x\}$ under conjugation. Then let's consider some $x \notin Z(G)$. The size of the orbit must be strictly greater than $1$ (since otherwise it would commute with everything). Consider that the sabalizer must have at least three elements ($1,r^2,x$), and so must have at least order four. It can't have order eight since the identity will not be in the centralizer, and so we know that $\abs{O_x} = 2$. This tells us that the orbit of \emph{any non-center} element has order two, which tells us that we can find the orbits of any elements \emph{super quickly} since we just need to conjugate it \emph{once} and get a new element and we are done! 17 | \end{enumerate} 18 | \end{example} 19 | 20 | \begin{theorem}[Class Equation] 21 | Let $G$ be a group. The center of the group contains all conjugacy classes of size $1$. List the classes of size greater than or equal to $2$ as $)_{x_1} ,\dotsc,O_{x_n}$. Then 22 | \[ \abs{G} = \abs{Z(G)} + \sum_{i=1}^n \abs{O_{x_i}} \] 23 | since the orbits partition $G$. 24 | \end{theorem} 25 | 26 | \begin{theorem} 27 | Let $\abs{G} = p^n$ where $p$ is prime. We know a few things about such a group. 28 | \begin{enumerate} 29 | \item $\abs{Z(G)} > 1$. 30 | \end{enumerate} 31 | \end{theorem} 32 | 33 | \begin{proof}[Proof of 1] 34 | We know from the class equation that \[ \abs{G} = \abs{Z(G)} + \sum_{i=1}^n \abs{O_{x_i}}. \] For all $i$ we know that $O_{x_1} \geqslant 2$ and we know that it divides the order of the group. Then $\abs{O_{x_1}} = p^k$ for some $1 \leqslant k \leqslant n$. Then $p$ divides the order of the sum of the orders of the orbits, and so $p$ must divide the order of the center. In fact, this tells us that $\abs{Z(G)}$ is at least $p$. 35 | \end{proof} 36 | 37 | \begin{proposition} 38 | For prime $p$, if $\abs{G} = p^2$ then 39 | \begin{enumerate} 40 | \item[(a)] $G$ is abelian, and 41 | \item[(b)] $G \cong C_{p^2}$ or $C_p \times C_p$. 42 | \end{enumerate} 43 | \end{proposition} 44 | 45 | \begin{proof}[Proof of~\textup{(a)}] 46 | Let $x \in G$, and assume by way of contradiction that $x \notin Z(G)$. Consider $H = \langle Z(G), x \rangle$. Since $G$ is a $p$ group, we know that the center cannot be one, and so $\abs{G} = p$ (since it must divide $p^2$ and if $\abs{G} = p^2$ then $x \in Z(G)$). And we know that $\abs{H} \geqslant \abs{Z(G)}$ so we know that $p \leqslant \abs{H} \leqslant p^2$ and the order must divide $p$, and so we know that $H = G$. But since $x$ commutes with everything in $H$ we know that $x \in Z(G)$. 47 | \end{proof} 48 | 49 | \begin{proof}[Proof of~\textup{(b)}] 50 | Left as an exercise to us. 51 | \end{proof} 52 | 53 | \subsection{Conjugation in \texorpdfstring{$S_n$}{Sn}} 54 | 55 | If you take an arbitrary element $\sigma$ of $S_n$, what can we reasonably expect the conjugacy class of $\sigma$ to look like? For example, consider $\sigma = (123)$. Well, $(14)(123)(14) = (423)$. We also know that $(256)(123)(652) = (153)$. Notice that we've found two $3$-cycles! We can hypothesis that $\abs{\tau \sigma \tau} = \abs{\sigma}$, and in fact we just replace the elements of the $3$-cycle with ``where the numbers in the conjugating cycles get sent.'' That is, if $\sigma = (a_1,\dotsc,a_k)$ then $\tau\sigma\tau^{-1} = (\tau(a_1)),\dotsc,\tau(a_k))$. We can infer a slightly stronger link here; in fact, $\sigma$ is conjugate to $\sigma'$ if and only if they have the same cycle structure. -------------------------------------------------------------------------------- /MATH-350/Lectures/18.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 12 October 2018} 2 | 3 | \epigraph{``We have hope. But hope doesn't mean much.''}{Miki} 4 | 5 | Let's return to the proposition we described last time, where we said that the equivalency classes in $S_n$ under conjugation are exactly the sets of permutations with the exact same cycle decomposition structure. That is, all elements of the form $(\cdot \cdot \cdot \cdot \cdot) \in S_n$ are conjugates with one another, and the same holds for $(\cdot \cdot \cdot)(\cdot \cdot)$ and all other cycle structures. 6 | 7 | \begin{proposition} \hfill 8 | \begin{parts} 9 | \part[prf:cda] If $\sigma \in S_n$ is a $k-cycle$ where $\sigma = (a_1,\cdots,a_k)$, and $\tau \in S_n$, then $\tau\sigma\tau^{-1} = (\tau(a_1), \dotsc, \tau(a_k))$. 10 | \part[prf:cdb] If $\sigma$ is a product of disjoint cycles $\sigma_i \cdots \sigma_r$ then $\tau\sigma\tau^{-1}$ is the product of disjoint cycles $\tau \sigma_i \tau^{-1}$. 11 | \part[prf:cdc] Cycles $\sigma, \sigma'$ are conjugate if and only if they have the same cycle structure. 12 | \end{parts} 13 | \end{proposition} 14 | 15 | \begin{proof}[Proof of\/~\ref{prf:cda}] 16 | Let $A = \{1, \dotsc, n\}$ so that $\tau A = A$. Then $A = \{\tau(1), \dotsc, \tau(n)\}$. \note{FINISH THIS} 17 | \end{proof} 18 | 19 | \begin{proof}[Proof of\/~\ref{prf:cdb}] 20 | Let $\sigma = \sigma_1 \cdots \sigma_r$. Then $\tau\sigma\tau^{-1}$ can be written as $\tau\sigma_1(\tau^{1}\tau) \cdots (\tau^{1}\tau) \sigma_r \tau^{-1}$, and by associativity the proposition holds. Since the cycles were disjoint to begin with, permuting each $\sigma_i$ under $\tau$ ensure that the products are still disjoint. 21 | \end{proof} 22 | 23 | \begin{proof}[Proof of\/~\ref{prf:cdc}] 24 | The forward direction follows immediately from the previous two proofs. 25 | Next, assume $\sigma, \sigma'$ have the same cycle structure. Then \[\sigma = (a_1^1 \cdots a_{k_1}^1)(a_1^2 \cdots a_{k_2}^2) \cdots (a_1^r \cdots a_{k_r}^r),\] and \[\sigma' = (b_1^1 \cdots b_{k_1}^1)(b_1^2 \cdots b_{k_2}^2) \cdots (b_1^r \cdots b_{k_r}^r).\] Then $A = \{1, \dotsc, n\} = \{a_i^j\} = \{b_i^j\}$. Then take $\tau(a_i^j) = b_i^j$, since this is just a permutation on the elements in $A$, so by \ref{prf:cda} and \ref{prf:cdb} this holds. 26 | \end{proof} 27 | 28 | \subsection{Proving the simplicity of \texorpdfstring{$A_5$}{A5}} 29 | 30 | This is a big deal. 31 | 32 | \begin{proof} 33 | We want to show that $A_5$ (or any $A_n$, for that matter) has no proper normal subgroups. Recall the orbit stabalizer theorem, where $\abs{G} = \abs{G_x} \cdot \abs{O_x}$ for any $x \in G$. Recall also that if $N \normal G$ then $N$ is the union of conjugacy classes. Let's start by finding the class equation for $A_5$. Since $A_5$ must have even sign, we know that the only cycles in $A_5$ are of the form $e$, $(\cdot \cdot)$, $(\cdot \cdot \cdot)$, and $(\cdot \cdot \cdot \cdot \cdot)$. Let $O_{x}^{S_5}$ be the orbit of an element $x$ in $S_5$ while $O_{x}^{A_5}$ is the orbit in $A_5$. Note that $\abs{O_x^{A_5}} \leq \abs{O_x^{S_5}}$. Similarly, anything in $A_5$ which fixes $x$ must also fix $x$ in $S_5$ so $\abs{(S_5)_x} \geq \abs{(A_5)_x}$. We also know by Orbit-Stabalizer that $\abs{(A_5)_x} \cdot \abs{O_x^{A_5}} = \abs{A_5} = 60$ while $\abs{(S_5)_x} \cdot \abs{O_x^{S_5}} = \abs{S_5} = 120$. Combining these inequalities with the Orbit-Stabalizer theorem (and recognizing that everything here is an integer), we are left with the option that either the orbits are the same size and the centralizer in $A_5$ is half of the centralizer in $S_5$, or that the centralizers are the same and the orbits in $A_5$ are half that of the orbits in $S_5$. 34 | 35 | Let's figure out which of these cases is true. Consider $x = (\cdot \cdot \cdot) = (123) \in S_5$ without a loss of generality. What is the size of the orbit of $x$ in $S_5$? Well, it's all three-cycles, so there are $2 \cdot {5 \choose 3} = 20$ elements in the orbit of $x$ in $S_5$. By Orbit-Stabalizer, the size of the sabalizer is then $120/20 = 6$. Note that $(45) \in (S_5)_x$ since it doesn't move $x$, but because $(45)$ is not in $A_5$ since it has the wrong sign, we know that it is the stablizer which has shrunk and the orbits have the same size. 36 | 37 | Let's do the same thing with $x = (\cdot \cdot)(\cdot\cdot) = (12)(34)$ without a loss of generality. Then $\abs{O_x^{S_5}} = {5 \choose 1} \cdot 3 = 15$ elements in the orbit of $x$ in $S_5$. Since this is odd, we know that the orbit can't shrink so it \emph{again} must be the case that the stabalizer has shrunk. 38 | 39 | Now let $x = ( \cdot \cdot \cdot \cdot \cdot)$. The orbit of $x$ is then of order $5!/5 = 4!$ while the stabalizer is of order $5!/4! = 5$. In this case, it is now the \emph{orbit} which has shrunk. 40 | 41 | Then $\abs{A_5} = 1 + 20 + 15 + 2\cdot 12$ where $20$ comes from the $3$-cycles, $15$ comes from the double $2$-cycles, and the $24$ comes from the two $5$-cycles. Now suppose that $N \normal A_5$. We know it is the union of conjugacy classes and it contains the identity, so $\abs{N} = 1 + \{\text{some of }12,12,15,20\}$, and it must divide $\abs{A_5} = 60$. Note that this can happen \emph{only} if $\abs{N} = 1$ or $\abs{N} = 60$, so $A_5$ contains no proper normal subgroups and is simple. 42 | \end{proof} -------------------------------------------------------------------------------- /MATH-350/Lectures/19.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 15 October 2018} 2 | 3 | \epigraph{``Perfectly balanced, as all things should be.'' (when referring to left and right actions)}{Miki} 4 | \epigraph{``Our theorem is gone! Oh no!''}{Miki} 5 | 6 | \begin{problem} 7 | Does right multiplication define an action of $S_4$ on itself? 8 | \end{problem} 9 | \begin{solution} 10 | No, since we can find two elements for which $g_1(g_2(x)) \not= x \cdot (g_1g_2)$. Consider $(12)$ and $(23)$ acting on the identity. In general, right multiplication is an action if and only if the group is commutative since we are ``switching the order of the multiplication.'' 11 | \end{solution} 12 | 13 | \subsection{Right Actions} 14 | In order to fix this ``unfairness,'' we often define something called a right action $A \times G \to A$, where the associativity of the action is specified as $a \cdot (gh) = (a \cdot g) \cdot h$. This turns right multiplication into a ``right action.'' There really isn't any distinction between the two, which is why we just speak of ``the action.'' 15 | 16 | How do we turn left actions into right actions? Suppose we have a left action $g \cdot a$. Define $a \cdot g = g^{-1} \cdot a$; that is, the right action of $g$ on $a$ is just the left action by the inverse of $g$. This works since $(gh)^{-1} = h^{-1}g^{-1}$ so the order follows the rules for right multiplication/action. 17 | 18 | \begin{problem} 19 | Consider $\Z$ acting on itself through left addition, where $m \cdot n \mapsto m + n$, and consider that when we turn this into a right action. Then $n \cdot m \mapsto -m + n = n - n$, and we've just invented subtraction. 20 | \end{problem} 21 | 22 | \begin{example} 23 | Consider $A_3 \normal S_3$, and consider conjugating $(123) \in A_3$ by something in $S_3$. We know we'll get either another three cycle or the identity, since we know that $gNg^{-1} = N$. Then if $g \in S_3$ there there exists a $\sigma_g : S_3 \to S_3$ which acts on $g$ by conjugation. The consider $\sigma_g |_N$ restricted to acting on $N$. Then we have a map from $N$ to itself. If $g \in N$ then we get the trivial map (since this is just $Z_3$), and otherwise we must not get the trivial map and so $(123) \mapsto (132)$ and \emph{vice versa}. In the latter case, we've created not just a random map but a homomorphism from $N$ to itself. This homomorphism $x \mapsto x^3$ in the group $\langle x \mid x^3 = 1 \rangle$, which is both injective and surjective and we know that this is a homomorphism since the generators satisfy the relations under the map since $x^6 = 1$. 24 | \end{example} 25 | 26 | \subsection{Group Automorphisms} 27 | 28 | \begin{definition}[Automorphism] 29 | A group automorphism is an isomorphism from $G$ to itself. 30 | \end{definition} 31 | 32 | For every group $G$ there is a group $\operatorname{Aut}(G)$ which is the group of all automorphism of $G$ under composition. Miki told us to prove for ourselves that this is actually a group. Now consider $G$ acting on itself through conjugation where $g \mapsto \sigma_g : x \mapsto gxg^{-1}$. For an normal subgroup of $G$ we know that $\sigma_g|_N : N \to N$, and so we have a homomorphism $\psi$ from $G$ to $\operatorname{Aut}(N)$ where $g \mapsto \sigma_g |_N$. The kernel of $\psi$ is the set of all elements in $G$ which commute with $N$, and so $\ker \psi = C_G(N)$. Then $G/C_G(N)$ is isomorphic to a subgroup of $\operatorname{Aut}(N)$ by the First Isomorphism Theorem. 33 | 34 | There are two things to unpack here. First, how to we know that $\psi$ is actually a homomorphism? That is, why is $\sigma_g|_N \in \operatorname{Aut}(N)$? Well, consider that $\sigma_g(nn') = gnn'g^{-1} = gng^{-1} \cdot gn'g^{-1} = \sigma_g(n)\sigma_g(n')$, and so $\sigma_g |_N$ preserves the group operation. Next, how to we know that the map $g \mapsto \sigma_g$ is a homomorphism? That is, why does $\sigma_{gg'} = \sigma_{g}\sigma_{g'}$. Well, since conjugation is a well-defined action on $G$, this forms a homomorphism. Note that the restriction to $N$ isn't important here, but the reason we require normality since we won't we able to compose the conjugations since $gHg^{-1} \not= H$. 35 | 36 | \begin{corollary} 37 | Take $G = N$. Then we get a homomorphism from $G$ to its own automorphism group, and so $G/C_G(G) = G/Z(G)$ is isomorphism to a subgroup of $\operatorname{Aut}(G)$. 38 | \end{corollary} 39 | 40 | \begin{corollary} 41 | Let $H \leq G$ be any subgroup of $G$. Then for all $g \in G$, $gHg^{-1} \cong H$, but they are not necessarily equal to one another. 42 | \end{corollary} 43 | 44 | \begin{corollary} 45 | Let $H \leq G$ be any subgroup of $G$. Then $N_G(H) / C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$, since the centralizer is always normal in the normalizer. This is really just a general case of the preceeding statements. 46 | \end{corollary} 47 | 48 | \begin{proof} 49 | Since $H \normal N_G(H)$, we just let $G' = N_G(H)$ and apply the result. 50 | \end{proof} -------------------------------------------------------------------------------- /MATH-350/Lectures/2.tex: -------------------------------------------------------------------------------- 1 | \section{August 31, 2018} 2 | 3 | As always, Miki began class at precicely 10:25 \textsc{am}. She wrote a review of last lecture on the bard, and then posed the following question as a warm up. She also talked about how the DUS department is arguing over whether money should be spent on T-shirts or chocolate (Miki thinks chocolate). 4 | 5 | \begin{problem}[Warm Up] 6 | Are these groups? 7 | \begin{parts} 8 | \part[w:1a] $(\Z/n\Z, \times)$; 9 | \part[w:1b] $(\Z/n\Z \setminus \{0\}, \times)$ 10 | \end{parts} 11 | \end{problem} 12 | 13 | \begin{solution} 14 | The solutions to the warm-up 15 | \begin{subproof}[Solution to~\ref{w:1a}] 16 | No, since $0$ has no inverse. 17 | \end{subproof} 18 | 19 | \begin{subproof}[Solution to~\ref{w:1b}] 20 | No, this only works when $n$ is prime. For any factors $a,b$ of $n$, $a \times b = 0$, which isn't in the group. We say that $(\Z/p\Z, \times)$ is a group for all prime $p$. 21 | \end{subproof} 22 | \end{solution} 23 | 24 | \begin{theorem}[Fermat's little theorem] 25 | For prime $p$ and composite $a = np$, then $a^{p-1} \equiv 1 \pmod{p}$. 26 | \end{theorem} 27 | 28 | \begin{lemma} 29 | If $\bar{a} \in \Z/p\Z \setminus \{0\}$, then $\bar{a}$ has an inverse in $(\Z/p\Z \setminus \{0\}, \times)^*$. 30 | \end{lemma} 31 | 32 | \begin{definition}[Units] 33 | A unit is something which has an inverse. The units of a group are denoted by putitng an asterisk after teh group, eg $(\Z/p\Z \setminus \{0\}, \times)^*$. 34 | \end{definition} 35 | 36 | \begin{example} 37 | For integers modulo 4, $(\Z/4\Z, \times)^* = \{\bar{1}, \bar{3}\}$. 38 | \end{example} 39 | 40 | \begin{problem}[On Homework] 41 | What are the conditions for determining the units of a group? We know it must have an inverse, but that's hard to check. Instead, we know that $a$ is a unit if and only if $\gcd(a,n) = 1$. Prove this. 42 | \end{problem} 43 | 44 | \subsection{Symmetries of a regular \texorpdfstring{$n$}{n}-gon} 45 | Miki is angry with the book because she doesn't like how it treats symmetries, I think because she wants $D_{2n}$ to be called $D_n$. 46 | 47 | Miki drew a triangle on the board, and began talking about the different operations we can preform on that triangle to preserve symmetries. She introduced $s$ to mean a reflection, and $r$ to mean a rotation. For a triangle, there are three distinct reflections, 48 | \[ s = \{s_1, s_2, s_3 \}, \] 49 | where $s_i$ is the reflection across the line $OA_1$. We can also rotate the triangle in two directions. 50 | 51 | We know that these are all the symmetries, since we can count the permutations of the triangle. We've exhauseted then, so we know that there can't be any more elements of the triangle-symmetry group $D_6$. In fact, because of the permutation fact, we konw that $\abs{D_{2n}} = 2n$. Some other observations about $D_{2n}$: 52 | \begin{itemize} 53 | \item $s^2 = e \implies s = s^{-1}$; 54 | \item rotating twice clockwise is the same as rotating counterclockwise, so these aren't unique elememnts; 55 | \item $r^n = e$ 56 | \item $rs = s_2$, so $s_n$ is just a combination of $r$ and $s$ --- then we can generate the entire group with just $r$ and $s$. 57 | \end{itemize} 58 | These things lead us to discover a new object. 59 | 60 | \begin{definition}[Generators] 61 | For a group $G$, the generators of $G$ is a set $ S = \{a,b,\dotsc : a,b,\dotsc \in G\}$ where $G$ is equal to all possible sombinations of elements of $S$. For $D_{2n}$, we could say that $D_{2n}$ is generated by $r$ and $s$. 62 | Usually there isn't a way to guess the generators of a group easily. 63 | \end{definition} 64 | 65 | \begin{definition}[Relations] 66 | A relation is a way of writing equivalent elements of groups. For example, in $D_{2n}$, 67 | \[ r^3 \equiv 1, \qquad s^2 \equiv 1, \qquad sr \equiv r^2s. \] 68 | Relations allow us to define how we can commute elements of the group. 69 | \end{definition} 70 | 71 | \begin{definition}[Presentation] 72 | A presentation of a group are the generators combined with the relations necessary to create the group. The largest group which is generated from the generators and which satisfies the relations, and has no other relations, is our group. A presentation is written as $\langle a,b \mid \text{relations between $a$ and $b$} \rangle$, where $a$ and $b$ are the generators of the group. 73 | \end{definition} 74 | 75 | Now Miki told us that the group of the symmetires of a regular $n$-gon is the dihedral group of order $2n$, written either as \{$D_{2n}$ or $D_{n}$\}, depending on if you are a representation theorist or not. 76 | 77 | \begin{problem}[HW] 78 | Why is the order of $D_{2n}$ always $2n$? 79 | \end{problem} 80 | 81 | \subsection{Symmetric group on \texorpdfstring{$n$}{n} elements} 82 | 83 | Miki defined the symmetric group on $n$ elements $S_n$, which is just the permutations of $n$ elemnts. Notice that $D_{2n}$ is a subgroup of $S_n$. We know that the order of $S_n = n!$ and the order of $D_{2n} = 2n$. 84 | 85 | [Insert diagrams of different ways to denote permuations, like the cycle notation] 86 | -------------------------------------------------------------------------------- /MATH-350/Lectures/20.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 22 October 2018} 2 | 3 | \epigraph{``You all look so unhappy.''}{Miki} 4 | \epigraph{``$p$ is going to be prime for \emph{at least} two more days.''}{Miki} 5 | 6 | \subsection{Clasifying Automorphisms} 7 | 8 | Let's talk automorphisms! 9 | 10 | \begin{definition}[Inner Automorphism] 11 | Let $g \in G$ and let $\sigma_g : G \to G : x \mapsto gxg^{-1}$ be an automorphism (i.e., an automorphism by conjugation). Then $\sigma_g$ is an \emph{inner automorphism}. The collection of all inner automorphisms forms a group $\Inn(G) \leq \Aut(G)$ which is isomorphic to $G/\mathrm{Z}(G)$ by the first isomorphism theorem. 12 | \end{definition} 13 | 14 | \begin{example} 15 | Let $G = \Z/n\Z$. We proved on homework that $\Aut(G) \cong (\Z/n\Z)^\times$, and so any $\sigma \in \Aut(G)$ is uniquely determined by the map which sends $1$ to $a$ for some unit $a$. Since $G$ is commutative, conjugation doesn't really do anything, so $\Inn(G) = \sigma_1$. Put another way, $\mathrm{Z}(G) = G$, so $\Inn(G)$ is as small as it could be. 16 | \end{example} 17 | 18 | \begin{example} 19 | Let $G = D_8$. The center of $D_8$ is $\mathrm{Z}(D_8) = \langle r^2 \rangle$. We know that $\Inn(G) \cong G/\langle r^2 \rangle \cong K_4$. 20 | \end{example} 21 | 22 | \begin{definition}[Characteristic] 23 | A subgroup $H \leq G$ is \emph{characteristic} if $\sigma(H) = H$ for any $\sigma \in \Aut(G)$. This is like a normal subgroup, except that a normal subgroup need only be preserved under \emph{inner automorphism} while a being characteristic subgroup is a stronger condition. 24 | \end{definition} 25 | 26 | \begin{example} 27 | Let $G = D_8$ and let $H = \langle r^2 \rangle$. Since $H$ is the center, this is characteristic (this is true in general). Next let $K = \langle r \rangle \leq G$. Since $\Im(r)$ is either $r$ or $r^3$ (check the order under isomorphism) then $\sigma(\langle r \rangle) = \langle r \rangle$ for any $\sigma \in \Aut(D_8)$ and so it is characteristic. 28 | \end{example} 29 | 30 | Just to make the point explicit, if $H$ is characteristic in $G$ then it must be normal in $G$, but the reverse is not true. Additionally, if $H$ is the unique subgroup of a particular order in $G$ then it must be characteristic since there's nothing else it could be sent to under an automorphism since it's image must be a subgroup of the same order. 31 | 32 | \subsection{Sylow \texorpdfstring{$p$}{p}-subgroups} 33 | 34 | \begin{definition} 35 | Let $p$ be prime. A $p$-subgroup is a subgroupd of order $p^n$ for $n \geq 0$. 36 | \end{definition} 37 | 38 | \begin{definition} 39 | Let $\abs{G} = p^am$ where $p$ does not divide $m$. If there is a subgroup of order $p^a$ (there is) then a subgroup of this order is called a Sylow $p$-subgroup. The set of all such groups is written as $\Syl_p(G)$. The number of such groups is written as $n_p(G) = \abs{\Syl_p(G)}$. 40 | \end{definition} 41 | 42 | \begin{example} 43 | If $p$ does not divide $\abs{G}$ the the only Sylow $p$-subgroup is the trivial subgroup. If $\abs{G} = p^a$ then the unique Sylow $p$-subgroup is $\Syl_p(G) = \{G\}$. 44 | \end{example} 45 | 46 | \begin{example} 47 | Let $G = S_3$ which has order $2 \cdot 3$. Let $p = 2$, $m = 3$, and $a = 1$. Then the largest $\Syl_p$ subgroup is $C_2$, of which there are three such subgroups (things generated by $2$-cycles). If we let $p = 3$, then there is one Sylow $p$-subgroup, generated by a $3$-cycle. 48 | \end{example} 49 | 50 | \subsection{Sylow Theorems} 51 | 52 | Throughout, let $p$ be prime and let $G$ be a group of order $p^am$ where $a > 0$ and $p$ does not divide $m$. 53 | 54 | \begin{theorem}[Sylow I] 55 | There exists a subgroup of $P \leq G$ where $\abs{P} = p^a$. 56 | \end{theorem} 57 | 58 | \begin{theorem}[Sylow II] 59 | For each $p$, the Sylow $p$-subgroups are conjugate to one another. 60 | \end{theorem} 61 | 62 | \begin{theorem}[Sylow III] 63 | The number of Sylow $p$-subgroups of $G$, written $n_p(G)$, divides $m$ and is congruent to $1 \bmod{p}$. 64 | \end{theorem} 65 | 66 | We'll prove these next time (with a lot of chocolate). Today we'll just talk about the implications of these theorems. 67 | 68 | \begin{corollary} 69 | There must exist an $x \in G$ whose order is $p$. 70 | \end{corollary} 71 | \begin{proof} 72 | Let $y \in P$ be not the identity. Then $\abs{y} = p$, so for some $0 < b \leq a$ we know that $x = y^{p^{b-1}}$. 73 | \end{proof} 74 | 75 | \begin{corollary} 76 | The Sylow $p$-subgroups are all conjugate. 77 | \end{corollary} -------------------------------------------------------------------------------- /MATH-350/Lectures/21.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 24 October 2018} 2 | 3 | Today was a presentation of the proof of the Sylow theorems found in the textbook. As such, notes are omitted in favor of reading the relevant section in the book (and I'm rather tired today and I don't want to type anything up). -------------------------------------------------------------------------------- /MATH-350/Lectures/22.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 26 October 2018} 2 | 3 | Didn't go to class today! Something about direct products I think. -------------------------------------------------------------------------------- /MATH-350/Lectures/23.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 20 October 2018} 2 | 3 | \epigraph{``Let's write down all finitely generated abelian groups. What fun.''}{Miki} 4 | 5 | We have two goals for today. 6 | \begin{enumerate} 7 | \item Is $Z_{20} \times Z_{18} \cong Z_{36} \times Z_{10}$? 8 | \item How do we classify \emph{all} finitely generatred abelian groups? 9 | \end{enumerate} 10 | 11 | To start answering these, we'll begin with a proposition. 12 | 13 | \begin{proposition} 14 | $Z_n \times Z_m \cong Z_{mn}$ if and only if $\gcd(m,n) = 1$. 15 | \end{proposition} 16 | 17 | \begin{proof} 18 | Let $d = \gcd(m,n)$, and let $Z_m = \langle x \rangle$ and let $Z_n = \langle y \rangle$. Consider $G = Z_m \times Z_n = \{(x^a, y^b)\}$. Consider $(c,f) \in G$. Then $\abs{(c,f)} = \lcm (\abs{c}, \abs{f})$. If $d = 1$ then $\abs{(x,y)} = \lcm (m,n) = mn$, so $Z_{mn} \cong \langle (x,y) \rangle \leq G$. Since the orders are the same, it is isomorphic to the whole thing. 19 | On the other hand, if $d > 1$, let $(c,f) \in G$, and consider $(c,f)^{mn/d} = (c^{mn/d}, f^{mn/d}) = (e,e)$, so every element has order strictly less than $mn$ since $d > 1$. Therefore $G \not\cong Z_{mn}$. 20 | \end{proof} 21 | 22 | \begin{example} 23 | Consider $Z_9 \times Z_6 \not\cong Z_{54}$. Note that $Z_9 \times Z_6 \cong Z_9 \times Z_3 \times Z_2 = Z_{18} \times Z_3$. 24 | \end{example} 25 | 26 | \begin{example} 27 | Use the proposition we just proved to ``factor'' the groups into the same decomposition. Ta-Da! 28 | \end{example} 29 | 30 | fdas 31 | 32 | \subsection{Classifying Finitely-Generated Abelian Groups} 33 | 34 | \begin{definition}[Free Abelian Group] 35 | Let $\Z^r = \Z \times \cdots \times \Z$ ($r$ times) be the free abelian group of rank $r$. 36 | \end{definition} 37 | 38 | \begin{theorem}[Classification Theorem for Finitely Genreated Abelian Groups] 39 | Let $G$ be a finitely generated abelian group. Then there is a unique decomposition of $G$ satisfying 40 | \begin{enumerate} 41 | \item $G \cong \Z^r \times Z_{n_1} \times \cdots Z_{n_s}$ for $r,n_i \in \Z$, 42 | \item $n_i > 2$ for all $i$, and 43 | \item $n_{i+1}$ must divide $n_i$ for all $1 \leq i \leq s-1$. 44 | \end{enumerate} 45 | \end{theorem} 46 | 47 | \subsection{Classifying Finitely-Generated Abelian Groups 2, Electric Boogaloo} 48 | 49 | \begin{example} 50 | Consider $Z_{60} \cong Z_{2^2} \times Z_{3} \times Z_{5}$. Notice now that all the components are $p$-subgroups. 51 | \end{example} 52 | 53 | \begin{theorem} 54 | Let $\abs{G} = n = \prod p_i^{a_i}$, where $a_i \geq 1$. Then we can write $G$ uniquely (up to order of primes) as $G \cong A_1 \times \cdots \times A_k$ where $\abs{A_i} = p_i^{a_i}$, and for all $A = A_i$ where $\abs{A} = p^a$, we know that $A \cong Z_{p^{b_1}} \times \cdots \times Z_{p^{b_\ell}}$ where $b_1 \geq b_2 \geq \cdots \geq b_\ell$, where the sum of all $b_i$ is $a$. 55 | \end{theorem} 56 | -------------------------------------------------------------------------------- /MATH-350/Lectures/24.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 31 October 2018} 2 | 3 | \epigraph{``Oh. You are a gamer.''}{Miki} 4 | 5 | \subsection{FGAGs for Dayyyyysssss} 6 | 7 | Miki started us off with some exercises to apply what we learned last lecture about classifying finite abelian groups. 8 | 9 | \begin{exercise} 10 | Conver $G = Z_{36} \times Z_{12}$ into elementary divisor notation. 11 | 12 | \begin{solution} 13 | $G \cong (Z_{2^2} \times Z_{3^2}) \times (Z_{2^2} \times Z_3) \cong (Z_{2^2} \times Z_{2^2}) \times (Z_{3^2} \times Z_3)$. 14 | \end{solution} 15 | \end{exercise} 16 | 17 | \begin{exercise} 18 | Convert $(Z_{16} \times Z_{4} \times Z_2) \times (Z_9 \times Z_3)$ into invariant factor notation. 19 | 20 | \begin{solution} 21 | Group up the $i$\textsuperscript{th} terms in each parenthetical term, so $G \cong (Z^{16} \times Z_9) \times (Z_4 \times Z_3) \times Z_2 \cong Z_{144} \times Z_{12} \times Z_{2}$. 22 | \end{solution} 23 | \end{exercise} 24 | 25 | \begin{exercise} 26 | Classify all abelian groups of order $24$. 27 | 28 | \begin{solution} 29 | Let's use the invariant factor notaion. If $p$ divides the order of $\abs{G}$ then $p$ divides $n_1$. The factorization of $24$ is $2^3 \times 3$. Then $n_1$ could be $24$, and $G_1 \cong Z_{24}$. It could be that $n_1 = 4 \cdot 3$, so $n_2 = 2$ and $G_{2} \cong Z_{12} \times Z_2$. It could be that $n_1 = 2 \cdot 3$, so $n_2 = 2$ and $n_3 = 2$, (can't be $n_2 = 4$ since $4$ doesn't divide $6$), so $G_3 \cong Z_6 \times Z_2 \times Z_2$. 30 | \end{solution} 31 | 32 | \begin{solution} 33 | Let's use the elementary divisor notation. Let $\abs{H} = 24$. Then $H \cong A_1 \times A_2$ where $\abs{A_1} = 2^3$ and $\abs{A_3} = 3$. Then $A_2 \cong Z_3$. For $A_1$, we must take all non-increasing partitions of $3$, so $3 = 3, 2 + 1, 1+1+1$. The the possibilities for $A_1$ are $Z_{2^3}$, $Z_{2^2} \times Z_2$, and $Z_2 \times Z_2 \times Z_2$. Then $H$ is either $Z_3 \times Z^{2^4} \cong Z_24$ or $Z_3 \times Z_{2^2} \times Z_2 \cong Z_{12} \times Z_2$ or $Z_3 \times Z_2 \times Z_2 \times Z_2 \cong Z_6 \times Z_2 \times Z_2$. 34 | \end{solution} 35 | \end{exercise} 36 | 37 | \subsection{The Shape of Things to Come} 38 | 39 | Over the next few lectures, we'll cover how to take the product of groups which aren't abelian, and understanding how we can ``factor'' non abelian groups in the same way that we now know how to factor abelian groups. 40 | Warning: it'll be the hardest single thing we do in this class. 41 | 42 | \subsection{Commutators} 43 | 44 | Let $G$ be a group with $x,y \in G$. We defined the \emph{commutator} to be $[x,y] = xyx^{-1}y^{-1}$. The commutator of any two elements of $G$ is one if and only if $xy = yx$. The commutator subgroup $G' = \langle [x,y] \mid x,y \in G \rangle$, which is normal in $G$ and the quotient $G/G'$ is abelian. 45 | 46 | Suppose we had a homomorphism $\phi$ from $G$ to $H$, where $H$ is abelian. Then it must be that $G' \leq \ker \phi$, since $\phi(x)\phi(y) = \phi(y)\phi(x)$, so $[\phi(x),\phi(y)] = \phi([x,y])$ for all $x,y \in G$. 47 | 48 | The quotient of $G$ by $G'$ is the largest abelian quotient of $G$, which means that the commutator group is the smallest subgroup for which the quotient is abelian. 49 | 50 | \begin{example} 51 | Let $G = S_3$. What is $G'$? Given the sign map $\pi : S_3 \to Z_2$, we know that $\ker \pi = A_3$, so we know that $G' \leq A_3$ since $Z_2$ is commutative. Then $G' = A_3$. 52 | \end{example} 53 | 54 | \begin{example} 55 | Let $G = D_{12}$. Is it a direct product of some proper subgroups? Consider $K = \langle r^3 \rangle \leq Z(G)$ and let $H = \langle s, r^2 \rangle$. Notice that $\langle H,K \rangle \leq G$ and $\langle H,K \rangle$ contains $s$ and $r$, so it is equal to the whole group. It's also true that $H$ and $K$ commute with one another. As it turns out (note quite a proof yet) $G \cong K \times H$. 56 | \end{example} 57 | 58 | \begin{theorem} 59 | Let $G$ be a group and let $H,K$ be subgroups of $G$ satisfying 60 | \begin{enumerate} 61 | \item $H \normal G$ and $K \normal G$, 62 | \item $H \cap K = \{e\}$, and 63 | \item $\langle H,K \rangle = HK = G$. 64 | \end{enumerate} 65 | Then $G \cong H \times K$. 66 | \end{theorem} 67 | 68 | \begin{proof} 69 | First, we show that $H$ and $K$ commute with one another. Consdier $[h,k]$. Notice that $(hkh^{-1})k^{-1} \in K$ and $h(kh^{-1}k^{-1}) \in H$, so it's in both $H$ and $K$, but the intersection of $H$ and $K$ is $\{e\}$ which means that the commutator is the identity, which happens if and only if $H$ and $K$ commute with one another. Next, consider that $\abs{HK} = \abs{H}\cdot\abs{K}/ \abs{\{e\}} = \abs{H} \cdot \abs{K}$. Next, create a map $\phi : H \times K \to G$ where $(h,k) \mapsto hk$. We show that this is a homomorphism (and an isomorphism). Notice that $\phi(h,k)\phi(h',k') = hh'kk' = \phi(hh',kk')$ so $\phi$ is a homormphism. It is also injective. Suppose that $\phi(hk) = 1$, which tells us that $h = k^{-1}$, which means that $h = k = 1$ since $h \in H \cap K$. It is surjective, since the sizes of the groups are the same. Then $G \cong H \times K$ through $\phi$. 70 | \end{proof} 71 | 72 | 73 | -------------------------------------------------------------------------------- /MATH-350/Lectures/25.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 2 November 2018} 2 | 3 | \epigraph{``You're not going to like this.''}{Miki} 4 | 5 | \subsection{Semidirect Products} 6 | We want to generalize our understanding of the direct product to non-commutativity. 7 | 8 | \begin{example} 9 | Let $G = D_6$, and let $H = \langle r \rangle \cong Z_3$ and let $K = \langle s \rangle \cong Z_2$. Notice $H \normal G$. Since their intersection is trivial, $HK = G$. Here, we know that $G \not\cong H \times K$ (since then it would be abelian) but it's really similar, since we can write any $g$ as a product of $h$ and $k$. 10 | \end{example} 11 | 12 | We kinda cheated with this example since we already know how $r$ and $s$ relate to one another, but what if we don't know how to ``conjugate'' them? What if we don't know how to multiply words? 13 | 14 | \begin{definition}[Semidirect Product] 15 | Let $H$ and $K$ be groups, and let $\phi : K \to \Aut(H)$ be a homomorphism (aside from the trivial one, this may not exist). Then $\phi$ defines an action of $K$ on $H$ where $k \cdot h = \phi(k)(h)$. Then $G = H \rtimes_\phi K$ is called the semidirect product, where $g \in G = (h,k)$ and $(h_1,k_1)(h_2,k_2) = (h_1(k_1 \cdot h_2), k_1,k_2)$. This is a generalization of conjugation. 16 | \end{definition} 17 | 18 | There are some important properties of this. First, $G$ is a group where $\abs{G} = \abs{H} \cdot \abs{K}$. Second, $H \leq G$ via $h \mapsto (h,1)$ and $K \leq G$ via $k \mapsto (1,k)$. Third, $H \cong \langle (h,1) \rangle \normal G$. Fourth, $H \cap K = \{e = (1,1)\}$. Finally, for all $k \in K$ and $h \in H$ we know that $k \cdot h = khk^{-1}$. 19 | 20 | \begin{example} 21 | Let $K = \langle k \rangle \cong Z_2$ and $H = \langle h \rangle \cong Z_3$. What could $H \rtimes K$ be? First, let's find $\Aut(H) \cong (\Z/(3))^\times$, so there are only two possible choices for our defining homorphism; it can be either the identity, or one other map. Then let $\phi : k \mapsto e$, so $h \cdot h = h$ and the groups commute, so $H \rtimes_\phi K \cong H \times K$. Or, we could take that $\psi : k \mapsto \chi$ where $k \cdot h = h^2$ and where $\chi$ is the other element of $\Aut(H)$, and so $H \rtimes_\varphi K \cong S_3$. Notice that $H \rtimes_\phi K \not\cong H \rtimes_\varphi K$. 22 | \end{example} 23 | 24 | \begin{example} 25 | What if we keep $H$ and $K$ but make $H \cong Z_2$ and $K \cong Z_3$ (i.e., we flip the group order)? Well, $k \cdot 1 = 1$ and then, by exhaustion, $k \cdot h = h$ (since that's all we can do with a homomorphism), so $H \rtimes K \cong H \times K$. This tells us that $H \rtimes K$ is, in general, not the same thing as $K \rtimes H$. 26 | \end{example} -------------------------------------------------------------------------------- /MATH-350/Lectures/26.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 5 November 2018} 2 | 3 | \epigraph{``Now everybody is happy. Or not, but it works.''}{Miki} 4 | 5 | \subsection{More Semidirect Products} 6 | 7 | A continuation of the discussion from last lecture. 8 | 9 | \begin{example} 10 | Suppose $H \cong Z_{20} \times Z_{45}$ (or any abelian group $\abs{H} > 2$). Since $H$ is abelian we know that the map $\gamma \in \Aut(H) : x \mapsto x^{-1}$ is a homomorphism. Then $\gamma$ is of order two. Let's take some semidirect products. Consider 11 | \begin{parts} 12 | \part[prt:sd:a] $H \rtimes Z_2$ where $\phi(k) = \gamma$. Then $k \cdot h = h^{-1}$ for all $h \in H$. 13 | \part[prt:sd:b] $H \rtimes Z_4$. Here we could also send $k$ to $\gamma$ since $\gamma^4 = 1$. Then $\phi(k^2) = 1$, so $k \cdot h = h^{-1}$, and $k^2 \cdot h = h$. This sends $Z_4$ to $Z_4 / \langle k^2 \rangle \cong Z_2$ to $\langle \gamma \rangle$. 14 | \part[prt:sd:c] $H \rtimes S_3$. We can quotient $S_3$ by $A_3$, which is isomorphic to $Z_2$, and then we map this into $\Aut(H)$. Composition of these maps yields a nontrivial homomorphism $\phi$. Here, $\psi : S_3 \to S_3/A_3$ is the sign map, so for all $\sigma in \S_3$, we say that $\psi :\sigma \mapsto \gamma$ if the sign of sigma is $-1$, and $\sigma \mapsto 1$ if the sign of sigma is positive. 15 | \end{parts} 16 | \end{example} 17 | 18 | \subsection{Identitfying Groups as Semidirect Products} 19 | 20 | Given a group $G$, how can we tell what its semidirect product decomposition could be? 21 | 22 | \begin{theorem} 23 | Let $G$ be a group with $H,K \leq G$ such that 24 | \begin{enumerate} 25 | \item $H \normal G$, 26 | \item $H \cap K = \{1\}$, and 27 | \item $G = HK$. 28 | \end{enumerate} 29 | Then $G \cong H \rtimes K$ where $k \cdot h = khk^{-1}$ for all $h \in H$ and for all $k \in K$. 30 | \end{theorem} 31 | 32 | This is really similar to the requirements for $G$ being the \emph{direct product} of $H$ and $K$; we just drop the requirement that $K \normal G$. 33 | 34 | \begin{proof} 35 | We know that $H \cap K = \{1\}$ so $\abs{H}\abs{K} = \abs{G}$, so for every $g \in G$ there is a unique way of writing it as $hk$. Then we can create a well-defined map $HK = G \to H \rtimes K$ via $\pi : g \mapsto (h,k)$. To show that this map is onto, observe that we can get all elements of the form $(1,k)$ and $(h,1)$, so any $(h,k)$ is in the image of $hk$ under our map. We know that $\abs{HK} = \abs{H \rtimes K}$ so they are isomorphic if we can show that $\pi$ is a homomorphism (turns out, it is). Proof of this is left as an exercise to the reader. 36 | \end{proof} 37 | 38 | \subsection{Classification} 39 | 40 | Let $\abs{G} = pq$ where $p \leq q$. We showed with the Sylow theorems that $P \in \Syl_p(G)$ and $Q \in \Syl_q(G)$ imply that $Q \normal G$. We also showed that if $p$ doesn't divide $q - 1$ then $n_p = 1$ and $P \normal G$. In these cases, $G \cong Z_{pq}$. What happens if $p$ \emph{does} divide $q-1$? In this case we don't need to create a $Z_{pq}$, and this is where semidirect products come in. Here, we can apply the previous theorem to construct groups of the form $Q \rtimes_\phi P$ for some $\phi$. 41 | 42 | Let's look at $\Aut(Q)$, which we know is isomorphic to $(\Z_q)^\times$ so the order is $q-1$. We also know\footnote{since $q$ is prime} that $\Aut(Q)$ is cyclic, so $\Aut(Q) \cong Z_{q-1}$. We know that $p$ divides $q-1$, so there is a unique subgroup $\langle \gamma \rangle \leq \Aut(Q)$ which is isomorphic to $Z_p$. We need to map a generator of $P$ into $\Aut(Q)$, so it must have order $1$ or $p$. Let $P = \langle x \rangle$. We have a few options. We know that $\phi(x) = \gamma^i$ for some $0 \leq i \leq p-1$ since we need $\phi(x)^p = 1$. 43 | \begin{enumerate} 44 | \item The case where $i = 0$, wherein have the trivial action. Then $P$ and $Q$ commute one another since $x \cdot y = y$, and so $G \cong Q \times P$. 45 | 46 | \item The case where $i = 1$. Then $G \cong Q \rtimes_{\phi_1} R$ where $x \cdot y = \gamma(y)$. This involves mapping $x$ to $\gamma$. This semidirect product is nonabelian. What exactly $\gamma$ is depends on $P$ and $Q$, but we konw that it does exist since $p$ divides $q-1$, and so a $p$-order subgroup of $\Aut(Q)$ must exist. 47 | 48 | \item The case where $i > 1$. We know that $x \mapsto \gamma^i$ since $P \cong \langle \gamma \rangle$. Then call $x'$ the element which is mapped to $\gamma$. Then we revert to the case where we defied $\phi_1$ by $x'$, so all cases where $i \not= 0$ are isomorphic, and the exact value depends on which $x'$ you choose. 49 | \end{enumerate} -------------------------------------------------------------------------------- /MATH-350/Lectures/27.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 7 November 2018} 2 | 3 | \subsection{Classify Groups of Order \texorpdfstring{$12$}{12}} 4 | Online. Look it up. 5 | 6 | \subsection{``A Simple Song''} 7 | 8 | This may be the best thing I have ever seen. Lyrics to follow shortly. 9 | 10 | \begin{raggedright} 11 | \raggedright\itshape\small 12 | What are the orders of all simple groups? \\ 13 | I speak of the honest ones, not of the loops. \\ 14 | It seems that old Burnside their orders has guessed. \\ 15 | Except for the cyclic ones, even the rest. \\ \hspace{1em} \\ 16 | 17 | Groups made up with permutes will produce 18 | some more: \\ 19 | For A\textsubscript{n} is simple, if n exceeds 4. \\ 20 | Then, there was Sir Matthew who came into 21 | view \\ 22 | Exhibiting groups of an order quite new. \\ \hspace{1em} \\ 23 | 24 | Still others have come on to study this thing. \\ 25 | Of Artin and Chevalley now we shall sing. \\ 26 | With matrices finite they made quite a list. \\ 27 | The question is: Could there be others they've missed? \\ \hspace{1em} \\ 28 | 29 | Suzuki and Ree then maintained it's the case \\ 30 | That these methods had not reached the end of 31 | the chase. \\ 32 | They wrote down some matrices, just four by 33 | four, \\ 34 | That made up a simple group. Why not make 35 | more? \\ \hspace{1em} \\ 36 | 37 | And then came the opus of Thompson and Feit, \\ 38 | Which shed on the problem remarkable light. \\ 39 | A group, when the order won't factor by two, \\ 40 | Is cyclic or solvable. That's what is true. \\ \hspace{1em} \\ 41 | 42 | Suzuki and Ree had caused eyebrows to raise, \\ 43 | But the theoreticians they just couldn't faze. \\ 44 | Their groups were not new: if you added a twist, \\ 45 | You could get them from old ones with a flick of the wrist. \\ \hspace{1em} \\ 46 | 47 | Still, some hardy souls felt a thorn in their side. \\ 48 | For the five groups of Mathieu all reason defied; \\ 49 | Not $A_n$, not twisted, and not Chevalley, \\ 50 | They called them sporadic and filed them away. \\ \hspace{1em} \\ 51 | 52 | Are Mathieu groups creatures of heaven or hell? \\ 53 | Zvonimir Janko determined to tell. \\ 54 | He found out what nobody wanted to know: \\ 55 | The masters had missed 1 7 5 5 6 0. \\ \hspace{1em} \\ 56 | 57 | The floodgates were opened! New groups were the rage! \\ 58 | (And twelve or more sprouted, to greet the new age.) \\ 59 | By Janko and Conway and Fischer and Held \\ 60 | McLaughlin, Suzuki, and Higman, and Sims. \\ \hspace{1em} \\ 61 | 62 | No doubt you noted the last lines don't rhyme. \\ 63 | Well, that is, quite simply, a sign of the time. \\ 64 | There's chaos, not order, among simple groups; \\ 65 | And maybe we'd better go back to the loops 66 | \end{raggedright} 67 | 68 | \subsection{Introductions to Rings} 69 | 70 | \begin{definition}[Ring] 71 | A \emph{ring} is a tuple $(R,+,\times)$ where $(R,+)$ is an abelian group, $\times$ is an associative binary operaation, and the distributive laws hold. 72 | \end{definition} -------------------------------------------------------------------------------- /MATH-350/Lectures/28.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 9 November 2018} 2 | 3 | \epigraph{``I'm an advertisement for Czech people.''}{Miki} 4 | 5 | \subsection{Warmups} 6 | 7 | \begin{exercise} 8 | Are the following objects rings? 9 | \begin{parts} 10 | \part The set of all even integers with the operations you expect. 11 | \part The set of all odd integers with the operations you expect. 12 | \end{parts} 13 | \end{exercise} 14 | 15 | \begin{solution} 16 | The even integers do form a ring (we don't require a ring to have identity) while the odd integers don't (there is no additive identity and it isn't even closed). 17 | \end{solution} 18 | 19 | \subsection{Quaternions, Division Rings, and Fields, Oh My!} 20 | 21 | \begin{exercise} 22 | What is $\C$? 23 | \end{exercise} 24 | 25 | \begin{solution} 26 | We can think of it as a vector space over $\R$ with a basis $\{1,i\}$, which gives us how to multiply and add things in $\C$ if we keep in mind that $i^2 = 1$. 27 | \end{solution} 28 | 29 | We can extend this idea of $\C$ into a higher-dimensional object $\mathbb{H}$, known as the \emph{Hamiltonian quaternions}. This forms a vector space over $\R$ with a basis $\{1,i,j,k\}$ where the ring multiplication is defined by the group $Q_8$. Some ``fun'' facts about $\mathbb{H}$: 30 | \begin{enumerate} 31 | \item $\mathbb{H}$ is not commutative; 32 | \item $\mathbb{H}$ is a ring with identity. For any $a + bi + cj + dk$, the inverse is $(a - bi - cj - dk)/(a^2+b^2+c^2+d^c)$. 33 | \end{enumerate} 34 | 35 | We can extend these definitions even further. 36 | 37 | \begin{definition}[Division Ring] 38 | A ring $R$ with identity $1$ is a \emph{division ring} if for all nonzero $a \in R$ there exists an $a^{-1} \in R$ 39 | \end{definition} 40 | 41 | \begin{definition}[Field] 42 | A \emph{field} is a commutative division ring. 43 | \end{definition} 44 | 45 | About all rings we can make some claims. 46 | 47 | \begin{proposition} 48 | Let $R$ be a ring. Then 49 | \begin{enumerate} 50 | \item $0a = a0 = 0$ for all $a \in R$; 51 | \item $(-a)b = a(-b) = -(ab)$ for all $a,b \in R$; 52 | \item $(-a)(-b) = ab$ for all $a,b \in R$; 53 | \item If $R$ has an identity the it is unique and satisfies $(-a)1 = 1(-a) = -a$ for all $a \in R$. 54 | \end{enumerate} 55 | \end{proposition} 56 | 57 | \begin{proof} 58 | If you took 230, you've already done this. If not, it's a good if somewhat tedious exercise. 59 | \end{proof} 60 | 61 | \begin{definition}[Zero Divisor] 62 | An nonzero $a \in R$ is a zero divisor if there exists a nonzero $b \in R$ for which $ab = 0$ or $ba = 0$. 63 | \end{definition} 64 | 65 | \begin{definition}[Unit] 66 | Let $R$ be a ring with identity. A unit is any element $x \in R$ if there exists a $u = x^{-1} \in R$ for which $xu = ux = 1$. We denote the set of units of $R$ by $R^\times$. 67 | \end{definition} 68 | 69 | \begin{example} 70 | Consider $R = \Z/6\Z$. Find all units and zero divisors. 71 | \end{example} 72 | 73 | \begin{solution} 74 | The units are $1,5$, and the zero divisors are $2,3,4$. 75 | \end{solution} 76 | 77 | \begin{proof}[Proof that $2$ is not a unit] 78 | Suppose by way of contradiction that there exist $x \in R$ such that $2x = 1$. Then $3(2x) = 3 = (3 \cdot 2)x = 0$, which is a contradiction. Generally, if $x \in R$ is a zero divisor then it is not a unit. However, the converse is not true (consider $2 \in \Z$, which is neither a unit nor a zero divisor). 79 | \end{proof} 80 | 81 | \begin{definition}[Integral Domain] 82 | A commutative ring $R$ with identity is an integral domain if it has no zero divisors. This gives us the cancellation law! [The proof is super short and intuitive.] 83 | \end{definition} 84 | 85 | \begin{proposition} 86 | Any finite integral domain is a field. The proof is pretty chill. Check it out some time! 87 | \end{proposition} 88 | 89 | \begin{definition}[Subring] 90 | A subring $S \subseteq R$ is an object where $(S,+) \leq (R,+)$ and $S$ is closed under multiplication. For example, $2\Z \subset \Z$, $\Z \subset \Q$, $\Q \subset \R$, $\R \subset \C$, the set of continuous functions from $\R$ to $\R$ is a subring of all functions from $\R$ to $\R$. 91 | \end{definition} 92 | 93 | \begin{example} 94 | Consider $\Q(i) = \{a + bi \mid a,b \in \Q\}$ where $i^2 = 1$. We know how to add and multiply this from middle school. This is actually a field equal to $\Q[i]$, where $\Q[i]$ is actually the field of rational polynomials over $1$ and $i$. Now consider $\Z[i]$ (this won't be a field), which is integral polynomials in $1$ and $i$. This isn't a field since $2$ has no inverse. 95 | \end{example} 96 | 97 | \begin{definition}[Norm] 98 | Let the norm of $\Q(i)$ be a map $N : \Q(i) \to \Q$ defined by $a + bi \mapsto (a+bi)(a-bi) = a^2+b^2$. This map is multiplicative. 99 | \end{definition} 100 | 101 | \begin{proposition} 102 | $N(x) \in \Z$ for all $x \in \Z[i]$, and $x$ is a unit if and only if $N(x) = \pm 1$. 103 | \end{proposition} -------------------------------------------------------------------------------- /MATH-350/Lectures/29.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 12 November 2018} 2 | 3 | \begin{example} 4 | Consider $\Z[\sqrt{5}] \subset \Q(\sqrt{5})$ where $N(a+b\sqrt{5}) = a^2 + 5b^2$. Then $(a+b\sqrt{5})^{-1} = (a-b\sqrt{5})/(a^2-5b^2)$ for anything where $a+b\sqrt{5} \not=0$. Then $x \in \Z[\sqrt{5}]$ is a unit if and only if the norm is $\pm 1$. Notice that this method depended on the fact that $5$ was square-free (no repeated primes). 5 | \end{example} 6 | 7 | \subsection{Polynomial Rings} 8 | 9 | Polynomial rings are the focus of Galois Theory which will be covered in depth next semester. Let $R$ be a commutative ring with identity, and form $R[x]$. This is the ring of polynomials with coefficients in $R$. Let $p(x) \in R[x]$ be a nonzero polynomail, and let ${\partial p} = n$ be the degree of the polynomial. We say that $p(x)$ is monic if $a_n = 1$. 10 | 11 | \begin{example} 12 | \begin{enumerate} 13 | \item Let $R = \Z/2\Z$. Then all polynomials have coefficient $1$ or $0$. Then there are $2^n$ distinct polynomials of degree $5$. 14 | 15 | \item Let $R = \Z/4\Z$. This ring has zero devisors, so $(2x)^2 = 0$. Also consider $(2x+1)^2 = 1$, so $2x + 1$ is a unit in $R[x]$. 16 | 17 | \item Notice that $R$ is a subring of $R[x]$ since it's just the ring of constant polynomails. 18 | \end{enumerate} 19 | \end{example} 20 | 21 | \begin{proposition} 22 | Let $R$ be a commutative unital ring which is an integral domain, and let $p,q \in R[x]$. Then the following hold. 23 | \begin{enumerate} 24 | \item $\partial (pq) = \partial p + \partial q$. 25 | \item $R[x]^\times = R^\times$. 26 | \item $R[x]$ is also an integral domain. 27 | \end{enumerate} 28 | \end{proposition} 29 | 30 | \begin{proof}[Proof of (1.) and (3.)] 31 | Let $p = a_0 + \cdots + a_nx^n$ and let $q = b_0 + \cdots + b_nx^m$. Notice that $pq = a_0b_0 + \cdots + a_nb_mx^{n+m}$, and since $a_nb_m \not= 0$ then $\partial(pq) = \partial p + \partial q$. 32 | \end{proof} 33 | 34 | \begin{proof}[Proof of (2.)] 35 | Notice that $R^\times \subset R[x]^\times$ trivially, since we didn't do anything to inverses. To show the other direction, we show that all units of $R[x]$ have degree zero. Suppose $p \in R[x]^\times$, so there exists a $q \in R[x]$ such that $pq = 1$. Since degrees are additive, it must be the case that $\partial(pq) = \partial p + \partial q = 0$ so $\partial p = 0$. 36 | \end{proof} 37 | 38 | \subsection{Matrix Rings} 39 | 40 | Pick a ring $R$, any ring! Also let $n \in \Z_{>0}$. Then $M_n(R)$ is the ring of matricies with entries in $R$. Notice that even in $R$ is commutative, $M_n(R)$ won't be for all $n \geq 2$. Fun fact, if $R \not= 0$ then this ring \emph{will have} zero divisors so it cannot be an integral domain. As an example, consider 41 | \[ 42 | \begin{pmatrix} 43 | 0 & 1 \\ 0 & 0 44 | \end{pmatrix}^2 = 0, 45 | \] 46 | for $n = 2$. Just as $R \subset R[x]$ is a subring, so to $R \subset M_n(R)$, where $r \in R$ is isomorphic to the diagonal matrix with all nonzero entries $r$. The units of $M_n(R)$ form a ring $\mathrm{GL}_n(R) = \left(M_n(R)\right)^\times$. 47 | 48 | \subsection{Group Rings} 49 | 50 | \begin{example} 51 | Let $G = Z_2$ be a group. Consider a ring $R = \Z$. We're gonna stick them together to form $RG$ by ``forming a vector space of $G$ over $R$'' (pretty sure this is actually a module). Then $RG = \{a1 + bx \mid z,b \in \Z\}$ where component addition is inherited from $R$, and components add only if they are the same ``variable,'' and multiplication is inherited from $G$. 52 | \end{example} 53 | 54 | Let $(G, \cdot) = \{g_1, \dotsc, g_n\}$ be any finite group, and let $R$ be any commutative ring with identity. Then we form $RG = \{a_1g_1 + \cdots + a_ng_n \mid a_i \in R\}$ where we add and multiply in the expected way. 55 | 56 | \begin{problem} 57 | What is the copy of $R$ inside of $RG$? It's $\{r1_G\}$, so exactly the constant terms just like in the polynomials. 58 | \end{problem} 59 | 60 | \begin{example} 61 | Let $G = D_6$ and let $R = \Z$. Consider $(3s + 2rs)(r^2 + s) = 3sr^2 + 3s^2 + 2rsr^2 +2rs^2 = 3rs + 4r + s + 3$. This actually has zero divisors! Consider $(1+s)(1-s) = 0$. 62 | \end{example} 63 | 64 | In general, for $g \in RG$ where $\abs{g} = m$, we have that $(1-g)(1 + g + \cdots + g^{m-1}) = 1 - g^{m} = 0$. Then if $m > 0$, $1-g$ is a zero divisor. 65 | 66 | I want to know that the analogous concept to a group action will be. Miki says to ask again in four lectures. 67 | 68 | \subsection{Why doesn't \texorpdfstring{$\mathbb{H} = \R Q_8$}{H = RQ8}?} 69 | 70 | Check the orders. Notice that $\abs{\R Q_8} = 8$ while $\abs{\mathbb{H}} = 4$. This problem arises because $\R Q_8$ treats $i$ and $-i$ as \emph{different variables}. Also, note that there are zero divisors in $\R Q_8$, but there are no zero divisors in $\mathbb{H}$ since it is a division ring. In fact, since we already proved that $RG$ contains zero divisors in $\abs{G} > 1$ then we know that $\mathbb{H} \not= RG$ for \emph{any} ring $R$ and group $G$. -------------------------------------------------------------------------------- /MATH-350/Lectures/3.tex: -------------------------------------------------------------------------------- 1 | \section{September 4, 2018} 2 | 3 | \begin{definition} 4 | For a set $\Omega$, the symmetric group on $\Omega$ is $S_\Omega = \{\text{bijective maps $\Omega \to \Omega$}\}$. For $n \in \N$, we say that $S_n = S_{\{1,\dotsc,n\}}$. This is usually called the symmetric group on $n$ letters. 5 | \end{definition} 6 | 7 | Let's consider this example for $S_4$ (warning, there's some cyclic decomposition for $g_1, g_2$?) 8 | 9 | \begin{example} 10 | Consider the following maps $g_1, g_2 \in S_4$, 11 | \[ 12 | \begin{array}{c} 13 | g_1 \\ 14 | 1 \to 2 \\ 15 | 2 \to 1 \\ 16 | 3 \to 4 \\ 17 | 4 \to 2 18 | \end{array} \qquad 19 | \begin{array}{c} 20 | g_2 \\ 21 | 1 \to 3 \\ 22 | 2 \to 1 \\ 23 | 3 \to 2 \\ 24 | 4 \to 4 25 | \end{array} 26 | \] 27 | We can also write these as $g_1 = (12)(34)$ and $g_2 = (132)(4)$. In this notation, how to we multiply things? E.g., what is $g_2g_1$? Well, we can write this naïvely as $(132)(4)(12)(34)$, but we don't want to repeat any numbers. Let's see what happens to $1$: 28 | \[ (132)(4)(12)(34) \cdot 1 = (132)(4)(12) \cdot 1 = (132) \cdot 2 = 1. \] For $2$, we get 29 | \[ (132)(4)(12)(34) \cdot 2 = 3. \] For $3$, this comes $g_2g_1 \cdot 3 = 4$, and for $4$ we have $g_2g_1 \cdot 4 = 2$. Then $g_2g_1 = (1)(234)$. Unfortunately, doing this sort of element-wise reduction is the fastest way to multiply anything. 30 | \end{example} 31 | 32 | \begin{problem} 33 | Someone asked the question ``does order matter?'' E.g., is it true that $(12)(34) = (34)(12)$ always? 34 | \end{problem} 35 | \begin{solution} 36 | No. They are the same. Also, $(abc) = (bca)$; as long as the sign of the permutation of the cycle elements is $+1$, it won't matter how you order the elements of a cycle. 37 | \end{solution} 38 | 39 | \begin{problem} 40 | Does order matter when there is a number repeated (when the cycles are not disjoint)? E.g. does $g_1g_2 = g_2g_1$? 41 | \end{problem} 42 | \begin{solution} 43 | Yeah, order does matter. Consider that $(12)(13) \not= (13)(12)$. This means that, in general, $S_n$ is not abelian. 44 | \end{solution} 45 | 46 | \begin{problem} 47 | Consider $S_5$, where $g = (123)(45)$ and $h = (12345)$. Find $g^2, g^{-1}, h^{-1}$. Fun fact, it's easy. 48 | \end{problem} 49 | 50 | These facts lead us to an interesting and useful conclusion. 51 | \begin{proposition} 52 | For any $g \in S_n$, we can write $g$ as a product of disjoint cycles. 53 | \end{proposition} 54 | 55 | This gives us an interesting observation for $S_n$. 56 | 57 | \begin{proposition} 58 | Let $g \in S_n$ be written as the product of disjoint cycles. Then the order of $g$ is the least common multiple of the orders of the disjoint cycles. 59 | \end{proposition} 60 | 61 | \subsection{Fields \texorpdfstring{$n$}{n} stuff} 62 | 63 | \begin{definition} 64 | A field $k$ is a triple $(F, +, \times)$ where $(F,+)$ and $(F \setminus \{0\},\times)$ are groups where $F^\times = F \setminus \{0\}$ and where multiplication distributes over addition. Some cannonical examples are $\Q$, $\R$, $\C$, $\F_p = \Z/p\Z$ for prime $p$. 65 | \end{definition} 66 | 67 | A brief note on finite fields: for a finite field $\F$, we know that $\abs{\F} = p^n$ for some prime $p$ and some $n \geq 1$. 68 | 69 | Now that we have fields, we can get matrices for free. Consider the cannonical matrix group $\mathrm{GL}_n(k)$ of invertible matrices with entries in $k$. 70 | 71 | \begin{example} 72 | Consider $\mathrm{GL}_2(\F_2)$ where $\F_2 = \{\bar{1}, \bar{2}\}$ (note that this is just $\Z / 2\Z$). What is the order of $\mathrm{GL}_2(\F_2)$? 73 | \end{example} 74 | 75 | \begin{proof} 76 | There are six. Any element cannot have three or four zeros in it, nor two zeros in the same row or column. Then just count the total possibilities. 77 | \end{proof} -------------------------------------------------------------------------------- /MATH-350/Lectures/30.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 16 November 2016} 2 | 3 | \epigraph{``No one really cares about left ideals.''}{Miki} 4 | 5 | \begin{problem}[Challenge Problem] 6 | What is $\Aut(Z_6 \times Z_2)$? 7 | \end{problem} 8 | 9 | \subsection{Ring Maps} 10 | 11 | \begin{definition}[Ring Homomorphisms] 12 | A \emph{ring homomorphism} s a map $\phi : R \to S$ where $\phi(a + b) = \phi(a) + \phi(b)$ and $\phi(ab) = \phi(a)\phi(b)$. The kernel of $\phi$ is the underlying group kernel; that is, $\ker \phi = \{a \in R \mid \phi(a) = 0\}$. If $\phi$ is bijective it is a ring isomorphism. 13 | \end{definition} 14 | 15 | \begin{example} 16 | \begin{enumerate} 17 | \item Consider $\Z \to \Z/n/Z$ via $a \mapsto \bar{a}$. Then $\bar{a}\bar{b} = \overline{ab}$ and $\bar{a} + \bar{b} = \overline{a + b}$. 18 | \item Consider $\Q[x] \to \Q$ via $f(x) \mapsto f(0)$. That is, we evaluate the function at zero. Multiplying and adding polynomials works so this map is a homomorphism. 19 | \end{enumerate} 20 | \end{example} 21 | 22 | \begin{proposition} 23 | If $\phi : R \to S$ is a homomorphism then $\Im(\phi)$ is a subring of $S$ and $\ker \phi$ is a subring of $S$. 24 | \end{proposition} 25 | 26 | \begin{corollary} 27 | The important fact about the kernel is that if $a \in \ker\phi$ then $\phi(a)\phi(x) = 0$ for \emph{any} $x \in R$. This is a really strong property. It means that the kernel isn't just closed under multiplication, its closed under multiplication by \emph{anything}. 28 | \end{corollary} 29 | 30 | \begin{definition}[Ideals] 31 | A left ideal $I \subseteq R$ is a subring of $R$ with the property that $rI \subseteq I$ for all $r \in R$. A similar definition exists for a right idea, where $Ir \subseteq I$ for all $r \in R$. If $I$ is both a left and right ideal, then it is just called an \emph{ideal}. 32 | \end{definition} 33 | 34 | \begin{example} 35 | For any ring $R$ and any homomorphism $\phi$, then $\ker \phi$ is an ideal in $R$. 36 | \end{example} 37 | 38 | \subsection{Quotient Rings} 39 | 40 | Consider $\phi : \Z \mapsto \Z/2\Z$. Then let $\ker\phi = I$, so $(I,+) \leq (\Z,+)$. We know that we have two cosets, $0 + I$ and $1 + I$ which are derived from our cosets of $\Z/2\Z$. Then we can form a new object, $\Z/I$. We know from group structure that the multiplication is well defined, and multiplication also works since $\textsc{odd} \times \textsc{even} = \textsc{even}$ and $\textsc{even} \times \textsc{even} = \textsc{even}$. 41 | 42 | \begin{example}[Example where this doesn't work] 43 | Let $R = \Z[x]$ and $S \subset R$ be the polynomials in $x^2$, so $s = 1 + x^2 + 3x^6$, for example. Notice that $R/S$ doesn't work. For example, we want that $\bar{1} \cdot \bar{x} = \bar{x}$. But $(1 + x^2) \cdot x = x + x^3$ which is not in $x + S$. 44 | \end{example} 45 | 46 | The punchline for this is that quotient rings only work when $S$ is an ideal. This is kind of like the condition for normality for groups. For any $S \subseteq R$ we know that $(R/S,+)$ is okay since $(S,+) \normal (R,+)$ since $R$ is commutative in addition, but the multiplication is where we get tripped up. For any $r_1,r_2 \in R$ and any $a \in S$ we want that $r_1(r_2 + a) \in r_1r_2 + S$. But $r_1(r_2 + a) = r_1r_2 + r_1a$ which means that $r_1a$ must be in $S$ for \emph{any} $r_1$ which happens only if $S$ is a left ideal. Similarly, we need that $(r_1 + a)r_2 = r_1r_2 + ar_2$, which tells us that $S$ must be a right idea, so it must be an idea. 47 | 48 | \begin{proposition} 49 | If $I \subseteq R$ is an ideal then $R/I$ is a ring and $(r + I) + (s + I) = (r+s) + I$ and $(r+I) + (s + I) = rs + I$. 50 | \end{proposition} 51 | 52 | \begin{example} 53 | \begin{parts} 54 | \part Consider $\Q[x] \to \Q$ again where $p(x) \mapsto p(0)$. The kernel of this is the set of polynomials with a zero constant term. Notice then that $\Q[x]/\ker\phi \cong \Q$ which looks a hell of a lot like the first isomorphism theorem for groups. 55 | \part Consdier $\Q[x]$ and $I = \{\text{polynomials of degree $\geq 2$}\}$. This doesn't work out well. 56 | \end{parts} 57 | \end{example} 58 | 59 | \subsection{Using Quotient Rings} 60 | 61 | Consider $\pi : \Z \to \Z/p\Z = \F_p$. This induces a map $\Z[x] \to \F_p[x]$ where $a_nx^n \mapsto \bar{a}_nx^n$ (looking at the coefficients modulo $p$). Suppose there exists an $a \in Z$ such that $f(a) = 0$. Then $\bar{f}(\bar{a}) = \bar{0}$. 62 | 63 | \begin{example} 64 | Consider $f(x) = 3x^{52} + 4x^{49} - 16x^{20} - 17x - 1755$. Let's look at it modulo $2$. Then $\bar{f}(x) = x^{52} + x + 1$. Then $\bar{f}(\bar{0}) \not= \bar{0}$ and $\bar{f}(\bar{1}) \not= \bar{0}$ so there are no integral roots. 65 | \end{example} 66 | 67 | \begin{problem} 68 | If we have a ring homomorphism $\phi : R \to S$, can we always induce a homomorphism $\psi : R[x] \to S[x]$. 69 | \end{problem} -------------------------------------------------------------------------------- /MATH-350/Lectures/31.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 26 November 2018} 2 | 3 | \epigraph{``I can hear your excitement all the way over hear.''}{Miki} 4 | 5 | \subsection{Warm-Up} 6 | 7 | Let $I$ and $J$ be ideals of $R$. Then 8 | \begin{itemize} 9 | \item $I +J = \{i+j \mid i \in I, j \in J\}$, 10 | \item $IJ = \{ij \mid i \in I, j \in J\}$, 11 | \item $I^n = I \cdots I$. 12 | \end{itemize} 13 | 14 | \begin{problem}[Warm-Up] 15 | Let $R = \Z$ and let $I = 2\Z$ and let $J = 3\Z$. 16 | What are the following objects? 17 | \begin{parts} 18 | \part $I+J = \Z$; 19 | \part $IJ = 6\Z$; 20 | \part $I^n = 2^n\Z$. 21 | \end{parts} 22 | \end{problem} 23 | 24 | \subsection{Ideals Generated by Subset} 25 | 26 | For all of today, let $R$ have $1 \not= 0$. 27 | 28 | \begin{definition}[Generated Ideal] 29 | Let $A \subset R$ be a subset. Then $(A)$ is the smallest ideal containing $A$, also called the \emph{ideal generated by $A$}. Then let $RA = \{r_1a_1 + \cdots r_na_n \mid r_i \in R, a_i \in A\}$, and analogously construct $AR$. These are the left and right ideals generated by $A$, respectively. Then $RAR$ is the ideal generated by $A$, so $(A) = \bigcap I$ where $A \subset I$. 30 | \end{definition} 31 | 32 | \begin{definition}[Principal Ideal] 33 | An ideal $I \subset R$ is finitely generated if there exists an $A \subset I$ with $I = (A)$. An ideal is principal if there exists an $a \in I$ such that $I = (a)$, so $I$ is finitely generated by a single element. 34 | \end{definition} 35 | 36 | \begin{example} 37 | \begin{parts} 38 | \part Let $R = \Z$. An ideal $I \subset R$ is an abelian subgroup, which means that $I = n\Z$ for some $n \in \Z$. Then all ideals of $\Z$ are of the form $(n)$, and so all are principal. 39 | 40 | \part Let $R = \Z[x]$, and let $I = (2,x)$. This generates any polynomial which is a multiple of $x$, and if it has a constant term it must be even, so $RA = 2 \cdot p(x) + x \cdot q(x)$ for any integral polynomials $p,q$. Does there exist a single generator of $(2,x)$? Suppose that $I = (a(x))$. Then $2 = a(x)b(x)$ for some $b(x) \in \Z[x]$. Since degrees add in an integral domain, it must be that $a(x)$ is constant and $\partial a = 0$. Then $a = \pm 1$ or $\pm 2$. Since $I \not= R$ $a \not= \pm 1$, and if $a = \pm 2$ then $x \notin (a)$, so $(2,x)$ is not principal. 41 | \end{parts} 42 | \end{example} 43 | 44 | \begin{definition}[Maximal Idea] 45 | An ideal $M \subset R$ is maximal if $M \not= R$ and the only ideals containing $M$ are $M$ and $R$. 46 | \end{definition} 47 | 48 | \begin{corollary} 49 | The only maximal ideals of $\Z$ are $(p)$ for some prime $p$. 50 | \end{corollary} 51 | 52 | \begin{proposition} 53 | If $I \subset R$ is a proper ideal then there exists a maximal ideal $M \subset R$ such that $I \subset M$. 54 | \end{proposition} 55 | 56 | \begin{proof}[Proof in $\Z$] 57 | Take any prime factor of $n$, so that $(n) \subset (p)$. 58 | \end{proof} 59 | 60 | \subsection{Fields} 61 | 62 | \begin{proposition} 63 | Let $R$ be a commutative ideal. Then $R$ is a field if and only if $R$ has exactly two ideals, $(0)$ and $R$. 64 | \end{proposition} 65 | 66 | \begin{proof} 67 | First, assume that $R$ is a field. Note that $0 \not= 1$ so $(0)$ and $R$ are two distinct ideals. Then let $I \not= (0)$ be an ideal of $R$. Then there exists some $a \not= 0 \in I$. Since $R$ is a field, every nonzero element has an inverse, so $a \in R \setminus \{0\} = R^\times$. Then $I = R$, since $aa^{-1} = 1$ which generates $R$. 68 | 69 | In the other direction, assume that $R$ has exactly two ideals. Let $a \in R\setminus \{0\}$, and let $I = (a)$. Since $a \not= 0$, it must be that $I = R$. Then $1 \in (a)$, which means that $1$ is a multiple of $a$, which means that $a$ has an inverse for any nonzero $a$. Then $R^\times = R \setminus \{0\}$, so $R$ is a field. 70 | \end{proof} 71 | 72 | \begin{corollary} 73 | If $\phi : R \to S$ is a homomorphism where $R$ is a field, then $\phi$ is the zero map or $\phi$ is injective. 74 | \end{corollary} 75 | 76 | \begin{proof} 77 | Notice that $\ker \phi$ is an ideal of $R$, so it is $(0)$ or $R$. 78 | \end{proof} 79 | 80 | \begin{proposition} 81 | Let $R$ be commutative and let $I \subset R$ be a proper ideal. Then $I$ is maximal if and only if $R/I$ is a field. 82 | \end{proposition} 83 | 84 | \begin{proof} 85 | $R/I$ is a field if and only if there are exactly two ideals. Then the ideals of $R/I$ correspond one to one to the ideals of $R$ which contain $I$, by the second/third? isomorphism theorem. Then there can only by two ideals of $R$ which contain $I$ which must be $R$ and $I$, so $I$ is maximal in $R$. 86 | \end{proof} 87 | 88 | -------------------------------------------------------------------------------- /MATH-350/Lectures/32.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 28 November 2018} 2 | 3 | \epigraph{``You have nothing.''}{Miki} 4 | 5 | \subsection{Prime Ideals} 6 | 7 | Recall the following results from last lecture about commutative unital rings. 8 | \begin{itemize} 9 | \item $R$ is a field if and only if $R$ has exactly two ideals, $0$ and $R$. 10 | \item $I \subset R$ is maximal if and only if $R/I$ is a field. 11 | \end{itemize} 12 | 13 | \begin{definition}[Prime Ideal] 14 | An ideal $I \subset R$ is \emph{prime} if $I \not= R$ and if $a,b \in R$ and $ab \in R$ then $a \in I$ or $b \in I$. 15 | \end{definition} 16 | 17 | \begin{example} 18 | \begin{parts} 19 | \part Consdier that $I = (6) \subset \Z$ is not prime since $2 \cdot 3 = 6 \in I$ but $2,3 \notin I$. 20 | \part Consider that $I = (2) \subset \Z$ is prime since if a number is even it is divisible by two (you can't ``factor'' two into anything). 21 | \end{parts} 22 | \end{example} 23 | 24 | There is a direct correspondence between integral domains and prime idals. For example, consider that $\Z/6\Z$ is not an integral domain since $\bar{2}\cdot\bar{3} = \bar{0}$ while $\Z/2\Z$ is an integral domain. 25 | 26 | \begin{proposition} 27 | Let $R$ be a commutative ring with unit, and let $I \subseteq R$ be an ideal. Then $I$ is prime if and only if $R/I$ is an integral domain. 28 | \end{proposition} 29 | 30 | \begin{proposition} 31 | Every maximal ideal of a nonzero commutative ring is prime. 32 | \end{proposition} 33 | 34 | \begin{proof} 35 | We proved last time that $M$ is maximal if and only if $R/M$ is a field, which is an integral domain. Then this follows from the above proposition. 36 | \end{proof} 37 | 38 | \subsection{The Chinese Remainder Theorem} 39 | 40 | \begin{theorem} 41 | If $m,n$ are relatively prime then $\Z/mn\Z$ is isomorphic to $\Z/m\Z \times \Z/n\Z$. 42 | \end{theorem} 43 | 44 | \subsection{Rings of Fractions} 45 | 46 | How do we make $\Q$ from $\Z$. Well, we can define $\Q$ as $\{(a,b) \mid a,b \in \Z, b \not= 0\}$ with an equivalence relation that $(a,b) \sim (c,d)$ if and only if $ad = bc$. 47 | 48 | Can we do this with any (commutative) ring? 49 | 50 | \begin{example} 51 | Suppose that $R = \Z/6\Z$, and we generate a new ring analogously by how we made $\Q$. Notice that $2/1 = 2 \cdot 3 / 3 = 0 / 3 = 0$ which is really terrible. Now Miki is sad. 52 | \end{example} 53 | 54 | We get around this problem by just not using any zero divisors. 55 | 56 | \begin{definition}[Multiplicative Subset] 57 | A subset $D \subseteq R$ is \emph{multiplicative} if 58 | \begin{itemize} 59 | \item $D$ is closed under multiplication; 60 | \item $D$ does not contain $0$; 61 | \item $D$ does not contain any zero divisors from $R$. 62 | \end{itemize} 63 | \end{definition} 64 | 65 | \begin{theorem} 66 | Let $R$ be a commutative ring and let $D \subset R$ be multiplicative and nonempty.Then there exists a commutative ring $Q = D^{-1}R$ which contains $R$ and for all $a \in D$ there exists an $a^{-1} \in Q$. We construct this as 67 | \[ D^{-1}R = \{(a,b) \mid a \in R, b \in D\}, \] 68 | along with the equivalence relation $(a,b) \sim (c,d)$ if and only if $ad = bc$. Then we define multiplication as $(a,b)(c,d) = (ac,bd)$ and addition as $(a,b) + (c,d) = (ad + bc, bd)$. 69 | \end{theorem} 70 | 71 | How do we construct an injection from $R$ to $D^{-1}R$? We let $d \in D$ be arbitrary, and use the map $r \mapsto (rd,d)$. This is a general case of $r \mapsto (r,1)$ with $d = 1$, which only works if $R$ is unital. 72 | 73 | \begin{definition}[Field of Fractions] 74 | If $R$ is an integral domain and $D = R \setminus \{0\}$ then $D^{-1}R$ is known as a \emph{field of fractions}, denoted $\operatorname{Frac}(R)$. 75 | \end{definition} 76 | 77 | \begin{corollary} 78 | $\operatorname{Frac}(R)$ is the smallest field containing $R$. 79 | \end{corollary} 80 | 81 | \begin{example} 82 | \begin{parts} 83 | \part $\operatorname{Frac}(\Z) = \Q$. 84 | \part $\operatorname{Frac}(\Q) = \Q$ since $\Q$ is already a field. 85 | \part Let $R = \Z[x]$. Then $\operatorname{Frac}(R) = p(x),q(x) \in \Z[x]$ where $q \not= 0$. 86 | \part Let $R = 2\Z$. Then $\operatorname{Frac}(2\Z) = \Q$. 87 | \part Let $R = \Q[x]$. Then $\operatorname{Frac}(\Q[x])$ is the same as $\operatorname{Frac}(\Z[x])$. 88 | \end{parts} 89 | \end{example} -------------------------------------------------------------------------------- /MATH-350/Lectures/33.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 30 November 2018} 2 | 3 | Didn't take notes :( -------------------------------------------------------------------------------- /MATH-350/Lectures/34.tex: -------------------------------------------------------------------------------- 1 | \section{Monday, 3 December 2018} 2 | 3 | \epigraph{``Proof: You just monkey around with it.''}{Miki} 4 | 5 | Suppose that $R$ is an integral domain. 6 | 7 | \begin{definition}[Reducible] 8 | An element $r \in R$, $r \not= 0$, $r \in R^\times$ is reducible if there exist $a,b \in R \setminus R^\times$ such that $ab = r$. Otherwise, the element is irreducible. 9 | \end{definition} 10 | 11 | \begin{definition} 12 | An element $r \in R$ is prime if $(r)$ is a prime idea. This means that if $r$ divides $xy$ then it must divide $x$ or $y$. 13 | \end{definition} 14 | 15 | \begin{definition} 16 | An elemenet $r$ is associate to $s$ if there exists a $u \in R^\times$ such that $r=us$. 17 | \end{definition} 18 | 19 | \begin{proposition} 20 | Let $R$ be an integral domain. If $p \in R$ is prime then $p$ is irreducible. 21 | \end{proposition} 22 | 23 | \begin{proof} 24 | Suppose that $(p)$ is a prime ideal. Take any $a,b \in R$ such that $p = ab$. We want to show that at least one is a unit. If $ab = p$ then $ab \in (p)$, which means that either $a \in (p)$ or $b \in (p)$ since $(p)$ is a prime ideal. Without a loss of generality, let $a \in (p)$. Then $a = px$ for some $x \in R$. Then $p = ab = pxb$. Since $R$ is an integral domain, so $p(1-vb) = 0$ implies that $xb = 1$ which implies that $b \in R^\times$. 25 | \end{proof} 26 | 27 | \begin{proposition} 28 | There exist irreducible elements which are not prime. 29 | \end{proposition} 30 | 31 | \begin{proof} 32 | Consider $\Z[\sqrt{-5}]$ with norm $N(a+b\sqrt{-5}) = a^2 + 5b^2$. First, we show that $3$ is irreducible. Suppose by way of contradiction that $3=ab$ for non-unit $ab$. Since this can't happen in $\Z$, we can assume that at least one of $a,b$ can be written as $x + y\sqrt{5}$ where $y \not=0$. Note that $N(3) = N(ab) = 9$, and since $b$ is not a unit its norm must be at least $2$, while the norm of $ab$ is then at least $10$, so $3$ is irreducible. Next, we show that $3$ is not prime. Note that $(1+\sqrt{-5})(1+\sqrt{-5}) = 6 \in (3)$, but $1 \pm \sqrt{-5} \notin (3)$ since $N(1 \pm \sqrt{-5}) = 6$ while $N((3)) = 9$. 33 | \end{proof} 34 | 35 | \begin{proposition} 36 | If $R$ is a PID then $p \in R^\times, p \not= 0$ is irreducible implies that $p$ is prime. 37 | \end{proposition} 38 | 39 | \begin{proof} 40 | Assume that $p$ is irreducible. We will show that $(p)$ is maximal which, in a PID, implies that it is prime. Let $I$ be an ideal such that $(p) \subset I \subset R$. Since $R$ is a PID, $I = (x)$ for some $x \in R$. If $p \in I$ then $p = rx$ for some $r \in R$. Since $p$ is irreducible, either $x$ or $r$ is a unit. Suppose that $r \in R^x$. Then $pr^{-1} = x \implies x \in (p)$, so $I \subset (p)$ and $I = (p)$. On the other hand, if $x \in R^\times$ and $(x) = I$ then $I = R$ since units generate the whole ring. 41 | \end{proof} 42 | 43 | \begin{definition}[Unique Factorization Domain] 44 | A Unique Factorization Domain (UFD) is an integral domain where for every nonzer0 element $a \in R \setminus R^\times$ can be written as a finite product of irreducible elements, and this decomposition is unique up to order and associates. 45 | \end{definition} 46 | 47 | \begin{example} 48 | \begin{parts} 49 | \part $\Q$ and $\Q[x]$ are both UFDs. 50 | \part $\Z[\sqrt{-5}]$ is \emph{not}, since $6 = 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5})$. 51 | \part $\Q[x_1, x_2, \dots]/(x_1-x_2^2, x_2-x_3^2, \dots)$ is \emph{not} a UFD since $x_1$ has \emph{no irreducible decomposition oh god why is this allowed to happen???} 52 | \end{parts} 53 | \end{example} 54 | 55 | \begin{proposition} 56 | If $R$ is a UFD then every irreducible element is prime. 57 | \end{proposition} 58 | 59 | \begin{proof} 60 | Suppose that $x$ is irreducible, and suppose that $x$ divides $ab$, where neither are units. We can factor $a$ and $b$ into irreducible $a = a_1 \cdots a_k$ and $b = b_1 \cdots b_\ell$, and $x$ divides $a_1 \cdots a_k b_1 \cdots a_\ell$. Since this is unique, and $x$ is irreducibe, $x$ must be associate to one of these elements, it must divide either $a$ or $b$. 61 | \end{proof} 62 | 63 | \begin{proposition} 64 | All PIDs are UFDs. 65 | \end{proposition} 66 | 67 | \begin{proposition} 68 | $\Z$ is a UFD. 69 | \end{proposition} 70 | 71 | \begin{proof} 72 | $\Z$ is a PID. 73 | \end{proof} 74 | 75 | \begin{proposition} 76 | GCDs don't imply a Euclidean Domain. 77 | \end{proposition} -------------------------------------------------------------------------------- /MATH-350/Lectures/35.tex: -------------------------------------------------------------------------------- 1 | \section{Wednesday, 5 December 2018} 2 | 3 | \epigraph{``I'm trying to think of how this is \emph{good} news...''}{Miki} 4 | 5 | \epigraph{``You cannot send your computer to have tea with the bank's computer and discuss encryption.''}{Miki} 6 | 7 | \subsection{A Sketch of Real Life} 8 | The fact that authors define rings differently can really screw things up. For example, if a subring doesn't need to have the same identity as $R$, we could get something like $x$ is the identity in $S = \Z[x]/(x^2-x)$ but $x$ is not the identity in $\Z[x]$. In Galois Theory, it's common to say that $2x+2$ is irreducible even though by our definition it is reducible into $2(x+1)$ since our definition coincides with definitions of what ``prime'' means. 9 | 10 | \textsc{tl;dr:} Real life is messy. 11 | 12 | \subsection{Cryptography and RSA} 13 | 14 | RSA is a security protocol that we use to communicate securely, and it hopefuly won't be broken any time soon (spoiler, it probably will). The advantage is that it allows us to choose our encryption scheme in a very public manner while still remaining secure. 15 | 16 | \begin{theorem} 17 | Let $m$ and $n$ be relatively prime to one another. Then $m^{\varphi(n)} = 1 \pmod{n}$. Recall that $\bar{m} \in (\Z/n\Z)^\times$. 18 | \end{theorem} 19 | 20 | \begin{proposition} 21 | If $n = p$ is prime, and you take $k \in \Z$, then $m^{k\varphi(n) + 1} = m \pmod{n}$. 22 | \end{proposition} 23 | 24 | \begin{example}[Non RSA Example] 25 | Let $n = p = 11$, and let $\lambda = \varphi(n) = 10$. Pick a unit $e \in (\Z/\lambda\Z)^\times = \{1,3,7,9\}$ (say $e = 7$). Calculate the inverse of $e$. sp $d = e^{-1} = 3$ in our example. 26 | 27 | \noindent 28 | Let $(n,e)$ be \textsc{public information} and keep $d$ as \textsc{private information}. 29 | 30 | \noindent 31 | Let your message be an integer $0 \leq m < n$, and encode it by $c = m^e$. 32 | 33 | \noindent 34 | Decode it by $c^d = m^{ed}$. Since $ed \equiv 1 \pmod{\varphi(n)}$, $c^d = m$. 35 | 36 | \noindent 37 | This is really stupid from a security perspective since we can easily calculate $d$ from $(n,e)$, but it illustrates how the encryption-decryption scheme works. 38 | \end{example} 39 | 40 | \begin{proposition} 41 | Let $n = pq$ where $p,q$ are prime, and let $k \in \Z$. Then $m^{k \varphi(n) + 1} \equiv m \pmod{n}$. This is very similar to the previous proposition except that now we don't care that $m$ and $n$ are relatively prime. 42 | \end{proposition} 43 | 44 | \begin{theorem}[RSA] 45 | Let $n = pq$ where $p,q$ are distinct primes with very many digits. Note that $\lambda = \varphi(n) = (p-1)(q-1)$, which is very easy to calculate if you know that $p$ and $q$ are but very hard otherwise. Then pick an $e \in (\Z/\lambda\Z)$. Picking this isn't always trivial since we don't know how to factor quickly, so we turn to the Euclidean algorithm to tell us if a chosen $e$ is actually relatively prime to $\lambda$. This is very efficient in the number of digits, so it's okay from a use perspective. Then, we need to find a $d = e^{-1} \in (\Z/n\Z)^\times$. Since we have the Euclidean Algorithm, we know that $1 = xe + y\lambda$, and modulo lambda we know that $d = x$. Then our \textsc{public information} is again $(n,e)$ and our \textsc{private information} is $d$, but now it is extremely hard to calculate $d$ from $(n,e)$. Then you can post your public key online and take your message $M$ and break it into $\{m\}$ where $0 \leq m < n$ for all $m$, and encrypt each part of the message in turn by $c = m^e$. Then we can decode it by $m = c^d$. 46 | \end{theorem} -------------------------------------------------------------------------------- /MATH-350/Lectures/36.tex: -------------------------------------------------------------------------------- 1 | \section{Friday, 7 December 2018} 2 | 3 | \epigraph{``$a$ is a bad person who lies to you.''}{Miki} 4 | 5 | The last day of class :( 6 | 7 | \subsection{RSA Continued} 8 | 9 | The real security of RSA is the fact that it is very hard to factor large numbers, so long as we can choose our initial primes in a good manner. This involves determining if a number is prime, which is suprisingly nontrivial. There are a couple of ways to do this. One is due to Fermat: 10 | \begin{enumerate} 11 | \item Take $n$ and hope it's prime. 12 | \item if $n$ is prime, then for any $a \in \Z$ where $(a,n) = 1$ then $a^{n-1} \equiv 1 \pmod{n}$. 13 | \item Suppose on the other hand that $n$ is composite. Take $0 \leq a \leq n$ and evaluate $a^{n-1} \pmod{n}$. If you get something other than $1$, you know that $n$ isn't prime, and you call $a$ a Fermat witness. If you \emph{do} get $1$, then you don't know that it's not prime, and you call $a$ a Fermat liar since it lies to you that $n$ is prime, like a little liar. 14 | \item Repeat this process a bunch of times. If you keep getting $1$, then maybe $n$ is prime after all. 15 | \end{enumerate} 16 | This is a probabalistic test for how likely it is that $n$ is prime. How good of an estimate this is relates to how many liars there could be. For ``most'' $n$, at most half of the numbers in the range $0 < a < n$ are liars, so the chance of picking $k$ liars is around $(1/2)^k$. We should qualify what ``most $n$'' actually means. In fact, there exist $n$ called \emph{Carmichael numbers}, which have the terrible property that any $a$ which is relatively prime to $n$ is a Fermat liar. This means that if you choose such an $n$ your kinda screwed if you use this test. 17 | 18 | \begin{theorem} 19 | A number $n$ is Carmichael if and only if it is square-free and if $p$ is a prime which divides $p-n$ then $p-1$ divides $n-1$. 20 | \end{theorem} 21 | 22 | This classification doesn't help you if you can't factor $n$, but it does give us an idea of how many Carmichael numbers there are. 23 | 24 | \subsection{Fermat's Theorem} 25 | 26 | \begin{theorem} 27 | Let $p$ be an odd prime. Then there exist integers $a,b$ such that $p = a^2 + b^2$ if and only if $p \equiv 1 \pmod{4}$. 28 | \end{theorem} 29 | 30 | \begin{proof}[Forward direction] 31 | Suppose that $p$ is a sum of squares. If $x \in \Z$ then $x^2 \equiv 1$ or $0$ modulo $4$, which we check by squaring all $\bar{x} \in \Z/4\Z$. Since $p$ is the sum of two squares, $p$ is either $0$, $1$, or $2$ modulo $4$, and since $p$ is odd it must be $1 \pmod{4}$. 32 | \end{proof} 33 | 34 | Brief interlude. Recall when we constructed numbers like $\Z[\sqrt{D}] = a + b\sqrt{D}$, and wiht then we had a norm $N(a + b\sqrt{D}) = (a + b\sqrt{D})(a - b\sqrt{D}) = a^2-b^2D$. We showed that $N$ might not be a norm since $N(x)$ could be less than $0$, but we can fix this by just taking the absolute value of this. We also showed that $N$ is multiplicative and that $N(x) = \pm 1$ if and only if $x$ is a unit. It's also worth remembering that not every norm makes $\Z[\sqrt{D}]$ a Euclidean domain. 35 | 36 | \begin{proof}[Reverse Direction] 37 | Assume that $p \equiv 1 \pmod{4}$. First, let's construct $R = \Z[i]$ and $N(a+bi) = a^2 + b^2$. We claim that $p$ is nor a prime element in $R$, so there exist $x,y$ such that $p$ divides $xy$ but neither one individually. We know that $p \equiv 1 \pmod{4}$ so $4$ divides $p-1$. Consdier $(\Z/p\Z)^\times \cong Z_{p-1}$. Then there exists an $n \in \Z$ such that $\bar{n}$ has order $4$ in $(\Z/p\Z)^\times$. Then if $\abs{\bar{n}} = 4 \pmod{p}$ then $\abs{\bar{n}^2} = 2 \pmod{p}$ so $\bar{n}^2 = -1 \pmod{4}$ and $p$ divides $n^2 + 1$. In the Gaussian integers, then we know that $p$ divides $n^2 + 1 = (n-i)(n+i)$ but $p$ does not divide $n \pm i$. 38 | 39 | Note that $R$ is a PID, which implies that all irreducible elements are prime, so $p$ is not irreducible in $R$, o there exist $x,y \in R$ such that $x,y \notin R^\times$ but $xy = p$. Recall that $N(p) = p^2 = N(x)N(y)$, and we know that $N(x),N(y) \not= \pm 1$. Then both are either $+p$ or $-p$. We can write $x$ as $a+bi$, so $N(x) = a^2 + b^2 = p$. 40 | \end{proof} 41 | 42 | \subsection{The End} 43 | 44 | That's all, folks. Thanks to Miki for a great semester! -------------------------------------------------------------------------------- /MATH-350/Lectures/4.tex: -------------------------------------------------------------------------------- 1 | \section{September 7, 2018} 2 | 3 | Some facts about finite fields. 4 | \begin{enumerate} 5 | \item For all prime $p$, there exists a field $\F_p$ where $\abs{\F_p} = p$; 6 | \item For all prime $p$ and $n > 0$, there exists a field $\F$ where $\abs{\F} = p^n$; 7 | \item Every finite field has order $p^n$ for some prime $p$ and some $n > 0$. 8 | \end{enumerate} 9 | 10 | From last time, we know for for all prime $p$ and all $n > 0$, there exists a field with $p^n$ elements. However, the naïve choice for this field isn't always right. 11 | 12 | \begin{example} 13 | Consider $\F_4$; what could it be? 14 | \end{example} 15 | 16 | \begin{solution} 17 | It can't be $\Z/4\Z$, since inverses aren't unique as $4$ isn't prime, and so $(\Z/4\Z \setminus \{0\}, \times)$ isn't a group. However, it could be the direct product of $\Z/2\Z$ with itself; i.e., $\F^4 \cong \Z_2 \cross \Z_2$. Whatever it is, we know it has elements $\{\bar{0},\bar{1}, x, x+1\}$ which satisfies $x^2 + x + 1 = 0$, $\bar{1} + \bar{1} = 0$, and $x + x = 0$. Then $x^2 = -x-1 = x+1$. 18 | \end{solution} 19 | 20 | \subsection{Homomorphisms} 21 | 22 | \begin{definition}[Homomorphism] 23 | A group homomorphism is a map $\varphi : (G, \ast) \to (H, \times)$ which preserves the operations between the groups, so $\varphi(a \ast b) = \varphi(a) \times \varphi(b)$. Usually, this is just abbreviated into $\varphi(ab) = \varphi(a)\varphi(b)$. 24 | \end{definition} 25 | 26 | \begin{lemma} 27 | Let $\varphi$ be a homomorphism. Then $\varphi(1_G) = 1_H$, and $\varphi(a^{-1}) = \varphi(a)^{-1}$. 28 | \end{lemma} 29 | 30 | \begin{proof} 31 | We know that $1 \cdot 1 = 1$. Then $\varphi(1 \cdot 1) = \varphi(1)\varphi(1) = \varphi(1)$. Then multiply by $\varphi(1)^{-1}$, and we have that $\varphi(1) = 1$. Consider then that $1 = aa^{-1}$, so $\varphi(1) = \varphi(a)\varphi(a^{-1}) = 1$ (by the previous result). Then $1 = \varphi(a)\varphi(a^{-1})$, so $\varphi(a^{-1}) = \varphi(a)^{-1}$ since inverses are unique. 32 | \end{proof} 33 | 34 | \begin{example}[Examples of Homomorphisms] 35 | \begin{enumerate} 36 | \item The identity map $g \mapsto g$; 37 | \item The determinant $\det : \mathrm{GL}_n(\R) \to (R^\times, \times)$; 38 | \item The map $(\Z, +) \to (\Z_n, +)$ where $a \mapsto \bar{a}$; 39 | \item Let $g \in G$. Then we have a map $(\Z, +) \to G$ where $n \mapsto g^n$. 40 | \end{enumerate} 41 | \end{example} 42 | 43 | \begin{definition}[Isomorphism] 44 | An isomorphism is a bijective homomorphism. Note that the inverse of an isomorphism is also a group isomorphism. 45 | \end{definition} 46 | 47 | What does it mean for two things to be isomorphic? Well, it means that anything you care about can be preserved under a sufficiently good map, so two isomorphic groups aren't the same, but they're ``the same.'' As an example of why they aren't actually the same, consider that $(\Z_2, +)$ and $(\Z_3^\times, \times)$ are isomorphic. These groups don't have the same elements or the same operations, but they are isomorphic to one another. 48 | 49 | \begin{lemma} 50 | Let $\phi : G \to H$ be a homomorphism where $g_i \mapsto h_i$. Then any relation on $\{g_i\}$ is satisfied by $\{h_i\}$. For example, if $G$ is abelian then $H$ is abelian as well. 51 | \end{lemma} 52 | 53 | \begin{corollary} 54 | If $G = \langle g_1,\dotsc,g_n \mid \text{relations } \rangle$, and if $h_1,\dotsc,h_n \in H$ satisfy the same relations, then there exists a homomorphism $\phi : G \to H$ where $g_i \mapsto h_i$. However, any map which does preserve these relations need not be surjective nor injective. This means that presentations aren't enough to determine group isometry. Worse, minimal generating sets may not even have the same size for distinct generators. For example $\{1\}$ and $\{2,3\}$ both generate $\Z_6$. 55 | \end{corollary} 56 | 57 | \begin{corollary} 58 | Homomorphisms don't actually preserve order, since if $g^n = 1$ then $\phi(g^n) = 1$, but the order of $\phi(g^n)$ might just be a divisor of $n$, not $n$ itself. 59 | \end{corollary} 60 | 61 | \begin{definition}[Subgroup] 62 | A subgroup $H$ of $G$ is a group where the set of $H$ is a subset of $G$ 63 | and $H$ inherits its operation from $G$. Formally, $H$ is a subgroup of $G$ if the following are satisfied: 64 | \begin{itemize} 65 | \item $e \in G$; 66 | \item $a \in H \implies a^{-1} \in H$; 67 | \item $a,b \in H \implies ab \in H$. 68 | \end{itemize} 69 | \end{definition} -------------------------------------------------------------------------------- /MATH-350/Lectures/5.tex: -------------------------------------------------------------------------------- 1 | \section{September 10, 2018} 2 | 3 | A useful fact about orders a generated subgroups. 4 | 5 | \begin{lemma} 6 | Let $x \in G$, and let $\langle x \rangle \subset G$. Then $\abs{x} = \abs{\langle x \rangle}$. 7 | \end{lemma} 8 | 9 | Today, we're gonna connect the notion of a homomorphism and the notion of a group. Let $\phi : H \to G$ be a homomorphism. Then the image $\phi(H)$ is a subgroup of $G$. Why is this true? Well, trivially, $\phi(H)$ is a subset of $G$. Since $\phi$ is a homomorphism, we know that $\phi(1) = 1$ and so $1 \in \phi(H)$. Since $\phi$ is multiplicative, $\phi(a),\phi(b) \in \phi(H) \implies \phi(a)\phi(b) \in \phi(H)$. And since $\phi(a)^{-1} = \phi(a^{-1})$, $\phi(H)$ contains inverses for all $\phi(a) \in \phi(H)$. However, there's not much that we can say about $\phi(H)$ in relation to $G$, other than the fact that it must not be larger than $G$. However, if $\phi : H \xhookrightarrow{} G$ is injective, then $H$ and $\phi(H)$ are isomorphic, so there's a copy of $H$ inside of $G$. 10 | 11 | \subsection{Representation Theory} 12 | 13 | Miki says that she's not supposed to talk about representation theory in this class but she can't resist mentioning it here when we discuss group actions. 14 | 15 | \begin{definition}[Group Action] 16 | Let $G$ be a group. A group action is a map $\phi : G \times A \to A$, where $A$ is a set on which $G$ is acting, which obeys the following axioms. 17 | \begin{enumerate} 18 | \item The identity in $G$ becomes the identity map, so $\phi(1_G, a) = a$ for all $a \in A$; 19 | \item The action $\phi$ is associative, so $\phi(g, \phi(h,a)) = \phi(gh, a)$. 20 | \end{enumerate} 21 | \end{definition} 22 | 23 | The simplest example of a group action is the \emph{trivial action}, which is simply the map $\phi(g,a) = a$ for any $a \in A$ and any $g \in G$. Another example is \emph{translation}, where we map each $a$ to $a + n$ from some $n$. \emph{Reflection} is where we map $a$ to $-a$. 24 | 25 | \begin{example} 26 | Some food for thought: the group operation is also an action on that group. 27 | \end{example} 28 | 29 | Some facts about group actions. 30 | \begin{lemma} 31 | For all $g \in G$, we get a map $\sigma_g : A \to A$ where $a \mapsto g\cdot a$; then $\sigma_g$ is bijective since it's just a permutation of $a$; then we have a map $\pi : G \to S_A : g \mapsto \sigma_g$. 32 | This map $\pi$ is a group homomorphism. However, we don't know that $\pi$ is necessarily injective. 33 | \end{lemma} 34 | \begin{proof} 35 | We know that $\sigma_g$ is bijective since it has an inverse in $\sigma_{g^{-1}}$. Since it's bijective, we know that $\sigma_g$ is a permutation, and so is an element of $S_A$. Then consider that $\pi(gh)(a) = (gh)(a) = g \cdot (h \cdot a) = \pi(g)\pi(h)(a)$, so $\pi$ is multiplicative. 36 | \end{proof} 37 | 38 | \begin{example} 39 | Let $A = G$, so our action is left multiplication $\ast : G \times G \to G : (g,a) \mapsto ga$. Since multiplication already fulfills the requirements for group actions, we know this forms a valid action. For this action, look at the map $\pi : G \to S_G$. We know $\pi(g) = \sigma_g$ is bijective. Then $\pi$ is injective. This fact gives us that every finite group is isomorphic to a subgroup of $S_n$ for some $n$, since $G \cong \pi(G) \subset S_n$ for $n = \abs{G}$. 40 | 41 | \begin{proof} 42 | Suppose $\sigma_g = \sigma_h$. Then $\sigma_g(a) = \sigma_h(a)$ for all $a \in A = G$. In particular, let $a = 1$. The $g = g \cdot 1 = h \cdot 1 = h$, so $g = h$. Then $\pi(g) = \pi(h)$ if and only if $g = h$. 43 | \end{proof} 44 | \end{example} 45 | 46 | \subsection{Isomorphisms and Equality} 47 | Why do we bother saying that groups are isomorphic instead of just saying that groups are ``equal.'' Consider $D_{8}$ acting on a square $\square$. There is a subgroup $H_1 = \langle r^2 \rangle = \{1, r^2\}$. We know that $H_1 \cong \Z/2\Z$. There is also the subgroup $H_2 = \langle s \rangle = \{1,s\}$. We know that $H_2$ is also isomorphic to $\Z/2\Z$. However, it's pretty clear that $H_1 \not= H_2$ even though $H_1 \cong H_2$. Then isomorphic groups can be distinguished by their group actions. -------------------------------------------------------------------------------- /MATH-350/Lectures/6.tex: -------------------------------------------------------------------------------- 1 | \section{September 12, 2018} 2 | 3 | \epigraph{``Oh, I erased my smiley face. How sad.'' (she did not sound sad)}{Miki} 4 | 5 | Today we'll officially state something we covered last time. 6 | 7 | \begin{theorem}[Caley's Theorem] 8 | Every finite group $G$ is isomorphic to a subgroup of $S_n$ for some $n$. 9 | \end{theorem} 10 | \begin{proof} 11 | Let $n = \abs{G}$. 12 | \end{proof} 13 | 14 | \subsection{Kernels} 15 | 16 | Let's discuss formally the idea of a kernel of a homomorphism and a kernel of a group action. 17 | 18 | \begin{definition}[Kernel] 19 | Let $\phi : G \to H$ be a homomorphism. Then the kernel of $\phi$, written $\ker \phi$, is the set of all elements in $G$ which are mapped to the identity in $H$; i.e., $\ker \phi = \{g \mid \phi(g) = 1_h \}$. 20 | \end{definition} 21 | 22 | \begin{definition} 23 | Suppose $G$ acts on $A$ by $\pi$. Then the kernel of the action is the set of all elements of $g$ which act trivially on $A$; i.e., $\ker \pi = \{g \mid ga = a \text{ for all $a \in A$}\}$. 24 | \end{definition} 25 | 26 | \begin{example} 27 | Consider the action $\phi : \mathrm{GL}_2(\R) \to (\R^\times, \times) : A \mapsto \det A$. Then the kernel of $\phi$ are all matricies with determinant $1$, called $\mathrm{SL}_2(\R)$. 28 | \end{example} 29 | 30 | \begin{definition}[Stabalizer] 31 | Let $\pi : G \times A \to A$ be a group action, and fix $a \in A$. The \emph{stabalizer} is $G_a = \{g \in G \mid ga = a\}$. By this definition, the kernel is contained within any stabalizer, and in fact is equal to the intersection of all stabalizers. 32 | \end{definition} 33 | 34 | \begin{example} 35 | Let $G = \mathrm{GL}_2(\R)$ and let $A = \R^2$ defined with the usual action (vector-matrix multiplication). What is the kernel of this action? Then let $c = (0,1)^\top \in \R^2$. What is the stabalizer of $c$? 36 | \end{example} 37 | 38 | \begin{corollary} 39 | The kernel of an action is a subgroup of $G$, and $G_a$ is a subgroup of $G$ for any fixed $a \in A$. 40 | \end{corollary} 41 | 42 | \begin{definition}[Orbit] 43 | Fix $a \in A$. The orbit of $a$ is the image of $a$ under the group action; i.e., $O_a = \{ga \mid g \in G\}$. Intuitively, it's everywhere $a$ can go under a specific group action. Notice that the orbits partition $A$, and so are equivalence classes in $A$. 44 | \end{definition} 45 | 46 | \begin{example} 47 | Let $G = \mathrm{GL}_2(\R)$ and let $A = \R^2$ defined with the usual action (vector-matrix multiplication). What is the orbit of $(1,0)^\top$? 48 | \end{example} 49 | 50 | \begin{definition}[Faithful] 51 | An action is faithful if the kernel is the identity. This means that the base element of the action must be the identity. This tells us that $G$ is injective into $S_A$. 52 | \end{definition} 53 | 54 | \begin{example} 55 | Consider $D_8$ acting on a square (technically the set $A = \{1,2,3,4\}$). The orbit $O_1$ is all possible vertices, since you can rotate any vertex to any position. The stabalizer is $\{1, s\}$. 56 | \end{example} 57 | 58 | \begin{lemma} 59 | As it turns out, for a fixed $a \in A$, we see that $\abs{O_a}\abs{G_a} = \abs{G}$. We'll prove this later. (Orbit-Stabalizer Theorem I think?) 60 | \end{lemma} 61 | 62 | \begin{definition}[Conjugation] 63 | Consider the action $\pi : G \times G \to G : (g,a) \mapsto gag^{-1}$. This action is known as \emph{conjugation}. 64 | \end{definition} 65 | 66 | \begin{definition}[Centralizer] 67 | Let $S \subset G$. The \emph{centralizer} of $S$ in $G$, written $C_G(S) = \{g \in G \mid gsg^{-1} = s \text{ for all $s \in S$}\}$. This is the set of things that fix $S$ in $G$ pointwise under conjucation. By definition, this is the set of elements in $G$ which commute with all elements in $S$. In the case that $S = \{s\}$ we see that $C_G(S) = G_S$. 68 | \end{definition} 69 | 70 | \begin{definition}[Normalizer] 71 | Let $S \subset G$. The \emph{normalizer} of $S$ in $G$ is $N_G(S) = \{g \in G \mid gSg^{-1} = S\}$. Essentially, this is just a centralizer on a set, except that it may permute the elements of $S$. Then $C_G(S) \subset N_G(S)$. 72 | \end{definition} 73 | 74 | \begin{example} 75 | Suppose that $G$ is abelian. For any $S \subset G$, we see that $C_G(S) = N_G(S) = G$. 76 | \end{example} 77 | 78 | \begin{example} 79 | Let $G = S_3$, and let $S = G$. What is the normalizer of $S$? (It's the whole thing since $G$ is closed under its operation.) What is the centralizer of $S$? (It's the identity.) 80 | \end{example} -------------------------------------------------------------------------------- /MATH-350/Lectures/8.tex: -------------------------------------------------------------------------------- 1 | \section{September 17, 2018} 2 | 3 | \epigraph{``The only thing I learned for years was how to count hedgehogs in a field.'' (In the midst of a wonderful and inspirational talk about being a mathematician.)}{Miki} 4 | 5 | \subsection{Finite Cyclic Groups} 6 | 7 | Today, we'll cover finite cyclic groups. This will be very similar to the previous lecture on infinite cyclic groups. As a reminder, here are the propositions for infinite cyclic groups: 8 | 9 | \begin{proposition}[Infinite Cyclic Groups] 10 | Let $G$ be an infinite cyclic group. 11 | \begin{enumerate} 12 | \item The order of $G$ is infinite, with $G = \{\dotsc,x^{-1},1,x,x^2,\dotsc\}$ all distinct. 13 | \item The group $G$ is isomorphic to $\Z$. 14 | \item The group $G$ is generated by $x^n$ if and only if $n = \pm 1$. 15 | \item Every subgroup $G$ is cyclic. 16 | \end{enumerate} 17 | \end{proposition} 18 | 19 | Now for the finite case. 20 | 21 | \begin{proposition}[Finite Cyclic Groups] 22 | Let $G = \langle x \rangle$ with $\abs{G} = n < \infty$. 23 | \begin{enumerate}[label=P\arabic*.] 24 | \item The group $G$ is exactly $\{1,x,\dotsc,x^{n-1}\}$. 25 | \item The group $G$ is isomorphic to $\Z/n\Z$. 26 | \item The group $G$ is generated by $x^k$ if and only if $\gcd(k,n) = 1$. 27 | \item Every subgroup of $G$ is also cyclic. That is, for all $k > 0$ where $k$ divides $n$ we get a subgroup $H$ of order $k$ generated by $x^{n/k}$ 28 | \end{enumerate} 29 | \end{proposition} 30 | \begin{proof}[Proof of~P1] 31 | We know that $1,\dotsc,x^{n-1}$ are all in $G$. Suppose that $x^a = x^b$ for some distinct $a,b$. Then $x^{b-a} = 1$ for $0 < b-a < n$, which is a contradiciton since $\abs{x} = n$. In fact, this set enumerates $G$. Suppose that $x^k \in G$ for some $k \in \Z$. We use the divison algorithm to write that $k = an + r$ for some $a,r \in \Z$. Then $x^k = x^{an+r} = (x^n)^ax^r = x^r$, so $x^k$ is in $G$. 32 | \end{proof} 33 | \begin{proof}[Proof of~P2] 34 | Let $\phi : \Z/n\Z \to G : \bar{k} \mapsto x^k$ where $k$ is any representative of $\bar{k} \in \Z$. To show that $\phi$ is well-defined, consider another representative $\ell$ of $\bar{k} \in \Z$. Then $\ell = k + an$, so $x^\ell = x^{k+an} = x^k(x^n)^a = x^k$. To show that $\phi$ is a homomorphism, consider that $\phi(\bar{m}+\bar{n}) = x^{m+n} = x^{m}x^n$, so $\phi$ is multiplicative. Finally, we know that $\phi$ is surjective and injective by P1. This tells us that, up to an isomorphism, there are only really two cyclic groups; $\Z$ if the group is of infinite order, or $\Z/n\Z$ if it is finite. 35 | \end{proof} 36 | \begin{proof}[Proof of~P3] 37 | This is more of a sketch. Recall that $\langle \abs{x^k}\rangle = \abs{x^k}$, and this is $n$ if and only if $\gcd(k,n) = 1$. In general, $\abs{x^k} = n / \gcd(k,n)$. 38 | \end{proof} 39 | \begin{proof}[Proof of~P4] 40 | Exactly the same as the infinite case. 41 | \end{proof} 42 | 43 | Now that we've covered cyclic groups, it's helpful to introduce some notation to represent them. 44 | \begin{notation}[$\Z_n, C_n$] 45 | We write the multiplicative cyclic group of order $n$ as $\Z_n$. The additive cyclic group of order $n$, which we've been writing as $\Z/n\Z$, is commonly written as $C_n$. 46 | \end{notation} 47 | 48 | \subsection{Subgroups} 49 | 50 | \cite{DF} uses the notation $S \subset G$ to mean that $S$ is a subset of $G$, and $H \leq G$ to mean that $H$ is a subgroup of $G$. 51 | 52 | \begin{definition}[Subgroup] 53 | Let $S \subset G$ be nonempty. Let $H = \{a_1^{\varepsilon_1}\cdots a_k^{\varepsilon_k}\}$ where $a_i \in S$ and $\varepsilon_i = \pm 1$ for $k \in \Z_{\geq 0}$. This sequence of $a_i^{\varepsilon_i}$ is called a \emph{word}. Note that $a_i$ need not be distinct. Then $H$ is a subgroup of $G$. 54 | \end{definition} 55 | 56 | \begin{proof} 57 | First, we know that $H \subset G$. We know that $1 \in H$. Since any concatenation of words is also a word, we know that $H$ is closed under multiplication. Finally, since $((ab)^n)^{-1} = b^{-n}a^{-n}$ we know that the inverses of a word are words themselves, and so $H$ is closed under inversion. 58 | \end{proof} -------------------------------------------------------------------------------- /MATH-350/Lectures/9.tex: -------------------------------------------------------------------------------- 1 | \section{September 19, 2018} 2 | 3 | \epigraph{``Funny things happen with groups, which is why they're fun!''}{Miki} 4 | 5 | Recall from last time how we defined a subgroup $H$ of $G$ in terms of words where the powers of each element was $\pm 1$. If $G$ is abelian we can combine elements of like bases to get powers which can be any integral value. If we assume that $\abs{a_i} = d_i$ is finite for all $a_i \in H$, then we know that $\abs{H} \leq d_1 \cdots d_k$. This gives us a limit on the order of a subgroup; if $G$ is abelian then the order of a subgroup is bounded above by the product of the orders of the generating elements. On the other hand, if $G$ is not abelian then this does not always hold. Consider $G = \langle a,b \mid a^2 = b^2 = 1 \rangle$. If $G$ isn't commutative, then $(ab)^n \not= a^nb^n$ for all $n$ and so we can just create infinitely many words by appending $ab$ to one another and so the order is infinite. 6 | 7 | \begin{lemma} 8 | Let $G = \{a_1^{n_1} \cdots a_k^{n_k}\}$ be abelian, and let each $a_i$ have finite order $d_i$. Then $\abs{G} \leq d_1 \cdots d_i$. 9 | \end{lemma} 10 | 11 | \begin{proposition} 12 | Let $G$ be a group and let $\mathcal{L}$ be a collection of subgroups of $G$. Then \[ K = \bigcap_{L \in \mathcal{L}} L \] 13 | is a subgroup of $G$. 14 | \end{proposition} 15 | 16 | \begin{definition}[Subgroup] 17 | Let $S \subset G$ and let $\mathcal{L} = \{L \leq G \mid s \subset L\}$. Then the subgroup generated by $S$ is \[ K = \bigcap_{L \in \mathcal{L}} L. \] 18 | \end{definition} 19 | 20 | What do we know from this definition? Well, $S \subset K$ and $K \leq G$. We want to say that $K \in \mathcal{L}$ is the minimal element, so $K = L_i$ for some $i$. 21 | 22 | \begin{definition}[Minimal Element] 23 | Let $\mathcal{M}$ be a collection of subsets of $G$. A minimal element is an element $M$ of $\mathcal{M}$ such that if $M' \in \mathcal{M}$ and $M' \subset M$ then $M = M'$. It's like ``the smallest element'' except there could be multiple minimal elements. 24 | \end{definition} 25 | 26 | We want to show that $K$ is \emph{the} minimal element of $\mathcal{L}$. 27 | 28 | \begin{proof}[Proof $K$ is minimal] 29 | Let $L \in \mathcal{L}$. Then $K \subset L$. Then either $K = L$ or $L$ is not minimal. 30 | \end{proof} 31 | 32 | \begin{proof}[Proof $K$ is \emph{the} minimal element] 33 | Suppose there is another minimal $M$ in $\mathcal{L}$. By definition $K \subset M$ so by minimality $M = K$. 34 | \end{proof} 35 | 36 | \begin{proposition} 37 | Our two definitions for subgroup (generated by words $H$ and minimal element of collection $K$) containing elements of $S \subset G$ are equivalent. 38 | \end{proposition} 39 | 40 | \begin{proof} 41 | $H \leq G$ and $S \subset H$ by the construction of $1$-letter words. Then $H \in \mathcal{L}$ so $K \subset H$. On the other hand, $S \subset K$ and $K$ is a group. Then $K$ contains all inverses and products of elements in $S$, so it contains all words and therefore contains $H$. Then $H \subset K$. Putting this together, we have that $K = H$. 42 | \end{proof} 43 | 44 | \begin{definition}[Lattice] 45 | Given a group $G$, a lattice is a diagram showing all subgroups of $G$ which shows containment between the subgroups. 46 | \end{definition} 47 | 48 | \begin{figure}[h] 49 | \caption{Lattice Diagram of $C_2$} 50 | \end{figure} 51 | 52 | Recall from last time that for $C_n$ the subgroups are paired with the divisors $k$ of $n$; then $\langle k \rangle$ generates subgroup of order $n / k$. 53 | 54 | \begin{figure}[ht] 55 | \caption{Lattice Diagram of $C_4$} 56 | \end{figure} 57 | 58 | \begin{figure}[ht] 59 | \caption{Lattice Diagram of $C_8$} 60 | \end{figure} 61 | 62 | \begin{figure}[ht] 63 | \caption{Lattice Diagram of $C_6$} 64 | \end{figure} 65 | 66 | \begin{figure}[ht] 67 | \caption{Lattice Diagram of $S_3$} 68 | \end{figure} -------------------------------------------------------------------------------- /MATH-350/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/MATH-350/notes.pdf -------------------------------------------------------------------------------- /MATH-350/notes.tex: -------------------------------------------------------------------------------- 1 | \documentclass{lnotes} 2 | 3 | % Bibliography 4 | \usepackage[ 5 | style = alphabetic , 6 | ]{biblatex} 7 | \addbibresource{references.bib} 8 | 9 | % Packages 10 | \usepackage{epigraph} 11 | \usepackage{wasysym} 12 | \usepackage[osf]{libertineRoman} 13 | \usepackage{tikz-cd} 14 | 15 | % Document Information 16 | \title{Introduction to Abstract~Algebra} 17 | \course{\textsc{math} 350} 18 | \place{Yale University} 19 | \term{Fall} 20 | \year{2018} 21 | 22 | \blurb{ 23 | These are lecture notes for \textsc{math} 350a, ``Introduction to Abstract Algebra,'' taught by Marketa Havlickova at Yale University during the fall of 2018. 24 | These notes are not official, and have not been proofread by the instructor for the course. 25 | These notes live in my lecture notes respository at 26 | \[\text{\url{https://github.com/jopetty/lecture-notes/tree/master/MATH-350}.}\] 27 | If you find any errors, please open a bug report describing the error, and label it with the course identifier, or open a pull request so I can correct it. 28 | } 29 | 30 | \begin{document} 31 | 32 | \section*{Syllabus} 33 | 34 | \begin{center} 35 | \begin{tabular}{@{}rp{10cm}@{}} 36 | \toprule 37 | \textbf{Instructor} & Marketa Havlickova, \url{miki.havlickova@yale.edu} \\ 38 | \textbf{Lecture} & MWF 10:30--11:20 \textsc{am}, LOM 205 \\ 39 | \textbf{Recitation} & TBA \\ 40 | \textbf{Textbook} & \fullcite{DF} \\ 41 | \textbf{Midterms} & Wednesday, October 10, 2018 \\ 42 | & Wednesday, November 14, 2018 \\ 43 | \textbf{Final} & Monday, December 17, 2018, 2:00--5:30 \textsc{pm} \\ 44 | \bottomrule 45 | \end{tabular} \\[3ex] 46 | \end{center} 47 | 48 | Abstract Algebra is the study of mathematical structures carrying notions of ``multiplication'' and/or ``addition''. Though the rules governing these structures seem familiar from our middle and high school training in algebra, they can manifest themselves in a beautiful variety of different ways. The notion of a group, a structure carrying only multiplication, has its classical origins in the study of roots of polynomial equations and in the study of symmetries of geometrical objects. Today, group theory plays a role in almost all aspects of higher mathematics and has important applications in chemistry, computer science, materials science, physics, and in the modern theory of communications security. The main topics covered will be (finite) group theory, homomorphisms and isomorphism theorems, subgroups and quotient groups, group actions, the Sylow theorems, ring theory, ideals and quotient rings, Euclidean domains, principal ideal domains, unique factorization domains, module theory, and vector space theory. Time permitting, we will investigate other topics. This will be a heavily proof-based course with homework requiring a significant investment of time and thought. The course is essential for all students interested in studying higher mathematics, and it would be helpful for those considering majors such as computer science and theoretical physics. 49 | 50 | Your final grade for the course will be determined by 51 | \[ \max\left\{ 52 | \begin{array}{cccc} 53 | 25\%\text{ homework} + 20\%\text{ exam 1} + 20\%\text{ exam 2} + 35\%\text{ final} \\ 54 | 25\%\text{ homework} + 10\%\text{ exam 1} + 20\%\text{ exam 2} + 45\%\text{ final} \\ 55 | 25\%\text{ homework} + 20\%\text{ exam 1} + 10\%\text{ exam 2} + 45\%\text{ final} 56 | \end{array} 57 | \right\}. \] 58 | 59 | \printbibliography 60 | 61 | \include{Lectures/1} 62 | \include{Lectures/2} 63 | \include{Lectures/3} 64 | \include{Lectures/4} 65 | \include{Lectures/5} 66 | \include{Lectures/6} 67 | \include{Lectures/7} 68 | \include{Lectures/8} 69 | \include{Lectures/9} 70 | \include{Lectures/10} 71 | \include{Lectures/11} 72 | \include{Lectures/12} 73 | \include{Lectures/13} 74 | \include{Lectures/14} 75 | \include{Lectures/15} 76 | \include{Lectures/16} 77 | \include{Lectures/17} 78 | \include{Lectures/18} 79 | \include{Lectures/19} 80 | \include{Lectures/20} 81 | \include{Lectures/21} 82 | \include{Lectures/22} 83 | \include{Lectures/23} 84 | \include{Lectures/24} 85 | \include{Lectures/25} 86 | \include{Lectures/26} 87 | \include{Lectures/27} 88 | \include{Lectures/28} 89 | \include{Lectures/29} 90 | \include{Lectures/30} 91 | \include{Lectures/31} 92 | \include{Lectures/32} 93 | \include{Lectures/33} 94 | \include{Lectures/34} 95 | \include{Lectures/35} 96 | \include{Lectures/36} 97 | 98 | \end{document} -------------------------------------------------------------------------------- /MATH-350/references.bib: -------------------------------------------------------------------------------- 1 | @book{ 2 | DF , 3 | title = { Abstract Algebra } , 4 | author = { Dummit and Foote } , 5 | edition = { 3 } , 6 | year = { 2004 } , 7 | publisher = { John Wiley \& Sons } 8 | } -------------------------------------------------------------------------------- /MATH-354/Lectures/1.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{January 14, 2019} 4 | 5 | Didn't go to lecture today. -------------------------------------------------------------------------------- /MATH-354/Lectures/12.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, 25 February 2019} -------------------------------------------------------------------------------- /MATH-354/Lectures/15.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, 4 March 2019} 4 | 5 | It's been a while. 6 | 7 | Let $p$ be prime. Recall the following definitions: 8 | \begin{itemize} 9 | \item $\Pi(x) = \sum_{p \leq x} 1$; 10 | \item $\vartheta(x) = \sum_{p \leq x} \log p$; 11 | \item $\psi(x) = \sum_{p^n \leq x} \log p = \sum_{n \leq x} \Lambda(x)$. 12 | \end{itemize} 13 | 14 | We had two important results from these definitions. 15 | 16 | \begin{theorem} 17 | $\abs{\vartheta(x) - \psi(x)} \leq O(\sqrt{x}\log^2x)$. 18 | \end{theorem} 19 | 20 | \begin{theorem} 21 | $\Pi(x) = \vartheta(x)/\log x - \int_1^x \vartheta(x)/t\log^2x \dd{t}$. 22 | \end{theorem} 23 | 24 | \begin{theorem} 25 | The following are equivalent: 26 | \begin{itemize} 27 | \item $\psi(x) \sim x$, 28 | \item $\vartheta(x) \sim x$, 29 | \item $\Pi(x) \sim x/\log x$. 30 | \end{itemize} 31 | \end{theorem} 32 | 33 | \begin{lemma} 34 | $M_{\log x} = \sum_{n \leq x} \log n = x\log x - x + O(\log x)$. 35 | \end{lemma} 36 | 37 | \begin{proof} 38 | Consider 39 | \[ \sum_{n \leq x} \log n \cdot 1. \] 40 | We apply Abel summation using $f(n) = 1$ and $\phi(x) = \log x$ to get that 41 | \[ \sum_{n \leq x} \log n \cdot 1 = \lfloor x \rfloor \log x - \int_1^x \frac{\lfloor t \rfloor}{t} \dd{t}, \] 42 | noting that $M_1 = \lfloor x \rfloor$. Notice that 43 | \[ \int_1^x \frac{\lfloor t \rfloor}{t} \dd{t} = \int_1^x 1 - \frac{t - \lfloor t \rfloor}{t} \dd{t} = x - 1 - O\qty(\int_1^x \frac{1}{t} \dd{t}) = x + O(\log x). \] 44 | This acheives the result we wanted. 45 | \end{proof} 46 | 47 | \begin{theorem}[Chebyshev] 48 | The inequality 49 | \[ x \log 2 + O(\log x) \leq \psi(x) \leq x\log 4 + O(\log^2x) \] holds. 50 | \end{theorem} 51 | 52 | \begin{proof} 53 | Recall that $M_{\log(x)} = \sum_{n \leq x} \log n$, and that 54 | \[ \log n = \sum_{d \mid n} \Lambda(d). \] 55 | Then 56 | \[ M_{\log(x)} = \sum_{n \leq x} \sum_{d \mid n} \Lambda(d) = \sum_{dq \leq x} \Lambda(d) = \sum_{q \leq x}\sum_{d \leq x/q} \Lambda(d) = \sum_{q \leq x} \psi\qty(\frac{x}{q}). \] 57 | Define the quantity 58 | \[ D(x) = M_{\log}(x) - 2M_{\log}\qty(\frac{x}{2}) = \log {x \choose x/2}. \] 59 | On one had, we can show that 60 | \[ D(x) = x\log(2) + O(\log x), \] 61 | and on the other had we know that 62 | \[ D(x) = \sum \psi(x/q) - 2\sum\psi(x/2q) = \psi(x) - \psi(x/2) + \psi(x/3) + \cdots. \] 63 | Note that $\psi(x)$ is ``sorta monotone increasing'' 64 | \end{proof} -------------------------------------------------------------------------------- /MATH-354/Lectures/3.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Lecture 3} 4 | 5 | Didn't take notes today. -------------------------------------------------------------------------------- /MATH-354/Lectures/4.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Wednesday, January 23} 4 | 5 | Recall the uniqueness of prime factorization, where for all $n \in \N$ we have a unique list of primes $p_1, \dotsc, p_k$ and $a_i, \dotsc, a_k \in \Z_{>0}$ such that $n = \prod_{i=1}^k p_i^{a_i}$. 6 | 7 | \subsection{Infinitude of Primes} 8 | \begin{problem} 9 | How many primes are there? 10 | \end{problem} 11 | 12 | \begin{theorem} 13 | There are infinitely many primes. 14 | \end{theorem} 15 | 16 | \begin{proof}[Euclid's Proof] 17 | Assume by way of contradiction we have a finite list of primes $p_1, \dotsc, p_k$ of all primes. Let $M = \prod p_i$, and consider $M+1$. By the existence of prime factorization, we know that $M+1 = \prod_{i=1}^k p_i^{a_i}$. Without a loss of generality assume that $a_1 \not= 0$. Then $p_1$ divide $M+1$ and since $p_1$ divides $M$ it must be the case that $p_1$ divides $1$ as well which presents a contradiction. 18 | \end{proof} 19 | 20 | \textbf{Fact:} Let $p_1, p_2, p_3, \dotsc$ be a list of primes in order. By the uniqueness of prime factorization, there is an injective correspondence between vectors $(a_1, a_2, \dotsc) \in (\Z_{\geq0})^\infty$ with finitely many nonzero entries and $\N$. The correspondence is $n = \prod p_i^{a_i}$ with a lot of $a_i$ being zero. 21 | 22 | \textbf{NOTE: THE BELOW IS BY CONTRADCTION and (*) ONLY HOLDS FOR $k=\infty$.} 23 | If we assume that the list of primes is finite, then we would have an injective correspondence between $(a_1, \dotsc, a_k) \in (\Z_{\geq 0})^k$ and $\N$. Therefore 24 | \[ \prod_{i=1}^k\qty(\sum_{j=0}^\infty \frac{1}{p_i}) = \sum_{n=1}^\infty \frac{1}{n}. \tag{*} \] 25 | Then by uniqueness of prime factorization for each $n \in \N$ we know that $1/n$ appears exactly once when you expand this product. This is Euler's product for the $\zeta$ function? 26 | 27 | \begin{proof}[Euler's Proof] 28 | Assume by way of contradiction that there are finitely many primes. Then 29 | \[ \sum_{n=1}^\infty \frac{1}{n} = \prod_{i=1}^k\qty(1 + \frac{1}{p_1} + \cdots) = \prod_{i=1}^k \qty(\frac{1}{1-1/p_i}) < \infty. \] 30 | Yet we know that $\sum 1/n$ diverges, which presents a contradiction. 31 | \end{proof} 32 | 33 | \begin{lemma} 34 | For any $n \in \Z$ there exists a unique $a,b \in \Z$ such that $a$ is square free (meaning that no square number divides it) and $n = ab^2$. 35 | \end{lemma} 36 | 37 | \begin{proof}[Erdős' Proof] 38 | Assume by way of contradiction that there are finitely many primes. Then any square-free number $n = \prod p_i^{a_i}$ where $a_i \in \{0,1\}$. Thus there are only $2^k$ square-free numbers. Now let's look at all numbers at most $N$ for some $N$. By the above lemma, they can be specified by $(a,b)$ where $a$ is square-free and $b^2$ is square. There are $2^k$ square-free numbers and at most $\sqrt{N}$ square numbers, so $N \leq 2^k\sqrt{N}$ for all $N$, so $2^k \geq \sqrt{N}$ for all $N$, which is very very false if $N > 2^{2k}$. 39 | \end{proof} 40 | 41 | \subsection{Congruence Equations \& Modular Arithmetic} 42 | 43 | \begin{definition}[Congruence] 44 | We say that $a \equiv b \pmod{m}$ if and only if $m$ divides $b-a$. Alternatively, we say that $a \equiv b \pmod{m}$ if and only if there exists some $k$ such that $a = b + mk$. 45 | \end{definition} 46 | 47 | \begin{theorem}[Some quick remarks] \hfill 48 | \begin{enumerate} 49 | \item Congruency is an equivalence relation on the integers (transitive, symmetric, and reflexive); 50 | \item For some fixed $m$, we define the congruence class $\bar{a}$ to be the set $\bar{a} = \{n \in \Z \mid n \equiv a \pmod{m}\}$; 51 | \item Arithmetic on these congruence classes holds; If $a \equiv b \pmod{m}$ and $c \equiv d \pmod{m}$ then $a+c \equiv b + d \pmod{m}$ and $ac \equiv bd \pmod{m}$. Thus $\bar{a} + \bar{b} = \overline{a + b}$ and $\bar{a}\bar{b} = \overline{ab}$. This forms the commutative ring $\Z/m\Z$. 52 | \end{enumerate} 53 | \end{theorem} 54 | 55 | \begin{problem} 56 | Let $a,b,m$ be fixed. 57 | When is the congruence $ax \equiv b \pmod{m}$ solvable? 58 | \end{problem} 59 | 60 | \begin{enumerate}[label=\textbf{Obs. \arabic*.}] 61 | \item If $(a,m)=1$ then we can use Bezout's theorem. This tells us that tehre exist some $X,Y$ such that $1 = aX + mY$. Then we multiply through by $b$ to get that $b = a(Xb) + m(Yb)$. Then $aXb \equiv b \pmod{m}$. 62 | \end{enumerate} 63 | 64 | \begin{lemma} 65 | The congruence $ax \equiv b \pmod{m}$ has solutions if and only if the \textsc{gcd} of $a$ and $m$ divides $b$. 66 | \end{lemma} 67 | 68 | \begin{proof} 69 | Let $d$ be the \textsc{gcd} of $a$ and $m$. By Bezout, there exists some $X_0, Y_0 \in \Z$ such that $d = aX_0 + bY_0$. Since $d$ divides $b$ there exists some $k$ such that $b = dk$. Then $b = aX_0k + mY_0k$ so $b \cong aX \pmod{m}$ for $X = X_0k$. In the other direction, just write it out. If there is a solution then $b \cong aX \pmod{m}$ so $b = aX + mY$. Since the \textsc{gcd} divides the right hand side it must divide the left as well, so $d$ divides $b$. 70 | \end{proof} 71 | 72 | \begin{problem} 73 | Can there be lots of different solutions? What do solutions look like? 74 | \end{problem} -------------------------------------------------------------------------------- /MATH-354/Lectures/5.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, January 28} 4 | 5 | \subsection{Solving \texorpdfstring{$ax \equiv b \pmod{m}$}{ax = b (m)}} 6 | 7 | Recall from last lecture that $ax \equiv b \pmod{m}$ is solvable if and only if the \textsc{gcd} divides $m$. If we let $m' = m/d$ then the solutions are unique modulo $m'$. 8 | 9 | \begin{proof} 10 | Let $x_1,x_2$ be solutions to $ax_1 \equiv b \pmod{m}$ and $ax_2 \equiv b \pmod{m}$. Consider then that $a(x_1-x_2) \equiv 0 \pmod{m}$. Let $a' = a/d$. Then $da'(x_1-x_2) = dm'k$. We know that $m'$ divides $a'(x_1-x_2)$, and since $(m',a') = 1$ we know that $m'$ divides $x_1-x_2$. 11 | \end{proof} 12 | 13 | \begin{corollary} 14 | If $(a,m) = 1$ then there is a unique solution to $ax \equiv b \pmod{m}$. 15 | \end{corollary} 16 | 17 | \begin{corollary} 18 | If $a \not\equiv 0 \pmod{p}$ for prime $p$ then there is a unique solution to $ax \equiv b \pmod{p}$ in $\Z/p\Z$. 19 | \end{corollary} 20 | 21 | \subsection*{Chinese Remainder Theorem} 22 | 23 | \begin{theorem}[Chinese Remainder Theorem] 24 | If we have $m_1, \dotsc, m_r$ all relatively prime and the system of equations 25 | \[ 26 | x \equiv a_1 \pmod{m_1}, \dotsc, a \equiv a_r \pmod{m_r}, 27 | \] 28 | then there is a unique solution modulo $M = m_1 \cdots m_r$. Alternatively, the rings 29 | \[ \Z/M\Z \cong \bigoplus_{i=1}^r \Z/m_i\Z \] 30 | are isomorphic. 31 | \end{theorem} 32 | 33 | \begin{lemma} 34 | If $a_1, \dotsc, a_r$ are pairwise relatively prime to $m$ then the product $a_1 \cdots a_r$ is also relatively prime to $m$ as well. 35 | \end{lemma} 36 | 37 | \begin{lemma} 38 | If $a_1, \dotsc, a_r$ all divide $m$ and are all pairwise relatively prime to $m$ then the product $a_1 \cdots a_r$ divides $m$. 39 | \end{lemma} 40 | 41 | \begin{proof}[Proof of CRT] 42 | Let $\hat{M}_i = M/m_i = \prod_{j \not= i} m_i$. We find a helper $y_i$ such that $y_i \equiv 0 \pmod{\hat{M}_i}$ and $y_i \equiv 1 \pmod{m_i}$. Then we'll have that $x = \sum a_iy_i$. Note that $(\hat{M_i}, m_i) = 1$ so we know that $1 = x_i\hat{M}_i + y_im_i$ has a solution. Let $y_i = x_i\hat{M}_i$. THis shows existence. To show uniqueness, just apply Lemma~5.2 above. 43 | \end{proof} 44 | 45 | \subsection{Algorithmic Speed for the Chinese Remainder Theorem} 46 | 47 | The Euclidean Algorithm runs in logarithmic time in the inputs $a,b$. The worst case is when we plug in two consecutive Fibonacci numbers since they are recursively defined in almost the exact opposite way that Euclid's algorithm reduces numbers. -------------------------------------------------------------------------------- /MATH-354/Lectures/6.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Wednesday, January 30} 4 | 5 | \subsection{Pollard-\texorpdfstring{$\rho$}{rho} Factorization} 6 | 7 | We want to factor $n$. If we can find $0 < a,b 0$ such that $r^a = e \in G$. 51 | \end{definition} 52 | 53 | \begin{theorem} 54 | $\qty(\Z/p\Z)^\times$ is cyclic for prime $p$. 55 | \end{theorem} 56 | 57 | \begin{lemma} 58 | It is always true that the order of $r$ modulo $p$ is a divisor of $p-1$. 59 | \end{lemma} 60 | 61 | \begin{proof} 62 | Lagrange's Theorem. 63 | \end{proof} -------------------------------------------------------------------------------- /MATH-354/Lectures/8.tex: -------------------------------------------------------------------------------- 1 | \section{No Notes} -------------------------------------------------------------------------------- /MATH-354/Lectures/9.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{No Notes} -------------------------------------------------------------------------------- /MATH-354/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/MATH-354/notes.pdf -------------------------------------------------------------------------------- /MATH-354/notes.tex: -------------------------------------------------------------------------------- 1 | % !TEX program = pdflatex 2 | 3 | \documentclass{lnotes} 4 | 5 | % Packages 6 | \usepackage{epigraph} 7 | 8 | % Formatting 9 | \let\field\mathbf 10 | 11 | % Metadata 12 | \title{\lsstyle Number Theory} 13 | \course{\textsc{math} 354} 14 | \place{yale university} 15 | \term{spring} 16 | \year{2019} 17 | \blurb{ 18 | These are lecture notes for \textsc{math} 354b, ``Number Theory,'' taught by 19 | Ross Berkowitz at Yale University during the spring of 2019. 20 | These notes are not official, and have not been proofread by the instructor 21 | for the course. 22 | They live in my lecture notes respository at 23 | \[ 24 | \text{\url{https://github.com/jopetty/lecture-notes/tree/master/MATH-354}.} 25 | \] 26 | If you find any errors, please open a bug report describing the error and 27 | label it with the course identifier, or open a pull request so I can correct 28 | it. 29 | } 30 | 31 | \begin{document} 32 | 33 | \include{Lectures/1} 34 | \include{Lectures/2} 35 | \include{Lectures/3} 36 | \include{Lectures/4} 37 | \include{Lectures/5} 38 | \include{Lectures/6} 39 | \include{Lectures/7} 40 | \include{Lectures/8} 41 | \include{Lectures/9} 42 | \include{Lectures/10} 43 | \include{Lectures/11} 44 | \include{Lectures/12} 45 | \include{Lectures/15} 46 | \include{Lectures/16} 47 | \end{document} -------------------------------------------------------------------------------- /MATH-370/15.tex: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/MATH-370/15.tex -------------------------------------------------------------------------------- /MATH-370/Lectures/11.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thursday, 21 February 2019} 4 | 5 | \epigraph{``You are right to be skeptical of this calculus business''}{Asher} 6 | 7 | Recall our proposition: 8 | 9 | \begin{proposition} 10 | Let $f \in F[x]$ and let $E/F$ be an extension generated by some subset of the roots of $f$. Suppose that $\Omega/F$ is the splitting field of $f$. Then $1 \leq \#\Hom_F(E,\Omega) \leq [E:F]$ with upper-bound equality when $f$ has all distinct roots. 11 | \end{proposition} 12 | 13 | \begin{corollary} 14 | If $E,E'$ are splitting fields of $f$ then $E \isom E'$. 15 | \end{corollary} 16 | 17 | \begin{proof} 18 | Let $E = \Omega$ as in the above proposition, and exhibit an $F$-homomorphism. Then let $E' = \Omega$, and do the same. Since $F$-homs are injective, and we have two finite dimensional vector-spaces with injective homs between them, then they are isomorphic. 19 | \end{proof} 20 | 21 | \begin{corollary} 22 | If $E/F$ is any finite extension and $L/F$ is any extension then $\#\Hom_F(E,L) \leq [E:F]$ and there always exists an extension $\Omega/L$ such that $\Hom_F(E,\Omega) \not= \varnothing$. 23 | \end{corollary} 24 | 25 | \subsection{Multiple Roots} 26 | 27 | Let $f \in F[x]$ be a polynomial, and let $\Omega$ be a field in which $f$ splits, and let $\alpha \in \Omega$ is a root of $f$. 28 | 29 | \begin{definition}[Multiple, Simple Roots; Seperable] 30 | We say that $\alpha$ is multiple root of $f$ if and only if $(x-\alpha)^k$ divides $f$ over $\Omega$ for multiplicity $k > 1$. Otherwise, we say that $\alpha$ is a simple root. The polynomial $f$ is seperable if every root is simple. 31 | \end{definition} 32 | 33 | Over $\C$, we can test for multiple roots. Suppose that $f(x) = (x-\alpha)^2g(x)$. Consider the derivative of $f$, which is 34 | \[ f'(x) = 2(x-\alpha)g(x) + (x-\alpha)^2g'(x) = (x-\alpha)h(x). \] 35 | Notice that we can factor out a term of $(x-\alpha)$. More generally, if $f$ has a root of multiplicity at least $m$, then $f'$ has the same root with multiplicity at least $m-1$. This gives us the formula 36 | \[ \operatorname{mult}_\alpha f = \max\{i < \deg f \mid f^{(i)}(\alpha) = 0\}. \] 37 | This implies that $(x-\alpha)$ divides $\gcd(f,f')$. 38 | 39 | \subsection{The Derivative} 40 | 41 | For any polynomial $f = \sum a_ix^i$, define $f' = \sum ia_ix^{i-1}$. This has some nice properties: 42 | \begin{itemize} 43 | \item $d/dx\colon F[x] \to F[x]$ is $F$-linear; 44 | \item $d/dx(a) = 0$ for constant $a$; 45 | \item The Chain Rule; 46 | \item The Product Rule; 47 | \end{itemize} 48 | 49 | \begin{proof} 50 | Induction. 51 | \end{proof} 52 | 53 | \begin{proposition} 54 | Let $f \in F[x]$. Then $f$ is seperable if and only if $f$ and $f'$ are relatively prime. 55 | \end{proposition} 56 | 57 | \begin{proof} 58 | ``$\implies$'' Let $f$ be seperable, and let $\Omega$ be a field where $f$ splits, and let $\alpha \in \Omega$ be a root of $f$. Then $\alpha$ is simple if and only if $f(x) = (x-\alpha)h(x)$ and $h(\alpha) \not= 0$. Consider $f'(x) = h(x) + (x-\alpha)h'(x)$ by the chain/product rules, and so $f'(\alpha) \not= 0$. Thus no root of $f$ is a root of $f'$, and hence they share no common factors and so are relatively prime in $F$. 59 | 60 | ``$impliedby$'' Proof by contrapositive. Assume that $f$ is not seperable, and let $\alpha$ be a multiple root. Hence $f(x) = (x-\alpha)^2g(x)$ over $\Omega$. Then $f' = 2(x-\alpha)g(x) + (x-\alpha)^2g'(x)$, and so $f'(\alpha) = 0$. Then $f$ and $f'$ have a common root, and so $m_\alpha(x)$ divides both $f$ and $f'$ over $F$. Thus their \textsc{gcd} is not $1$, and so they are not relatively prime. 61 | \end{proof} 62 | 63 | \begin{exercise} 64 | Let $D\colon F[x] \to F[x]$ be an $F$-linear map satisfying the product rule and $D(x) = 1$. Prove that $D$ is exactly the derivative. (Hint: Try Induction.) 65 | \end{exercise} 66 | 67 | \begin{definition}[Seperable Extension] 68 | An algebraic etension $K/F$ is called seperable if every element $\alpha \in K$ satisfies a seperable polynomial over $F$. Equivalently, the minimal polynomial over $F$ of every $\alpha \in K$ is seperable. 69 | \end{definition} 70 | 71 | \begin{proposition} Some nice facts: 72 | \begin{enumerate} 73 | \item Let $f \in F$ be irreducible. Then $f$ is seperable if and only if the derivative doesn't vanish. 74 | \item If $\operatorname{char} F = 0$ (so $\Q \subseteq F$) then every irreducible polynomial is seperable. 75 | \item If $\operatorname{char} F = p$ (so $p=0 \in F$) then $f$ is not seperable if and only if $f(x)$ is a polynomial in $x^p$, so $f(x) = g(x^p)$. 76 | \end{enumerate} 77 | \end{proposition} -------------------------------------------------------------------------------- /MATH-370/Lectures/12.tex: -------------------------------------------------------------------------------- 1 | \section{No Notes} -------------------------------------------------------------------------------- /MATH-370/Lectures/13.tex: -------------------------------------------------------------------------------- 1 | \section{No Notes} -------------------------------------------------------------------------------- /MATH-370/Lectures/16.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Tuesday, 26 March 2019} 4 | 5 | Today's lecture was just random questions. 6 | Also, Grace Hopper is Awesome. -------------------------------------------------------------------------------- /MATH-370/Lectures/17.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thursday, 28 March 2019} 4 | 5 | \subsection{Grace Hopper's Thesis} 6 | 7 | -------------------------------------------------------------------------------- /MATH-370/Lectures/4.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thursday, January 24} 4 | 5 | \begin{definition}[Extension] 6 | Let $k$ be a field. Then $F \subseteq k$ is a subfield if $F \subseteq k$ is a unital subring and $F$ is a field. 7 | We say that $k$ is an extension of $F$ and write $k/F$. 8 | \end{definition} 9 | 10 | \begin{example} 11 | If $k$ is an extension of $F$ then $k$ has the structure of an $F$-vector space. The cannonical example for this is thinking of $\C$ as a two-dimensional $\R$-vector space with basis $\{1, i\}$. 12 | \end{example} 13 | 14 | \begin{definition} 15 | An extension $k/F$ is finite if $k$ is a finite dimensional $F$-vector space. The degree of $k/F$, written as $[k : F] = \dim_Fk$ is the $F$ dimension of $k$. 16 | \end{definition} 17 | 18 | \begin{example}[Examples of degrees] 19 | \begin{enumerate} 20 | \item $\C/\R$ is finite and $[\C : \R] = 2$. 21 | \item $\R/\Q$ is infinite since $n$-tuples of $\Q$ are countable and $\R$ is uncountable. 22 | \item Let $\Q(i) = \{x + iy \in \C \mid x,y \in \Q\}$. Then $\Q(i)/\Q$ is finite with degree $2$. 23 | \end{enumerate} 24 | \end{example} 25 | 26 | \subsection{Field extensions generated by elements} 27 | 28 | Let $F$ be a field contained in some field $\Omega$. For $\alpha_1, \dotsc, \alpha_n \in \Omega$ we can consider two objects: 29 | \begin{enumerate} 30 | \item $F[\alpha_1, \dotsc, \alpha_n] \subseteq \Omega$ is a subring 31 | \item $F(\alpha_1, \dotsc, \alpha_n) \subseteq \Omega$ is a subfield 32 | \end{enumerate} 33 | 34 | We define these equantities in the following way. 35 | \[ F[\alpha_1, \dotsc, \alpha_n] = \bigcap_{\substack{R \subseteq \Omega \\ F \subseteq R, \alpha_i \in R}} R = \left\{\sum a_{i_1} \cdots a_{i_n}\alpha_1^{i_1} \cdots \alpha_n^{i_n} \mid a_{i_1,\dotsc,i_n} \in F\right\} \] 36 | and 37 | \[ 38 | F(\alpha_1, \dotsc, \alpha_n) = \bigcap_{\substack{k \subseteq \Omega \\ F \subseteq k, \alpha_i \in k}} k = 39 | \left\{ 40 | \frac{\alpha}{\beta} \mid \alpha,\beta \in F[\alpha_1, \dotsc, \alpha_n], \beta \not= 0 41 | \right\}. 42 | \] 43 | We say that $F[\alpha_1, \dotsc, \alpha_n]$ is the subring of $\Omega$ generated by $\alpha_1, \dotsc, \alpha_n$ and $F(\alpha_1, \dotsc, \alpha_n)$ is the subfield of $\Omega$ generated by $\alpha_1, \dotsc, \alpha_n$. 44 | 45 | \begin{example} 46 | Consider $\Q[i]$ and $\Q(i)$. We say that $\Q[i]$ are rational polynomials in $i$ and $\Q(i)$ are quotients of these polynomials, understanding that even powers of $i$ and odd powers of $i$ to collapse it into rational and imaginary components. These are equal to one another. However, $\Q[\pi] \subset \Q(\pi)$ but they are not equal (since the field extension is not finite as $\pi$ is not algebraic). 47 | \end{example} 48 | 49 | The Laurent series $F((x))$ is important for Complex Analysis and is analogous to the formal power series $R[[x]]$. 50 | 51 | \begin{theorem}[Tower Law] 52 | Let $L/k$ and $k/F$ be field extensions. Then $[L : F] = [L : k] \cdot [k : F]$. 53 | \end{theorem} 54 | 55 | \begin{proof} 56 | If $L$ is a finite dimensional $F$-vector space then $L$ is a finite dimensional $k$-vector space since if $z_1,\dotsc,z_p$ is an $F$-basis for $L$ then in turn $\alpha = \sum a_iz_i \in F $ for all $\alpha \in L$ so $z_1,\dotsc,z_p$ is a generating set. On the other hand, $k \subseteq L$ is an $F$-vector space. Assume that $[L : k] = n$ with $x_i$ a $k$-basis for $L$ and $[k : F] = m$ with $y_i$ and $F$-basis for $k$. Then we first show that $\{x_iy_j\}$ is an $F$-basis of $L$ so $[L:F] = nm$. First, linear independence: assume that $\sum a_{i,j}x_iy_j = 0$ for $a_{i,j} \in F$. Rewrite this as $\sum\qty(\sum a_{i,j}y_i)x_i = 0$, so $\sum a_{i,j}y_i = 0$ so $a_{i,j} = 0$ and the set is linearly independent. Next, we show this generates $L$ over $F$. Let $\alpha \in L$ be written as $\alpha = \sum \beta_ix_i$ for $\beta_i \in k$. But for each $i$ we have $\beta_i = \sum a_{i,j}y_i$ so $\alpha = \sum\qty(\sum a_{i,j}y_i)x_i = \sum a_{i,j}x_iy_i$, so these generate. Then this is a basis, and the proposition holds. 57 | \end{proof} -------------------------------------------------------------------------------- /MATH-370/Lectures/5.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Tuesday, January 29} 4 | 5 | \begin{definition}[Algebraic] 6 | Let $k/F$ be a field extension and let $\alpha \in k$. Then $\alpha$ is algebraic over $F$ if there exists a nonzero $f \in F[x]$ such that $f(\alpha) = 0$. Otherwise, we say that $\alpha$ is trancendental over $F$. 7 | \begin{itemize} 8 | \item $\sqrt{2} \in \R$ is algebraic over $\Q$ since it satisfies $f(x) = x^2-2$; 9 | \item $\pi \in \R$ is trancendental over $\Q$. 10 | \end{itemize} 11 | \end{definition} 12 | 13 | \begin{definition}[Simple Field Extension] 14 | Let $F$ be a field. Then $k/F$ is a \emph{simple} field extension if $k = F(\alpha)$ for some single $\alpha \in k$. 15 | \end{definition} 16 | 17 | \begin{theorem} 18 | Let $F$ be a field, let $\Omega/F$ be a field extension, let $\alpha \in \Omega$, and let $k = F(\alpha)$. Then either 19 | \begin{enumerate} 20 | \item $\alpha$ is trancendental, in which case $k/F$ is not finite; or 21 | \item $\alpha$ is algebraic, in which case $k/F$ has finite degree $n$, $F[\alpha] = F(\alpha)$, and there exists a unique monic irreducible polynomial $m_\alpha(x) \in F[x]$ of degree $n$ such that $m_\alpha(\alpha) = 0$. This $m_\alpha$ is called the minimal polynomial of $\alpha$ over $F$. 22 | \end{enumerate} 23 | \end{theorem} 24 | 25 | \begin{lemma} 26 | If $\alpha$ is trancendental over $F$ then $k = F(\alpha)$ is infinite over $F$ and $F[\alpha] \cong F[x]$. 27 | \end{lemma} 28 | 29 | \begin{lemma}\label{lemma:minimal} 30 | The polynomial $m_\alpha$ is the unique monic polynomial of minimal degree which has $\alpha$ as a root; consequently, it is irreducible. 31 | \end{lemma} 32 | 33 | \begin{proof}[Proof of~\ref{lemma:minimal}] 34 | By the proof of $\textsc{euclidean} \implies \textsc{pid}$ (i.e., any element of minimal degree in an ideal of $F[x]$ generates the ideal). Hence $m_\alpha(x)$ generates the idea, and so it is the unique monic element of minimal degree. To show it is irreducible, suppose that $m_\alpha = fg$; then $m_\alpha(\alpha) = f(\alpha)g(\alpha) = 0$. Since $K$ has no zero divisors we know that $f(\alpha) = 0$ or $g(\alpha)= 0$. Assume without a loss of generality that $f(\alpha) = 0$. Assume (to get a contradiction) that $\partial f, \partial g < \partial m_\alpha$. However, if this is the case then we have a polynomial with $\alpha$ as a root with a smaller degree than $m_\alpha$, which we had already shown to be minimal. This presents a contradiction, and so $m_\alpha$ is irreducible. 35 | \end{proof} 36 | 37 | \begin{proof}[Proof of part 1] 38 | If $\alpha$ is trancendental then $f(\alpha) \not= 0$ for any $f \in F[x]$. 39 | There is a unique $F$-algebra homomorphism from $F[x] \to k$ such that $x \mapsto \alpha$. Since the kernel of this map is trivial (since nonzero polynomials don't satisfy $\alpha$) then this $F$-algebra homomorphism is injective. The image of this homomorphism is precisely $F[\alpha] \subseteq k$. Then by the First Isomorphism Theorem for Rings, we know that $F[x] \cong F[\alpha]$ since the kernel is trivial. But $F[\alpha]$ is an $F$-vector space with infinite $F$-dimension. Hence $F(\alpha)$ is also of inifinite dimension since $F[\alpha] \subset F(\alpha)$ is an $F$-subspace. Then the degree of $F(\alpha)$ over $F$ is infinite. 40 | \end{proof} 41 | 42 | \begin{proof}[Proof of part 2] 43 | Since $\alpha$ is alegebraic, we know that the homomorphism is not injective. Then its kernel is a nontrivial ideal $I \subseteq F[x]$, so by \textsc{pid}-ness $I = (g(x))$ for some nonzero $g \in F[x]$. We can call this polynomial $g = m_\alpha$. 44 | \end{proof} -------------------------------------------------------------------------------- /MATH-370/Lectures/6.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thursday, January 31} 4 | 5 | Recall from last lecture: We have a field $F \subseteq \Omega$ and some $\alpha \in \Omega$ with an evaluation homomoprhism $F[x] \to \Omega$ where $x \mapsto \alpha$ and the image is $F[\alpha]$. Then either this is injective, which means that $\alpha$ is trancendental, or we have a nontrivial kernel equal to $(m_\alpha(x)) \subseteq F[x]$ where $m_\alpha(x)$ is the minimal polynomial of $\alpha$ over $F$. 6 | 7 | Recall also from \textsc{math 350} that if $R$ is a \textsc{pid} but not a field and $I \subseteq R$ is a prime ideal then $I \subseteq R$ is a maximal ideal. 8 | 9 | \begin{corollary} 10 | If $\alpha$ is algebraic over $F$ then $F[x]/(m_\alpha(x))$ is a field which is isomorphic to $F[\alpha]$ by the first isomorphism theorem. 11 | \end{corollary} 12 | 13 | \begin{proof} 14 | Recall that the minimal polynomial is irreducible, and in a \textsc{pid} irreducibility implies that it is prime. Then $(m_\alpha)$ is a prime ideal and so maximal. Then $F[x]/(m_\alpha)$ is a field. 15 | \end{proof} 16 | 17 | \begin{corollary} 18 | If $\alpha$ is algebraic over $F$ then $F[\alpha] = F(\alpha)$. 19 | \end{corollary} 20 | 21 | \begin{proof} 22 | By definition $F[\alpha] \subseteq F(\alpha)$. Since $F[\alpha]$ is a field containing $F,\alpha$ and $F(\alpha)$ is the smallest extension of $F$ containing $\alpha$ then $F(\alpha) \subseteq F[\alpha]$, and so they are equal by dual containment. 23 | \end{proof} 24 | 25 | \begin{theorem} 26 | Let $f(X) \in F[x]$ be irreducible of degree $n$. Then $F[x]/(f)$ is a field extension of $F$ of degree $n$ with an $F$-basis $\{\bar{1}, \bar{x}, \bar{x}^2, \dotsc, \bar{x}^{n-1}\}$ where $\bar{x}$ is the coset in $F[x]/(f)$ represented by $x$. 27 | \end{theorem} 28 | 29 | \begin{proof} 30 | By the corollaries, $K = F[x]/(f)$ is a field extension of $F$. Let $f(x) = a_0 + a_1x + \cdots + a_nx^n$. We claim then taht $1, \bar{x}, \dotsc, \bar{x}^{n-1}$ is generates $K$ over $F$. We know that $K$ is generated by the infinite set $\bar{1}, \bar{x}, \bar{x}^2,\dotsc$. However we see that $\bar{x}^n = 1\frac{1}{a_n}(a_0 + a_1\bar{x} + \cdots + a_{n-1}\bar{x}^{n-1})$, and so on for $\bar{x}^{k}$ for $k \geq n$. All of these expressions are linear combinations of $\bar{1}, \bar{x}, \dotsc, \bar{x}^{n-1}$. Then we must show that these are all linearly independent. Suppose that 31 | \[b + b_1\bar{x} + \cdots + b_{n-1}\bar{x}^{n-1} = 0. \] 32 | Notice that $\bar{x} \in K$ satisfies $f(\bar{x}) = 0$ and $f$ is irreducible ofer $F$. Then $\frac{1}{a_n}f(x)$ is the minimal polynomial of $\bar{x} \in K$. Then no polynomial of smaller degree can have $\bar{x}$ as a root, yet we just found $\star$ to have $\bar{x}$ as a root, and so $b_i = 0$ for all $i$ and this set is linearly independent. 33 | \end{proof} 34 | 35 | \begin{corollary} 36 | Let $\alpha, \beta \in \Omega$ have the same minimal polynomial over $F$. Then $F(\alpha) \cong F(\beta)$. 37 | \end{corollary} 38 | 39 | \begin{definition} 40 | $F(\alpha_1, \dotsc, \alpha_n)$ is a field extension of $F$ which is finitely generated. 41 | \end{definition} 42 | 43 | \begin{theorem} 44 | Let $F$ be a field. 45 | \begin{enumerate} 46 | \item Any finite extension $k/F$ is algebraic, where every elemetn $\alpha \in k$ is algebraic over $F$. 47 | \item Any finitely-generated and algebraic extension $k/F$ is finite. 48 | \end{enumerate} 49 | \end{theorem} -------------------------------------------------------------------------------- /MATH-370/Lectures/7.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Tuesday, February 5} 4 | 5 | \subsection{Ruler and Compass Construction} 6 | 7 | Euclid had this book which tried to lay the foundations for geometry. Among these are \emph{constructions}. 8 | 9 | \begin{enumerate} 10 | \item You have a set $S, \mathcal{P} \subseteq \C$ which contain $0$ and $1$. 11 | \item Given any two points $p,q$ in $S$, you can draw a line $\overline{PQ}$ through $p$ and $q$. 12 | \item Given any two points $p,q$ in $S$ you can draw the circle $C_p(q)$ centered at $p$ going through $q$. 13 | \item Any intersections of lines or circles which can be drawn are now points in $\mathcal{P}$. 14 | \end{enumerate} 15 | 16 | We call $\mathcal{P} \subseteq \C$ the set of constructable numbers which contains $0,1$ and is closed under this method of constructions. We can also create this idea algorithmicly. We proceed with induction for $n \geq 0$. Let $\mathcal{P}_0 = \{0,1\}$ and let $\mathcal{L}_0 = \mathcal{C}_0 = \emptyset$. Let $\mathcal{L}_{n+1} = \{\ell_{p,q} \mid p,q \in \mathcal{P}_n\}$ and let $\mathcal{C}_{n+1} = \{C_p(q) \mid p,q \in \mathcal{P}_{n}\}$. Then \[\mathcal{P}_{n+1} = \{z \in \C \mid z \in L \cap L', C \cap L, C \cap C', L,L' \in \mathcal{L}_{n+1},C,C' \in \mathcal{C}_{n+1}\}.\] 17 | Hence $\mathcal{P}_0 \subseteq \mathcal{P}_1 \subseteq \cdots \subseteq \mathcal{P}$, and each $\mathcal{P}_n$ are finite. Then $\bigcup_{n \geq 0} \mathcal{P}_n = \varinjlim \mathcal{P}_n = \mathcal{P}$. Then $\mathcal{P}$ is countable. 18 | 19 | \begin{theorem} 20 | The set $\mathcal{P} \subset \C$ is a field called Pythagorean closure of $\Q$ in $\Q$. 21 | \end{theorem} 22 | 23 | \subsection{Some Basic Constructions} 24 | 25 | \begin{enumerate} 26 | \item Given a `line segment' $\overline{PQ}$ we can bisect it by drawing the circles $C_p(q)$ and $C_q(p)$ and then drawing the line connecting the intersections of these circles. 27 | 28 | \item Given a line $\ell$ and a $p \in \ell$ we can draw the perpendicular line to $\ell$ at $p$. Draw a circle $C_p(q)$ where $q$ is any other point on $\ell$. Then draw two circles of twice the radius centered at the intersections of the smaller circle with $\ell$, and then connect their intersections. 29 | 30 | \item Give a line $\ell$ and a point $p \notin \ell$ we can draw the parallel line to $\ell$ through $p$. Draw a circle centered at $p$ intersecting $\ell$. Draw the perpendicular bisector of $\ell$ through $p$. 31 | \end{enumerate} -------------------------------------------------------------------------------- /MATH-370/Lectures/8.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Wednesday, 7 January} 4 | 5 | More constructions. -------------------------------------------------------------------------------- /MATH-370/Lectures/9.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Tuesday, 12 January} 4 | 5 | Recall that $\mathfrak{p} \subseteq \C$ is the subfield of constructible numbers, and $\Q^{\mathrm{py}} \subseteq \Q$ is the Pythagorean closure of the rationals. We had a theorem that these were equal to one another. A corollary to this is the fact that $\alpha \in \mathfrak{p}$ if and only if there exists a tower of fields of relative degree two ending at $\Q(\alpha)$, which tells us that $[\Q(\alpha) : \Q] = 2^n$. This proves that we cannot double the cube since $[\Q(\sqrt[3]{2}) : \Q] = 3$. 6 | 7 | \subsection{Squaring the Circle} 8 | There is also the problem of squaring the circle, which is also impossible since if a circle has area $\pi$ then the sides of the square would need length $\sqrt{\pi}$, which is not algebraic in $\Q$. This implicitly assumes that all constructible numbers are algebraic, which isn't necessarily a trivial observation. However, part of proving that $\mathfrak{p}$ is a field is showing that $\mathfrak{p}/Q$ is algebraic. 9 | 10 | \subsection{Trisecting an Arbitrary Angle} 11 | The general statement of this problem is on the homework. Consider $\theta = 2\pi/3$. Then $\theta/3 = 2\pi/9$. The intersection of $\theta$ with the unit circle is $\omega$ and the intersection with $\theta/3$ with the unit circle is $\zeta_9$. Then the constructibility $\theta/3$ is then equivalent to the constructibility of $\zeta_9$. Since the minimal polynomial of $\cos(\zeta_9)$ has degree three, it is not constructible. 12 | 13 | \begin{lemma} 14 | The angle $2\pi/n$ is constructible if and only if $\zeta_n$ is constructible if and only if $\cos(2\pi/n)$ and $\sin(2\pi/n)$ are constructible. 15 | \end{lemma} 16 | 17 | \subsection{Splitting Fields} 18 | Let $F \subseteq \Omega$ be a field and let $f(x) \in F[x]$ have degree $n$. We say that $f$ splits in $\Omega$ if $f(x) = a \prod (x-\alpha_i)$ for $\alpha_i \in \Omega$, so it separates into linear factors. 19 | 20 | \begin{example} 21 | Consider $f(x) = x^3 - 2 \in \Q$. We know that $\sqrt[3]{2}$ is a real root. How does $f(x)$ split in $\Q\qty(\sqrt[3]{2})$. Since it has a root, we know it factors as $f = (x-\sqrt[3]{2})g(x)$ where $g \in \Q\qty(\sqrt[3]{2})[x]$ has degree two. We can calculate what $g$ is through polynomial division to get that 22 | \[ f(x) = \qty(x-\sqrt[3]{2})\qty(x^2 + \sqrt[3]{2}x + \sqrt[3]{2}^2). \] 23 | We can check the discriminant is negative to see that the remaining quadratic has complex roots, and so cannot split further over $\Q\qty(\sqrt[3]{2})$. 24 | \end{example} 25 | 26 | \begin{definition}[Splitting field] 27 | A splitting field for $f(x) \in F[x]$ is an extension $E/F$ such that 28 | \begin{itemize} 29 | \item $f(x)$ splits in $E$, so $f = a\prod (x-\alpha_i)$ 30 | \item $E = F(\alpha_1, \dotsc, \alpha_n)$. 31 | \end{itemize} 32 | In this way, it is kinda like the ``minimal field'' over which $f$ splits. 33 | \end{definition} 34 | 35 | \begin{theorem} 36 | Let $f \in F[x]$. Then a splitting field $E/F$ for $f$ exists, and moreover if $\deg(f) = n$ then $[E:F] \leq n!$. 37 | \end{theorem} 38 | 39 | \begin{example} 40 | Recall the Cyclotomic polynomial $\Phi_p(x) = (x^p-1)/(x-1)$ for prime $p$. Any $p$\textsuperscript{th} root of $1$ is a root since $\Phi_p(\zeta_p) = (\zeta_p^p-1)/(\zeta_p-1) = 0$. But $\zeta_p^k$ is also a root, and there are $p-1$ of them, which is the degree of $\Phi_p(x)$. So in fact $\Phi_p(x)$ splits over $\Q(\zeta_p)$. 41 | \end{example} 42 | 43 | \begin{proof}[Proof of the existence of a finite splitting field] 44 | We can find the splitting field by adjoining all the roots to $\Q$. Then its finitely generated and algebraic, so it is finite. 45 | \end{proof} 46 | 47 | \begin{proof}[Proof of the bound on the degree] 48 | Just divide the polynomial by the one root we do have to get a polynomial of degree $n-1$. Then repeat the procedure until we get linear terms. Then we have a tower where the relative degree of the $i$\textsuperscript{th} is $n-i$, and so the degree of the full extension is bounded by $n \cdot (n-1) \cdots 1 = n!$. 49 | \end{proof} 50 | 51 | \begin{theorem} 52 | Splitting fields are unique up to an $F$-isomorphism. 53 | \end{theorem} 54 | 55 | \begin{theorem} 56 | Let $f \in F[x]$ be a monic polynomial. Let $E/F$ be generated by some subset of the roots of $f$. Let $\Omega/F$ be a field extension where $f$ splits. Then 57 | \begin{enumerate}[label={\textup{(\alph*)}}] 58 | \item There exist $F$-homomorphisms $\phi_i : E \to \Omega$, and the number of distinct $\phi_i$ is at most $[E : F]$ with equality if $f$ has distinct roots; 59 | \item If $E$ and $\Omega$ are splitting fields then they are isomorphic. 60 | \end{enumerate} 61 | \end{theorem} -------------------------------------------------------------------------------- /MATH-370/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/MATH-370/notes.pdf -------------------------------------------------------------------------------- /MATH-370/notes.tex: -------------------------------------------------------------------------------- 1 | % !TEX program = pdflatex 2 | 3 | \documentclass{lnotes} 4 | 5 | % Packages 6 | % \usepackage[osf]{libertineRoman} 7 | \usepackage{epigraph} 8 | \usepackage{tikz-cd} 9 | 10 | % Customization 11 | \let\mathbb\mathbf 12 | \def\Hom{\operatorname{Hom}} 13 | 14 | % Metadata 15 | \title{Fields and Galois Theory} 16 | \course{\textsc{math} 370} 17 | \place{yale university} 18 | \term{spring} 19 | \year{2019} 20 | \blurb{ 21 | These are lecture notes for \textsc{math} 370b, ``Fields and Galois Theory,'' taught by Asher Auel at Yale University during the spring of 2019. 22 | These notes are not official, and have not been proofread by the instructor for the course. 23 | They live in my lecture notes respository at 24 | \[\text{\url{https://github.com/jopetty/lecture-notes/tree/master/MATH-370}.}\] 25 | If you find any errors, please open a bug report describing the error and label it with the course identifier, or open a pull request so I can correct it. 26 | } 27 | 28 | % Bibliography 29 | \usepackage[ 30 | style = alphabetic, 31 | ]{biblatex} 32 | \addbibresource{references.bib} 33 | 34 | \begin{document} 35 | 36 | \section*{Syllabus} 37 | 38 | \begin{tabularx}{\textwidth}{rX} 39 | \toprule 40 | \textbf{Instructor} & Asher Auel, \url{asher.auel@yale.edu} \\ 41 | \textbf{Lecture} & \textsc{tr} 11:35 \textsc{am} -- 12:50 \textsc{pm} in \textsc{lom} 215 \\ 42 | \textbf{Peer Tutor} & Arthur Azvolinsky, \url{arthur.azvolinsky@yale.edu} \\ 43 | \textbf{Section} & Math Lounge \\ 44 | \textbf{Exams} & Midterm 1: Febuary 19; \quad Midterm 2: April 9; \quad Final: May 3. \\ 45 | \textbf{Textbook} & \fullcite{textbook} \\ 46 | \bottomrule 47 | \end{tabularx} \\ 48 | 49 | The main object of study in Galois theory are roots of single variable polynomials. 50 | Many ancient civilizations (Babylonian, Egyptian, Greek, Chinese, Indian, Persian) knew about the importance of solving quadratic equations. 51 | Today, most middle schoolers memorize the ``quadratic formula'' by heart. 52 | While various incomplete methods for solving cubic equations were developed in the ancient world, a general ``cubic formula'' (as well as a ``quartic formula'') was not known until the 16th century Italian school. 53 | It was conjectured by Gauss, and nearly proven by Ruffini, and then finally by Abel, that the roots of the general quintic polynomial could not be solvable in terms of nested roots. 54 | Galois theory provides a satisfactory explanation for this, as well as to the unsolvability (proved independently in the 19th century) of several classical problems concerning compass and straight-edge constructions (e.g., trisecting the angle, doubling the cube, squaring the circle). 55 | More generally, Galois theory is all about symmetries of the roots of polynomials. 56 | An essential concept is the field extension generated by the roots of a polynomial. 57 | The philosophy of Galois theory has also impacted other branches of higher mathematics (Lie groups, topology, number theory, algebraic geometry, differential equations). 58 | 59 | This course will provide a rigorous proof-based modern treatment of the main results of field theory and Galois theory. 60 | The main topics covered will be irreducibility of polynomials, Gauss's lemma, field extensions, minimal polynomials, separability, field automorphisms, Galois groups and correspondence, constructions with ruler and straight-edge, theory of finite fields. 61 | The grading in Math 370 is very focused on precision and correct details. 62 | Problem sets will consist of a mix of computational and proof-based problems. 63 | 64 | Your final grade for the course will be determined by 65 | \[ \max\left\{ 66 | \begin{array}{cccc} 67 | \text{20\% homework} + \text{25\% midterm 1} + \text{25\% midterm 2} + \text{30\% final} \\ 68 | \text{20\% homework} + \text{25\% midterm 1} + \text{15\% midterm 2} + \text{40\% final} \\ 69 | \text{20\% homework} + \text{15\% midterm 1} + \text{25\% midterm 2} + \text{40\% final} 70 | \end{array} 71 | \right\}. \] 72 | 73 | \printbibliography 74 | 75 | \include{Lectures/1} 76 | \include{Lectures/2} 77 | \include{Lectures/3} 78 | \include{Lectures/4} 79 | \include{Lectures/5} 80 | \include{Lectures/6} 81 | \include{Lectures/7} 82 | \include{Lectures/8} 83 | \include{Lectures/9} 84 | \include{Lectures/10} 85 | \include{Lectures/11} 86 | \include{Lectures/12} 87 | \include{Lectures/13} 88 | \include{Lectures/14} 89 | \include{Lectures/15} 90 | \include{Lectures/16} 91 | \include{Lectures/17} 92 | 93 | \end{document} -------------------------------------------------------------------------------- /MATH-370/references.bib: -------------------------------------------------------------------------------- 1 | @misc{textbook, 2 | author = { Milne, James S. }, 3 | title = { Fields and Galois Theory (v4.60) }, 4 | year = { 2018 }, 5 | pages = { 138 } 6 | } -------------------------------------------------------------------------------- /PHIL-126/Lectures/1.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, January 14} 4 | 5 | \epigraph{``Do you know what Pop-tarts are?''}{Michael Della Rocca} 6 | 7 | \subsection{Lecture Outline} 8 | \begin{multicols}{2} 9 | \begin{enumerate} 10 | \item 10 Philosophers 11 | \item Our Problems 12 | \item Dialogue 13 | \item Mind-Body Problems 14 | \item Naturalism 15 | \item Freedom 16 | \item Causation 17 | \item Skepticism 18 | \item Idealism 19 | \item God 20 | \item Metaphysics \& Epistemology 21 | \item PSR \& The Finest Hour 22 | \item My Motto 23 | \item Copernicus 24 | \item Mechanism 25 | \item Old Science 26 | \begin{enumerate} 27 | \item Final Causes 28 | \item Substatial Forms 29 | \item Trust the senses 30 | \end{enumerate} 31 | \end{enumerate} 32 | \end{multicols} 33 | 34 | This class will cover 17\textsuperscript{th} and 18\textsuperscript{th} century philosophers, including René Descartes, Nicolas Malebranch, Benedict de Spinoza, Anne Conway, Gottfried Leibniz, Emilie du Châtelet, John Locke, George Berkeley, David Hume, and Immanuel Kant. These philosophers defined the themes and methods of modern philosophy today. They tried to understand what the world is like and what our place in it is. 35 | 36 | Philosophy is not teleological; older philosophers may have a \emph{better understanding} of some ideas than contemporary philosophers do --- this is one reason why we study older ideas. There is also a rich tradition of dialogue with older philosophers. Much of philosopher is thinkers responding to and criticizing older philosophers. 37 | 38 | \begin{problem}[The Mind-Body Problem] 39 | What is the relation, if any, between the physical body and the mind? How do mental states relate to bodily states? Are the two distinct or the same? Is the mind destroyed if the body is destroyed? Are we the same person we were in previous years? How to the mind and body interact? 40 | \end{problem} 41 | 42 | \begin{definition}[Naturalism] 43 | The belief that everything plays by the same rules, and that the laws of nature apply to everything (including immaterial things, like the mind). Advocated by Spinoza and Hume. 44 | \end{definition} 45 | 46 | \begin{problem}[Freedom] 47 | Does free will exist? Are we ever truly free? How are our choices actually made? How should we be held responsible for our actions? If determinism is true, how can we be free in any sense? 48 | \end{problem} 49 | 50 | \begin{definition}[Determinism] 51 | Everything that takes place now was determined in the past by earlier events. Past states necessitate the current state of the world. If you know a given state and all the natural laws, you can predict with certainty how the system will progress. 52 | \end{definition} 53 | 54 | \begin{problem}[Causation] 55 | What does it mean for one thing to determine another thing? How does one billiard ball cause another to move when they strike one another? When a rock breaks a window, does the rock cause the breakage, or does God? 56 | \end{problem} 57 | 58 | \begin{problem}[Skepticism] 59 | Do we really know that things exist or the state of things? What does knowing something actually mean? How do we know that we aren't all dreaming? Is there a deceiving God which invents a false reality for use to perceive? Do we really know future events, like the sun will rise tomorrow or if the eraser is let go then it will fall to the ground? 60 | \end{problem} 61 | 62 | \begin{definition}[Idealism] 63 | Physical objects which exist in the world (objective things) are dependent on the mind, and the perception of these objects by the mind is what instantiates them. 64 | \end{definition} 65 | 66 | \begin{problem}[God] 67 | Does God exist? How do we know? Can we prove it? Is God beholden to natural laws? Is God nature itself? 68 | \end{problem} 69 | 70 | \begin{definition}[Metaphysics] 71 | The study of what exists. 72 | \end{definition} 73 | 74 | \begin{definition}[Epistemology] 75 | The study of how do we know that something exists. 76 | \end{definition} 77 | 78 | \begin{definition}[Principle of Sufficient Reason] 79 | The claim that for everything that exists and every occurance, there is an explanation for it. There is always a way to understand something even if we don't understand it yet. Invoked in a philosophers' \emph{Finest Hour}. 80 | \end{definition} -------------------------------------------------------------------------------- /PHIL-126/Lectures/3.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Friday, January 18} 4 | 5 | \begin{multicols}{2} 6 | \begin{enumerate} 7 | \item Reasons for doubt 8 | \item Principles 9 | \item Piece of paper 10 | \item Dream argument 11 | \item Coherence 12 | \item Beliefs left standing 13 | \item The big guns 14 | \item Make my day 15 | \item Coming and going 16 | \item Rock bottom 17 | \item Archimedes 18 | \item \emph{Sum} 19 | \item \emph{Cogito ergo sum} 20 | \end{enumerate} 21 | \end{multicols} 22 | 23 | Descartes arrives at a radical skepticism, through which he attempts to further his three main aims. He wants to get rid of the skeptical doubts by the end of his meditations. Some original belifes come back, but his (old Aristotelian) views on physical objects do not. He doesn't want to doubt for the sake of doubt, but wants to have \emph{reasons} for doubt. He also knows he has infinitely many beliefs, so instead of going through each he goes after the ``fundamental pinciples'' of his beliefs. In particular, he scruitinizes beliefs which arise from his senses. Our eyes can be tricked (oasis, liquid on a hot road, square tower in the distance may look round, Dick Cheeny thinking his friend's face was a bird). However, this isn't a general reason to doubt the senses since there are always specific nonoptimal situations which give rise to the doubt, but under optimal circumstances we have no reason for doubt; for example, being here sitting by the fire with a piece of paper in my hand. To doubt this would make one insane. HOWEVER, what if I were dreaming? This does give rise to a valid reason for doubt. 24 | 25 | \begin{definition}[Valid Argument] 26 | A argument in which the conclusion necessarily follows from the premises. 27 | \end{definition} 28 | 29 | \begin{definition}[Sound Argument] 30 | A valid argument in which the premises are true. 31 | \end{definition} 32 | 33 | \begin{enumerate} 34 | \item[\textbf{Premise 1}]\label{des:p1} In order to be certain that I'm sitting, I must \emph{first} be certain that I'm not dreaming. 35 | \item[\textbf{Premise 2}]\label{des:p2} I can't be certain that I'm not dreaming. 36 | \item[\textbf{Conclusion}]\label{des:conc} I can't be certain that I'm sitting. 37 | \end{enumerate} 38 | 39 | This calls into doubt Physics, Astronomy, Medicine, and Empirical Science. Mathematics and Logic, however, he feels are safe from this doubt. This is a valid argument since \textbf{Premise 1} and \textbf{Premise 2} necessarily imply \textbf{Conclusion}. Descartes will eventually argue that it isn't sound though since it would imply that God is a deceiver, a premise which he later refutes in \emph{Meditations 3 \& 4}. -------------------------------------------------------------------------------- /PHIL-126/Lectures/4.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Wednesday, January 13} 4 | 5 | \begin{multicols}{2} 6 | \begin{enumerate} 7 | \item Cogito ergo sum 8 | \item Ambulo ergo sum 9 | \item Transparency 10 | \item Cogitatur 11 | \item Quid igitur sum? 12 | \item Sum res cogitant 13 | \item The essence of the mind 14 | \item Truth rule 15 | \item Ugly head 16 | \item The great chain of being 17 | \item Degrees of reality 18 | \end{enumerate} 19 | \end{multicols} 20 | 21 | Doubt presupposes existence, so the existence of doubt is an argument for the existence of `I.' Something like \emph{walking} wouldn't work since we can only be certain that we \emph{think} that we're walking, but we can be sure that we are thinking. In this way, Descartes believes that he is certain of the content of his own mind. This certainty opens a gap between the mind and the world, everything the mind percieves. The mind is transparent to itself, however. 22 | 23 | \begin{problem} 24 | How can Descartes be sure that the `I' which thinks is preserved between thoughts? Likewise, how can't we really only be sure that thought is going on? 25 | \[ \text{thinking exists} \implies \text{there is a thinking thing} \implies \text{I am that thinking thing} \] 26 | Descartes first argues that thought must have a bearer in some substance, and then argues that there is an inherent quality of self about thoughts (you can't be aware of other people's thoughts, so any thoughts one percieve must be one's own). \note{How does Descartes address the idea that the I which thinks each thought might be different each time, just with different false memories of other beings? Kant considers this, but Descartes likely views this as an implicit part of a thinking substance.} 27 | \end{problem} 28 | 29 | Descartes makes some claims about the nature of a thinking mind. Here are some possible claims one could make about the essence of the mind/self: The first two claims are addressed in the second meditation, the third is addressed in the sixth meditation. 30 | \begin{enumerate} 31 | \item I am essentially (necessarily) a thinking thing; Descartes said ``A thought alone is not seperable from me.'' 32 | \item I am not essentially (necessarily) extended; 33 | \item I am not an extended thing; \note{What exactly does \emph{extended} mean?} 34 | \item I am not identical to my body; 35 | \end{enumerate} 36 | 37 | \subsection{The Truth Rule} 38 | 39 | Descartes reflects on his certainty that he is a thinking thing, and seeks to extrapolate to understand what is a sufficient condition for being sure of something. He ends up with \emph{whatever I perceive very clearly and distinctly is true.} What constitutes clarity and distinctness is not really something Descartes addresses. Descartes also worries that a deceiving god could construct a clear and distinct argument which is actually false, and we could not be sure that we do not fall victim to our own assumtions about this claim. Thus Descartes seeks to first establish that God exists and that God is not a deceiver, and then the truth rule can necessarily follow. If he cannot then even the \emph{cogito ergo sum} argument is called into doubt once again. The proof Descartes proposed is almost universally taken to be wrong (but that doesn't mean that it's not valuable or not worth studying). 40 | 41 | \subsection{Descartes' Proof of God's Existence} 42 | 43 | There are many different possible kinds of beings which exist in the world, and they have different degrees of reality. All beings are made of substance, meaning that they have properties but are not themselves properties of anything else. \note{Need substances be material? Is the mind a substance?} Furthermore, substances have a greater degree of reality than their properties, states, or modes. This is because the properties of a substance depend on the substance yet the substance doesn't depend on the properties: You can have a table which isn't rectangular, but you can't just have `rectangular' without it being a rectangular \emph{something}. The highest degree of being would be an infinite substance since it would be independent of anything else, uncreated and permanent. Below this are created, finite substances like a table or a mind. Below even this are the properties of finite substances. Since God is superlative, if God exists then he must be an infinite substance. The kind of reality which substances have is called \emph{formal reality}, reality by the virtue of existance. This is contrasted with objective reality. -------------------------------------------------------------------------------- /PHIL-126/Lectures/5.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, January 28} 4 | 5 | \begin{multicols}{2} 6 | \begin{enumerate} 7 | \item Degrees of Reality; 8 | \item Formal Reality \& Objective Reality; 9 | \item Mister Ed; 10 | \item Two causal principles \& and the PSR; 11 | \item Whodunit; 12 | \item D.'s finest hour; 13 | \item Not a deceiver; 14 | \item Cartesian Circle; 15 | \item Intellectual Problem of Evil; 16 | \item Belief \& the Will. 17 | \end{enumerate} 18 | \end{multicols} 19 | 20 | \subsection{Degrees of Reality} 21 | 22 | Descartes' proof of God's existence hinges on different degrees of reality, which in some sense is a measure of independence; in Descartes' view, God (an infinite substance) ought to be independent on anything else while everything else ought to be dependent on God. Finite and/or extended substances (table, mind, Mr.\ Ed, etc.) all depend on God, and in turn the shape of a table depends on the table itself. 23 | 24 | \begin{definition}[Formal Reality] 25 | The reality something has by virtue of its existence. Usually a measure of how independent a thing is. Often also derives from the complexity of the thing; a machine would have more formal reality than a rock. 26 | \end{definition} 27 | 28 | \begin{definition}[Objective Reality] 29 | The objective reality of an idea is equal to the formal reality of the object of the idea has, or would have if it existed. Only ideas can have objective reality. 30 | \end{definition} 31 | 32 | \begin{example} 33 | Consider the idea of Mr.\ Ed. This has relatively low formal reality since it's just a though in a mind, but has relatively high objective reality since Mr.\ Ed himself has a relatively high formal reality. The idea of god has infinite objective reality. 34 | \end{example} 35 | 36 | \subsection{Two Causal Principles} 37 | 38 | \begin{proposition} 39 | The formal reality of a cause is greater than or equal to the formal reality of the effect. 40 | \end{proposition} 41 | 42 | \begin{proposition} 43 | The formal reality of the cause is greater than or equal to the objective reality of the effect. 44 | \end{proposition} 45 | 46 | From this, God asks ``What causes my idea of God?'' -------------------------------------------------------------------------------- /PHIL-126/Lectures/6.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Wednesday, January 30} 4 | 5 | \begin{multicols}{2} 6 | \begin{enumerate} 7 | \item Getting out of the circle; 8 | \item Intellectual Problem of evil; 9 | \item Belief \& the Will; 10 | \item Do bodies exist? 11 | \item Whodunit again? 12 | \item The Real distinction 13 | \item Leibniz's Law 14 | \item Mind-body problem 15 | \item Elisabeth (1618--1680) 16 | \item E's finest hour and D's darkest hour 17 | \item Mind-body union 18 | \end{enumerate} 19 | \end{multicols} 20 | 21 | \subsection{Theodicy and Free Will} 22 | 23 | It sure seems like Descartes' thinking involves circular reasoning, so much so that there's a name for it: `The Cartesian Circle.' Do try and get out of this, he somewhat restricts the scope of his doubt to no longer be skeptical of his current clear and distinct ideas. 24 | 25 | \begin{problem} 26 | Why can't God deceive our current clear and distinct ideas? Descartes doesn't really have an answer for that. 27 | \end{problem} 28 | 29 | \begin{problem} 30 | Why should a good God allow me to make any mistakes at all? 31 | \end{problem} 32 | 33 | Answering these problems is a form of \emph{Theodicy} --- trying to get God off the hook for the problems in the world. Descartes does this by saying that the problems are our fault entirly. Being mistaken is a function of a disconnect between the \emph{intellect} and the \emph{will}. When the will supercedes the intellect and makes judgements for which it has no justification then we get a false belief. But we shouldn't fault God for giving us extra will, so he is therefore off the hook for the problems of our intellectual errors. 34 | 35 | \begin{problem} 36 | I can't assent to the idea that I can fly. 37 | 38 | \begin{solution}[Response] 39 | Descartes says that you freely don't assent to this idea. You could if you want to, but you choose to weigh your past experiences against your current beliefs. The only things the will doesn't have control over are clear and distinct ideas, like $2+2 = 4$. Descartes ardently defends the freedom of will in almost all circumstances. 40 | \end{solution} 41 | \end{problem} 42 | 43 | \subsection{The Nature of Bodies (Meditation 5)} 44 | 45 | A body is something which is extended --- it exists in three dimensions --- and has size, shape, and motion. We don't have a clear and distinct idea about the existence of bodies, so we cannot use this to prove their existence. Descartes does argue that the mind must be distinct from the body, if we are to have a body in the first place. However, the fact that we have ideas of anything means that they must be caused by something. Furthermore, our sensory ideas of bodies come to us against our will so they cannot originate in the mind. They can't come from God since this would mean that God is a deciever for two reasons: first, if God does this we have no recourse to figure out the truth against our will, and this is a deception. Second, God has given us a propensity to believe our senses, and if these were fully false then God would be saddling us with a mistake which we have no means of correcting and a great propensity to give our assent to these ideas. This is not the action of a good and undeceptive God. Thus the only remaining possibility is that the cause of our sensory ideas of bodies must be the bodies themselves. 46 | 47 | \subsection{Mind-Body Distinction} 48 | 49 | Descartes' strategy for the distinguishing the mind and the body is extremely influential, and is still used by contemporary philosophers today. In Aristotelian tradition, the mind and body were intimately related; the mind was just the form of the body. For Descartes, they are very distinct things. Descaret's argues that 50 | \begin{enumerate} 51 | \item The mind is essentailly thinking; one cannot separate thoughts from the mind itself. 52 | \item The mind is not essentially extended; we don't need a physical conception of the mind to understand its function and for it to do its job. 53 | \item Unlike the mind, the body must be extended. 54 | \item The body is not essentailly thinking; there are many kinds of bodies which have no thoughts. 55 | \end{enumerate} 56 | This argument relies on something which will come to be called \emph{Leibniz's Law}. 57 | 58 | \begin{theorem}[Leibniz's Law] 59 | If $A$ and $B$ have different properties then $A$ and $B$ are not identical. 60 | \end{theorem} -------------------------------------------------------------------------------- /PHIL-126/Lectures/7.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, February 4} 4 | 5 | \begin{multicols}{2} 6 | \begin{enumerate} 7 | \item Mind-Body Interaction 8 | \item Pineal gland 9 | \item Intelligibility 10 | \item Princess E.\ 1618--1680 11 | \item Bohemian Rhapsody 12 | \item E.'s Finest Hour 13 | \item Mind-Body Union 14 | \item Malebranche 1638--1715 15 | \item What's the occasion? 16 | \item Not just mind-body, but body-body too 17 | \item No substantial forms 18 | \item Causation + conceptual connection 19 | \item M.'s finest hour 20 | \item Spinoza 1632--1677 21 | \end{enumerate} 22 | \end{multicols} 23 | 24 | \subsection{Bohemian Rhapsody} 25 | 26 | Descartes was also a scientist in his day, and he conjectured about the physical relation between the mind and the body. He thought that the \emph{pineal gland} was the locus of this interaction. Elisabeth of Bohemia doubted Descartes account of the mind interacts with the body; in particular she wondered how a nonphysical, nonextended thing like the mind could cause interactions with a physical, extended thing like the body. She thought that contact and extension were necessary to induce movement or feeling in a body. Descartes' response was to doubt that these were actually necessary for interaction. He thinks that two things of different natures can interact. This ability is attributed to God, ever the helpful fellow that he is. These causal connections are in our interest, and so that is the reason for their existence. Elisabeth doubts this, and Descartes comes back to say ``Don't think about it too much, it wouldn't make sense for use to have a good idea of the connection anyways.'' Descartes calls this a \emph{primitive notion} which seems like a way for Descartes explain away the flaws in his reason. 27 | 28 | \begin{problem} 29 | How does this not violate the principle of sufficient reason? 30 | 31 | \begin{solution}[Descarte's Response] 32 | Yeah, it does. Oops. But I only use the PSR to justify God, not to justify anything else. 33 | \end{solution} 34 | \end{problem} 35 | 36 | \subsection{Malebranche (1638--1715)} 37 | 38 | Malebranche was a Cartesian philosopher who came a bit after Descartes. He was mainly concerned with Theodicy. He believed in the Mind-Body distinction but he had a hard time accepting Descartes' account for how the two interact. Malebranche supposes that, on the occasion of physical interaction with our bodies, a feeling appears in our mind due to God. When we intend to move our body, God moves our body in response to this intention. This divine intervention occurs in a regular fashion. For Malebranche, this is a general theory that also explains body-body interaction; all interactions are resultant from God's will. This theory of intervention by God when certain things happen is called \emph{occasionalism}. 39 | 40 | \begin{problem} 41 | Doesn't this just mean that God is the ultimate cause of all suffering and pain? 42 | 43 | \begin{proof}[Malebranche's Response] 44 | God is only acting on our intentions; they are the ultimate cause of these things. The only causal power finite objects have is the will to accept the good which God presents to us or to turn away from it. 45 | \end{proof} 46 | \end{problem} 47 | 48 | \begin{problem} 49 | What about the phantom limb problem? 50 | 51 | \begin{proof}[Malebranche's Response] 52 | God set up the system in the simplest way possible. Extra checks to ensure that the effects are valid are superfluous to the ultimate design of the system. 53 | \end{proof} 54 | \end{problem} 55 | 56 | Malebranche, like Descartes, wants to get rid of the substantial forms of Plato. To do this, they stripped bodies of all causal power and delegate causal power to God, or maybe to the mind in the case of will. For Malebranche, a cause is something which by its nature is connected to the effect. -------------------------------------------------------------------------------- /PHIL-126/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/PHIL-126/notes.pdf -------------------------------------------------------------------------------- /PHIL-126/notes.tex: -------------------------------------------------------------------------------- 1 | % !TEX program = xelatex 2 | 3 | \documentclass{lnotes} 4 | 5 | % Packages 6 | \usepackage{multicol} 7 | \usepackage{epigraph} 8 | \usepackage{fontspec} 9 | 10 | % Fonts 11 | \setmainfont[ 12 | Ligatures = {Common, Discretionary}, 13 | Numbers = {OldStyle} 14 | ]{Libertinus Serif} 15 | 16 | % \setsansfont{Libertinus Sans} 17 | 18 | % Document Metadata 19 | \title{Introduction to Modern~Philosophy} 20 | \year{2019} 21 | \term{spring} 22 | \place{yale university} 23 | \course{phil 126} 24 | \year{2019} 25 | \blurb{ 26 | These are lecture notes for \textsc{phil} 126b, ``Introduction to Modern Philosophy,'' taught by Michael Della Rocca at Yale University during the spring of 2019. 27 | These notes are not official, and have not been proofread by the instructor for the course. 28 | They live in my lecture notes respository at 29 | \[\text{\url{https://github.com/jopetty/lecture-notes/tree/master/PHIL-126}.}\] 30 | If you find any errors, please open a bug report describing the error and label it with the course identifier, or open a pull request so I can correct it. 31 | } 32 | 33 | \begin{document} 34 | 35 | \include{Lectures/1} 36 | \include{Lectures/2} 37 | \include{Lectures/3} 38 | \include{Lectures/4} 39 | \include{Lectures/5} 40 | \include{Lectures/6} 41 | \include{Lectures/7} 42 | 43 | \end{document} -------------------------------------------------------------------------------- /PHIL-175/Lectures/1.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{January 15, 2019} 4 | 5 | \epigraph{``You are welcome to sit adoringly at my feet if you so wish.''}{Shelly} 6 | 7 | Shelly (he asked us to call him that) talked about how there really isn't a good definition of what Philosophy is, perhaps unlike many other academic disciplines. Rather, it is defined by the types of things we ask questions about and the types of thoughts we study. Perhaps it's a ``Roller-coaster for your mind.'' 8 | 9 | \begin{definition}[Ethics] 10 | Moral Philosophy or Ethics is the attempt to answer the question ``How should I live?'' In the course of answering this, various other questions will arise and must be answered along the way. 11 | \end{definition} 12 | 13 | This doesn't deal with the empirics of ``How \emph{do} we live.'' That is a realm for sociologists. We're also not really interested in people's opinions on justified actions --- we're concerned with what the truth of the matter actually is. There will be no ``comparative ethics'' here. We're also not doing a history of philosophical ideas here. 14 | 15 | Philosophy thrives on rival views, not on proven results. There isn't a philosophical answer to ``what is well being?'' --- you have to decide for youself how to answer that question. -------------------------------------------------------------------------------- /PHIL-175/Lectures/2.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{January 17, 2019} 4 | 5 | \epigraph{``The blood gurgled out of his neck as he gasped his last breath.''}{Shelly} 6 | 7 | \subsection*{Outline of Lecture} 8 | 9 | \begin{enumerate} 10 | \item Application of Morality --- Applied/Practical Ethics 11 | \item Content of Morality --- Normative Ethics 12 | \item Foundation of Morality --- Normative \& Metaethics 13 | \item Nature of Morality --- Normative and Metaethics 14 | \end{enumerate} 15 | 16 | \subsection*{Questions of Moral Philosophy} 17 | 18 | \begin{example}[Questions of Moral Philosophy] 19 | \begin{parts} 20 | \part What are the various moral rules? What are the specific details? For example, if ``Don't kill'' is the rule, is it every permissible to kill people? When and why? If a child picks up a loaded gun and the only way to stop them from killing people is to shoot the child, is that moral? \emph{(This is called an innocent threat)} Are there proportionality considerations? 21 | 22 | \part Do we have obligations to keep our promises? What if we have conflicting promises? How should we weight our promises? Are there ever a valid set of rules we can follow? Can we break our promises if a more pressing situation arises where we could do more good (saving someone in cardiac arrest but skipping a meeting to do so)? 23 | 24 | \part Can war ever be moral? What about collateral damage and civilian deaths? 25 | 26 | \part Is abortion moral or immoral? Does killing a fetus break the ``don't kill'' command? When do we have rights to life? 27 | \end{parts} 28 | \end{example} 29 | 30 | The subfields of morality (although there really isn't a clear line between these areas): 31 | 32 | \begin{enumerate} 33 | \item \textbf{Content of Morality} --- What are the rules of a just society? What factors make an action right or wrong? 34 | \item \textbf{Application of Morality} --- How can we apply the rules of morality to real-life situations? What happens when situations are ambiguous? What conflicts arise? Abortion and Capital Punishment? This class doesn't really touch on this. 35 | \item \textbf{Foundation of Morality} --- Can our rules be derived from one another? Is a prohibition on killing entirely separate from ``Do not harm people'' or is it just a specific case one the more general rule? How to we systematize ethical rules? Is there a single basic rule from which all others derive (like the Golden Rule or Categorical Imperative or Utilitarianism)? What makes the true rules true? Is true even the right word? Why are moral systems just? By virtue of what is morality founded? Divine commandment, contractarianism, etc. 36 | \item \textbf{Nature of Morality} --- What's ethics all about? What's the point of ethics? What does it even mean to be moral? Why are some things more valid than others? Why do values even exist? How do we include morality in a scientific worldview? Are there ``morality facts''? How could we ever hope to know whether something is right or wrong? What do ``right'' and ``wrong'' even mean? We probably won't touch on this here either (Take Moral Skepticism with Shelly in the fall). 37 | \end{enumerate} -------------------------------------------------------------------------------- /PHIL-175/Lectures/3.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Tuesday, January 22} 4 | 5 | \epigraph{``If you planted a bomb in a school yard, I guess that would be a pretty bad thing''}{Shelly} 6 | 7 | A lot of what drives moral philosophy is ``why be moral,'' but to answer this we first must have an idea of what morality actually is. Our first framework will be \emph{Utilitarianism}, popularized by Mill and Bentham, which will we will build up to. Shelly made a note that although we find objections to this or any view, this doesn't mean that they aren't worth considering, and since we can find objections about any view there isn't a ``perfect'' answer to these questions. Our goal is to understand the flaws and strengths of the arguments and gain a better understanding of morality and ethics through this. 8 | 9 | Mill does not own Utilitarianism, he merely wrote arguments to support it. Mill's words on the matter are not God's words on it (the same holds for Kant and \emph{Kantianism}), so it's important not to take this perspective. 10 | 11 | Philosophy isn't like physics; this isn't a ``learn it or fail'' kind of course. You will need to decide for youself what the right answers are and justify why you think that. There aren't any \emph{knock-down} arguments in Philosophy but this doesn't mean that there aren't right answers. To compare to Theology, there are disagreements about whether or not god exists but that doesn't mean there is a truth value to that statement. Shelly wants us to know that he has bias himself and so do we, so it's important to be aware of this. He'll try to keep his cards close to his chest, so to speak. 12 | 13 | \begin{problem} 14 | If we believe something but there isn't a set of arguments which will pursuade any rational person of it, does that mean that our belief comes from a source outside of reason? 15 | \end{problem} 16 | 17 | Shelly used the analogy of a jury deliberating to illustrate how there can be rational conflicts. 18 | 19 | \subsection{Why have morality?} 20 | 21 | There are many possible answers. One possible answer is that different actions have different results, and we are not indifferent to the results of actions. We care about what happens in the world. This makes possible the thought that one should consider the results of one's actions; in a moral sense, we should consider whether the results are good or bad when deciding what one ought to do. Shelly calls this idea \emph{Resultism}. 22 | 23 | \begin{definition}[Resultism] 24 | The results of an action are morally relevant to determine the moral value of the action. 25 | \end{definition} 26 | 27 | The modest statement of this is that the results are \emph{one of the things} which are relevent to this question; the strong statement says that results are the \emph{only} things one should consider. This \textbf{bold resultism} is an agreement in Utilitarianism. Pretty much every moral theory has the modest version at least. 28 | 29 | \begin{example}[Weak Resultism] 30 | If you read the paper and saw that many people died, almost everyone would agree that the results of that earthquake were bad (having been given no other information). 31 | \end{example} 32 | 33 | In order to use the results as a basis for morality, we need some way to measure the goodness or badness of results. This forms what's called \textbf{a theory of the good}. This lets us rank outcomes from better to worse. Here are some elements we may want to include in such a theory (intuitively plausible ideas, acceptible \emph{prima facie}): 34 | \begin{itemize} 35 | \item What are the short term results? 36 | \item What are the long term results? 37 | \item What are the effects on \emph{everybody}? (constasts with the idea that morality should only consider how one's actions affect oneself, known as \emph{ethical egoism}) 38 | \end{itemize} 39 | 40 | \begin{example}[Qualities we should care about?] 41 | \begin{enumerate} 42 | \item Is there a difference between planting a bomb in a school yard which will go off today, tomorrow, next year, in 100 years? Would any length of time be acceptible? [Long Term Results] 43 | \item If we store nuclear waste in a cavern in Utah which won't leak for a thousand years, is that still acceptible if it kills 10,000 Utahans \emph{eventually}? [Long Term Results] 44 | \item Can we put our nuclear waste in Kenya so that no Americans are killed? [Who is affected] 45 | \end{enumerate} 46 | \end{example} 47 | 48 | The various flavors of ethical egoism: 49 | \begin{enumerate} 50 | \item Rational egoism: Asking ``what should I do'' as a matter of rationality; this is the in-house religion of the department of economics. 51 | \item Ethical egoism: Should I consider how my actions affect someone else, or just consider what is good for me. 52 | \item Psychological egoism: Our actions are entirely determined by what we think is good for ourselves individually. 53 | \end{enumerate} 54 | You don't need to believe one to believe the other. Bentham believed rational but not ethical. 55 | 56 | \begin{example} 57 | Suppose there is a lever which you can pull to get a candy bar. However, when you pull the lever 10,000 people in France are electrocuted. For an ethical egoist, this isn't a moral dilemma: you should just pull the lever. 58 | \end{example} -------------------------------------------------------------------------------- /PHIL-175/Lectures/4.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thursday, January 24} 4 | 5 | \epigraph{``Ice cream is valuable because of the pleasurable sensations.''}{Shelly} 6 | 7 | \epigraph{``I just don't know what to say about masochists.''}{Shelly} 8 | 9 | \subsection{What is the correct theory of the good?} 10 | 11 | Generally we want to have a theory which is impartial and considers everyone equally. Sometimes we don't want to single out some people over others. But that still doesn't tell us how to pick one set of results over another. 12 | 13 | \begin{proposition} 14 | Let's measure how well-off people are after the different results. The modest way says that this is a relevant factor, the bolder way says it is the only thing thats relevant. 15 | \end{proposition} 16 | 17 | \begin{definition}[Welfarism] 18 | The view that people's welfare is the only relevant criterion in determining which actions are morally better than another. 19 | \end{definition} 20 | 21 | This still doesn't give us a good account of well-being, and it doesn't tell us how to aggregate individual well-being to understand how good or bad an outcome is as a whole. We have to answer both of these questions. Should it be based on material or personal goods, like money or vacations or respect or sex or one's books to be published? In turn we need to understand why each of these good (or bad) things is valuable in its own right. It seems like the root of all these things is ``pleasure'' --- everything else is an instrument to acquire pleasure. 22 | 23 | \begin{problem} 24 | Can pleasure be both inherently good and instrumentally good? Yes, cf.\ reproduction of genes through pleasurable activities. 25 | \end{problem} 26 | 27 | \begin{example} 28 | The pain you get at the dentist is intrinsically bad (because pain is bad) but instrumentally good (since it helps you avoid further pain and live a healthier life). 29 | \end{example} 30 | 31 | \begin{problem} 32 | Can something be both intrinsically good and intrinsically bad? What about masochism? 33 | \end{problem} 34 | 35 | \begin{problem} 36 | How do we know if our list of inherently good and bad things complete? Are all pleasures equally good? Are all pains equally bad? 37 | \end{problem} 38 | 39 | For this first question, Welfarism contends that pleasure and pain are the only good and bad things on that list. 40 | 41 | \begin{definition}[Hedonism] 42 | The view that the value of a life is the sum difference of pleasures and pains. 43 | \end{definition} 44 | 45 | \begin{example}[Painless Death] 46 | If you life is more good than bad, death would probably be bad. If your life is worse than it is good, death might be good. If you believe that the afterlife is amazing then death wouldn't really be a problem. 47 | \end{example} 48 | 49 | \begin{problem} 50 | Do Hedonists consider the pleasure of other people? 51 | \end{problem} 52 | 53 | \begin{definition}[Quantitative Hedonists] 54 | A hedonist who wants to measure the amount of pain or pleasure in an experience to rank them. 55 | \end{definition} -------------------------------------------------------------------------------- /PHIL-175/Lectures/5.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Tuesday, January 29} 4 | 5 | \epigraph{``I've never been a pig.''}{Shelly} 6 | 7 | Mill differentiates himself from Bentham by accepting that the quality of pleasures is important, even moreso than the quantity of the pleasures. This leads to a hierarchy of pleasures which we should prefer. The social and intellectual and spiritual and aesthetic pleasures are `higher' than the more base pleasures. In order to distinguish which are better than others, Mill proposes that if we ask those who understand and experience both and a decided majority support one over the other then that is the higher good. 8 | 9 | \begin{problem} 10 | How do we actually know that this is the answer people would give? Mill seems to priveledge these particular `highbrow pleasures' but then \emph{justify} that choice by saying that `people will naturally choose that.' But what if people don't choose that --- would Mill still hold those higher virtues above the lower ones? 11 | \end{problem} 12 | 13 | \begin{problem} 14 | If we accepted Qualitative Hedonism, wouldn't that imply that we should force people to do these higher pleasures over the lower ones? 15 | 16 | \begin{solution}[Response] 17 | Right now we're discussing a theory of well-being, not a theory of what one ought to do. We might accept that someone might be happier if we forced them to do certain things, but we don't say that we must force people to be happer, only recognize that this is something which would make them happier. Furthermore, \emph{how} we experience the world is just as important as the thing we percieve; if we don't like classical music then we aren't experiencing any pleasure regardless of the supposed benefit of seeing it. 18 | \end{solution} 19 | \end{problem} 20 | 21 | \begin{proposition} 22 | In order to understand and appreciate the higher virtues, we must train them to appreciate them through education. 23 | \end{proposition} 24 | 25 | \begin{problem} 26 | Five-year-olds seem to be a lot happier than adults. 27 | 28 | \begin{solution}[Response] 29 | Would you really choose to be a five-year-old for the rest of your life? Most people (I think) wouldn't. 30 | \end{solution} 31 | \end{problem} 32 | 33 | \begin{problem} 34 | Is getting the right mental state all that matters? Imagine that scientists can stimulate your brain directly to create experiences identical to those you would have while doing anything else. For example, you could fully experience what it would feel like to climb mount everest while inside the machine the whole time. But while you are in this machine, you don't know that you are in the machine since this knowledge would change your perspective. Do you want to be hooked up to the machine? 35 | 36 | \begin{solution}[Response (Personal)] 37 | I don't think so. 38 | \end{solution} 39 | \end{problem} 40 | -------------------------------------------------------------------------------- /PHIL-175/Lectures/6.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thursday, January 31} 4 | 5 | \epigraph{``I went to the pound since Tuesday's class and bought a stray cat. Brought it home, doused it in gasoline, and lit it on fire while it shreiked and died an agonizing death....and that seems wrong.''}{Shelly} 6 | 7 | The results of the experience machine experiment tell us that most people doen't really buy Hedonism --- there's more to well-being than just the experience. There are many different theories which propose ideas about what the missing pieces are. One alternative is that we must have the things we want, and when we accomplish this our life is better than when we don't. This theory explains why the experience machine fails; we want the real things, not merely the experience. These Preference Theories are the in-house philosophy of the department of economics. 8 | 9 | It also seems like it ought to matter how welfare is aggregated. In some cases this is easy; if everybody is better off in world $A$ than in world $B$ its clear than $A$ presents a better existence. But it's often not so clear cut. Perhaps we ought to use the maximum happiness in the world, or perhaps we should use the average amount of happiness. There are scenarios which make either look bad. This has been going on for forty years and philosophers still don't have a consensus. These theories all require us to have interpersonal comparisons of happiness or welfare, which may not even be possible. 10 | 11 | \begin{problem} 12 | Do animals count in the welfare aggregation? 13 | \end{problem} -------------------------------------------------------------------------------- /PHIL-175/Lectures/7.tex: -------------------------------------------------------------------------------- 1 | \section{No Notes} -------------------------------------------------------------------------------- /PHIL-175/Lectures/8.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Thusrday, February 7} 4 | 5 | \epigraph{``They're graduate students in philosophy, they won't contribute \emph{anything} to society.''}{Shelly} 6 | 7 | \subsection*{Does Utilitarianism Give Plausible Answers?} 8 | 9 | \begin{problem} 10 | Why is murder wrong? 11 | 12 | \begin{solution}[Utilitarianism Response] 13 | Because you had other options, and the option to murder someone decreased happiness/welfare. 14 | \end{solution} 15 | \end{problem} 16 | 17 | \begin{problem} 18 | Should we steal from the blind man in the subway? 19 | 20 | \begin{solution}[Utilitarianism Response] 21 | No. 22 | \end{solution} 23 | \end{problem} 24 | 25 | \begin{problem} 26 | We are sailing a boat when we see someone drowning; we turn to rescue them, but then we notice five people drowning somewhere else. We don't have time to save everyone. What do we do? 27 | 28 | \begin{solution}[Utilitarianism Response] 29 | It's better to save five than to save one. 30 | \end{solution} 31 | 32 | What if we tweak the scenario? What if the one was about the find the cure for cancer, and the five were all graduate students in philosophy. What about the impacts of what these people contribute to society? 33 | 34 | \begin{solution}[Utilitarianism Response] 35 | Those things matter. We can't say that everything is equal, as before. 36 | \end{solution} 37 | \end{problem} 38 | 39 | What if we aren't sure of the outcomes? We can't know everything, after all. What should a utilitarianism (or a consequentialist) do in light of this? What if you accidentally save Hitler from drowning? 40 | 41 | \begin{example} 42 | There's a distinction here in what we're asking: Did you do the right thing or are you a morally good person? Utilitarianism and consequentialism are concerned with actions, and so in and of themselves do not concern themselves with whether or not people are good. We can develop a utilitarian theory of this, but it is supplemental to the core of the philosophy. 43 | \end{example} -------------------------------------------------------------------------------- /PHIL-175/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/PHIL-175/notes.pdf -------------------------------------------------------------------------------- /PHIL-175/notes.tex: -------------------------------------------------------------------------------- 1 | % !TEX program = xelatex 2 | 3 | \documentclass{lnotes} 4 | 5 | % Packages 6 | \usepackage{multicol} 7 | \usepackage{epigraph} 8 | 9 | % Fonts 10 | \setmainfont[ 11 | Ligatures = {Common, Discretionary}, 12 | Contextuals = {Alternate}, 13 | Numbers = {OldStyle} 14 | ]{Libertinus Serif} 15 | 16 | % Document Metadata 17 | \title{Introduction to Ethics} 18 | \year{2019} 19 | \term{spring} 20 | \place{yale university} 21 | \course{phil 126} 22 | \year{2019} 23 | \blurb{ 24 | These are lecture notes for \textsc{phil} 175b, ``Introduction to Ethics,'' taught by Shelly Kagan at Yale University during the spring of 2019. 25 | These notes are not official, and have not been proofread by the instructor for the course. 26 | They live in my lecture notes respository at 27 | \[\text{\url{https://github.com/jopetty/lecture-notes/tree/master/PHIL-175}.}\] 28 | If you find any errors, please open a bug report describing the error and label it with the course identifier, or open a pull request so I can correct it. 29 | } 30 | 31 | \begin{document} 32 | 33 | \include{Lectures/1} 34 | \include{Lectures/2} 35 | \include{Lectures/3} 36 | \include{Lectures/4} 37 | \include{Lectures/5} 38 | \include{Lectures/6} 39 | \include{Lectures/7} 40 | \include{Lectures/8} 41 | 42 | \end{document} -------------------------------------------------------------------------------- /PHIL-427/Lectures/1.tex: -------------------------------------------------------------------------------- 1 | \section{Lecture 1} 2 | 3 | Didn't attend class. -------------------------------------------------------------------------------- /PHIL-427/Lectures/2.tex: -------------------------------------------------------------------------------- 1 | \section{Lecture 2} 2 | 3 | Didn't attend class. -------------------------------------------------------------------------------- /PHIL-427/Lectures/3.tex: -------------------------------------------------------------------------------- 1 | % !TEX root = ../notes.tex 2 | 3 | \section{Monday, January 28} 4 | 5 | -------------------------------------------------------------------------------- /PHIL-427/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/jopetty/lecture-notes/068ba5af3233663aa1368fc43ffc8f64ba1c0b9e/PHIL-427/notes.pdf -------------------------------------------------------------------------------- /PHIL-427/notes.tex: -------------------------------------------------------------------------------- 1 | % !TEX program = pdflatex 2 | 3 | \documentclass{lnotes} 4 | 5 | % Packages 6 | \usepackage[osf]{libertineRoman} 7 | 8 | % Metadata 9 | \title{Computability and Logic} 10 | \course{phil 427} 11 | \term{spring} 12 | \year{2019} 13 | \place{yale university} 14 | 15 | \begin{document} 16 | 17 | \include{Lectures/1} 18 | \include{Lectures/2} 19 | \include{Lectures/3} 20 | 21 | \end{document} -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Lecture Notes 2 | 3 | This repository contains lecture notes and study material for various classes that I have taken. 4 | I type up these notes as a means of learning the material and figuring out where I have gaps in my understanding of the course content. These notes are not TeX'ed up directly during the lecture, but are rather reconstructed from much more abbreviated notes that I jot down while I'm seeing the material for the first time. 5 | 6 | These notes are not official, and haven't been proofread by the instructor for each course. As such, they're liable to contain errors or typos. If you find such an error, I'd greatly appreciate it if you could open a pull request to fix the issue (if you are proficient in LaTeX) or open an issue (so that I can fix it). 7 | 8 | If you like the formatting of the notes, you are free to use my `notes` class from my [`coursework-latex`](https://github.com/jopetty/coursework-latex) repository. 9 | 10 | ## Courses 11 | | Course | Course ID | Instructor | Term | 12 | |--------|-----------|------------|------| 13 | |Abstract Algebra | MATH 350 | Miki Havlickova | Fall 2018 | 14 | | Galois Theory | MATH 370 | Asher Auel | Spring 2019 | 15 | | Number Theory | MATH 305 | Ross Berkowitz | Spring 2019 | 16 | | Modern Philosophy | PHIL 126 | Michael Della Rocca | Spring 2019 | 17 | | Introduction to Ethics | PHIL 175 | Shelly Kagan | Spring 2019 | --------------------------------------------------------------------------------