├── BY-NC-SA.pdf ├── .gitignore ├── mybibtex.bib ├── .github └── workflows │ ├── pull_request_check.yml │ ├── generate_release_pdf.yml │ └── dev_branch_check.yml ├── README.md ├── chapter4.tex ├── hmcpset.cls ├── chapter8.tex ├── Solution_to_Algebra_Chapter_0.tex ├── chapter6.tex ├── chapter5.tex ├── LICENSE ├── chapter7.tex └── chapter1.tex /BY-NC-SA.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/hooyuser/Solution-to-Algebra-Chapter-0/HEAD/BY-NC-SA.pdf -------------------------------------------------------------------------------- /.gitignore: -------------------------------------------------------------------------------- 1 | *.aux 2 | *.log 3 | *.out 4 | *.synctex.gz 5 | *.toc 6 | *.thm 7 | .history/* 8 | *.synctex(busy) 9 | 10 | Solution_to_Algebra_Chapter_0.pdf -------------------------------------------------------------------------------- /mybibtex.bib: -------------------------------------------------------------------------------- 1 | @book{aluffi2009algebra, 2 | title={Algebra: chapter 0}, 3 | author={Aluffi, Paolo}, 4 | volume={104}, 5 | year={2009}, 6 | publisher={American Mathematical Soc.} 7 | } 8 | 9 | 10 | -------------------------------------------------------------------------------- /.github/workflows/pull_request_check.yml: -------------------------------------------------------------------------------- 1 | name: Pull Request Check 2 | on: [pull_request, workflow_dispatch] 3 | jobs: 4 | build_latex: 5 | runs-on: ubuntu-latest 6 | steps: 7 | - name: Set up Git repository 8 | uses: actions/checkout@v4 9 | - name: Compile LaTeX document 10 | uses: xu-cheng/latex-action@v2 11 | with: 12 | root_file: Solution_to_Algebra_Chapter_0.tex 13 | -------------------------------------------------------------------------------- /.github/workflows/generate_release_pdf.yml: -------------------------------------------------------------------------------- 1 | name: Generate Release PDF 2 | 3 | on: 4 | push: 5 | branches: 6 | - master 7 | paths-ignore: 8 | - '**.md' 9 | - '.gitignore' 10 | pull_request: 11 | branches: 12 | - master 13 | paths-ignore: 14 | - '**.md' 15 | - '.gitignore' 16 | workflow_dispatch: 17 | 18 | jobs: 19 | release_pdf: 20 | uses: hooyuser/pdf-ci-templates/.github/workflows/latex-pdf-release.yml@main 21 | with: 22 | TEX_FILE_NAME: Solution_to_Algebra_Chapter_0 23 | -------------------------------------------------------------------------------- /.github/workflows/dev_branch_check.yml: -------------------------------------------------------------------------------- 1 | name: Dev Branch Check 2 | on: 3 | push: 4 | branches: 5 | - dev 6 | paths-ignore: 7 | - '.github/**' 8 | - '.gitignore' 9 | - '**.md' 10 | env: 11 | TEX_FILE_NAME: Solution_to_Algebra_Chapter_0 12 | jobs: 13 | build_latex: 14 | runs-on: ubuntu-latest 15 | steps: 16 | - name: Set up Git repository 17 | uses: actions/checkout@v2 18 | - name: Compile LaTeX document 19 | uses: xu-cheng/latex-action@v2 20 | with: 21 | root_file: ${{ env.TEX_FILE_NAME }}.tex 22 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Solution to Algebra Chapter 0 2 | ![Lines of Code](https://img.shields.io/endpoint?url=https://raw.githubusercontent.com/hooyuser/github-badges/main/badges/hooyuser-Solution-to-Algebra-Chapter-0.json) 3 | ![GitHub repo size](https://img.shields.io/github/repo-size/hooyuser/Solution-to-Algebra-Chapter-0) 4 | 5 | **Preview Here**: [Solution to Algebra, Chapter 0.pdf](https://hooyuser.github.io/Solution-to-Algebra-Chapter-0/Solution_to_Algebra_Chapter_0.pdf) 6 | 7 | **Download Here**: [Solution to Algebra, Chapter 0.pdf](https://github.com/hooyuser/Solution-to-Algebra-Chapter-0/releases/latest/download/Solution_to_Algebra_Chapter_0.pdf) 8 | 9 | This is a solution manual for Paolo Aluffi's **ALGEBRA, Chapter 0**, a textbook that introduces abstract algebra using the language of categories. 10 | I find this approach to be particularly appealing. So I'm enthusiastic to share my solutions to the exercises in this book for others to reference. 11 | 12 | As a beginner, I'm trying to make my proof explict and detailed, but unfortunately it seems verbose and tedious sometimes. 13 | Anyway, I would appreciate it if you can point out my mistakes, big or small. 14 | 15 |
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20 | -------------------------------------------------------------------------------- /chapter4.tex: -------------------------------------------------------------------------------- 1 | \section{Chapter IV.\hspace{0.2em} Groups, second encounter} 2 | \subsection{\textsection1. The conjugation action} 3 | \begin{problem}[1.1] 4 | $\triangleright$ Let $p$ be a prime integer, let $G$ be a $p$-group, and let $S$ be a set such that $|S| \not \equiv 0 \bmod p .$ If $G$ acts on $S,$ prove that the action must have fixed points. $[\S 1.1$ $\S 2.3]$ 5 | \end{problem} 6 | \begin{solution} 7 | let $Z$ be the fixed-point set of the action. We have 8 | \[ 9 | |Z| \equiv|S| \quad \bmod p. 10 | \] 11 | Since $|S| \not \equiv 0 \bmod p$, there must be $|Z|\ne0$, which implies the action must have fixed points. 12 | \end{solution} 13 | 14 | \begin{problem}[1.2] 15 | Find the center of $D_{2 n}$. (The answer depends on the parity of $n$. You have actually done this already: \hyperlink{Exercise II.2.7}{Exercise II.2.7}. This time, use a presentation.) 16 | \end{problem} 17 | \begin{solution} 18 | \[ 19 | D_{2n}=\left\langle r, s \mid r^{n}=s^{2}=(sr)^{2}=e\right\rangle 20 | \] 21 | It is clear that $r^is^j\in Z(D_{2n})$ if and only if $r^is^j$ commutes with $r$ and $s$. That is, 22 | \[ 23 | (r^is^j)r=r(r^is^j)\iff s^jr=rs^j\iff j=0 24 | \] 25 | and 26 | \[ 27 | (r^is^j)s=s(r^is^j)\iff r^is=sr^i\iff i=0\text{ or }n. 28 | \] 29 | Therefore, 30 | \[ 31 | Z(D_{2n})=\{e,r^n\}. 32 | \] 33 | \end{solution} 34 | 35 | \begin{problem}[1.4] 36 | $\triangleright$ Let $G$ be a group, and let $N$ be a subgroup of $Z(G)$. Prove that $N$ is normal in $G .[\S 2.2]$ 37 | \end{problem} 38 | \begin{solution} 39 | Since for all $g\in G$, $a\in N$, 40 | \[ 41 | gag^{-1}=agg^{-1}=a\in N, 42 | \] 43 | we see $N$ is normal in $G$. 44 | 45 | \end{solution} 46 | 47 | 48 | \begin{problem}[1.5] 49 | $\triangleright$ Let $G$ be a group. Prove that $G / Z(G)$ is isomorphic to the group $\operatorname{Inn}(G)$ of inner automorphisms of $G .$ (Cf. \hyperlink{Exercise II.4.8}{Exercise II.4.8}.) Then prove Lemma 1.5 again by using the result of \hyperlink{Exercise II.6.7}{Exercise II.6.7}. $[\S 1.2]$ 50 | \end{problem} 51 | \begin{solution} 52 | Define 53 | \begin{align*} 54 | f:G/Z(G)&\longrightarrow \operatorname{Inn}(G)\\ 55 | gZ(G)&\longmapsto (\gamma_g:a\longmapsto gag^{-1}). 56 | \end{align*} 57 | We can check that $f$ is a homomorphism 58 | \[ 59 | f(g_1g_2Z(G))=\gamma_{g_1g_2}=\gamma_{g_1}\gamma_{g_2}=f(g_1Z(G))f(g_2Z(G)). 60 | \] 61 | Since 62 | \[ 63 | gZ(G)\in\ker f\iff \gamma_g=\mathrm{id}_G\iff\forall a\in G, gag^{-1}=a\iff g\in Z(G), 64 | \] 65 | we have 66 | \[ 67 | \ker f=Z(G), 68 | \] 69 | which means $f$ is injective. It is clear that $f$ is surjective. Thus we show $f$ is an isomorphism and $G / Z(G)\cong\operatorname{Inn}(G)$. 70 | 71 | 72 | \end{solution} 73 | 74 | \begin{problem}[1.6] 75 | $\triangleright$ Let $p, q$ be prime integers, and let $G$ be a group of order $p q .$ Prove that either $G$ is commutative, or the center of $G$ is trivial. Conclude (using Corollary 1.9$)$ that every group of order $p^{2}$, for a prime $p$, is commutative. $[\S 1.3]$ 76 | \end{problem} 77 | \begin{solution} 78 | Since $Z(G)\operatorname{\triangleleft} G$, it follows that $|Z(G)|\in\{1,p,q,pq\}$. Suppose $|Z(G)|=p\text{ or }q$, we have $|G/Z(G)|=p\text{ or }q$. Since groups of prime orders are cyclic, $G/Z(G)$ is cyclic. According to Lemma 1.5, it implies $G / Z(G)=\{e_G\}$, which yields a contradiction. Hence $|Z(G)|=1\text{ or }pq$, that is, 79 | \[ 80 | Z(G)=\{e_G\}\text{ or }G. 81 | \] 82 | Therefore, we prove that either $G$ is commutative, or the center of $G$ is trivial. 83 | 84 | Let $G$ be a group of order $p^{2}$. According to Corollary 1.9, the center of the nontrivial $p$-group $G$ is nontrivial. Therefore, $G$ must be commutative. 85 | 86 | \end{solution} -------------------------------------------------------------------------------- /hmcpset.cls: -------------------------------------------------------------------------------- 1 | % HMC Math dept HW class file 2 | % v0.04 by Eric J. Malm, 10 Mar 2005 3 | %%% IDENTIFICATION -------------------------------------------------------- 4 | \NeedsTeXFormat{LaTeX2e}[1995/01/01] 5 | \ProvidesClass{hmcpset} 6 | [2005/03/10 v0.04 HMC Math Dept problem set class] 7 | 8 | %%% INITIAL CODE ---------------------------------------------------------- 9 | 10 | % test whether the document is being compiled with PDFTeX 11 | \RequirePackage{ifpdf} 12 | 13 | 14 | %%% DECLARATION OF OPTIONS ------------------------------------------------ 15 | %% Header Options: header*, no header 16 | \newif\ifhmcpset@header 17 | 18 | % no header block in upper right hand corner 19 | \DeclareOption{noheader}{% 20 | \hmcpset@headerfalse% 21 | } 22 | 23 | % do print header block 24 | \DeclareOption{header}{% 25 | \hmcpset@headertrue% 26 | } 27 | 28 | %% Font Options: palatino*, cm 29 | \newif\ifhmcpset@palatino 30 | 31 | % use palatino fonts 32 | %\DeclareOption{palatino}{% 33 | % \hmcpset@palatinotrue% 34 | %} 35 | 36 | % use compuer modern fonts 37 | \DeclareOption{cm}{% 38 | \hmcpset@palatinofalse% 39 | } 40 | 41 | %% Problem Boxing: boxed*, unboxed 42 | \newif\ifhmcpset@boxed 43 | 44 | % box problem statements 45 | \DeclareOption{boxed}{% 46 | \hmcpset@boxedtrue% 47 | } 48 | 49 | % don't box problem statements 50 | \DeclareOption{unboxed}{% 51 | \hmcpset@boxedfalse% 52 | } 53 | 54 | % pass remaining options to article class 55 | \DeclareOption*{\PassOptionsToClass{\CurrentOption}{article}} 56 | 57 | %%% EXECUTION OF OPTIONS -------------------------------------------------- 58 | %% default to: 59 | % including header, 60 | % loading mathpazo package for palatino fonts, 61 | % boxing problem statements 62 | \ExecuteOptions{header,palatino,boxed} 63 | 64 | \ProcessOptions 65 | 66 | %%% PACKAGE LOADING ------------------------------------------------------- 67 | %% based on std article class 68 | \LoadClass{article} 69 | 70 | 71 | %% Font loading: Palatino text/math fonts 72 | \ifhmcpset@palatino 73 | \RequirePackage{mathpazo} 74 | \fi 75 | 76 | %% AMSLaTeX math environments and symbols 77 | \RequirePackage{amsmath} 78 | \RequirePackage{amssymb} 79 | 80 | %% boxed minipage for boxed problem environment 81 | \RequirePackage{boxedminipage} 82 | 83 | %%% MAIN CODE ------------------------------------------------------------- 84 | %% Tell dvips/pdflatex correct page size 85 | \ifpdf 86 | \AtBeginDocument{% 87 | \setlength{\pdfpageheight}{\paperheight}% 88 | \setlength{\pdfpagewidth}{\paperwidth}% 89 | } 90 | \else 91 | \AtBeginDvi{\special{papersize=\the\paperwidth,\the\paperheight}}% 92 | \fi 93 | 94 | 95 | %% Problem set environments 96 | % boxed problem environment 97 | \newenvironment{problem}[1][]{% 98 | \ifhmcpset@boxed\def\hmcpset@probenv{boxed}\else\def\hmcpset@probenv{}\fi% 99 | \bigskip% put space before problem statement box % 100 | \noindent\begin{\hmcpset@probenv minipage}{\columnwidth}% 101 | \def\@tempa{#1}% 102 | \ifx\@tempa\empty\else% 103 | \hmcpset@probformat{#1}\hspace{0.5em}% 104 | \fi% 105 | }{% 106 | \end{\hmcpset@probenv minipage}% 107 | } 108 | % display optional argument to problem in bold 109 | \let\hmcpset@probformat\textbf 110 | 111 | % solution environment with endmark and optional argument 112 | \newenvironment{solution}[1][]{% 113 | \begin{trivlist}% 114 | \def\@tempa{#1}% 115 | \ifx\@tempa\empty% 116 | \item[]% 117 | \else% 118 | \item[\hskip\labelsep\relax #1]% 119 | \fi% 120 | }{% 121 | \mbox{}\penalty10000\hfill\hmcpset@endmark% 122 | \end{trivlist}% 123 | } 124 | 125 | % default endmark is small black square 126 | \def\hmcpset@endmark{\ensuremath{\scriptscriptstyle\blacksquare}} 127 | 128 | %% Problem set list, for top of document 129 | \newcommand{\problemlist}[1]{\begin{center}\large\sffamily{#1}\end{center}} 130 | 131 | %% commands for upper-right id header block 132 | \newcommand{\headerblock}{% 133 | \begin{flushright} 134 | \mbox{\hmcpset@name}\protect\\ 135 | \mbox{\hmcpset@assignment}\protect\\ 136 | \hmcpset@updatedate% 137 | \ifx\hmcpset@extraline\empty\else\protect\\\hmcpset@extraline\fi% 138 | \end{flushright}% 139 | } 140 | 141 | % put id header block at start of document 142 | \ifhmcpset@header\AtBeginDocument{\headerblock}\fi 143 | 144 | % internal state for headerblock 145 | \def\hmcpset@name{} 146 | \def\hmcpset@assignment{} 147 | \def\hmcpset@updatedate{} 148 | \def\hmcpset@extraline{} 149 | 150 | % commands to set header block info 151 | \newcommand{\name}[1]{\def\hmcpset@name{#1}} 152 | 153 | \newcommand{\assignment}[1]{\def\hmcpset@assignment{#1}} 154 | \newcommand{\updatedate}[1]{\def\hmcpset@updatedate{#1}} 155 | \newcommand{\extraline}[1]{\def\hmcpset@extraline{#1}} 156 | -------------------------------------------------------------------------------- /chapter8.tex: -------------------------------------------------------------------------------- 1 | \section{Chapter VIII.\hspace{0.2em} Linear Algebra, reprise} 2 | \subsection{\textsection1. Preliminaries, reprise} 3 | \begin{problem}[1.1] 4 | Let $\mathscr{F}: \mathsf{C} \rightarrow \mathsf{D}$ be a covariant functor, and assume that both $\mathsf{C}$ and $\mathsf{D}$ have products. Prove that for all objects $A, B$ of $C$, there is a unique morphism $\mathscr{F}(A \times B) \rightarrow \mathscr{F}(A) \times \mathscr{F}(B)$ such that the relevant diagram involving natural projections commutes. 5 | 6 | \hspace*{2em}If $D$ has coproducts (denoted II) and $\mathscr{G}: C \rightarrow D$ is contravariant, prove that there is a unique morphism $\mathscr{G}(A)\amalg\mathscr{G}(B) \rightarrow \mathscr{G}(A \amalg B)$ (again, such that an appropriate diagram commutes). 7 | \end{problem} 8 | \begin{solution} 9 | According to the universal property of $\mathscr{F}(A \times B)$ in $\mathsf{D}$, we have 10 | \[\xymatrix{ 11 | &\mathscr{F}(A \times B) 12 | \ar@{->}[d]_{\exists !} 13 | \ar@{->}[rd]^{\mathscr{F}(\pi_{\mathsf{C}_2})} 14 | \ar@{->}[ld]_{\mathscr{F}(\pi_{\mathsf{C}_1})} 15 | & \\ 16 | \mathscr{F}(A) 17 | &\mathscr{F}(A) \times \mathscr{F}(B) 18 | \ar@{->}[l]^{\pi_{\mathsf{D}_1}\hspace{1.2em}} 19 | \ar@{->}[r]_{\hspace{1.5em}\pi_{\mathsf{D}_2}} 20 | &\mathscr{F}(B) 21 | }\] 22 | Similarly, according to the universal property of $\mathscr{G}(A)\amalg\mathscr{G}(B)$ in $\mathsf{D}$, we have 23 | \[\xymatrix{ 24 | &\mathscr{G}(A \amalg B) 25 | & \\ 26 | \mathscr{G}(A) 27 | \ar@{->}[ru]^{\mathscr{G}(i_{\mathsf{C}_2})} 28 | \ar@{->}[r]_{i_{\mathsf{D}_1}\hspace{1.5em}} 29 | &\mathscr{G}(A) \amalg \mathscr{G}(B) 30 | \ar@{->}[u]^{\exists !} 31 | &\mathscr{G}(B) 32 | \ar@{->}[lu]_{\mathscr{G}(i_{\mathsf{C}_1})} 33 | \ar@{->}[l]^{\hspace{2em}i_{\mathsf{D}_2}} 34 | }\] 35 | \end{solution} 36 | 37 | \begin{problem}[1.2] 38 | $\triangleright$ Let $\mathscr{F}: \mathsf{C} \rightarrow \mathsf{D}$ be a fully faithful functor. If $A, B$ are objects in $\mathsf{C}$, prove that $A \cong B$ in $\mathsf{C}$ if and only if $\mathscr{F}(A) \cong \mathscr{F}(B)$ in $\mathsf{D}$. [\S1.3] 39 | \end{problem} 40 | \begin{solution} 41 | If $A \cong B$ in $\mathsf{C}$ , then there exists $f:A\to B$ and $g:B\to A$ such that $f\circ g=\mathrm{id}_B$ and $g\circ f=\mathrm{id}_A$. Since $\mathscr{F}$ is a functor, we have $\mathscr{F}(f)\circ\mathscr{F}(g)=\mathrm{id}_{\mathscr{F}(B)}$ 42 | and $\mathscr{F}(g)\circ\mathscr{F}(f)=\mathrm{id}_{\mathscr{F}(A)}$. Hence $\mathscr{F}(A) \cong \mathscr{F}(B)$ in $\mathsf{D}$. 43 | 44 | If $\mathscr{F}(A) \cong \mathscr{F}(B)$ in $\mathsf{D}$, there exists $f':\mathscr{F}(A)\to\mathscr{F}(B)$ and $g':\mathscr{F}(B)\to\mathscr{F}(A)$ such that $f'\circ g'=\mathrm{id}_{\mathscr{F}(B)}$ and $g'\circ f'=\mathrm{id}_{\mathscr{F}(A)}$. Since $\mathscr{F}$ is fully faithful, there exists $f:A\to B$ and $g:B\to A$ such that $\mathscr{F}(f)=f'$ and $\mathscr{F}(g)=g'$. Now we have 45 | \begin{align*} 46 | \mathscr{F}\left(f\circ g\right)&=\mathscr{F}(f)\circ\mathscr{F}(g)=f'\circ g'=\mathrm{id}_{\mathscr{F}(B)},\\ 47 | \mathscr{F}\left(g\circ f\right)&=\mathscr{F}(g)\circ\mathscr{F}(f)=g'\circ f'=\mathrm{id}_{\mathscr{F}(A)}, 48 | \end{align*} 49 | which implies $f\circ g=\mathrm{id}_B$ and $g\circ f=\mathrm{id}_A$. Hence $A \cong B$ in $\mathsf{C}$. 50 | \end{solution} 51 | 52 | 53 | \begin{problem}[1.3] 54 | Recall (\S II.1) that a group $G$ may be thought of as a groupoid $\mathsf{G}$ with a single object. Prove that defining the action of $G$ on an object of a category $\mathsf{C}$ is equivalent to defining a functor $\mathsf{G} \rightarrow \mathsf{C}$. 55 | \end{problem} 56 | \begin{solution} 57 | Suppose a group $G$ acts on an object $B$ of a category $\mathsf{C}$ and $\mathsf{G}$ is the groupoid with a single object $A$ with $\mathrm{End}_{\mathsf{G}}(A)=G$ . Then the functor $\mathsf{G} \rightarrow \mathsf{C}$ is defined by $\mathsf{G}(A) \rightarrow B$ and $\mathsf{G}(g) \rightarrow g$ for all $g \in G$. 58 | 59 | Conversely, suppose $\mathsf{G}$ is a groupoid with a single object $A$ and $F:\mathsf{G} \rightarrow \mathsf{C}$ is a functor that maps $A$ to $B$ in $\mathsf{C}$. Then the group action on $B$ is defined by 60 | \begin{align*} 61 | \mathrm{Aut}_{\mathsf{G}}(A) &\longrightarrow \mathrm{Aut}_{\mathsf{C}}(B),\\ 62 | g &\longmapsto F(g). 63 | \end{align*} 64 | \end{solution} 65 | 66 | 67 | \begin{problem}[1.4] 68 | $\neg$ Let $R$ be a commutative ring, and let $S \subseteq R$ be a multiplicative subset in the sense of \hyperlink{Exercise V.4.7}{Exercise V.4.7} . Prove that `localization is a functor': associating with every $R$-module $M$ the localization $S^{-1} M$ (\hyperlink{Exercise V.4.8}{Exercise V.4.8}) and with every $R$-module homomorphism $\varphi: M \rightarrow N$ the naturally induced homomorphism $S^{-1} M \rightarrow S^{-1} N$ defines a covariant functor from the category of $R$-modules to the category of $S^{-1} R$ modules. [1.25] 69 | \end{problem} 70 | \begin{solution} 71 | 72 | The functor $\mathscr{F}: R\text{-}\mathsf{Mod} \rightarrow S^{-1} R\text{-}\mathsf{Mod}$ is defined as follows: 73 | \begin{itemize} 74 | \item object mapping: $M \mapsto S^{-1} M$; 75 | \item morphism mapping $\varphi\mapsto F(\varphi)$ is given by the following commutative diagram 76 | \[\xymatrix{ 77 | S^{-1}M 78 | \ar@{-->}[r]^{\mathscr{F}(\varphi)} & S^{-1}N\\ 79 | M \ar[u]^{\ell}\ar[r]_{\varphi} & N \ar[u]^{\ell} 80 | }\] 81 | \end{itemize} 82 | We can check that $\mathscr{F}$ is a functor by checking the functoriality of $\mathscr{F}$ 83 | \begin{align*} 84 | \mathscr{F}\left(\varphi\circ\varphi'\right)&=\mathscr{F}(\varphi)\circ\mathscr{F}(\varphi'), 85 | \end{align*} 86 | which can be derived from the following commutative diagram. 87 | \[\xymatrix{ 88 | S^{-1}M 89 | \ar@{-->}[r]^{\mathscr{F}(\varphi)} & S^{-1}N\ar@{-->}[r]^{\mathscr{F}(\varphi')}& S^{-1}L\\ 90 | M \ar[u]^{\ell}\ar[r]_{\varphi} & N \ar[u]^{\ell}\ar[r]_{\varphi'}& L \ar[u]^{\ell} 91 | }\] 92 | 93 | \end{solution} -------------------------------------------------------------------------------- /Solution_to_Algebra_Chapter_0.tex: -------------------------------------------------------------------------------- 1 | % Compile with PdfLaTeX 2 | \documentclass[12pt,letterpaper,boxed]{hmcpset} 3 | 4 | % set 1-inch margins in the document 5 | \usepackage[margin=1in]{geometry} 6 | 7 | % include this if you want to import graphics files with /includegraphics 8 | \usepackage{enumerate} 9 | \usepackage{graphicx} 10 | \usepackage{amsmath,amssymb,amsthm} 11 | \usepackage[all]{xy} 12 | 13 | \usepackage{fancyhdr} 14 | \pagestyle{fancy} 15 | \fancyhf{} 16 | \rhead{} 17 | \lhead{\small\leftmark} 18 | \rfoot{-\hspace{4pt}\thepage\hspace{4pt}- } 19 | 20 | \usepackage[usenames,dvipsnames]{color} 21 | \usepackage{xcolor} 22 | \definecolor{myRoyalBlue}{RGB}{50,60,150} 23 | \definecolor{myGreen}{RGB}{45,100,45} 24 | \usepackage[colorlinks,linkcolor=myRoyalBlue,urlcolor=myGreen,citecolor=BurntOrange]{hyperref} 25 | \usepackage{mathrsfs} 26 | \usepackage{cancel} 27 | \usepackage{cite} 28 | \usepackage{setspace} 29 | \usepackage{tikz} 30 | \usepackage{tikz-cd} 31 | 32 | 33 | \usepackage{mathtools} 34 | \DeclarePairedDelimiter\ceil{\lceil}{\rceil} 35 | \DeclarePairedDelimiter\floor{\lfloor}{\rfloor} 36 | 37 | \renewcommand{\baselinestretch}{1.05} 38 | \renewcommand\appendix{\setcounter{secnumdepth}{-2}} 39 | 40 | \newcommand{\Obj}{\mathrm{Obj}} 41 | \newcommand{\Hom}{\mathrm{Hom}} 42 | \newcommand{\GL}{\mathrm{GL}} 43 | \newcommand{\SL}{\mathrm{SL}} 44 | \newcommand{\Aut}{\mathrm{Aut}} 45 | \newcommand{\Inn}{\mathrm{Inn}} 46 | \newcommand{\im}{\mathrm{im}\hspace{1.5pt}} 47 | \newcommand{\Grp}{\mathsf{Grp}} 48 | \newcommand{\Set}{\mathsf{Set}} 49 | \newcommand{\Ab}{\mathsf{Ab}} 50 | \newcommand{\Mod}{\mathsf{Mod}} 51 | \newcommand{\R}{\mathbb{R}} 52 | \newcommand{\C}{\mathbb{C}} 53 | \newcommand{\Q}{\mathbb{Q}} 54 | \newcommand{\Z}{\mathbb{Z}} 55 | \newcommand{\N}{\mathbb{N}} 56 | \newcommand{\id}{\mathrm{id}} 57 | \newcommand{\from}{\leftarrow} 58 | 59 | \newcommand{\set}[1]{\left\{\,#1\,\right\}} 60 | \newcommand{\abs}[1]{\left|#1\right|} 61 | \newcommand{\quotuniv}[1]{\overline{#1}} 62 | \newcommand{\quot}[2]{#1/\!\!#2\,\,} 63 | \newcommand{\quotntws}[2]{#1/\!\!#2} 64 | 65 | \newcommand{\inv}[1]{#1^{-1}} 66 | \newcommand{\cycl}[1]{\mathbb{Z}/#1\mathbb{Z}} 67 | 68 | \DeclareMathOperator{\lcm}{lcm} 69 | 70 | % info for header block in upper right hand corner 71 | 72 | \begin{document} 73 | \thispagestyle{empty} 74 | \vspace{-5em} 75 | \rightline{Hooy} 76 | \rightline{\today} 77 | %\name{Hooy} 78 | %\updatedate{\vspace*{1pt}\today} 79 | \vspace*{1.5em} 80 | \problemlist{\textbf{\LARGE Algebra, Chapter 0}\\\vspace*{0.6em}By Paolo Aluffi} 81 | 82 | \tableofcontents 83 | \thispagestyle{empty} 84 | \appendix 85 | 86 | \newpage 87 | \thispagestyle{empty} 88 | \section*{Notation for Problems} 89 | $\vartriangleright$: those problems that are directly referenced from the text.\\ 90 | 91 | \noindent $\neg$: those problems that are referenced from other 92 | problems.\\ 93 | 94 | \noindent [\textsection II.8.1]: related to the text in II.8.1 (Chapter II Section 8 Subsection 1).\\ 95 | 96 | \noindent [II.8.10]: related to the Definition/Example/Proposition/Lemma/Corollary/Claim 8.10 in Chapter II (the 10th Definition/Example/Proposition/Lemma/Corollary/Claim in Chapter II Section 8). 97 | 98 | \section*{Acknoledgement} 99 | It is kind of Shane Creighton-Young to share his solutions to Paolo Aluffi's ``Algebra: Chapter 0"\cite{aluffi2009algebra} on the Github website \href{https://github.com/srcreigh/aluffi}{\texttt{https://github.com/srcreigh/aluffi}}. He takes the credit for some parts of the first two chapters of this manuscript. 100 | 101 | \section*{Contact} 102 | Github: 103 | \href{https://github.com/hooyuser/Solution-to-Algebra-Chapter-0}{\texttt{https://github.com/hooyuser/Solution-to-Algebra-Chapter-0}} 104 | 105 | \noindent E-mail: \href{mailto:hooyuser@outlook.com}{\texttt{hooyuser@outlook.com}} 106 | 107 | 108 | \section*{License} 109 | This work is licensed under a \href{http://creativecommons.org/licenses/by-nc-sa/4.0/}{Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License}. 110 | \vfill 111 | \begin{figure}[h] 112 | \centering\includegraphics[width=0.38\textwidth]{./BY-NC-SA} 113 | \end{figure} 114 | 115 | \newpage 116 | \include{chapter1.tex} 117 | 118 | \newpage 119 | \include{chapter2.tex} 120 | 121 | \newpage 122 | \include{chapter3.tex} 123 | 124 | \newpage 125 | \include{chapter4.tex} 126 | 127 | \newpage 128 | \include{chapter5.tex} 129 | 130 | \newpage 131 | \include{chapter6.tex} 132 | 133 | \newpage 134 | \include{chapter7.tex} 135 | 136 | \newpage 137 | \include{chapter8.tex} 138 | 139 | \newpage 140 | \section{Appendix} 141 | \hypertarget{Lemma II.1}{} 142 | \textbf{Lemma II.1} (von Dyck) 143 | Given a presentation $(A|\mathscr{R})=F(A)/R$, where $A$ is the set of generators, $\mathscr{R}\in F(A)$ is the set of relators and $R$ is the smallest normal subgroup of $F(A)$ containing $\mathscr{R}$. Define inclusion mapping $i:A\to F(A)$ and projection $\pi:F(A)\to F(A)/R$. If $f$ is a mapping from $A$ to a group $G$, and every relations in $\mathscr{R}$ holds in $G$ via $f$, that is, $\mathscr{R}\subset\ker\varphi$ where $\varphi$ is the unique homomorphism induced by the universal property of free group, then there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $f=\psi\circ\pi\circ i$. If $G$ is generated by $f(A)$, then $\psi$ is surjective. 144 | \[\xymatrix{ 145 | F(A)/R\ar@{-->}[rd]^{\exists!\psi}\\ 146 | F(A)\ar@{-->}[r]^{\varphi}\ar[u]^{\pi} &G\\ 147 | A\ar[ru]_{f}\ar[u]^{i}& 148 | }\] 149 | \textbf{Proof of the lemma.} Since $R$ is the smallest normal subgroup of $F(A)$ containing $\mathscr{R}$ and the normal subgroup $\ker\varphi$ contains $\mathscr{R}$, we must have $R\subset\ker\varphi$. Then according to Theorem 7.12, there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $\varphi=\psi\circ\pi$, which means the whole diagram commutes. If there exists a homomorphism $\zeta:F(A)/R\to G$ such that $f=\zeta\circ\pi\circ i$, then we have $\varphi\circ i=\zeta\circ\pi\circ i$, which implies $\varphi(t)= \zeta(\pi(t))$ for all $t\in A$. Note that a homomorphism defined on $F(A)$ can be specified only by its valuation on the set of generators $A$, we can assert that $\varphi=\zeta\circ\pi$. Since there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $\varphi=\psi\circ\pi$, we have $\zeta=\psi$. Thus we show that there exists a unique homomorphism $\psi:F(A)/R\to G$ such that $f=\psi\circ\pi\circ i$. 150 | 151 | Moreover, if $G$ is generated by $f(A)$, then $\mathrm{im}\psi=G$, since $f(A)=\psi(\pi( i(A)))\subset\mathrm{im}\psi$ implies $G\subset\mathrm{im}\psi$.\hfill$\lrcorner$ 152 | 153 | 154 | \newpage 155 | 156 | \bibliographystyle{plain} 157 | \bibliography{mybibtex} 158 | \end{document} 159 | -------------------------------------------------------------------------------- /chapter6.tex: -------------------------------------------------------------------------------- 1 | \section{Chapter VI.\hspace{0.2em} Linear algebra} 2 | \subsection{\textsection1. Free modules revisited} 3 | \hypertarget{Exercise VI.1.1}{} 4 | \begin{problem}[1.1] 5 | $\neg$ Prove that $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as $\mathbb{Q}$-vector spaces. (In particular, $(\mathbb{R},+$ ) and $(\mathbb{C},+)$ are isomorphic as groups.) [II.4.4] 6 | \end{problem} 7 | \begin{solution} 8 | The $\mathbb{Q}$-vector space $\mathbb{R}$ is a free $\mathbb{Q}$-module, it adimits a basis $B$. We can show that $B$ is a infinite set by showing that for any positive integer $n$, there exists a linearly independent subset $B_n$ such that $|B_n|=n$. An example is $B_n=\left\{1,\pi,\cdots,\pi^n\right\}$, given the fact that $\pi$ is trascendent over $\mathbb{Q}$. Let 9 | \begin{align*} 10 | B'=B\bigcup \left\{bi\mid b\in B\right\}. 11 | \end{align*} 12 | Then we see $B'$ is a basis of $\mathbb{C}$ and $|B'|=|B|$. Therefore, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as $\mathbb{Q}$-vector spaces. 13 | \end{solution} 14 | 15 | \hypertarget{Exercise VI.1.4}{} 16 | \begin{problem}[1.4] 17 | Let $V$ be a vector space over a field $k$. A \emph{Lie bracket} on $V$ is an operation 18 | $[\cdot, \cdot]: V \times V \rightarrow V$ such that 19 | \begin{itemize} 20 | \item $(\forall u, v, w \in V),(\forall a, b \in k)$, 21 | $$ 22 | [a u+b v, w]=a[u, w]+b[v, w], \quad[w, a u+b v]=a[w, u]+b[w, v], 23 | $$ 24 | \item $(\forall v \in V),[v, v]=0$, 25 | \item and $(\forall u, v, w \in V),[[u, v], w]+[[v, w], u]+[[w, u], v]=0$. 26 | \end{itemize} 27 | (This axiom is called the \emph{Jacobi identity}.) A vector space endowed with a Lie bracket is called a \emph{Lie algebra}. Define a category of Lie algebras over a given field. Prove the following: 28 | \begin{itemize} 29 | \item In a Lie algebra $V,[u, v]=-[v, u]$ for all $u, v \in V$. 30 | \item If $V$ is a $k$-algebra (Definition III.5.7), then $[v, w]:=v w-w v$ defines a Lie bracket on $V$, so that $V$ is a Lie algebra in a natural way. 31 | \item This makes $\mathfrak{g l}_n(\mathbb{R}), \mathfrak{g l}_n(\mathbb{C})$ into Lie algebras. The sets listed in Exercise III.1.4 are all Lie algebras, with respect to a Lie bracket induced from $\mathfrak{g l}$. 32 | \item $\mathfrak{s u}{ }_2(\mathbb{C})$ and $\mathfrak{s o}_3(\mathbb{R})$ are isomorphic as Lie algebras over $\mathbb{R}$. 33 | \end{itemize} 34 | \end{problem} 35 | \begin{solution} 36 | 37 | \end{solution} 38 | 39 | \begin{problem}[1.5] 40 | $\triangleright$ Let $R$ be an integral domain. Prove or disprove the following: 41 | \begin{itemize} 42 | \item Every linearly independent subset of a free $R$-module may be completed to a basis. 43 | \item Every generating subset of a free $R$-module contains a basis. 44 | \end{itemize} 45 | [\S 1.3] 46 | \end{problem} 47 | \begin{solution} 48 | \begin{itemize} 49 | \item False. Given the $\mathbb{Z}$-module $\mathbb{Z}$, $\left\{2\right\}$ is a linearly independent subset of $\mathbb{Z}$, but it cannot be completed to a basis because the dimension of $\mathbb{Z}$ is $1$. 50 | \item False. Given the $\mathbb{Z}$-module $\mathbb{Z}$, $\left\{2, 3\right\}$ is a generating subset of $\mathbb{Z}$, but it does not contain a basis because neither $\left\{2\right\}$ and $\left\{3\right\}$ can generate $\mathbb{Z}$. 51 | \end{itemize} 52 | \end{solution} 53 | 54 | \subsection{\textsection4. Presentations and resolutions} 55 | \begin{problem}[4.1] 56 | $\triangleright$ Prove that if $R$ is an integral domain and $M$ is an $R$-module, then $\operatorname{Tor}(M)$ is a submodule of $M$. Give an example showing that the hypothesis that $R$ is an integral domain is necessary. $[\S 4.1]$ 57 | \end{problem} 58 | \begin{solution} 59 | Given any $m_1,m_2\in\operatorname{Tor}(M)$, we can suppose there exist $r_1,r_2\in R$ such that $r_1m_1=0(r_1\ne0)$ and $r_2m_2=0(r_2\ne0)$. Since $R$ is an integral domain, we have $r_1r_2\ne0$. Thus, there exist nonzero element $r_1r_2\in R$ such that 60 | $$ 61 | r_1r_2(m_1+m_2)=r_2(r_1m_1)+r_1(r_2m_2)=0, 62 | $$ 63 | which implies $m_1+m_2\in\operatorname{Tor}(M)$. 64 | 65 | Given any $r\in R$, $m\in\operatorname{Tor}(M)$, we can suppose 66 | \[ 67 | r_0m=0 68 | \] 69 | where $r_0\in R$ and $r_0\ne0$. Thus we have 70 | \[ 71 | r_0(rm)=r(r_0m)=0, 72 | \] 73 | which implies $rm\in\operatorname{Tor}(M)$. Therefore, we show that $\operatorname{Tor}(M)$ is a submodule of $M$. 74 | 75 | $\mathbb{Z}/6\mathbb{Z}$ is not an integral domain and $\mathbb{Z}/6\mathbb{Z}$ is a $\mathbb{Z}/6\mathbb{Z}$-module. Since 76 | \[ 77 | \operatorname{Tor}(\mathbb{Z}/6\mathbb{Z})=\{[0]_6,[2]_6,[3]_6\}, 78 | \] 79 | we have $[2]_6+[3]_6=[5]_6\notin \operatorname{Tor}(\mathbb{Z}/6\mathbb{Z})$, which implies $\operatorname{Tor}(\mathbb{Z}/6\mathbb{Z})$ is not a submodule of $\mathbb{Z}/6\mathbb{Z} 80 | ++$. 81 | \end{solution} 82 | 83 | 84 | 85 | 86 | 87 | \begin{problem}[4.4] 88 | $\triangleright$ Let $R$ be a commutative ring, and $M$ an $R$-module. 89 | \begin{itemize} 90 | \item Prove that $\operatorname{Ann}(M)$ is an ideal of $R$. 91 | \item If $R$ is an integral domain and $M$ is finitely generated, prove that $M$ is torsion if and only if $\operatorname{Ann}(M) \neq 0$. 92 | \item Give an example of a torsion module $M$ over an integral domain, such that $\operatorname{Ann}(M)=0 .$ (Of course this example cannot be finitely generated!) 93 | \end{itemize} 94 | $[\S 4.2, \S 5.3]$ 95 | \end{problem} 96 | \begin{solution} 97 | \begin{itemize} 98 | \item $\operatorname{Ann}(M)$ is defined as 99 | \[ 100 | \operatorname{Ann}(M)=\{r \in R \mid \forall m \in M, r m=0\}. 101 | \] 102 | Given any $r_1,r_2\in\operatorname{Ann}(M)$, we have 103 | \[ 104 | \forall m\in M, (r_1+r_2)m=r_1m+r_2m=0, 105 | \] 106 | which implies $r_1+r_2\in\operatorname{Ann}(M)$. 107 | 108 | Given any $r\in R$, $s\in\operatorname{Ann}(M)$, we have 109 | \[ 110 | \forall m\in M, (rs)m=r(sm)=0, 111 | \] 112 | which implies $rs\in\operatorname{Ann}(M)$. 113 | \item If $\operatorname{Ann}(M) \neq 0$, there exists nonzero $r\in R$ such that and $rm=0$ for any $m\in M$, which implies $M$ is a torsion module. 114 | 115 | Assume that $M$ is torsion and is generated by $\{m_1,m_2,\cdots,m_n\}$. There exist nonzero elements $r_1,r_2,\cdots,r_n\in R$ such that 116 | \[ 117 | r_1m_1=r_2m_2=\cdots=r_nm_n=0. 118 | \] 119 | Let $r_0=r_1r_2\cdots r_n$. Since $R$ is an integral domain, $r_0\ne0$. Given any $m\in M$, we have 120 | \[ 121 | m=k_1m_1+k_2m_2+\cdots+k_nm_n 122 | \] 123 | and 124 | \[ 125 | r_0m=k_1r_2\cdots r_n(r_1m_1)+\cdots+k_nr_1\cdots r_{n-1} (r_nm_n)=0. 126 | \] 127 | Thus we show $r_0\in\operatorname{Ann}(M)$ and $\operatorname{Ann}(M)\ne0$. 128 | \item $\mathbb{Q}/\mathbb{Z}$ is a module over the integral domain $\mathbb{Z}$. Since 129 | \[ 130 | \forall \left[\frac{p}{q}\right]\in \mathbb{Q}/\mathbb{Z},\ \exists q\in\mathbb{Z},\ q\left[\frac{p}{q}\right]=[0], 131 | \] 132 | we see $\mathbb{Q}/\mathbb{Z}$ is a torsion module. If $r\in\operatorname{Ann}(M)$, then for all $q\in\mathbb{Z}$, 133 | \[ 134 | r\left[\frac{1}{q}\right]=0\iff q|r\implies r=0\text{ or } r\ge q. 135 | \] 136 | Therefore, there must be $r=0$, which means $\operatorname{Ann}(M)=0$. 137 | 138 | \end{itemize} 139 | \end{solution} -------------------------------------------------------------------------------- /chapter5.tex: -------------------------------------------------------------------------------- 1 | \section{Chapter V.\hspace{0.2em} Irreducibility and factorization in integral domains} 2 | \subsection{\textsection1. Chain conditions and existence of factorizations} 3 | Remember that in this section all rings are taken to be \emph{commutative}. 4 | 5 | \hypertarget{Exercise V.1.1}{} 6 | \begin{problem}[1.1] 7 | $\triangleright$ Let $R$ be a Noetherian ring, and let $I$ be an ideal of $R$. Prove that $R / I$ is a Noetherian ring. [\textsection1.1] 8 | \end{problem} 9 | \begin{solution} 10 | Let $\pi:R\to R / I$ be the projection. According to \hyperlink{Exercise III.4.2}{Exercise III.4.2}, the homomorphic image of the Noetherian ring $R$ is Noetherian. That is, $\pi(R)=R / I$ is Noetherian. 11 | \end{solution} 12 | 13 | \begin{problem}[1.2] 14 | Prove that if $R[x]$ is Noetherian, so is $R$. (This is a `converse' to Hilbert's basis theorem.) 15 | \end{problem} 16 | \begin{solution} 17 | According to \hyperlink{Exercise V.1.1}{Exercise V.1.1}, $R[x]$ is Noetherian implies $R\left[x\right]/\left(x\right)$ is Noetherian. Since in \hyperlink{Exercise III.4.12}{Exercise III.4.12} we have shown that 18 | \[ 19 | R\cong R\left[x\right]/\left(x\right), 20 | \] 21 | we see $R$ is Noetherian. 22 | \end{solution} 23 | 24 | \subsection{\textsection4. Unique factorization in polynomial rings} 25 | \hypertarget{Exercise V.4.7}{} 26 | \begin{problem}[4.7] 27 | $\triangleright$ A subset $S$ of a commutative ring $R$ is a \emph{multiplicative subset} (or \emph{multiplicatively closed}) if (i) $1 \in S$ and (ii) $s, t \in S \Longrightarrow s t \in S$. Define a relation on the set of pairs $(a, s)$ with $a \in R, s \in S$ as follows: 28 | $$ 29 | (a, s) \sim\left(a^{\prime}, s^{\prime}\right) \Longleftrightarrow(\exists t \in S), t\left(s^{\prime} a-s a^{\prime}\right)=0 . 30 | $$ 31 | Note that if $R$ is an integral domain and $S=R \backslash\{0\}$, then $S$ is a multiplicative subset, and the relation agrees with the relation introduced in $\S 4.2$. 32 | \begin{itemize} 33 | \item Prove that the relation $\sim$ is an equivalence relation. 34 | \item Denote by $\frac{a}{s}$ the equivalence class of $(a, s)$, and define the same operations $+$, $\cdot$ on such `fractions' as the ones introduced in the special case of $\S 4.2$. Prove that these operations are well-defined. 35 | \item The set $S^{-1} R$ of fractions, endowed with the operations $+$, $\cdot$ is the \emph{localization} of $R$ at the multiplicative subset $S$. Prove that $S^{-1} R$ is a commutative ring and that the function $a \mapsto \frac{a}{1}$ defines a ring homomorphism $\ell: R \rightarrow S^{-1} R$. 36 | \item Prove that $\ell(s)$ is invertible for every $s \in S$. 37 | \item Prove that $R \rightarrow S^{-1} R$ is initial among ring homomorphisms $f: R \rightarrow R^{\prime}$ such that $f(s)$ is invertible in $R^{\prime}$ for every $s \in S$. 38 | \item Prove that $S^{-1} R$ is an integral domain if $R$ is an integral domain. 39 | \item Prove that $S^{-1} R$ is the zero-ring if and only if $0 \in S$. 40 | \end{itemize} 41 | $[4.8,4.9,4.11,4.15$, VII.2.16, VIII.1.4, VIII.2.5, VIII.2.6, VIII.2.12, §IX.9.1] 42 | \end{problem} 43 | \begin{solution} 44 | \begin{itemize} 45 | \item Reflexivity. 46 | \begin{align*} 47 | 1(sa-sa)=0\implies (a,s)\sim(a,s). 48 | \end{align*} 49 | Symmetry. 50 | \begin{align*} 51 | &(a, s) \sim\left(a^{\prime}, s^{\prime}\right) \Longleftrightarrow\left(\exists t \in S\right), t\left(s^{\prime} a-s a^{\prime}\right)=0 \\ 52 | \iff& (\exists t \in S), t\left(s a^{\prime}-s^{\prime} a\right)=0\iff \left(a^{\prime}, s^{\prime}\right) \sim (a, s). 53 | \end{align*} 54 | Transitivity. 55 | \begin{align*} 56 | &(a, s) \sim\left(a^{\prime}, s^{\prime}\right) ,\left(a^{\prime}, s^{\prime}\right) \sim \left(a'', s''\right)\\ 57 | \implies &\left(\exists t_1,t_2 \in S\right), t_1\left(s^{\prime} a-s a^{\prime}\right)=t_2\left(s'' a'-s' a''\right)=0 \\ 58 | \implies & (\exists t_1t_2s' \in S), t_1t_2s'\left(s'' a-s a''\right)=(t_2s'')(t_1s'a)-(t_1s)(t_2s'a'')\\ 59 | &\hspace{13.25em}=(t_2s'')(t_1sa')-(t_1s')(t_2s'' a')\\ 60 | &\hspace{13.25em}=0\\ 61 | \implies & (a, s)\sim \left(a'', s''\right) . 62 | \end{align*} 63 | \item Define 64 | \begin{align*} 65 | \frac{a_1}{s_1}+\frac{a_2}{s_2}&=\frac{a_1s_2+a_2s_1}{s_1s_2},\qquad \frac{a_1}{s_1}\cdot\frac{a_2}{s_2}=\frac{a_1a_2}{s_1s_2}. 66 | \end{align*} 67 | If $\frac{a_1}{s_1} = \frac{a_1^\prime}{s_1^\prime}$, then there exists $t\in S$ such that $ts_1^\prime a_1= ts_1 a_1^\prime$. Hence 68 | \begin{align*} 69 | &\;t(s_1^\prime s_2)(a_1s_2+a_2s_1)-t(s_1s_2)(a_1^\prime s_2+a_2s_1^\prime)\\ 70 | =&\;t\left(s_1^\prime s_2a_1s_2+s_1^\prime s_2a_2s_1-s_1s_2a_1^\prime s_2-s_1s_2a_2s_1^\prime\right)\\ 71 | =&\;t\left(s_1^\prime s_2a_1s_2+s_1^\prime s_2a_2s_1-s_1^\prime s_2a_1 s_2-s_1^\prime s_2a_2s_1\right)a\\ 72 | =&\;0, 73 | \end{align*} 74 | which implies 75 | \begin{align*} 76 | \frac{a_1}{s_1}+\frac{a_2}{s_2}=\frac{a_1s_2+a_2s_1}{s_1s_2}= \frac{a_1^\prime s_2+a_2s_1^\prime}{s_1^\prime s_2}=\frac{a_1^\prime}{s_1^\prime}+\frac{a_2}{s_2}. 77 | \end{align*} 78 | If $\frac{a_2}{s_2} = \frac{a_2^\prime}{s_2^\prime}$, in a similar way, we can prove 79 | \begin{align*} 80 | \frac{a_1}{s_1}+\frac{a_2}{s_2}=\frac{a_1s_2+a_2s_1}{s_1s_2}= \frac{a_1s_2^\prime+a_2^\prime s_1}{s_1s_2^\prime}=\frac{a_1}{s_1}+\frac{a_2^\prime}{s_2^\prime}. 81 | \end{align*} 82 | Therefore, $+$ is well-defined. 83 | 84 | If $\frac{a_1}{s_1} = \frac{a_1^\prime}{s_1^\prime}$ and $\frac{a_2}{s_2} = \frac{a_2^\prime}{s_2^\prime}$, then there exists $t_1,t_2\in S$ such that $t_1s_1^\prime a_1= t_1s_1 a_1^\prime$ and $t_2s_2^\prime a_2= t_2s_2 a_2^\prime$. Hence we have 85 | \begin{align*} 86 | (t_1t_2)(s_1^\prime s_2^\prime a_1a_2 - s_1 s_2 a_1^\prime a_2^\prime)&=(t_1s_1^\prime a_1 )( t_2s_2^\prime a_2) -(t_1 s_1a_1^\prime ) (t_2s_2a_2^\prime )\\ 87 | &=(t_1s_1 a_1^\prime )( t_2s_2 a_2^\prime) -(t_1 s_1a_1^\prime ) (t_2s_2a_2^\prime )\\ 88 | &=0, 89 | \end{align*} 90 | which implies 91 | \begin{align*} 92 | \frac{a_1}{s_1}\cdot \frac{a_2}{s_2}=\frac{a_1a_2}{s_1s_2}= \frac{a_1^\prime a_2^\prime}{s_1^\prime s_2^\prime}=\frac{a_1^\prime}{s_1^\prime}\cdot \frac{a_2^\prime}{s_2^\prime}. 93 | \end{align*} 94 | Therefore, $\cdot$ is well-defined. 95 | \item It is straightforward to check that $S^{-1} R$ is a commutative ring. Define 96 | \begin{align*} 97 | \ell: R &\longrightarrow S^{-1} R\\ 98 | a &\longmapsto \frac{a}{1}. 99 | \end{align*} 100 | Then 101 | \begin{align*} 102 | \ell(a+b)&=\frac{a+b}{1}=\frac{a}{1}+\frac{b}{1}= \ell(a)+\ell(b),\\ 103 | \ell(ab)&=\frac{ab}{1}=\frac{a}{1}\cdot\frac{b}{1}=\ell(a)\ell(b). 104 | \end{align*} 105 | Hence $\ell$ is a ring homomorphism. 106 | \item Since 107 | \begin{align*} 108 | \frac{\ell(s)}{1}\frac{1}{s}=\frac{\ell(s)}{s}=\frac{s}{s}=\frac{1}{1}, 109 | \end{align*} 110 | $\ell(s)$ is invertible in $S^{-1}R$ for every $s \in S$. 111 | \item 112 | \[\xymatrix{ 113 | S^{-1}R 114 | \ar@{-->}[r]^{g} & R'\\ 115 | R\ar[ru]_{f}\ar[u]^{\ell} & 116 | }\] 117 | Suppose $f: R \rightarrow R^{\prime}$ is a ring homomorphism such that $f(s)$ is invertible in $R^{\prime}$ for every $s \in S$. Define 118 | \begin{align*} 119 | g: S^{-1}R &\longrightarrow R^{\prime}\\ 120 | \frac{a}{s} &\longmapsto f(s)^{-1}f(a). 121 | \end{align*} 122 | We can check that $g$ is a ring homomorphism as follows: 123 | \begin{align*} 124 | g\left(\frac{a_1}{s_1}+\frac{a_2}{s_2}\right)&=g\left(\frac{s_1a_2+s_2a_1}{s_1s_2}\right)\\ 125 | &=f(s_1s_2)^{-1}f(s_1a_2+s_2a_1)\\ 126 | &=f(s_1)^{-1}f(s_2)^{-1}f(s_1)f(a_2)+f(s_1)^{-1}f(s_2)^{-1}f(s_2)f(a_1)\\ 127 | &=f(s_1)^{-1}f(a_1)+f(s_2)^{-1}f(a_2)\\ 128 | &=g\left(\frac{a_1}{s_1}\right)+g\left(\frac{a_2}{s_2}\right), 129 | \end{align*} 130 | \begin{align*} 131 | g\left(\frac{a_1}{s_1}\cdot \frac{a_2}{s_2}\right)&=g\left(\frac{a_1a_2}{s_1s_2}\right)\\ 132 | &=f(s_1s_2)^{-1}f(a_1a_2)\\ 133 | &=f(s_1)^{-1}f(a_1)f(s_2)^{-1}f(a_2)\\ 134 | &=g\left(\frac{a_1}{s_1}\right) g\left(\frac{a_2}{s_2}\right), 135 | \end{align*} 136 | \begin{align*} 137 | g\left(\frac{1}{1}\right)&=f(1)^{-1}f(1)=1. 138 | \end{align*} 139 | Since 140 | \begin{align*} 141 | g(\ell(a))&=g\left(\frac{a}{1}\right)=f(1)^{-1}f(a)=f(a), 142 | \end{align*} 143 | we see $g$ gives a commutative diagram. 144 | 145 | Suppose $g'$ is another ring homomorphism such that $g'\circ \ell=f$. Then we have 146 | \begin{align*} 147 | g'\left(\frac{a}{s}\right)&= g'\left(\frac{1}{s}\frac{a}{1}\right)\\ 148 | &=g'\left(\frac{1}{s}\right)g'\left(\frac{a}{1}\right)\\ 149 | &= \left(g'\left(\frac{s}{1}\right)\right)^{-1}g'\left(\frac{a}{1}\right)\\ 150 | &=f(s)^{-1}f(a)\\ 151 | &=g\left(\frac{a}{s}\right), 152 | \end{align*} 153 | which implies $g=g'$. Therefore, $\ell:R \rightarrow S^{-1} R$ is initial among ring homomorphisms $f: R \rightarrow R^{\prime}$ such that $f(s)$ is invertible in $R^{\prime}$ for every $s \in S$. 154 | \item If $R$ is an integral domain, then for any $\frac{a_1}{s_1}, \frac{a_2}{s_2} \in S^{-1} R$, we have 155 | \begin{align*} 156 | \frac{a_1}{s_1}\cdot\frac{a_2}{s_2}=\frac{a_1a_2}{s_1s_2}=\frac{0}{1}\implies a_1a_2=0\implies a_1=0\text{ or }a_2=0. 157 | \end{align*} 158 | That means $S^{-1} R$ is an integral domain. 159 | \item If $0 \in S$, then for any $a,a'\in R$, $s,s' \in S$, there exists $0\in S$ such that $0(s'a-sa')=0$. Therefore, $S^{-1} R$ only contains 1 element, which implies $S^{-1} R$ is a zero-ring. 160 | \end{itemize} 161 | \end{solution} 162 | 163 | \hypertarget{Exercise V.4.8}{} 164 | \begin{problem}[4.8] 165 | $\neg$ Let $S$ be a multiplicative subset of a commutative ring $R$, as in \hyperlink{Exercise V.4.7}{Exercise V.4.7}. For every $R$-module $M$, define a relation $\sim$ on the set of pairs $(m, s)$, where $m \in M$ and $s \in S$: 166 | $$ 167 | (m, s) \sim\left(m^{\prime}, s^{\prime}\right) \Longleftrightarrow(\exists t \in S), t\left(s^{\prime} m-s m^{\prime}\right)=0 . 168 | $$ 169 | Prove that this is an equivalence relation, and define an $S^{-1} R$-module structure on the set $S^{-1} M$ of equivalence classes, compatible with the $R$-module structure on $M$. The module $S^{-1} M$ is the localization of $M$ at $S$. [4.9, 4.11, 4.14, VIII.1.4, VIII.2.5, VIII.2.6] 170 | \end{problem} 171 | \begin{solution} 172 | We can check that $\sim$ is an equivalence relation as follows:\\ 173 | Reflexivity. 174 | \begin{align*} 175 | (\exists 1 \in S), 1\cdot\left(s m-s m\right)=0\implies (m, s) \sim\left(m, s\right). 176 | \end{align*} 177 | Symmetry. 178 | \begin{align*} 179 | &(m, s) \sim\left(m^{\prime}, s^{\prime}\right) \Longleftrightarrow\left(\exists t \in S\right), t\left(s^{\prime} m-s a^{\prime}\right)=0 \\ 180 | \iff& (\exists t \in S), t\left(sm^{\prime}-s^{\prime} m\right)=0\iff \left(m^{\prime}, s^{\prime}\right) \sim (m, s). 181 | \end{align*} 182 | Transitivity. 183 | \begin{align*} 184 | &(m, s) \sim\left(m^{\prime}, s^{\prime}\right) ,\left(m^{\prime}, s^{\prime}\right) \sim \left(m'', s''\right)\\ 185 | \implies &\left(\exists t_1,t_2 \in S\right), t_1\left(s^{\prime} m-s m^{\prime}\right)=t_2\left(s'' m'-s' m''\right)=0 \\ 186 | \implies & (\exists t_1t_2s' \in S), t_1t_2s'\left(s'' m-s m''\right)=(t_2s'')(t_1s'm)-(t_1s)(t_2s'm'')\\ 187 | &\hspace{13.25em}=(t_2s'')(t_1sm')-(t_1s')(t_2s'' m')\\ 188 | &\hspace{13.25em}=0\\ 189 | \implies & (m, s)\sim \left(m'', s''\right) . 190 | \end{align*} 191 | Denote by $\frac{m}{s}$ the equivalence class of $(m, s)$. The addtion of $S^{-1} M$ is defined as follows: 192 | \begin{align*} 193 | \frac{m_1}{s_1}+\frac{m_2}{s_2}&=\frac{s_2m_1+s_1m_2}{s_1s_2}. 194 | \end{align*} 195 | We can show that the addition is well-defined and makes $S^{-1} M$ an abelian group.\\ 196 | Define the scalar multiplication of $S^{-1}R$ on $S^{-1} M$ as follows: 197 | \begin{align*} 198 | \frac{a}{s_1}\cdot\frac{m}{s_2}&=\frac{am}{s_1s_2}. 199 | \end{align*} 200 | We can show that the scalar multiplication is well-defined as follows: 201 | \begin{align*} 202 | &\frac{a}{s_1}=\frac{a'}{s_1'},\frac{m}{s_2}=\frac{m'}{s_2'}\\ 203 | \implies& (\exists t_1,t_2 \in S), t_1\left(s_1a^{\prime}-s_1^{\prime} a\right)=0,t_2\left(s_2m^{\prime}-s_2^{\prime} m\right)=0\\ 204 | \implies& (\exists t_1t_2 \in S), t_1t_2\left(s_1's_2'am-s_1s_2a^{\prime}m^{\prime}\right)=0\\ 205 | \implies& \frac{am}{s_1s_2}=\frac{a'm'}{s_1's_2'}. 206 | \end{align*} 207 | We can check that the scalar multiplication is compatible with the addition as follows: 208 | \begin{align*} 209 | \frac{a}{s_1}\left(\frac{m_1}{s_2}+\frac{m_2}{s_3}\right)&= \frac{a}{s_1}\frac{s_3m_2+s_2m_1}{s_2s_3}=\frac{as_3m_2+as_2m_1}{s_1s_2s_3}=\frac{am_1}{s_1s_2}+\frac{am_2}{s_1s_3}=\frac{a}{s_1}\frac{m_1}{s_2}+\frac{a}{s_1}\frac{m_2}{s_3},\\ 210 | \left(\frac{a_1}{s_1}+\frac{a_2}{s_2}\right)\frac{m}{s_3}&=\frac{s_2a_1+s_1a_2}{s_1s_2}\frac{m}{s_3}=\frac{s_2a_1m+s_1a_2m}{s_1s_2s_3}=\frac{a_1m}{s_1s_3}+\frac{a_2m}{s_2s_3}=\frac{a_1}{s_1}\frac{m}{s_3}+\frac{a_2}{s_2}\frac{m}{s_3},\\ 211 | \left(\frac{a_1}{s_1}\frac{a_2}{s_2}\right)\frac{m}{s_3}&=\frac{a_1a_2}{s_1s_2}\frac{m}{s_3}=\frac{(a_1a_2)m}{(s_1s_2)s_3}=\frac{a_1(a_2m)}{s_1(s_2s_3)}=\frac{a_1}{s_1}\frac{a_2m}{s_2s_3}=\frac{a_1}{s_1}\left(\frac{a_2}{s_2}\frac{m}{s_3}\right),\\ 212 | \frac{1}{1}\frac{m}{s}&=\frac{1m}{1s}=\frac{m}{s}. 213 | \end{align*} 214 | \end{solution} 215 | 216 | 217 | 218 | 219 | 220 | 221 | 222 | 223 | 224 | 225 | 226 | 227 | 228 | 229 | 230 | 231 | 232 | 233 | -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | Attribution-NonCommercial-ShareAlike 4.0 International 2 | 3 | ======================================================================= 4 | 5 | Creative Commons Corporation ("Creative Commons") is not a law firm and 6 | does not provide legal services or legal advice. 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For 434 | the avoidance of doubt, this paragraph does not form part of the 435 | public licenses. 436 | 437 | Creative Commons may be contacted at creativecommons.org. 438 | -------------------------------------------------------------------------------- /chapter7.tex: -------------------------------------------------------------------------------- 1 | \section{Chapter VII.\hspace{0.2em} Fields} 2 | \subsection{\textsection1. Field extensions, I} 3 | \begin{problem}[1.1] 4 | $\triangleright$ Prove that if $k \subseteq K$ is a field extension, then $\mathrm{char}\,k=\mathrm{char}\,K$. Prove that the category $\mathsf{Fld}$ has no initial object. $[\S 1.1]$ 5 | \end{problem} 6 | \begin{solution} 7 | Since $\mathbb{Z}$ is initial in $\mathsf{Ring}$, we have unique ring homomorphisms $i_k: \mathbb{Z} \rightarrow k$ and $i_K: \mathbb{Z} \rightarrow K$. Note that the inclusion $j:k\rightarrow K$ is a ring homomorphism, we have $i_K=j\circ i_k$. Since 8 | \[ 9 | \ker i_K=\ker(j\circ i_k)=i_{k}^{-1}(\ker j)=i_{k}^{-1}(\{0\})=\ker i_k, 10 | \] 11 | we have $\mathrm{char}\,k=\mathrm{char}\,K$. If there exists one initial object $A$ in $\mathsf{Fld}$, all the objects in $\mathsf{Fld}$ will have the same characteristic as $A$. This contradicts with the fact that $\mathrm{char}\,{\mathbb{Z}}_2 \ne \mathrm{char}\,\mathbb{Z}_3$. 12 | \end{solution} 13 | 14 | \begin{problem}[1.3] 15 | $\triangleright$ Let $k \subseteq F$ be a field extension, and let $\alpha \in F .$ Prove that the field $k(\alpha)$ consists of all the elements of $F$ which may be written as rational functions in $\alpha$, with coefficients in $k$. Why does this not give (in general) an onto homomorphism $k(t) \rightarrow k(\alpha) ?[\S 1.2, \S 1.3]$ 16 | \end{problem} 17 | \begin{solution} 18 | Since $k(\alpha)$ is smallest subfield of $F$ containing both $k$ and $\alpha$, for any $g(t)\in k(t)$ we have $g(\alpha)\in k(\alpha)$. Thus we can define the evaluation mapping 19 | \begin{align*} 20 | \mathrm{ev}_{\alpha}:k(t) &\longrightarrow k(\alpha)\\ 21 | g(t)&\longmapsto g(\alpha). 22 | \end{align*} 23 | It is clear to see $\mathrm{ev}_{\alpha}(k(t))\subseteq k(\alpha)$ and $k(\alpha)\subseteq \mathrm{ev}_{\alpha}(k(t))$, which means $k(\alpha)= \mathrm{ev}_{\alpha}(k(t))$. 24 | If $\mathrm{ev}_{\alpha}:k(t) \rightarrow k(\alpha)$ is an onto field homomorphism, then $\mathrm{ev}_{\alpha}$ must be a field isomorphism. If we consider the simple extension $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2})$, we will find that 25 | \[ 26 | \mathrm{ev}_{\sqrt{2}}(0)=\mathrm{ev}_{\sqrt{2}}(t^2-1)=0, 27 | \] 28 | which contradicts the fact that $\mathrm{ev}_{\sqrt{2}}$ is an isomorphism. 29 | \end{solution} 30 | 31 | 32 | \begin{problem}[1.4] 33 | Let $k \subseteq k(\alpha)$ be a simple extension, with $\alpha$ transcendental over $k$. Let $E$ be a subfield of $k(\alpha)$ properly containing $k$. Prove that $k(\alpha)$ is a finite extension of $E$. 34 | \end{problem} 35 | \begin{solution} 36 | The field extension $E\subseteq k(\alpha)$ is finitely generated because $E(\alpha)= k(\alpha)$. Since $E$ be a subfield of $k(\alpha)$ properly containing $k$, we can suppose that there exist $f,g\in k[t]$ such that $g\ne0$, $\deg f>0$ and 37 | \[ 38 | \frac{f(\alpha)}{g(\alpha)}\in E. 39 | \] 40 | Then let 41 | \[ 42 | h(t)=f(t)-\frac{f(\alpha)}{g(\alpha)}g(t)\in E[t]. 43 | \] 44 | It is immediate that $h(\alpha)=0$, which means $\alpha$ is algebraic over $E$. Thus we show that $E\subseteq E(\alpha)$ is a finite extension, or equivalently $k(\alpha)$ is a finite extension of $E$. 45 | \end{solution} 46 | 47 | \begin{problem}[1.5] 48 | $\triangleright$ (Cf. Example 1.4.) 49 | \begin{itemize} 50 | \item Prove that there is exactly one subfield of $\mathbb{R}$ isomorphic to $\mathbb{Q}[t] /\left(t^{2}-2\right)$. 51 | \item Prove that there are exactly three subfields of $\mathbb{C}$ isomorphic to $\mathbb{Q}[t] /\left(t^{3}-2\right)$. 52 | \end{itemize} 53 | From a `topological' point of view, one of these copies of $\mathbb{Q}[t] /\left(t^{3}-2\right)$ looks very different from the other two: it is not dense in $\mathbb{C}$, but the others are. $[\S 1.2]$ 54 | \end{problem} 55 | \begin{solution} 56 | \begin{itemize} 57 | \item We will show that $\mathbb{Q}(\sqrt{2})$ is the unique subfield of $\mathbb{R}$ isomorphic to $\mathbb{Q}[t] /\left(t^{2}-2\right)$. 58 | 59 | First we assert that $\mathbb{Q}[t] /\left(t^{2}-2\right)\cong\mathbb{Q}(\sqrt{2})$. To prove this, notice that $\mathbb{Q}[t]$ is a PID and $t^2-2$ is irreducible over $\mathbb{Q}[t]$, which implies $\left(t^{2}-2\right)$ is a maximal ideal of $\mathbb{Q}[t]$ and $\mathbb{Q}[t] /\left(t^{2}-2\right)$ is a field. Then we can check that 60 | \begin{align*} 61 | \varphi:\mathbb{Q}[t] /\left(t^{2}-2\right)&\longrightarrow \mathbb{Q}(\sqrt{2}) \\ 62 | p(t)+(t^2-2)&\longmapsto p(\sqrt{2}) 63 | \end{align*} 64 | is an isomorphism. 65 | 66 | Suppose that $F$ is a subfield of $\mathbb{R}$ and $\psi:\mathbb{Q}[t] /\left(t^{2}-2\right)\to F$ is an isomorphism. Let $\bar{t}=t+(t^2-2)\in \mathbb{Q}[t] /\left(t^{2}-2\right)$. Since $\psi(\bar{t})\in F\subseteq\mathbb{R}$, we have 67 | \[ 68 | \psi\left(\bar{t}\right)^2-2=\psi\left(\overline{t}^2-2\right)=0\implies \psi(\bar{t})=\sqrt{2}\text{ or }-\sqrt{2}\implies \mathbb{Q}(\sqrt{2})\subseteq F. 69 | \] 70 | For any $x\in F$, there exists $g=q_1t+q_2+\left(t^{2}-2\right)\in\mathbb{Q}[t] /\left(t^{2}-2\right)$ such that $\psi(g)=x$, where $q_1,q_2\in\mathbb{Q}$. If $q_1=0$, we have 71 | \[ 72 | x-q_2=\psi(g-q_2)=0\implies x=q_2\in \mathbb{Q}(\sqrt{2}). 73 | \] 74 | If $q_1\ne0$, we have 75 | \[ 76 | (x-q_2)^2-2q_1^2=\psi((g-q_2)^2-2q_1^2)=\psi(q_1^2t^2-2q_1^2+(t^2-2))=0, 77 | \] 78 | which implies $x=\pm q_1\sqrt{2}+q_2\in \mathbb{Q}(\sqrt{2})$. Thus we have $F\subseteq\mathbb{Q}(\sqrt{2})$. Therefore $F=\mathbb{Q}(\sqrt{2})$, which guarantees the uniqueness of $\mathbb{Q}(\sqrt{2})$. 79 | \item We can show that 80 | \begin{align*} 81 | \mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}\left(\frac{-1+\sqrt{3}i}{2}\sqrt[3]{2}\right),\mathbb{Q}\left(\frac{-1-\sqrt{3}i}{2}\sqrt[3]{2}\right) 82 | \end{align*} 83 | are all isomorphic to $\mathbb{Q}[t] /\left(t^{3}-2\right)$. $\mathbb{Q}(\sqrt[3]{2})$ is not dense in $\mathbb{C}$. 84 | \end{itemize} 85 | \end{solution} 86 | 87 | \begin{problem}[1.6] 88 | $\triangleright$ Let $k \subseteq F$ be a field extension, and let $f(x) \in k[x]$ be a polynomial. Prove that $\operatorname{Aut}_{k}(F)$ acts on the set of roots of $f(x)$ contained in $F$. Provide examples showing that this action need not be transitive or faithful. [\S1.2, \S1.3] 89 | \end{problem} 90 | \begin{solution} 91 | Let $S_f$ be the set of roots of $f(x)$ contained in $F$. Note that for any $\sigma\in\operatorname{Aut}_{k}(F)$ and any $\alpha\in S_f$, $\sigma$ can be extend to $F[x]$ and $\sigma(f)=f$. Since 92 | \[ 93 | f(\sigma(\alpha))=\sigma(f)(\sigma(\alpha))=\sigma(f(\alpha))=\sigma(0)=0\implies\sigma(\alpha)\in S_f, 94 | \] 95 | we can define the mapping 96 | \begin{align*} 97 | \cdot:\operatorname{Aut}_{k}(F)\times S_f &\longrightarrow S_f\\ 98 | (\sigma, \alpha) &\longmapsto \sigma\cdot\alpha:=\sigma(\alpha). 99 | \end{align*} 100 | Since for any $\sigma_1,\sigma_2\in\operatorname{Aut}_{k}(F)$ and any root $\alpha$ of $f(x)$, 101 | \[ 102 | \sigma_1\cdot(\sigma_2\cdot\alpha)=\sigma_1\cdot\sigma_2(\alpha)=\sigma_1(\sigma_2(\alpha))=(\sigma_1\circ\sigma_2)(\alpha)=(\sigma_1\circ\sigma_2)\cdot\alpha, 103 | \] 104 | we see that $\operatorname{Aut}_{k}(F)$ acts on the set of roots of $f(x)$ contained in $F$. 105 | \end{solution} 106 | 107 | 108 | \hypertarget{Exercise VII.1.7}{} 109 | \begin{problem}[1.7] 110 | Let $k \subseteq F$ be a field extension, and let $\alpha \in F$ be algebraic over $k$. 111 | \begin{itemize} 112 | \item Suppose $p(x) \in k[x]$ is an irreducible monic polynomial such that $p(\alpha)=0$; prove that $p(x)$ is the minimal polynomial of $\alpha$ over $k$, in the sense of Proposition 1.3. 113 | \item Let $f(x) \in k[x]$. Prove that $f(\alpha)=0$ if and only if $p(x) \mid f(x)$. 114 | \item Show that the minimal polynomial of $\alpha$ is the minimal polynomial of a certain $k$-linear transformation of $F$, in the sense of Definition $\mathrm{VI}.6.12$. 115 | \end{itemize} 116 | \end{problem} 117 | \begin{solution} 118 | \begin{itemize} 119 | \item Suppose $q(x)$ is the minimal polynomial of $\alpha$ over $k$. Since $\deg q(x)\le \deg p(x)$, we have Euclidean division in the Euclidean domain $k[x]$ as follows 120 | \[ 121 | p(x) = m(x)q(x) + r(x),\quad \deg r(x)<\deg q(x). 122 | \] 123 | Taking $x=\alpha$, there must be $r(\alpha)=0$. Since $q(x)$ is the minimal polynomial, we can assert $r(x)=0$ and accordingly $q(x)\mid p(x)$. The irreducibility of $p(x)$ means $p(x) = q(x)$. 124 | \item It is clear that $p(x) \mid f(x)\implies f(\alpha)=p(\alpha)h(\alpha)=0$. To show $f(\alpha)=0\implies p(x) \mid f(x)$, we can just follow the same procedure as the first problem. 125 | \item Suppose $p(x) \in k[x]$ is the minimal polynomial of $\alpha$. We can easily check that 126 | \begin{align*} 127 | T_\alpha:F&\longrightarrow F,\\ 128 | m&\longmapsto\alpha m 129 | \end{align*} 130 | is a $k$-linear transformation. Since for any $m\in F$, 131 | \begin{align*} 132 | p(T_\alpha)(m)&=c_0+c_1T_\alpha m+c_2T_\alpha^2 m+\cdots+c_nT_\alpha^n m\\ 133 | &=c_0+c_1\alpha m+c_2\alpha^2 m+\cdots+c_n\alpha^n m\\ 134 | &=\left(c_0+c_1\alpha +c_2\alpha^2 +\cdots+c_n\alpha^n\right) m=0, 135 | \end{align*} 136 | we have $p(T_\alpha)=0$. Since $p(x)$ is an irreducible monic polynomial, $p(x)$ is exactly the minimal polynomial of $T_\alpha$. 137 | \end{itemize} 138 | \end{solution} 139 | 140 | \begin{problem}[1.8] 141 | $\neg$ Let $f(x) \in k[x]$ be a polynomial over a field $k$ of degree $d$, and let $\alpha_{1}, \ldots, \alpha_{d}$ be the roots of $f(x)$ in an extension of $k$ where the polynomial factors completely. For a subset $I \subseteq\{1, \ldots, d\}$, denote by $\alpha_{I}$ the sum $\sum_{i \in I} \alpha_{i}$. Assume that $\alpha_{I} \in k$ only for $I=\emptyset$ and $I=\{1, \ldots, d\}$. Prove that $f(x)$ is irreducible over $k$. [7.14] 142 | \end{problem} 143 | \begin{solution} 144 | Suppose $f(x)=(x-\alpha_1)\cdots(x-\alpha_d)$ is reducible over $k$ and $g(x)=\prod_{i\in I} (x-\alpha_{i})\in k[x]$ is a factor of $f$, where $I\subsetneq \{1, \ldots, d\}$. Note the coefficient of the $n-1$-th degree term of $g(x)$ is 145 | \[ 146 | -\sum_{i \in I} \alpha_{i}\notin k. 147 | \] 148 | We derive a contradiction. Hence we can conclude that $f(x)$ is irreducible over $k$. 149 | \end{solution} 150 | 151 | \begin{problem}[1.10] 152 | $\neg$ Let $k$ be a field. Prove that the ring of square $n \times n$ matrices $\mathcal{M}_{n}(k)$ contains an isomorphic copy of every extension of $k$ of degree $\leq n$. (Hint: If $k \subseteq F$ is an extension of degree $n$ and $\alpha \in F$, then `multiplication by $\alpha$' is a $k$-linear transformation of $F$.) [5.20] 153 | \end{problem} 154 | \begin{solution} 155 | Suppose $k\subseteq F$ is an extension of degree $n$ and $\alpha \in F$, then 156 | \begin{align*} 157 | T_\alpha:F&\longrightarrow F,\\ 158 | m&\longmapsto\alpha m 159 | \end{align*} 160 | is a $k$-linear transformation of $F$, which is isomorphic to a matrix $A_{T_\alpha}\in \mathcal{M}_{n}(k)$ in $\mathsf{Vect}_k$. What is left to do is to show that $A_{T_{\alpha+\beta}}=A_{T_{\alpha}}+A_{T_{\beta}}$, $A_{T_{\alpha\beta}}=A_{T_{\alpha}}A_{T_{\beta}}$ and $A_{T_{1_F}}=I_{n\times n}$, which amounts to $T_{\alpha\beta}=T_{\alpha}+T_{\beta}$, $T_{\alpha\beta}=T_{\alpha}T_{\beta}$ and $T_{1_F}=\mathrm{id}_F$. The verification is straightforward. 161 | \end{solution} 162 | 163 | \hypertarget{Exercise VII.1.11}{} 164 | \begin{problem}[1.11] 165 | $\neg$ Let $k \subseteq F$ be a finite field extension, and let $p(x)$ be the characteristic polynomial of the $k$-linear transformation of $F$ given by multiplication by $\alpha$. Prove that $p(\alpha)=0$. 166 | 167 | \hspace{2em}This gives an effective way to find a polynomial satisfied by an element of an extension. Use it to find a polynomial satisfied by $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$, and compare this method with the one used in Example 1.19. [1.12] 168 | \end{problem} 169 | \begin{solution} 170 | In \hyperlink{Exercise VII.1.7}{Exercise VII.1.7} we show that the minimal polynomial of $\alpha$ coincides with the minimal polynomial of $T_\alpha$. Denote the minimal polynomial as $f(x)$. Then we have $f(\alpha)=0$ and $f(x)\mid p(x)$, which implies $p(\alpha)=0$. 171 | 172 | Consider the field extension $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2},\sqrt{3})$ and the algebraic element $\alpha=\sqrt{2}+\sqrt{3}$. Since $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ is a basis of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, we can represent the linear transformation by a matrix $A$ as follows 173 | \begin{align*} 174 | &\hspace{15pt}\left(T_\alpha 1,T_\alpha\sqrt{2},T_\alpha\sqrt{3},T_\alpha\sqrt{6}\right)\\ 175 | &=\left(\sqrt{2}+\sqrt{3},2+\sqrt{6},3+\sqrt{6},3\sqrt{2}+2\sqrt{3}\right)\\ 176 | &=\left(1,\sqrt{2},\sqrt{3},\sqrt{6}\right) 177 | \begin{pmatrix} 178 | 0&2&3&0\\ 179 | 1&0&0&3\\ 180 | 1&0&0&2\\ 181 | 0&1&1&0 182 | \end{pmatrix}\\ 183 | &=\left(1,\sqrt{2},\sqrt{3},\sqrt{6}\right)A. 184 | \end{align*} 185 | The characteristic polynomial of $A$ is 186 | \[ 187 | p(x)=\det(xI-A)=x^4-10x^2+1. 188 | \] 189 | \end{solution} 190 | 191 | \hypertarget{Exercise VII.1.12}{} 192 | \begin{problem}[1.12] 193 | $\neg$ Let $k \subseteq F$ be a finite field extension, and let $\alpha \in F$. The norm of $\alpha$, $N_{k \subseteq F}(\alpha)$, is the determinant of the linear transformation of $F$ given by multiplication by $\alpha$ (cf. \hyperlink{Exercise VII.1.11}{Exercise VII.1.11}, Definition VI.6.4). 194 | Prove that the norm is multiplicative: for $\alpha, \beta \in F$, 195 | $$ 196 | N_{k \subseteq F}(\alpha \beta)=N_{k \subseteq F}(\alpha) N_{k \subseteq F}(\beta) . 197 | $$ 198 | Compute the norm of a complex number viewed as an element of the extension $\mathbb{R} \subseteq \mathbb{C}$ (and marvel at the excellent choice of terminology). Do the same for elements of an extension $\mathbb{Q}(\sqrt{d})$ of $\mathbb{Q}$, where $d$ is an integer that is not a square, and compare the result with \hyperlink{Exercise III.4.10}{Exercise III.4.10}. [1.13, 1.14, 1.15, 4.19, 6.18, VIII.1.5] 199 | \end{problem} 200 | \begin{solution} 201 | consider the field extension $\mathbb{R} \subseteq \mathbb{C}$ and a complex number $w=a+bi\in \mathbb{C}$. Given a basis $1$, $i$ of $\mathbb{C}$, denote the matrix representation of the linear transformation $F:z\mapsto wz$ as $A$. Then we have 202 | \begin{align*} 203 | &\hspace{15pt}\left(F(1),F(i)\right)\\ 204 | &=\left(a+bi,-b+ai\right)\\ 205 | &=\left(1,i\right) 206 | \begin{pmatrix} 207 | a&-b\\ 208 | b&a\\ 209 | \end{pmatrix}\\ 210 | &=\left(1,i\right)A. 211 | \end{align*} 212 | The norm $N_{\mathbb{R} \subseteq \mathbb{C}}(w)=\det(A)=a^2+b^2$. 213 | 214 | consider the field extension $\mathbb{Q} \subseteq \mathbb{Q}\left(\sqrt{d}\right)$ and a number $c=a+b\sqrt{d}\in \mathbb{Q}\left(\sqrt{d}\right)$. Given a basis $1$, $\sqrt{d}$ of $ \mathbb{Q}\left(\sqrt{d}\right)$, denote the matrix representation of the linear transformation $F:x\mapsto cx$ as $B$. Then we have 215 | \begin{align*} 216 | &\hspace{15pt}\left(F(1),F\left(\sqrt{d}\right)\right)\\ 217 | &=\left(a+b\sqrt{d},bd+a\sqrt{d}\right)\\ 218 | &=\left(1,\sqrt{d}\right) 219 | \begin{pmatrix} 220 | a&bd\\ 221 | b&a\\ 222 | \end{pmatrix}\\ 223 | &=\left(1,\sqrt{d}\right)B. 224 | \end{align*} 225 | The norm $N_{\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{d})}(c)=\det(B)=a^2-db^2$. 226 | 227 | \end{solution} 228 | 229 | \hypertarget{Exercise VII.1.13}{} 230 | \begin{problem}[1.13] 231 | $\neg$ Define the trace $\operatorname{tr}_{k \subseteq F}(\alpha)$ of an element $\alpha$ of a finite extension $F$ of a field $k$ by following the lead of \hyperlink{Exercise VII.1.12}{Exercise VII.1.12}. Prove that the trace is additive: 232 | $$ 233 | \operatorname{tr}_{k \subseteq F}(\alpha+\beta)=\operatorname{tr}_{k \subseteq F}(\alpha)+\operatorname{tr}_{k \subseteq F}(\beta) 234 | $$ 235 | for $\alpha, \beta \in F$. Compute the trace of an element of an extension $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{d})$, for $d$ an integer that is not a square. [1.14, 1.15, 4.19, VIII.1.5] 236 | \end{problem} 237 | \begin{solution} 238 | Given $c=a+b\sqrt{d}\in \mathbb{Q}\left(\sqrt{d}\right)$, the trace $\mathrm{tr}_{\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{d})}(c)=\operatorname{tr}(B)=2a$. 239 | \end{solution} 240 | 241 | \begin{problem}[1.14] 242 | $\neg$ Let $k \subseteq k(\alpha)$ be a simple algebraic extension, and let $x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$ be the minimal polynomial of $\alpha$ over $k$. Prove that 243 | $$ 244 | \operatorname{tr}_{k \subseteq k(\alpha)}(\alpha)=-a_{d-1} \quad \text { and } \quad N_{k \subseteq k(\alpha)}(\alpha)=(-1)^{d} a_{0} . 245 | $$ 246 | (Cf. \hyperlink{Exercise VII.1.12}{Exercise VII.1.12} and \hyperlink{Exercise VII.1.13}{Exercise VII.1.13}). [4.19] 247 | \end{problem} 248 | \begin{solution} 249 | Since $p(x)=x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$ is the minimal polynomial of $\alpha$, we see $1,\alpha,\cdots,\alpha^{d-1}$ is the basis of the $k$-vector space $k(\alpha)$ and 250 | \begin{align*} 251 | p(\alpha)=\alpha^{d}+a_{d-1} \alpha^{d-1}+\cdots+a_{0}=0 252 | \end{align*} 253 | The matrix representation of the linear map $F:z\mapsto \alpha z$ is 254 | \begin{align*} 255 | &\hspace{15pt}\left(F(1),F\left(\alpha\right),\cdots,F\left(\alpha^{d-1}\right)\right)\\ 256 | &=\left(1,\alpha,\alpha^2,\cdots,\alpha^{d-1}\right) 257 | \begin{pmatrix} 258 | 0&0&0&\cdots&0&-a_0\\ 259 | 1&0&0&\cdots&0&-a_1\\ 260 | 0&1&0&\cdots&0&-a_2\\ 261 | \vdots&\vdots&\vdots&&\vdots&\vdots\\ 262 | 0&0&0&\cdots&0&-a_{d-2}\\ 263 | 0&0&0&\cdots&1&-a_{d-1} 264 | \end{pmatrix}\\ 265 | &=\left(1,\alpha\right)A. 266 | \end{align*} 267 | Thus we have 268 | \begin{align*} 269 | \operatorname{tr}_{k \subseteq k(\alpha)}(\alpha)=\operatorname{tr}(A)=-a_{d-1} 270 | \end{align*} 271 | and 272 | \begin{align*} 273 | N_{k \subseteq k(\alpha)}(\alpha)=\det(A)=(-1)^{d-1} (-a_{0})= (-1)^{d} a_{0}. 274 | \end{align*} 275 | \end{solution} 276 | 277 | 278 | \begin{problem}[1.16] 279 | $\triangleright$ Let $k \subseteq L \subseteq F$ be fields, and let $\alpha \in F$. If $k \subseteq k(\alpha)$ is a finite extension, then $L \subseteq L(\alpha)$ is finite and $[L(\alpha): L] \leq[k(\alpha): k]$. [§1.3] 280 | \end{problem} 281 | \begin{solution} 282 | Since $k \subseteq k(\alpha)$ is a finite extension, it is a simple algebraic extension and there exists a minimal polynomial $p(t)$ of $\alpha$ over $k$. Since $p(t)$ can be seen as a polynomial in $L$ and $p(\alpha)=0$, we have $L \subseteq L(\alpha)$ is finite. Since the degree of the minimal polynomial of $\alpha$ over $L$ is not greater than the degree of $p(t)$, we have $[L(\alpha): L] \leq[k(\alpha): k]$. 283 | \end{solution} 284 | 285 | \begin{problem}[1.20] 286 | Let $p$ be a prime integer, and let $\alpha=\sqrt[p]{2} \in \mathbb{R}$. Let $g(x) \in \mathbb{Q}[x]$ be any nonconstant polynomial of degree $ 1$; so $a\diamond b$ and 127 | $b\diamond c$, but not $a\diamond c$. 128 | 129 | When we try to build a partition of $\mathbb{Z}$ using $\diamond$, we get 130 | "equivalence classes" that are not disjoint. For example, $[2]_{\diamond} = 131 | \{1,2,3\}$, but $[3]_{\diamond} = \{2,3,4\}$. Hence $\mathscr{P}_{\diamond}$ is 132 | not a partition of $\mathbb{Z}$. 133 | \end{solution} 134 | 135 | \hypertarget{Exercise I.1.6}{} 136 | \begin{problem}[1.6] 137 | Define a relation $\sim$ on the set $\mathbb{R}$ of real numbers, by setting $a\sim b\iff b-a \in\mathbb{Z}$. Prove that this is an equivalence relation, and find a \textquoteleft compelling' description for $\mathbb{R}/\sim$. Do the same for the relation $\approx$ on the plane $\mathbb{R}\times\mathbb{R}$ defined by declaring $(a_1, a_2)\approx(b_1, b_2)\iff b_1-a_1 \in\mathbb{Z}$ and $b_2-a_2 \in\mathbb{Z}$. [\textsection II.8.1, II.8.10] 138 | \end{problem} 139 | \begin{solution} 140 | Suppose $a,b,c\in\mathbb{R}$. We have that $a-a=0\in\mathbb{Z}$, so $\sim$ is reflexive. If $a\sim b$, then $b-a=k$ for some $k\in\mathbb{Z}$, so $a-b=-k\in\mathbb{Z}$, hence $b\sim a$. So $\sim$ is symmetric. Now, suppose that $a\sim b$ and $b\sim c$, in particular that $b-a=k\in\mathbb{Z}$ and $c-b=l\in\mathbb{Z}$. Then $c-a=(c-b) + (b-a) = l+k\in\mathbb{Z}$, so $a\sim c$. So $\sim$ is transitive. 141 | 142 | An equivalence class $[a]_{\sim}\in\mathbb{R}\,/\!\sim$ is the set of integers $\mathbb{Z}$ transposed by some real number $\epsilon\in[0,1)$. That is, for every set $X\in\mathbb{R}\,/\!\sim$, there is a real number $\epsilon\in[0,1)$ such that every $x\in X$ is of the form $k+\epsilon$ for some integer $k$. 143 | 144 | Now we will show that $\approx$ is an equivalence relation over $\mathbb{R}\times\mathbb{R}$. Supposing $a_1,a_2\in\mathbb{R}\times\mathbb{R}$, we have $a_1-a_1=a_2-a_2=0\in\mathbb{Z}$, so $(a_1,a_2)\approx(a_1,a_2)$. If we also suppose that $b_1,b_2,c_1,c_2\in\mathbb{R}\times\mathbb{R}$, then symmetry and transitivity can be shown as well: $(a_1,a_2)\approx(b_1,b_2)\implies b_1-a_1=k$ for some integer $k$ and $b_2-a_2=l$ for some integer $l$, hence $a_1-b_1=-k\in\mathbb{Z}$ and $a_2-b_2=-l\in\mathbb{Z}$, so 145 | $(b_1,b_2)\approx(a_1,a_2)$; also if $(a_1,a_2)\approx(b_1,b_2)$ and $(b_1,b_2)\approx(c_1,c_2)$, then 146 | $(b_1,b_2)-(a_1,a_2)=(k_1,k_2)\in\mathbb{Z}\times\mathbb{Z}$ as well as $(c_1,c_2)-(b_1,b_2)=(l_1,l_2)\in\mathbb{Z}\times\mathbb{Z}$, so $(c_1,c_2) - (a_1,a_2) = (c_1,c_2) - (b_1,b_2) + (b_1,b_2) - (a_1,a_2) = (k_1+l_1, k_2+l_2\in\mathbb{Z}\times\mathbb{Z}$. Thus $\approx$ is an equivalence relation. 147 | 148 | The interpretation of $\approx$ is similar to $\sim$. An equivalence class $X\in\mathbb{R}\times\mathbb{R}\,/\approx$ is just the 2-dimensional integer lattice $\mathbb{Z}\times\mathbb{Z}$ transposed by some pair of values $(\epsilon_1,\epsilon_2)\in[0,1)\times[0,1)$. 149 | 150 | Imaginatively, $\mathbb{R}/\sim$ can be viewed as a ring of length 1 by bending the real line $\mathbb{R}$ and gluing the points in the same equivalence class. And $\mathbb{R}\times\mathbb{R}/\approx$ can be viewed as a torus in a similar way. 151 | \end{solution} 152 | 153 | 154 | 155 | 156 | 157 | \subsection{\textsection2. Functions between sets} 158 | 159 | \begin{problem}[2.1] 160 | How many different bijections are there between a set $S$ with $n$ elements 161 | and itself? [\textsection II.2.1] 162 | \end{problem} 163 | \begin{solution} 164 | A function $f:S\to S$ is a graph $\Gamma_f\subseteq S\times S$. Since $f$ is 165 | bijective, then for all $y\in S$ there exists a unique $x\in S$ such that 166 | $(x,y)\in\Gamma_f$. We can see that $\abs{\Gamma_f} = n$. Since each $x$ must be 167 | unique, all the elements $x\in S$ must be present in the first component of 168 | exactly one pair in $\Gamma_f$. Furthermore, if we order the elements $(x,y)$ in 169 | $\Gamma_f$ by the first component, we can see that $\Gamma_f$ is just a 170 | permutation on the $n$ elements in $S$. For example, for $S=\{1,2,3\}$ one such 171 | $\Gamma_f$ is: 172 | % 173 | \[ \set{ (1,3), (2,2), (3,1) } \] 174 | % 175 | Since $\abs{S} = n$, the number of permutations of $S$ is $n!$. Hence there are $n!$ different bijections between $S$ and itself. 176 | \end{solution} 177 | 178 | 179 | % Problem 2.2 180 | \begin{problem}[2.2] 181 | $\rhd$ Prove statement (2) in Proposition 2.1. You may assume that given a 182 | family of disjoint subsets of a set, there is a way to choose one element in 183 | each member of the family. [\S2.5, V3.3] 184 | 185 | 186 | \end{problem} 187 | \begin{quote} 188 | \textbf{Proposition 2.1.} Assume $A\neq\varnothing$, and let $f:A\to 189 | B$ be a function. Then 190 | 191 | (1) $f$ has a left-inverse if and only if $f$ is injective; and \\ 192 | (2) $f$ has a right-inverse if and only if $f$ is surjective. 193 | \end{quote} 194 | \begin{solution} 195 | 196 | Let $A\neq\varnothing$ and suppose $f:A\to B$ is a function. 197 | 198 | ($\implies$) Suppose there exists a function $g$ that is a right-inverse of $f$. 199 | Then $f\circ g = \id_B$. Let $b\in B$. We have that $f(g(b)) = b$, so there 200 | exists an $a = g(b)$ such that $f(a) = b$. Hence $f$ is surjective. 201 | 202 | ($\impliedby$) Suppose that $f$ is surjective. We want to construct a function 203 | $g:B\to A$ such that $f(g(b)) = b$ for all $b\in B$. Since $f$ is surjective, 204 | for all $b\in B$ there is an $a\in A$ such that $f(a) = b$. For each $b\in B$ 205 | construct a set $\Lambda_b$ of such pairs: 206 | % 207 | \[ \Lambda_b = \set{ (a,b) \mid a \in A, f(a) = b } \] 208 | % 209 | Note that $\Lambda_b$ is non-empty for all $b\in B$. So that we can choose one 210 | pair $(a,b)$ ($a$ not necessarily unique) from each set in $\Lambda = 211 | \set{\Lambda_b\mid b\in B}$ to define $g:B\to A$: 212 | % 213 | \[ g(b) = a, \text{ where $a$ is in some $(a,b)\in\Lambda_b$} \] 214 | % 215 | Now, $g$ is a right-inverse of $f$. To show this, let $b\in B$. Since $f$ in 216 | surjective, $g$ has been defined such that when $a=g(b)$, $f(a)=b$, so we get 217 | that $f(g(b)) = (f\circ g)(b) = b$, thus $g$ is a right-inverse of $f$. 218 | \end{solution} 219 | 220 | 221 | % Problem 2.3 222 | \begin{problem}[2.3] 223 | Prove that the inverse of a bijection is a bijection and that the 224 | composition of two bijections is a bijection. 225 | \end{problem} 226 | 227 | \begin{enumerate} 228 | \item Suppose $f:A\to B$ is a bijection, and that $f^{-1}:B\to A$ is its inverse. 229 | We have that $f\circ f^{-1} = \id_B$ and $f^{-1}\circ f = \id_A$. Hence $f$ is 230 | the left- and right-inverse of $f^{-1}$, so $f^{-1}$ must be a bijection. 231 | 232 | \item Let $f:B\to C$ and $g:A\to B$ be bijections, and consider $f\circ g$. To 233 | show that $f$ is injective, let $a, a'\in A$ such that $(f\circ g)(a) = (f\circ 234 | g)(a')$. Since $f$ is a bijection, $f(g(a)) = f(g(a')) \implies g(a) = g(a')$. 235 | Also, since $g$ is a bijection, $g(a) = g(a') \implies a=a'$. Hence $f\circ g$ 236 | is injective. Now, let $c\in C$. Since $f$ is surjective, there is a $b\in B$ 237 | such that $f(b) = c$. Also, since $g$ is surjective, there is an $a\in A$ such 238 | that $g(a) = b$; this means that there is an $a\in A$ such that $(f\circ g)(a) = 239 | c$. So $f\circ g$ is bijective. 240 | \end{enumerate} 241 | 242 | 243 | % Problem 2.4 244 | \begin{problem}[2.4] 245 | $\rhd$ Prove that `isomorphism' is an equivalence relation (on any set 246 | of sets.) [\S4.1] 247 | \end{problem} 248 | 249 | \begin{solution} 250 | Let $S$ be a set. Then $\id_S$ is a bijection from $S$ to itself, so $S\cong S$. 251 | Let $T$ be another set with $S\cong T$, i.e. that there exists a bijection 252 | $f:S\to T$. Since $f$ is a bijection, it has an inverse $f^{-1}:T\to S$, so 253 | $T\cong S$. Finally, let $U$ also be a set, and assume that there exists 254 | bijections $f:S\to T$ and $g:T\to U$, i.e. that $S\cong T$ and $T\cong U$. From 255 | exercise \textbf{I.2.3} we know that the composition of bijections is itself a 256 | bijection. This means that $g\circ f: S\to U$ is a bijection, so $S\cong U$. 257 | Hence $\cong$ is an equivalence relation. 258 | \end{solution} 259 | 260 | 261 | % Problem 2.5 262 | \begin{problem}[2.5] 263 | $\rhd$ Formulate a notion of \textit{epimorphism}, in the style 264 | of the notion of \textit{monomorphism} seen in \S 2.6, and prove a result 265 | analogous to Proposition 2.3, for epimorphisms and surjections. 266 | \end{problem} 267 | 268 | \begin{solution} 269 | A function $f:A\to B$ is an \textit{epimorphism} if and only if for all sets $Z$ 270 | and all functions $\beta',\beta'':B\to Z$, if $\beta'\circ f = \beta''\circ f$, 271 | then $\beta' = \beta''$. Now we will show that $f$ is a surjection if and only if 272 | it is an epimorphism. 273 | 274 | ($\implies$) Suppose that $f:A\to B$ is surjective. Let $Z$ be a set and 275 | $\beta',\beta'':B\to Z$ be functions such that $\beta'\circ f = \beta''\circ f$. 276 | We need to show that $\beta' = \beta''$. Let $b\in B$. Since $f$ is surjective, 277 | there exists an $a\in A$ such that $f(a) = b$. Then $\beta'(b) = \beta'(f(a)) = 278 | (\beta'\circ f)(a) = (\beta''\circ f)(a) = \beta''(f(a)) = \beta''(b)$. Since 279 | $b\in B$ was arbitrary, $\beta' = \beta''$. Hence $f$ is an epimorphism. 280 | 281 | ($\impliedby$) Suppose that $f$ is an epimorphism. We need to show that $f$ is 282 | surjective. Let $Z = \{0,1\}$ be a two-element set. Define $\beta':B\to Z$ by 283 | $\beta'(b) = \begin{cases} 0 & \text{if } b\in \im(f) \\ 1 & \text{if } b\notin 284 | \im(f) \end{cases}$ and define $\beta'':B\to Z$ by $\beta''(b) = 0$ for all $b\in 285 | B$. Note that for any $a\in A$, we have $f(a)\in\im(f)$, so $\beta'(f(a)) = 0 = 286 | \beta''(f(a))$. Thus $\beta'\circ f = \beta''\circ f$. Since $f$ is an 287 | epimorphism, we have $\beta' = \beta''$. This means that $\beta'(b) = 0$ for all 288 | $b\in B$, which implies that $b\in\im(f)$ for all $b\in B$. Therefore $\im(f) = 289 | B$, so $f$ is surjective. 290 | \end{solution} 291 | 292 | 293 | % Problem 2.6 294 | \begin{problem}[2.6] 295 | With notation as in Example 2.4, explain how any function $f:A\to B$ determines 296 | a section of $\pi_A$. 297 | \end{problem} 298 | 299 | \begin{solution} 300 | Let $f:A\to B$ and let $\pi_A:A\times B\to A$ be such that $\pi_A(a,b) = a$ for 301 | all $(a,b)\in A\times B$. Construct $g:A\to A\times B$ defined as $g(a) = (a, 302 | f(a))$ for all $a\in A$. The function $g$ can be thought of as `determined by' 303 | $f$. Now, since $(\pi_A\circ g)(a) = \pi_A(g(a)) = \pi_A(a, f(a)) = a$ for all 304 | $a\in A$, $g$ is a right inverse of $\pi_A$, i.e. $g$ is a section of $\pi_A$ as 305 | required. 306 | \end{solution} 307 | 308 | 309 | % Problem 2.7 310 | \begin{problem}[2.7] 311 | Let $f:A\to B$ be any function. Prove that the graph $\Gamma_f$ of $f$ is 312 | isomorphic to $A$. 313 | \end{problem} 314 | 315 | \begin{solution} 316 | Recall that sets $\Gamma_A$ and $A$ are \textit{isomorphic}, written 317 | $\Gamma_A\cong A$, if and only if there exists a bijection $g:\Gamma_A\to A$. 318 | Let's construct such a function $g$, defined to be $g(a,b) = a$. Keep in mind 319 | that here $(a,b)\in\Gamma_f\subseteq A\times B$. 320 | 321 | Let $(a',b'),(a'',b'')\in\Gamma_f$ such that $g(a',b') = g(a'',b'')$. For 322 | contradiction, suppose that $(a',b')\neq (a'',b'')$. Since $g(a',b') = a' = a'' 323 | = g(a'',b'')$, it must be that $b'\neq b''$. However, this would mean that both 324 | $(a',b')$ and $(a',b'')$ are in $\Gamma_f$; this would mean that $f(a') = b' 325 | \neq b'' = f(a')$, which is impossible since $f$ is a function. Hence $g$ is 326 | injective. 327 | 328 | Let $a'\in A$. Since $f$ is a well-defined function with $A$ as its domain, 329 | there must exists a pair $(a',b')\in\Gamma_f$ for some $b'\in B$, in particular 330 | that $g(a',b') = a'$; thus $g$ is surjective, so it is a bijection. 331 | \end{solution} 332 | 333 | 334 | % Problem 2.8. 335 | \begin{problem}[2.8] 336 | Describe as explicitly as you can all terms in the canonical decomposition (cf. 337 | \S2.8) of the function $\mathbb{R}\to\mathbb{C}$ defined by $r\mapsto e^{2\pi 338 | ir}$. (This exercise matches one previously. Which one?) 339 | \end{problem} 340 | 341 | \begin{solution} 342 | Let $f:\mathbb{R}\to\mathbb{C}$ be as above. The first piece in the canonical 343 | decomposition is the equivalence relation $\sim$ defined as $x \sim x' \iff f(x) = 344 | f(x')$, i.e. $[x]_{\sim}$ is the set of all elements in $\mathbb{R}$ that get 345 | mapped to the same element in $\mathbb{C}$ by $f$ as $x$. 346 | 347 | The second piece is the set $\mathscr{P}_{\sim}$. This set is the set of all 348 | equivalence classes of $\mathbb{R}$ over equality up to $f$. Note that, since 349 | $f(x) = e^{2\pi i x} = \cos(2\pi x) + i\sin(2\pi x)$, $f$ is periodic with 350 | period $1$. That is, $f(x) = e^{2\pi i x} = e^{2\pi i x + 2\pi} = e^{2\pi i (x + 351 | 1)} = f(x+1)$. In other words, we can write $\mathscr{P}_{\sim}$ as, 352 | % 353 | \[ \mathscr{P}_{\sim} = \set{\set{r + k\mid k\in\mathbb{Z}}\mid 354 | r\in[0,1)\subseteq\mathbb{R}}, \] 355 | % 356 | and it is here when we notice uncanny similarities to \hyperlink{Exercise I.1.6}{Exercise I.1.6} 357 | where $x\sim y$, for $x,y\in\mathbb{R}$, if and only if $x-y\in\mathbb{Z}$, in 358 | which we could have written $\mathscr{P}_{\sim}$ in the same way. 359 | 360 | Now we will explain the mysterious $\tilde{f}:\mathscr{P}_{\sim}\to\im f$. This 361 | function is taking each \textit{equivalence class} $[x]_{\sim}$ over the reals 362 | w.r.t. $\sim$ and mapping it to the element in $\mathbb{C}$ that $f$ maps each 363 | element $x'\in[x]_{\sim}$ to; indeed, since $x\sim x'$ is true for 364 | $x,x'\in\mathbb{R}$ if and only if $f(x)=f(x')$, we can see that for any 365 | $x\in\mathbb{R}$, for all $x'\in[x]_{\sim}$, there exists a $c\in\mathbb{C}$ 366 | such that $f(x') = c$. To illustrate with the equivalence class over 367 | $\mathbb{R}$ w.r.t. $\sim$ corresponding to the element $0\in\mathbb{R}$, we 368 | have $[0]_{\sim} = \set{\dots, -2, -1, 0, 1, 2, \dots}$. We can see that 369 | $e^{-4\pi i} = e^{-2\pi i} = e^{0\pi i} = 1 = e^{2\pi i} = e^{4\pi i}$, etc; so 370 | the function would map $[0]_{\sim}\mapsto1\in\mathbb{C}$, and so on. 371 | Furthermore, we can see that $\tilde{f}$ is surjective, since for $y$ to be in 372 | $\im f$ is to say that there is an $x\in\mathbb{R}$ such that $f(x) = y$; so 373 | there must be an equivalence class $[x]_{\sim}$ which is mapped to $y$ by 374 | $\tilde{f}$. 375 | 376 | Finally, the simple map from $\im f\to\mathbb{C}$ that simply takes $c\mapsto 377 | c$. This can be thought of as a potential ``expansion'' of the domain of 378 | $\tilde{f}$. It is obviously injective, since (trivially) $c\neq c'\implies 379 | c\neq c'$. However, it may not be surjective: for example, $2\in\mathbb{C}$ is 380 | not in $\im f$ as it is defined above. 381 | \end{solution} 382 | 383 | 384 | % Problem 2.9 385 | \hypertarget{Exercise I.2.9}{} 386 | \begin{problem}[2.9] 387 | $\rhd$ Show that if $A'\cong A''$ and $B'\cong B''$, and further 388 | $A'\cap B'=\varnothing$ and $A''\cap B''=\varnothing$, then $A'\cup B'\cong A''\cup 389 | B''$. Conclude that the operation $A\amalg B$ is well-defined up to 390 | \textit{isomorphism} (cf. \S2.9) [\S2.9, 5.7] 391 | \end{problem} 392 | 393 | \begin{solution} 394 | Let $A',A'',B',B''$ be sets as described above. Since $A'\cong A''$ and $B'\cong 395 | B''$, we know there exists respective bijections $f:A'\to A''$ and $g:B'\to 396 | B''$. Now, we wish to show that $A'\cup B'\cong A''\cup B''$. Define a function 397 | $h:A'\cup B'\to A''\cup B''$ such that $h(x) = f(x)$ if $x\in A'$ and $g(x)$ if 398 | $x\in B'$. 399 | 400 | We will now show that $h$ is a bijection. Let $y\in A''\cup B''$. Then, since 401 | $A''\cap B''=\varnothing$, either $y\in A''$ or $y\in B''$. Without loss of 402 | generality suppose that $y\in A''$. Then, since $f:A'\to A''$ is a bijection, it 403 | is \textit{surjective}, so there exists an $x\in A'\subseteq A'\cup B'$ such 404 | that $h(x) = f(x) = y$. So $h$ is surjective. Now, suppose that $x\neq x'$, for 405 | $x,x'\in A'\cup B'$. If $x,x'\in A'$, then since $f$ is injective and $h(x) = 406 | f(x)$ for all $x\in A'$, then $h(x)\neq h(x')$. Similarly for if $x,x'\in B'$. 407 | Now, without loss of generality if $x\in A'$ and $x'\in B'$, then $h(x) = f(x) 408 | \neq g(x') = h(x')$ since $A''\cap B''=\varnothing$. Hence $h$ is a bijection, so 409 | $A'\cup B'\cong A''\cup B''$. 410 | 411 | Since these constructions of $A',A'',B',B''$ correspond to creating ``copies'' 412 | of sets $A$ and $B$ for use in the disjoint union operation, we have that 413 | disjoint union is a well-defined function \textit{up to isomorphism}. In 414 | particular, since $\cong$ is an equivalence relation, we can consider $\amalg$ 415 | to be well-defined from $\mathscr{P}_{\cong}$ to $A'\cup B'$. 416 | \end{solution} 417 | 418 | 419 | % Problem 2.10 420 | \hypertarget{Exercise I.2.10}{} 421 | \begin{problem}[2.10] 422 | $\rhd$ Show that if $A$ and $B$ are finite sets, then $\abs{B^A} = 423 | \abs{B}^{\abs{A}}$. [\S2.1, 2.11, I.4.1] 424 | \end{problem} 425 | 426 | \begin{solution} 427 | Let $A$ and $B$ be sets with $\abs{A}=n$ and $\abs{B}=m$, with $n,m$ being 428 | non-negative integers. Recall that $B^A$ denotes the set of functions $f:A\to 429 | B$. Now, if $A=B=\varnothing$ or $A=\varnothing$ and $\abs{B}=1$, we get one 430 | function, the empty function $\Gamma_f = \varnothing$, and $0^0 = 1^0 = 1$. If 431 | $\abs{A} = \abs{B} = 1$, then we get the singleton function 432 | $\Gamma_f=\{(a,b)\}$, and $1^1 = 1$. If $A\neq\varnothing$ and $B=\varnothing$, then 433 | no well-defined function can exist from $A$ to $B$ since there will be no value 434 | for the elements in $A$ to take; this explains $\abs{B^A} = \abs{B}^{\abs{A}} = 435 | 0^{\abs{A}} = 0$. 436 | 437 | Suppose that $B\neq\varnothing$ and $B$ is finite. We will show inductively that 438 | $\abs{B^A} = \abs{B}^{\abs{A}}$. First, suppose that $\abs{A} = 1$. Then there 439 | are exactly $\abs{B}$ functions from $A$ to $B$: if $B=\set{b_1,b_2,\dots,b_m}$, 440 | then the functions are $\{(a,b_1)\}, \{(a,b_2)\}$, etc. Hence $\abs{B^A} = 441 | \abs{B}^{\abs{A}} = \abs{B}$. Now, fix $k\geq 2$, and assume that $\abs{B^A} = 442 | \abs{B}^{\abs{A}}$ for all sets $A$ such that $\abs{A}=k-1$. Suppose that 443 | $\abs{A}=k$. Let $a\in A$. (We can do this since $\abs{A}=k\geq 2$.) Then, by 444 | the inductive hypothesis, since $\abs{A\backslash\{a\}}=k-1$, 445 | $\abs{B^{(A\backslash\{a\})}} = \abs{B}^{\abs{A}-1}$. Let $F$ be the set of 446 | functions from $A\backslash\{a\}$ to $B$. Then, for each of those functions 447 | $f\in F$, there is $\abs{B}$ ``choices'' of where to assign $a$: one choice for 448 | each element in $B$. Hence, $\abs{B^A} = \abs{B}\abs{B}^{\abs{A}-1} = 449 | \abs{B}^{\abs{A}}$ as required. 450 | \end{solution} 451 | 452 | 453 | % Problem 2.11 454 | \begin{problem}[2.11] 455 | $\rhd$ In view of Exercise 2.10, it is not unreasonable to use $2^A$ to denote 456 | the set of functions from an arbitrary set $A$ to a set with $2$ elements (say 457 | $\{0,1\}$). Prove that there is a bijection between $2^A$ and the \textit{power 458 | set} of $A$ (cf. \S1.2). [\S1.2, III.2.3] 459 | \end{problem} 460 | 461 | \begin{solution} 462 | Let $S = \{0,1\}$, and consider $f:\mathcal{P}(A)\to 2^A$, defined as 463 | % 464 | \[ f(X) = \set{(a,1) \text{ if $a\in X$, and }(a,0) \text{ otherwise}} \] 465 | % 466 | We will show that $f$ is bijective. Let $g\in 2^A$. Then $f$ is a 467 | function from $A$ to $S$. Let $A_1 = \set{a\in A\mid g(a) = 1}$. Then $A_1$ is a 468 | set such that $A_1\in\mathcal{P}(A)$, and $f(A_1)=g$. Hence $f$ is surjective. 469 | 470 | Now, suppose that $X,Y\subseteq A$ and $f(X) = f(Y)$. Then, for all $a\in A$, 471 | $a\in X \iff f(X)(a) = 1 \iff f(Y)(a) = 1 \iff a\in Y$. Hence $f$ is injective, 472 | so $2^A\cong\mathcal{P}(A)$. 473 | \end{solution} 474 | 475 | \subsection{\textsection3. Categories} 476 | \hypertarget{Exercise I.3.1}{} 477 | \begin{problem}[3.1] 478 | Let $\mathsf{C}$ be a category. Consider a structure $\mathsf{C}^{op}$ with: 479 | \begin{itemize} 480 | \item $\Obj(\mathsf{C}^{op}) := \Obj(\mathsf{C})$; 481 | \item for $A$, $B$ objects of $\mathsf{C}^{op}$ (hence, objects of $\mathsf{C}$), $\Hom_{\mathsf{C}^{op}} (A,B) := \Hom_\mathsf{C}(B,A)$ 482 | \end{itemize} 483 | Show how to make this into a category (that is, define composition of morphisms 484 | in $\mathsf{C}^{op}$ and verify the properties listed in \textsection3.1). 485 | Intuitively, the `opposite' category $\mathsf{C}^{op}$ is simply obtained by `reversing all the 486 | arrows' in C. [5.1, \textsection VIII.1.1, \textsection IX.1.2, IX.1.10] 487 | \end{problem} 488 | \begin{solution} 489 | \begin{itemize} 490 | \item For every object $A$ of $\mathsf{C}$, there exists one identity morphism $1_A\in\Hom_\mathsf{C}(A,A)$. Since $\Obj(\mathsf{C}^{op}) := \Obj(\mathsf{C})$ and $\Hom_{\mathsf{C}^{op}} (A,A) := \Hom_\mathsf{C}(A,A)$, for every object $A$ of $\mathsf{C}^{op}$, the identity on $A$ coincides with $1_A\in\mathsf{C}$. 491 | \item For $A$, $B$, $C$ objects of $\mathsf{C}^{op}$ and $f\in\Hom_{\mathsf{C}^{op}} (A,B)=\Hom_\mathsf{C}(B,A)$, $g\in\Hom_{\mathsf{C}^{op}} (B,C)=\Hom_\mathsf{C}(C,B)$, the composition laws in $\mathsf{C}$ determines a morphism $f*g$ in $\Hom_{\mathsf{C}} (C,A)$, which deduces the composition defined on $\mathsf{C}^{op}$: 492 | \[ 493 | \begin{aligned} 494 | \Hom_{\mathsf{C}^{op}} (A,B)\times\Hom_{\mathsf{C}^{op}} (B,C)&\longrightarrow \Hom_{\mathsf{C}^{op}} (A,C)\\ 495 | (f,g)&\longmapsto g\circ f:=f*g 496 | \end{aligned} 497 | \] 498 | \item Associativity. If $f\in\Hom_{\mathsf{C}^{op}} (A,B)$, $g\in\Hom_{\mathsf{C}^{op}} (B,C)$, $h\in\Hom_{\mathsf{C}^{op}} (C,D)$, then 499 | \[ 500 | f\circ(g\circ h)=f\circ(h*g)=(h*g)*f=h*(g*f)=(g*f)\circ h=(f\circ g)\circ h. 501 | \] 502 | \item Identity. For all $f\in\Hom_{\mathsf{C}^{op}} (A,B)$, we have 503 | \[ 504 | f\circ 1_A=1_A*f=f,\quad 1_B\circ f=f*1_B=f. 505 | \] 506 | \end{itemize} 507 | Thus we get the full construction of $\mathsf{C}^{op}$. 508 | \end{solution} 509 | 510 | % Problem 3.2 511 | \begin{problem}[3.2] 512 | If $A$ is a finite set, how large is $\mathrm{End}_{\mathsf{Set}}(A)$? 513 | \end{problem} 514 | \begin{solution} 515 | The set $\mathrm{End}_{\mathsf{Set}}(A)$ is the set of functions $f:A\to A$. 516 | Since $A$ is finite, write $\abs{A} = n$ for some $n\in\mathbb{Z}$. By \hyperlink{Exercise I.2.10}{Exercise I.2.10}, we know that $\abs{A^A} = \abs{A}^{\abs{A}} = n^n$. So the the set 517 | $\mathrm{End}_{\mathsf{Set}}(A)$ has size $n^n$. 518 | \end{solution} 519 | 520 | 521 | 522 | \begin{problem}[3.3] 523 | $\vartriangleright$ Formulate precisely what it means to say that $1_a$ is an identity with respect to composition in Example 3.3, and prove this assertion. [\textsection3.2] 524 | \end{problem} 525 | \begin{solution} 526 | Suppose $S$ is a set, and $\sim$ is a relation on $S$ satisfying the reflexive and transitive property. Then we can encode this data into a category $\mathsf{C}$: 527 | \begin{itemize} 528 | \item Objects: the elements of $S$; 529 | \item Morphisms: if $a, b$ are objects (that is: if $a, b \in S$) then let $\Hom(a, b)$ be the set consisting of the element $(a, b) \in S \times S$ if $a \sim b$, and $\Hom(a, b) = \varnothing$. 530 | otherwise. 531 | \end{itemize} 532 | Given the composition of two morphisms 533 | \begin{align*} 534 | \Hom_\mathsf{C}(A,B) \times\Hom_\mathsf{C}(B,C)&\longrightarrow\Hom_\mathsf{C}(A,C)\\ 535 | (a,b)\circ(b,c)&\longmapsto(a,c) 536 | \end{align*} 537 | we are asked to check $1_a = (a, a)$ is an identity with respect 538 | to this composition. 539 | \end{solution} 540 | 541 | % Problem 3.4 542 | \begin{problem}[3.4] 543 | Can we define a category in the style of Example 3.3 using the relation $<$ on 544 | the set $\mathbb{Z}$? 545 | \end{problem} 546 | \begin{solution} 547 | No, we can't. This is because $<$ isn't reflexive: $x\not