├── README.md ├── Atiyah-Macdonald ├── 1 Rings and Ideals │ ├── SpecZ.pdf │ └── SpecZx.pdf ├── 5 Integral Dependence and Valuations │ └── 5 Integral Dependence and Valuations.tex └── 3 Rings and Modules of Fractions │ └── 3 Rings and Modules of Fractions.tex ├── .gitignore ├── Milnor └── Milnor.tex ├── Hartshorne ├── II Schemes │ ├── 4 Separated and Proper Morphisms │ │ └── II.4 Separated and Proper Morphisms.tex │ └── 6 Divisors │ │ └── II.6 Divisors.tex └── I Varieties │ ├── 3 Morphisms │ └── I.3 Morphisms.tex │ └── 1 Affine Varieties │ └── I.1 Affine Varieties.tex └── References.bib /README.md: -------------------------------------------------------------------------------- 1 | math-solutions 2 | ============== 3 | 4 | Solutions to various math textbooks 5 | -------------------------------------------------------------------------------- /Atiyah-Macdonald/1 Rings and Ideals/SpecZ.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/awasthi/math-solutions/HEAD/Atiyah-Macdonald/1 Rings and Ideals/SpecZ.pdf -------------------------------------------------------------------------------- /Atiyah-Macdonald/1 Rings and Ideals/SpecZx.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/awasthi/math-solutions/HEAD/Atiyah-Macdonald/1 Rings and Ideals/SpecZx.pdf -------------------------------------------------------------------------------- /.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | 10 | ## Intermediate documents: 11 | *.dvi 12 | *-converted-to.* 13 | # these rules might exclude image files for figures etc. 14 | # *.ps 15 | # *.eps 16 | # *.pdf 17 | 18 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 19 | *.bbl 20 | *.bcf 21 | *.blg 22 | *-blx.aux 23 | *-blx.bib 24 | *.brf 25 | *.run.xml 26 | 27 | ## Build tool auxiliary files: 28 | *.fdb_latexmk 29 | *.synctex.gz 30 | *.synctex.gz(busy) 31 | *.pdfsync 32 | 33 | ## Auxiliary and intermediate files from other packages: 34 | 35 | # algorithms 36 | *.alg 37 | *.loa 38 | 39 | # amsthm 40 | *.thm 41 | 42 | # beamer 43 | *.nav 44 | *.snm 45 | *.vrb 46 | 47 | #(e)ledmac/(e)ledpar 48 | *.end 49 | *.[1-9] 50 | *.[1-9][0-9] 51 | *.[1-9][0-9][0-9] 52 | *.[1-9]R 53 | *.[1-9][0-9]R 54 | *.[1-9][0-9][0-9]R 55 | *.eledsec[1-9] 56 | *.eledsec[1-9]R 57 | *.eledsec[1-9][0-9] 58 | *.eledsec[1-9][0-9]R 59 | *.eledsec[1-9][0-9][0-9] 60 | *.eledsec[1-9][0-9][0-9]R 61 | 62 | # glossaries 63 | *.acn 64 | *.acr 65 | *.glg 66 | *.glo 67 | *.gls 68 | 69 | # hyperref 70 | *.brf 71 | 72 | # listings 73 | *.lol 74 | 75 | # makeidx 76 | *.idx 77 | *.ilg 78 | *.ind 79 | *.ist 80 | 81 | # minitoc 82 | *.maf 83 | *.mtc 84 | *.mtc0 85 | 86 | # minted 87 | *.pyg 88 | 89 | # morewrites 90 | *.mw 91 | 92 | # nomencl 93 | *.nlo 94 | 95 | # sagetex 96 | *.sagetex.sage 97 | *.sagetex.py 98 | *.sagetex.scmd 99 | 100 | # sympy 101 | *.sout 102 | *.sympy 103 | sympy-plots-for-*.tex/ 104 | 105 | # todonotes 106 | *.tdo 107 | 108 | # xindy 109 | *.xdy 110 | Hartshorne/I Varieties/1 Affine Varieties/I.1 Affine Varieties.pdf 111 | *.pdf 112 | 113 | *.swp 114 | *.swo 115 | -------------------------------------------------------------------------------- /Milnor/Milnor.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt,letterpaper]{article} 2 | \usepackage{geometry} 3 | \geometry{letterpaper} 4 | \usepackage{amsmath,amssymb,amsthm,mathrsfs} 5 | \usepackage{mathtools} 6 | \usepackage{ifpdf} 7 | \ifpdf 8 | \setlength{\pdfpagewidth}{8.5in} 9 | \setlength{\pdfpageheight}{11in} 10 | \else 11 | \fi 12 | \usepackage{hyperref} 13 | 14 | \usepackage[utf8]{inputenc} 15 | \usepackage{csquotes} 16 | \usepackage[american]{babel} 17 | %\usepackage[style=alphabetic,firstinits=true,backend=biber,texencoding=utf8,bibencoding=utf8]{biblatex} 18 | %\bibliography{../../References} 19 | %\AtEveryBibitem{\clearfield{url}} 20 | %\AtEveryBibitem{\clearfield{doi}} 21 | %\AtEveryBibitem{\clearfield{issn}} 22 | %\AtEveryBibitem{\clearfield{isbn}} 23 | %\renewbibmacro{in:}{} 24 | %\DeclareFieldFormat{postnote}{#1} 25 | %\DeclareFieldFormat{multipostnote}{#1} 26 | 27 | \newtheorem{problem}{Problem}[section] 28 | \renewcommand{\theproblem}{\thesection-\Alph{problem}} 29 | \newtheorem{lemma}{Lemma}%[section] 30 | 31 | \numberwithin{equation}{section} 32 | 33 | %Put Macros here 34 | \newcommand{\RR}{\ensuremath \mathbf{R}} 35 | 36 | \title{Solutions to Milnor's \emph{Characteristic Classes}} 37 | \author{Takumi Murayama, Matt Stevenson, Feng Zhu} 38 | 39 | \begin{document} 40 | \maketitle 41 | \section{Smooth Manifolds} 42 | \begin{problem} 43 | Let $M_1 \subset \RR^A$ and $M_2 \subset \RR^B$ be smooth manifolds. Show that 44 | $M_1 \times M_2 \subset \RR^A \times \RR^B$ is a smooth manifold, and that the 45 | tangent manifold $D(M_1 \times M_2)$ is canonically diffeomorphic to the 46 | product $DM_1 \times DM_2$. Note that a function $x \mapsto (f_1(x),f_2(x))$ 47 | from $M$ to $M_1 \times M_2$ is smooth if and only if both $f_1 \colon M \to 48 | M_1$ and $f_2 \colon M \to M_2$ are smooth. 49 | \end{problem} 50 | \begin{proof} 51 | \textbf{Proof would go here}. 52 | \end{proof} 53 | 54 | \section{Vector Bundles} 55 | 56 | \begin{problem} 57 | Show that the unit sphere $S^n$ admits a vector field which is nowhere zero, providing that $n$ is odd. Show that the normal bundle of $S^n \subset \RR^{n+1}$ is trivial for all $n$. 58 | \end{problem} 59 | 60 | \begin{proof} 61 | Let $S^n \subset \RR^{n+1}$ be the standard embedding and let $(x^1,\ldots,x^{n+1})$ be the standard coordinates on $\RR^{n+1}$. For each $\vec{x} = (x^1,\ldots,x^{n+1}) \in S^n$, define $v(\vec{x}) := (x^2, -x^1, \ldots, x^{n+1}, -x^n) \in \RR^{n+1}$. This function $v \colon S^n \to TS^n$ is a vector field, since 62 | \[ 63 | v(\vec{x}) \cdot \vec{x} = (x^1 x^2 - x^2 x^1 ) + \ldots + (x^{n} x^{n+1} - x^{n+1} x^n) = 0 \textrm{ for all $\vec{x} \in S^n$.} 64 | \] 65 | It is clear that $v$ does not vanish. 66 | 67 | To each $\vec{x} \in S^n$, let $s(\vec{x}) = (\vec{x},\vec{x})$, where the first coordinate denotes the basepoint and the second coordinate denotes the point on the line through the origin and $\vec{x}$. As the normal bundle of $S^n \subset \RR^{n+1}$ has rank 1, the existence of the nowhere vanishing section $s$ implies that the bundle is trivial, by Theorem 2.2. 68 | 69 | \end{proof} 70 | 71 | \begin{problem} 72 | If $S^n$ admits a vector field which is nowhere zero, show that the identity map of $S^n$ is homotopic to the antipodal map. For $n$ even, show that the antipodal map of $S^n$ is homotopic to the reflection 73 | \[ 74 | r(x^1,\ldots,x^{n+1}) = (-x^1,x^2,\ldots,x^{n+1}), 75 | \] 76 | and therefore has degree $-1$. Combining these facts, show that $S^n$ is not parallelizable for $n$ even, $n \geq 2$. 77 | \end{problem} 78 | 79 | \begin{proof} 80 | Let $v \colon S^n \to \RR^{n+1}$ be a nonwhere zero vector field, then for $t \in [0,\pi]$, 81 | \[ 82 | (\vec{x},t) \mapsto H(\vec{x},t) := \cos(t) \vec{x} + \sin(t) \frac{v(\vec{x})}{\| v(\vec{x}) \|^2 } 83 | \] 84 | defines a homotopy from the identity map to the antipodal map (remark that $v$ is needed so that the homotopy stays on $S^n$: indeed, for any fixed $t$, $H(\vec{x},t) \cdot H(\vec{x},t) = 1$ since $\vec{x} \cdot v(\vec{x}) = 0$.) 85 | 86 | Middle proof still to be added. 87 | 88 | Let $n \geq 2$ be even and assume that $S^n$ is parallelizable, then there exist $n$ nowhere dependent (hence, nowhere zero) sections of $S^n$. Thus, the identity map is homotopic to the antipodal map, which is in turn homotopic to the above reflection; taking degrees, we find that $1= \deg(id) = \deg(-id) = \deg(r) = -1$, a contradiction. 89 | \end{proof} 90 | 91 | \begin{problem} 92 | (Existence theorem for Euclidean metrics) Using a partition of unity, show that any vector bundle over a paracompact base can be given a Euclidean metric. 93 | \end{problem} 94 | 95 | \begin{proof} 96 | Let $\{ U_{\alpha} \}$ be an open cover of the base space $M$ such that the vector bundle $(E,M,\pi)$ is trivial over each $U_{\alpha}$. Over each $U_{\alpha}$, we can equip $\pi^{-1}(U_{\alpha}) \simeq U_{\alpha} \times \RR^{\textrm{rank}(E)}$ with a Euclidean metric $\langle \cdot, \cdot \rangle_{\alpha}$. By paracompactness, there is a partition of unity $\{ \psi_{\alpha} \}$ subordinate to $\{ U_{\alpha} \}$, so $\langle \cdot , \cdot \rangle := \sum \psi_{\alpha} \langle \cdot , \cdot \rangle_{\alpha}$ is a Euclidean metric on $E$. 97 | \end{proof} 98 | 99 | \begin{problem} 100 | The Alexandroff line $L$ (also called the ``long line'') is a smooth, connected, 1-dimensional manifold which is not paracompact. Show that $L$ cannot be given a Riemannian metric. 101 | \end{problem} 102 | 103 | \begin{proof} 104 | A Riemannian metric induces a metric by taking the infimum of the lengths of paths connecting 2 points, but $L$ is not metrizable (proof needed). 105 | \end{proof} 106 | \end{document} 107 | -------------------------------------------------------------------------------- /Hartshorne/II Schemes/4 Separated and Proper Morphisms/II.4 Separated and Proper Morphisms.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt,letterpaper]{article} 2 | \usepackage{geometry} 3 | \geometry{letterpaper} 4 | \usepackage{microtype} 5 | \usepackage{amsmath,amssymb,amsthm,mathrsfs} 6 | \usepackage{mathtools} 7 | \usepackage{ifpdf} 8 | \ifpdf 9 | \setlength{\pdfpagewidth}{8.5in} 10 | \setlength{\pdfpageheight}{11in} 11 | \else 12 | \fi 13 | \usepackage{hyperref} 14 | 15 | \usepackage{tikz} 16 | \usepackage{tikz-cd} 17 | \usetikzlibrary{decorations.markings} 18 | \tikzset{ 19 | open/.style = {decoration = {markings, mark = at position 0.5 with { \node[transform shape] {\tikz\draw[fill=white] (0,0) circle (.3ex);}; } }, postaction = {decorate} }, 20 | closed/.style = {decoration = {markings, mark = at position 0.5 with { \node[transform shape, xscale = .8, yscale=.4] {\upshape{/}}; } }, postaction = {decorate} }, 21 | imm/.style = {decoration = {markings, mark = at position 0.3 with { \node[transform shape, xscale = .8, yscale=.4] {\upshape{/}}; }, mark = at position 0.6 with { \node[transform shape] {\tikz\draw[fill=white] (0,0) circle (.3ex);}; } }, postaction = {decorate} } 22 | } 23 | 24 | \usepackage{braket} 25 | 26 | \usepackage[utf8]{inputenc} 27 | \usepackage{csquotes} 28 | \usepackage[american]{babel} 29 | \usepackage[style=alphabetic,firstinits=true,backend=biber,texencoding=utf8,bibencoding=utf8]{biblatex} 30 | \bibliography{../Hartshorne} 31 | \AtEveryBibitem{\clearfield{url}} 32 | \AtEveryBibitem{\clearfield{doi}} 33 | \AtEveryBibitem{\clearfield{issn}} 34 | \AtEveryBibitem{\clearfield{isbn}} 35 | \renewbibmacro{in:}{} 36 | \DeclareFieldFormat{postnote}{#1} 37 | \DeclareFieldFormat{multipostnote}{#1} 38 | 39 | \renewcommand{\theenumi}{$(\alph{enumi})$} 40 | \renewcommand{\labelenumi}{\theenumi} 41 | 42 | \newcounter{enumacounter} 43 | \newenvironment{enuma} 44 | {\begin{list}{$(\alph{enumacounter})$}{\usecounter{enumacounter} \parsep=0em \itemsep=0em \leftmargin=2.75em \labelwidth=1.5em \topsep=0em}} 45 | {\end{list}} 46 | \newcounter{enumdcounter} 47 | \newenvironment{enumd} 48 | {\begin{list}{$(\arabic{enumdcounter})$}{\usecounter{enumdcounter} \parsep=0em \itemsep=0em \leftmargin=1.75em \labelwidth=1.5em \topsep=0em}} 49 | {\end{list}} 50 | \newtheorem*{theorem}{Theorem} 51 | \newtheorem*{universalproperty}{Universal Property} 52 | \newtheorem{problem}{Exercise}[section] 53 | \newtheorem{subproblem}{Exercise}[problem] 54 | \newtheorem{lemma}{Lemma}%[section] 55 | \newtheorem{proposition}{Proposition} 56 | \newtheorem{property}{Property}[problem] 57 | \newtheorem*{lemma*}{Lemma} 58 | \theoremstyle{definition} 59 | \newtheorem*{definition}{Definition} 60 | \newtheorem*{claim}{Claim} 61 | \theoremstyle{remark} 62 | \newtheorem*{remark}{Remark} 63 | 64 | \numberwithin{equation}{section} 65 | \numberwithin{figure}{problem} 66 | \renewcommand{\theequation}{\arabic{section}.\arabic{equation}} 67 | 68 | \DeclareMathOperator{\Ann}{Ann} 69 | \DeclareMathOperator{\Ass}{Ass} 70 | \DeclareMathOperator{\Supp}{Supp} 71 | \DeclareMathOperator{\WeakAss}{\widetilde{Ass}} 72 | \let\Im\relax 73 | \DeclareMathOperator{\Im}{im} 74 | \DeclareMathOperator{\Spec}{Spec} 75 | \DeclareMathOperator{\SPEC}{\mathbf{Spec}} 76 | \DeclareMathOperator{\Sp}{sp} 77 | \DeclareMathOperator{\maxSpec}{maxSpec} 78 | \DeclareMathOperator{\Hom}{Hom} 79 | \DeclareMathOperator{\Soc}{Soc} 80 | \DeclareMathOperator{\Ht}{ht} 81 | \DeclareMathOperator{\A}{\mathcal{A}} 82 | \DeclareMathOperator{\V}{\mathbf{V}} 83 | \DeclareMathOperator{\Aut}{Aut} 84 | \DeclareMathOperator{\Char}{char} 85 | \DeclareMathOperator{\Frac}{Frac} 86 | \DeclareMathOperator{\Proj}{Proj} 87 | \DeclareMathOperator{\stimes}{\text{\footnotesize\textcircled{s}}} 88 | \DeclareMathOperator{\End}{End} 89 | \DeclareMathOperator{\Ker}{Ker} 90 | \DeclareMathOperator{\Coker}{coker} 91 | \DeclareMathOperator{\LCM}{LCM} 92 | \DeclareMathOperator{\Div}{Div} 93 | \DeclareMathOperator{\id}{id} 94 | \DeclareMathOperator{\Cl}{Cl} 95 | \DeclareMathOperator{\dv}{div} 96 | \DeclareMathOperator{\Gr}{Gr} 97 | \DeclareMathOperator{\pr}{pr} 98 | \DeclareMathOperator{\trd}{tr.d.} 99 | \DeclareMathOperator{\rank}{rank} 100 | \DeclareMathOperator{\codim}{codim} 101 | \DeclareMathOperator{\sgn}{sgn} 102 | \DeclareMathOperator{\GL}{GL} 103 | \newcommand{\GR}{\mathbb{G}\mathrm{r}} 104 | \newcommand{\gR}{\mathrm{Gr}} 105 | \newcommand{\EE}{\mathscr{E}} 106 | \newcommand{\FF}{\mathscr{F}} 107 | \newcommand{\GG}{\mathscr{G}} 108 | \newcommand{\HH}{\mathscr{H}} 109 | \newcommand{\II}{\mathscr{I}} 110 | \newcommand{\LL}{\mathscr{L}} 111 | \newcommand{\MM}{\mathscr{M}} 112 | \newcommand{\OO}{\mathcal{O}} 113 | \newcommand{\Ss}{\mathscr{S}} 114 | \newcommand{\Af}{\mathfrak{A}} 115 | \newcommand{\Aa}{\mathscr{A}} 116 | \newcommand{\PP}{\mathcal{P}} 117 | \newcommand{\red}{\mathrm{red}} 118 | \newcommand{\Sh}{\mathfrak{Sh}} 119 | \newcommand{\Psh}{\mathfrak{Psh}} 120 | \newcommand{\LRS}{\mathsf{LRS}} 121 | \newcommand{\Sch}{\mathfrak{Sch}} 122 | \newcommand{\Var}{\mathfrak{Var}} 123 | \newcommand{\Rings}{\mathfrak{Rings}} 124 | \DeclareMathOperator{\In}{in} 125 | \DeclareMathOperator{\Ext}{Ext} 126 | \DeclareMathOperator{\Spe}{Sp\acute{e}} 127 | \DeclareMathOperator{\HHom}{\mathscr{H}\!\mathit{om}} 128 | \newcommand{\isoto}{\overset{\sim}{\to}} 129 | \newcommand{\isolongto}{\overset{\sim}{\longrightarrow}} 130 | \newcommand{\Mod}{\mathsf{mod}\mathchar`-} 131 | \newcommand{\MOD}{\mathsf{Mod}\mathchar`-} 132 | \newcommand{\gr}{\mathsf{gr}\mathchar`-} 133 | \newcommand{\qgr}{\mathsf{qgr}\mathchar`-} 134 | \newcommand{\uqgr}{\underline{\mathsf{qgr}}\mathchar`-} 135 | \newcommand{\qcoh}{\mathsf{qcoh}\mathchar`-} 136 | \newcommand{\Alg}{\mathsf{Alg}\mathchar`-} 137 | \newcommand{\coh}{\mathsf{coh}\mathchar`-} 138 | \newcommand{\vect}{\mathsf{vect}\mathchar`-} 139 | \newcommand{\imm}[1][imm]{\hspace{0.75ex}\raisebox{0.58ex}{% 140 | \begin{tikzpicture}[commutative diagrams/every diagram] 141 | \draw[commutative diagrams/.cd, every arrow, every label,hook,{#1}] (0,0ex) -- (2.25ex,0ex); 142 | \end{tikzpicture}}\hspace{0.75ex}} 143 | \newcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{tikzcd}[column sep=small,ampersand replacement=\&] {#1} \rar[dashed] \& {#2} \end{tikzcd}\hspace{-0.7em}}} 144 | 145 | \usepackage{todonotes} 146 | %\usepackage[notref,notcite]{showkeys} 147 | 148 | \title{Hartshorne Ch.~II,\\\S4 Separated and Proper Morphisms} 149 | \author{Takumi Murayama} 150 | 151 | \begin{document} 152 | \maketitle 153 | \setcounter{section}{4} 154 | \begin{problem} 155 | Show that a finite morphism is proper. 156 | \end{problem} 157 | \begin{lemma}\label{lem:finitebasechange} 158 | Finite morphisms are stable under base extension. 159 | \end{lemma} 160 | \begin{proof}[Proof of Lemma] 161 | We want to show that if $f\colon X \to S$ is finite, and $g\colon S' \to S$ is any morphism, then $f'\colon X \times_S S' \to S'$ is also finite. 162 | \par Cover $S$ with open affines $V_i = \Spec B_i$; then $g^{-1}(V_i)$ is a 163 | cover for $S'$. Now cover these $g^{-1}(V_i)$ with open affines $V_{ij}' = 164 | \Spec B_{ij}'$; then, $f^{\prime-1}(V_{ij}') = X \times_S V_{ij}' \cong f^{-1}(V_i) 165 | \times_{V_i} V_{ij}'$ by Thm.~3.3 Step 7. But for each $i$, $f^{-1}(V_i) = 166 | \Spec A_i \subset X$ for $A_{i}$ that are $B_i$-algebras, finitely 167 | generated as modules over $B_i$. But since the three 168 | schemes in this fibre product are affine, we see $f^{-1}(V_i) \times_{V_i} 169 | V_{ij}' = \Spec (A_i \otimes_{B_i} B_{ij}')$. Since $A_i$ are $B_i$-algebras 170 | finitely generated as modules over $B_i$, we have a surjection $B_i^{\oplus n} \to 171 | A_{i}$, and tensoring with $B_{ij}'$ over $B_i$ gives a surjection 172 | $B_{ij}^{\prime\oplus n} \to A_{i} \otimes_{B_i} B_{ij}'$ by the right-exactness of 173 | the tensor product, and so $A_{i} \otimes_{B_i} B_{ij}'$ is a 174 | $B_{ij}'$-algebra, finitely generated as a module over $B_{ij}'$. 175 | \end{proof} 176 | \begin{proof}[Main Proof] 177 | Let $f\colon X \to Y$ be a finite morphism. A finite morphism is automatically 178 | of finite type, and is universally closed since finite morphisms are stable 179 | under base extension by Lemma \ref{lem:finitebasechange}, and finite morphisms 180 | are closed by Exercise $3.5(b)$. 181 | \par Thus, it suffices to show $f$ is separated. Let $U_i = \Spec A_i \coloneqq 182 | f^{-1}(V_i)$ where $V_i = \Spec B_i$ is an affine open cover of $Y$, and $A_i$ 183 | is a $B_i$-algebra finitely generated as a module over $B_i$. Consider the diagonal 184 | morphism $\Delta \colon X \to X \times_Y X$. By Thm.\ 3.3, $X \times_Y X$ is 185 | covered by open affines $U_i \times_{V_i} U_i$. Now $U_i = \Delta^{-1}(U_i 186 | \times_{V_i} U_i)$, and the property of being a closed immersion is local on 187 | target by our proof of Exercise $3.11(a)$, hence $\Delta$ is a closed 188 | immersion by Prop.\ $4.1$. 189 | \end{proof} 190 | 191 | \begin{problem} 192 | Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let $Y$ be a 193 | separated scheme over $S$. Let $f$ and $g$ be two $S$-morphisms of $X$ to $Y$ 194 | which agree on an open dense subset of $X$. Show that $f = g$. Give examples 195 | to show that this result fails if either $(a)$ $X$ is nonreduced, or $(b)$ $Y$ 196 | is nonseparated. 197 | \end{problem} 198 | \begin{proof} 199 | Consider the map $h \colon X \to Y \times_S Y$ obtained by the universal 200 | property of fibre products from $f$ and $g$, and let $U$ be the open set in 201 | $X$ on which $f$ and $g$ agree. Since $X$ is irreducible by Prop.\ $3.1$, $U$ 202 | contains the generic point of $X$. 203 | \end{proof} 204 | 205 | \printbibliography 206 | \end{document} 207 | -------------------------------------------------------------------------------- /Hartshorne/I Varieties/3 Morphisms/I.3 Morphisms.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt,letterpaper]{article} 2 | \usepackage{geometry} 3 | \geometry{letterpaper} 4 | \usepackage{amsmath,amssymb,amsthm,mathrsfs} 5 | \usepackage{mathtools} 6 | \usepackage{ifpdf} 7 | \ifpdf 8 | \setlength{\pdfpagewidth}{8.5in} 9 | \setlength{\pdfpageheight}{11in} 10 | \else 11 | \fi 12 | \usepackage{hyperref} 13 | 14 | \usepackage{tikz} 15 | \usepackage{tikz-cd} 16 | \usetikzlibrary{decorations.markings} 17 | \tikzset{ 18 | open/.style = {decoration = {markings, mark = at position 0.5 with { \node[transform shape] {\tikz\draw[fill=white] (0,0) circle (.3ex);}; } }, postaction = {decorate} }, 19 | closed/.style = {decoration = {markings, mark = at position 0.5 with { \node[transform shape, xscale = .8, yscale=.4] {\upshape{/}}; } }, postaction = {decorate} }, 20 | imm/.style = {decoration = {markings, mark = at position 0.3 with { \node[transform shape, xscale = .8, yscale=.4] {\upshape{/}}; }, mark = at position 0.6 with { \node[transform shape] {\tikz\draw[fill=white] (0,0) circle (.3ex);}; } }, postaction = {decorate} } 21 | } 22 | 23 | \usepackage{braket} 24 | 25 | \usepackage{csquotes} 26 | \usepackage[american]{babel} 27 | \usepackage[style=alphabetic,firstinits=true,backend=biber,texencoding=utf8,bibencoding=utf8]{biblatex} 28 | \bibliography{../../../References} 29 | \AtEveryBibitem{\clearfield{url}} 30 | \AtEveryBibitem{\clearfield{doi}} 31 | \AtEveryBibitem{\clearfield{issn}} 32 | \AtEveryBibitem{\clearfield{isbn}} 33 | \renewbibmacro{in:}{} 34 | \DeclareFieldFormat{postnote}{#1} 35 | \DeclareFieldFormat{multipostnote}{#1} 36 | 37 | \renewcommand{\theenumi}{$(\alph{enumi})$} 38 | \renewcommand{\labelenumi}{\theenumi} 39 | 40 | \newcounter{enumacounter} 41 | \newenvironment{enuma} 42 | {\begin{list}{$(\alph{enumacounter})$}{\usecounter{enumacounter} \parsep=0em \itemsep=0em \leftmargin=2.75em \labelwidth=1.5em \topsep=0em}} 43 | {\end{list}} 44 | \newcounter{enumicounter} 45 | \newenvironment{enumi} 46 | {\begin{list}{$(\roman{enumicounter})$}{\usecounter{enumicounter} \parsep=0em 47 | \itemsep=0em \leftmargin=2.5em \labelwidth=1.75em \topsep=0em}} 48 | {\end{list}} 49 | \newcounter{enumdcounter} 50 | \newenvironment{enumd} 51 | {\begin{list}{$(\arabic{enumdcounter})$}{\usecounter{enumdcounter} \parsep=0em \itemsep=0em \leftmargin=1.75em \labelwidth=1.5em \topsep=0em}} 52 | {\end{list}} 53 | \newtheorem*{theorem}{Theorem} 54 | \newtheorem*{universalproperty}{Universal Property} 55 | \newtheorem{problem}{Exercise}[section] 56 | \newtheorem{subproblem}{Problem}[problem] 57 | \newtheorem{lemma}{Lemma}%[section] 58 | \newtheorem{proposition}{Proposition} 59 | \newtheorem{property}{Property}[problem] 60 | \newtheorem*{lemma*}{Lemma} 61 | \theoremstyle{definition} 62 | \newtheorem*{definition}{Definition} 63 | \newtheorem*{claim}{Claim} 64 | \theoremstyle{remark} 65 | \newtheorem*{remark}{Remark} 66 | 67 | \numberwithin{equation}{section} 68 | \numberwithin{figure}{problem} 69 | \renewcommand{\theequation}{\arabic{section}.\arabic{equation}} 70 | 71 | \DeclareMathOperator{\Ann}{Ann} 72 | \DeclareMathOperator{\Ass}{Ass} 73 | \DeclareMathOperator{\Supp}{Supp} 74 | \DeclareMathOperator{\WeakAss}{\widetilde{Ass}} 75 | \let\Im\relax 76 | \DeclareMathOperator{\Im}{im} 77 | \DeclareMathOperator{\Spec}{Spec} 78 | \DeclareMathOperator{\SPEC}{\mathbf{Spec}} 79 | \DeclareMathOperator{\Sp}{sp} 80 | \DeclareMathOperator{\maxSpec}{maxSpec} 81 | \DeclareMathOperator{\Hom}{Hom} 82 | \DeclareMathOperator{\Soc}{Soc} 83 | \DeclareMathOperator{\Ht}{ht} 84 | \let\AA\relax 85 | \DeclareMathOperator{\AA}{\mathbb{A}} 86 | 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\newcommand{\MOD}{\mathsf{Mod}\mathchar`-} 136 | \newcommand{\gr}{\mathsf{gr}\mathchar`-} 137 | \newcommand{\qgr}{\mathsf{qgr}\mathchar`-} 138 | \newcommand{\uqgr}{\underline{\mathsf{qgr}}\mathchar`-} 139 | \newcommand{\qcoh}{\mathsf{qcoh}\mathchar`-} 140 | \newcommand{\Alg}{\mathsf{Alg}\mathchar`-} 141 | \newcommand{\coh}{\mathsf{coh}\mathchar`-} 142 | \newcommand{\vect}{\mathsf{vect}\mathchar`-} 143 | \newcommand{\imm}[1][imm]{\hspace{0.75ex}\raisebox{0.58ex}{% 144 | \begin{tikzpicture}[commutative diagrams/every diagram] 145 | \draw[commutative diagrams/.cd, every arrow, every label,hook,{#1}] (0,0ex) -- (2.25ex,0ex); 146 | \end{tikzpicture}}\hspace{0.75ex}} 147 | \newcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{tikzcd}[column sep=small,ampersand replacement=\&] {#1} \rar[dashed] \& {#2} \end{tikzcd}\hspace{-0.7em}}} 148 | 149 | \usepackage{todonotes} 150 | 151 | \title{Hartshorne Ch.~I, \S3 Morphisms} 152 | \author{Takumi Murayama} 153 | 154 | \begin{document} 155 | \maketitle 156 | \setcounter{section}{3} 157 | \begin{problem}\mbox{} 158 | \begin{enuma} 159 | \item Show that any conic in $\AA^2$ is isomorphic either to $\AA^1$ or 160 | $\AA^1 - \{0\}$ \emph{(cf.\ Ex.~$1.1$)}. 161 | \item Show that $\AA^1$ is \emph{not} isomorphic to any proper open subset 162 | of itself. (This result is generalized by \emph{(Ex.~$6.7$)} below.) 163 | \item Any conic in $\PP^2$ is isomorphic to $\PP^1$. 164 | \item We will see later \emph{(Ex.\ $4.8$)} that any two curves are 165 | homeomorphic. But show now that $\AA^2$ is not even homeomorphic to 166 | $\PP^2$. 167 | \item If an affine variety is isomorphic to a projective variety, then it 168 | consists of only one point. 169 | \end{enuma} 170 | \end{problem} 171 | \begin{proof}[Proof of $(a)$] 172 | By Exercise $1.1$, any conic $W$ has $A(W) \cong k[t]$ or $k[s,s^{-1}]$. But 173 | $k[t] = A(\AA^1)$ and $k[s,s^{-1}] = A(\AA^1 - \{0\})$, so we are done by 174 | Cor.~3.7. 175 | \end{proof} 176 | \begin{proof}[Proof of $(b)$] 177 | By Ex.\ $1.1.1$, the proper open subsets of $\AA^1$ are the empty set and the 178 | complements of finite subsets. $\OO(\emptyset)$ has one element, the unique 179 | function $\emptyset \to k$, which is vacuously regular, so $\OO(\emptyset) 180 | \not\cong k[x]$, and $\emptyset \not\cong \AA^1$. 181 | \par Now suppose $U \subset \AA^1 \setminus \{p\} \subsetneq \AA^1$. Then, any 182 | function of the form $f/(x-p)$ for $f \in k[x]$ is regular on $U$, hence 183 | $k[x,(x-p)^{-1}] \subset \OO(U)$. But then $\OO(U) \not\cong k[x]$, since any 184 | $k$-algebra homomorphism $\OO(U) \to k[x]$ must have $(x-p) \mapsto a \in k 185 | \subset k[x]$, but the image of $a$ in $\OO(U)$ must be $a$. Hence $\OO(U) 186 | \not\cong k[x]$, and so $U \not\cong \AA^1$. 187 | \end{proof} 188 | \begin{proof}[Proof of $(c)$] 189 | Denote our conic by $C = Z(f) \subset \PP^2$ for $f\in k[x,y,z]$ an 190 | irreducible homogeneous degree $2$ polynomial. The dehomogenization 191 | $\alpha(f)$ in the chart $z \ne 0$ is an irreducible quadratic polynomial in 192 | $x,y$. By our analysis in Exercise $1.1$, 193 | \begin{equation*} 194 | \alpha(f) = \begin{cases} 195 | x^2 - y\ \text{or}\ x^2 - y^2 - 1 & \text{if}\ \Char k \ne 2\\ 196 | x^2 - y\ \text{or}\ xy - 1 & \text{if}\ \Char k = 2 197 | \end{cases} 198 | \end{equation*} 199 | after affine transformation in the chart $z \ne 0$; since any affine 200 | transformation in $z \ne 0$ extends to a projective transformation of $\PP^2$, 201 | we can therefore assume 202 | \begin{equation*} 203 | f = \begin{cases} 204 | x^2 - yz\ \text{or}\ x^2 - y^2 - z^2 & \text{if}\ \Char k \ne 2\\ 205 | x^2 - yz\ \text{or}\ xy - z^2 & \text{if}\ \Char k = 2 206 | \end{cases} 207 | \end{equation*} 208 | 209 | Assume $\Char k \ne 2$. Then, by the structure theorem for symmetric 210 | bilinear forms \cite[Ch.\ XV, Thm.\ 3.1]{Lan02}, and then by scaling, we can 211 | assume $f$ is one of $x^2$, $x^2 + y^2$, or $x^2 + y^2 + z^2$. In the first 212 | case, $Z(x^2) = Z(x) \cong \PP^1$ via the map $[0:y:z] \mapsto [y:z]$. In the 213 | second case, $x^2 + y^2 = (x+iy)(x-iy)$, so $C$ is reducible, a contradiction. 214 | Finally, if $f = x^2 + y^2 + z^2$, then 215 | \end{proof} 216 | 217 | \printbibliography 218 | \end{document} 219 | -------------------------------------------------------------------------------- /Atiyah-Macdonald/5 Integral Dependence and Valuations/5 Integral Dependence and Valuations.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt,letterpaper]{article} 2 | \usepackage{geometry} 3 | \geometry{letterpaper} 4 | \usepackage{amsmath,amssymb,amsthm,mathrsfs} 5 | \usepackage{mathtools} 6 | \usepackage{ifpdf} 7 | \ifpdf 8 | \setlength{\pdfpagewidth}{8.5in} 9 | \setlength{\pdfpageheight}{11in} 10 | \else 11 | \fi 12 | \usepackage{hyperref} 13 | 14 | \usepackage{tikz} 15 | \usepackage{tikz-cd} 16 | \usetikzlibrary{decorations.markings} 17 | \tikzset{ 18 | open/.style = {decoration = {markings, mark = at position 0.5 with { 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\DeclareMathOperator{\GL}{GL} 109 | \DeclareMathOperator{\lt}{lt} 110 | \DeclareMathOperator{\lc}{lc} 111 | \newcommand{\GR}{\mathbb{G}\mathrm{r}} 112 | \newcommand{\gR}{\mathrm{Gr}} 113 | \newcommand{\EE}{\mathscr{E}} 114 | \newcommand{\FF}{\mathscr{F}} 115 | \newcommand{\GG}{\mathscr{G}} 116 | \newcommand{\HH}{\mathscr{H}} 117 | \newcommand{\II}{\mathscr{I}} 118 | \newcommand{\LL}{\mathscr{L}} 119 | \newcommand{\MM}{\mathscr{M}} 120 | \newcommand{\OO}{\mathcal{O}} 121 | \newcommand{\Ss}{\mathscr{S}} 122 | \newcommand{\Af}{\mathfrak{A}} 123 | \newcommand{\Aa}{\mathscr{A}} 124 | \newcommand{\PP}{\mathcal{P}} 125 | \newcommand{\red}{\mathrm{red}} 126 | \newcommand{\Sh}{\mathfrak{Sh}} 127 | \newcommand{\Psh}{\mathfrak{Psh}} 128 | \newcommand{\LRS}{\mathsf{LRS}} 129 | \newcommand{\Sch}{\mathfrak{Sch}} 130 | \newcommand{\Var}{\mathfrak{Var}} 131 | \newcommand{\Rings}{\mathfrak{Rings}} 132 | \DeclareMathOperator{\In}{in} 133 | \DeclareMathOperator{\Ext}{Ext} 134 | \DeclareMathOperator{\Spe}{Sp\acute{e}} 135 | \DeclareMathOperator{\HHom}{\mathscr{H}\!\mathit{om}} 136 | \newcommand{\isoto}{\overset{\sim}{\to}} 137 | \newcommand{\isolongto}{\overset{\sim}{\longrightarrow}} 138 | \newcommand{\Mod}{\mathsf{mod}\mathchar`-} 139 | \newcommand{\MOD}{\mathsf{Mod}\mathchar`-} 140 | \newcommand{\gr}{\mathsf{gr}\mathchar`-} 141 | \newcommand{\qgr}{\mathsf{qgr}\mathchar`-} 142 | \newcommand{\uqgr}{\underline{\mathsf{qgr}}\mathchar`-} 143 | \newcommand{\qcoh}{\mathsf{qcoh}\mathchar`-} 144 | \newcommand{\Alg}{\mathsf{Alg}\mathchar`-} 145 | \newcommand{\coh}{\mathsf{coh}\mathchar`-} 146 | \newcommand{\vect}{\mathsf{vect}\mathchar`-} 147 | \newcommand{\imm}[1][imm]{\hspace{0.75ex}\raisebox{0.58ex}{% 148 | \begin{tikzpicture}[commutative diagrams/every diagram] 149 | \draw[commutative diagrams/.cd, every arrow, every label,hook,{#1}] (0,0ex) -- (2.25ex,0ex); 150 | \end{tikzpicture}}\hspace{0.75ex}} 151 | \newcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{tikzcd}[column sep=small,ampersand replacement=\&] {#1} \rar[dashed] \& {#2} \end{tikzcd}\hspace{-0.7em}}} 152 | 153 | %\usepackage{todonotes} 154 | %\usepackage[notref,notcite]{showkeys} 155 | 156 | \title{Atiyah-Macdonald Ch.~5 Integral Dependence and Valuations} 157 | \author{Takumi Murayama and Kyu Jun} 158 | 159 | \begin{document} 160 | \maketitle 161 | \setcounter{section}{5} 162 | \begin{problem}[AM Ch 5 \# 16] 163 | Let $k$ be a field and let $A \ne 0$ be a finitely generated $k$-algebra. Then there exist elements $y_1,\ldots,y_r \in A$ which are algebraically independent over $k$ and such that $A$ is integral over $k[y_1,\ldots,y_r]$. 164 | \end{problem} 165 | \begin{proof} 166 | We assume $k$ is infinite. Let $x_1,\ldots,x_n$ generate $A$ as a $k$-algebra. We can rearrange the $x_i$ such that $x_1,\ldots,x_r$ are algebraically independent over $k$ and each of $x_{r+1},\ldots,x_n$ is algebraic over $k[x_1,\ldots,x_r]$. Proceed by induction on $n$. If $n = r$ there is nothing to do, and so suppose $n > r$ and that the result is true for $n-1$ generators. In this case, the generator $x_n$ is algebraic over $k[x_1,\ldots,x_{n-1}]$, i.e., there is a polynomial $f \ne 0$ in $n$ variables such that $f(x_1,\ldots,x_{n-1},x_n) = 0$. Let $F$ be the homogeneous part of highest degree in $f$. Since $k$ is infinite, there exist $\lambda_1,\ldots,\lambda_{n-1} \in k$ such that $F(\lambda_1,\ldots,\lambda_{n-1},1) \ne 0$, since $F$ is a non-zero polynomial in $n-1$ variables, and so cannot induce the zero function on $k^{n-1}$ when $k$ is infinite. Now let $x_i' = x_i - \lambda_ix_n$ for $1 \le i \le n-1$, and let $A' = k[x_1',\ldots,x'_{n-1}]$. We claim that $x_n$ is integral over $A'$. 167 | \par We let $d = \deg(F)$ and choose $G_j$ in $n-1$ variables such that 168 | \begin{equation*} 169 | F(z_1,\ldots,z_n) = \sum_{j=0}^d z_n^jG_j(z_1,\ldots,z_{n-1}). 170 | \end{equation*} 171 | Each $G_j$ is a homogeneous polynomial of degree $d-j$. Letting $z_i' = z_i - \lambda_iz_n$, we compute 172 | \begin{align*} 173 | F(z_1,\ldots,z_n) &= \sum_{j=0}^d z_n^j G_j(z_1' + \lambda_1z_n,\ldots,z_{n-1}' + \lambda_{n-1}z_n)\\ 174 | &= \sum_{j=0}^d z_n^j [z_n^{d-j}G_J(\lambda_1,\ldots,\lambda_{n-1},1)+H_j(z'_1,\ldots,z'_{n-1},z_n)]\\ 175 | &= z_n^dF(\lambda_1,\ldots,\lambda_{n-1},1)+\sum_{j=0}^dz_n^jH_j(z'_1,\ldots,z'_{n-1},z_n), 176 | \end{align*} 177 | where each $H_j$ is a polynomial in $z'_1,\ldots,z'_{n-1},z_n$ with degree $< d-j$ in $z_n$, with coefficients in $k$. Defining a new polynomial 178 | \begin{equation*} 179 | \hat{F}(z) = z^d + \frac{1}{F(\lambda_1,\ldots,\lambda_{n-1},1)} \sum_{j=0}^d z^jH_j(x'_1,\ldots,x'_{n-1},z), 180 | \end{equation*} 181 | we see $\hat{F}$ is monic in $z$ with coefficients in $A'$ such that $\hat{F}(x_n) = F(x_1,\ldots,x_{n-1},x_n) = 0$. Thus, $x_n$ is integral over $A'$, i.e., $A = k[x_1,\ldots,x_n] = k[x_1',\ldots,x_{n-1}',x_n]$ is integral over $A'$. By the induction hypothesis, there are $y_1,\ldots,y_{n-1} \in A'$ algebraically independent over $k$ such that $A'$ is integral over $A'[y_1,\ldots,y_{n-1}]$. Now $y_1,\ldots,y_n \in A$ are algebraically independent over $k$ and $A$ is integral over $A[y_1,\ldots,y_n]$, and we are done. 182 | \end{proof} 183 | 184 | \begin{problem}[AM Exercise $5.10i$] 185 | A ring homomorphism $f : A \to B$ is said to have the \emph{going-up property} \emph{(}resp.~the \emph{going-down property)} if the conclusion of the going-up theorem $(5.11)$ \emph{(}resp.~the \emph{going-down theorem (5.16))} holds for $B$ and its subring $f(A)$. 186 | \par Let $f^* : \Spec(B) \to \Spec(A)$ be the mapping associated with $f$. 187 | \par Consider the following three statements: 188 | \begin{enum} 189 | \item $f^*$ is a closed mapping. 190 | \item $f$ has the going-up property. 191 | \item Let $\mathfrak{q}$ be any prime ideal of $B$ and let $\mathfrak{p} = \mathfrak{q}^c$. Then $f^* : \Spec(B/\mathfrak{q}) \to \Spec(A/\mathfrak{p})$ is surjective. 192 | \end{enum} 193 | Prove that $(a) \Rightarrow (b) \Leftrightarrow (c)$. 194 | \end{problem} 195 | \begin{proof} 196 | $(a) \Rightarrow (b)$. Suppose $\mathfrak{p}_1 \subseteq \mathfrak{p}_2$ a chain of prime ideals in $f(A)$ and $\mathfrak{q}_1 \subseteq B$ lies over $\mathfrak{p}_1$. Since $f^*(\mathfrak{q}_1) = f^{-1}(\mathfrak{p}_1) \subseteq f^{-1}(\mathfrak{p}_2)$, we have $f^{-1}(\mathfrak{p}_2) \in \V(f^*(\mathfrak{q}_1))$. Note $\V(f^*(\mathfrak{q}_1)) = \V(f^{-1}(\mathfrak{q}_1)) = \overline{f^*(\V(\mathfrak{q}_1))} = f^*(\V(\mathfrak{q}_1))$ by the AM Exercise $1.21iii)$ and the fact that $f^*$ is closed. Thus, $f^{-1}(\mathfrak{p}_2) \in f^*(\V(\mathfrak{q}_1))$, and so there exists $\mathfrak{q}_2 \in \V(\mathfrak{q}_1)$ such that $f^*(\mathfrak{q}_2) = f^{-1}(\mathfrak{p}_2)$. Since $f^*(\mathfrak{q}_2) = f^{-1}(\mathfrak{q}_2 \cap f(A))$, we then see that $\mathfrak{p}_2 = \mathfrak{q}_2 \cap f(A)$, and so $\mathfrak{q}_2$ lies over $\mathfrak{p}_2$. By induction as in the original proof of Going Up, we are done. 197 | \par $(b) \Rightarrow (c)$. Suppose $\mathfrak{q} \in \Spec(B)$ and let $\mathfrak{p} = f^{-1}(\mathfrak{q}) = f^{-1}(\mathfrak{q} \cap f(A))$. Consider $\mathfrak{p}' \in \V(\mathfrak{p})$; note that $\V(\mathfrak{p}) \xleftrightarrow{1-1\:} \Spec(A/\mathfrak{p})$, and so showing that $\mathfrak{p}' \in \Im f^*$ suffices. Then, $\mathfrak{p} \subseteq \mathfrak{p}'$, and so $f(\mathfrak{p}) \subseteq f(\mathfrak{p}')$ is a chain of prime ideals in $f(A)$ and $\mathfrak{q} \cap f(A) = f(\mathfrak{p})$. By $(b)$, the Going Up property, there exists $\mathfrak{q}' \subseteq B$ such that $\mathfrak{q}' \cap f(A) = f(\mathfrak{p}')$. Finally, $f^{-1}(\mathfrak{q}' \cap f(A)) = f^*(\mathfrak{q}') = f^{-1}(f(\mathfrak{p}')) = \mathfrak{p}'$, and so $f^*$ is surjective. 198 | \par $(c) \Rightarrow (b)$. Suppose $\mathfrak{p}_1 \subseteq \mathfrak{p}_2$ is a chain of prime ideals in $f(A)$ and $\mathfrak{q}_1 \subseteq B$ lies over $\mathfrak{p}_1$. Note that $f^{-1}(\mathfrak{p}_1)$ is prime since it is a contraction of a prime ideal, and also $f^{-1}(\mathfrak{p}_1) = f^{-1}(\mathfrak{q}_1 \cap f(A)) = (\mathfrak{q}_1)^c$. By $(c)$, $f^* : \V(\mathfrak{q}_1) \to \V(f^{-1}(\mathfrak{p}_1))$ is surjective, and so $f^{-1}(\mathfrak{p}_2) \in \V(f^{-1}(\mathfrak{p}_1))$ has preimage $\mathfrak{q}_2 = (f^*)^{-1}(f^{-1}(\mathfrak{p}_2)) \in \V(\mathfrak{q}_1)$. $\mathfrak{q}_2 \in \V(\mathfrak{q}_1) \implies \mathfrak{q}_1 \subseteq \mathfrak{q}_2$, and $f^{-1}(\mathfrak{p}_2) = f^*(\mathfrak{q}_2) = f^{-1}(\mathfrak{q}_2 \cap f(A)) \implies \mathfrak{p}_2 = \mathfrak{q}_2 \cap f(A)$, i.e., $\mathfrak{q}_2$ lies over $\mathfrak{p}_2$. By induction as in the original proof of Going Up, we are done. 199 | \end{proof} 200 | 201 | \begin{problem}[AM Exercise 5.1] 202 | Let $f: A \to B$ be an integral homomorphism of rings. Show that $f^* : \Spec(B) \to \Spec(A)$ is a \emph{closed} mapping, i.e., that it maps closed sets to closed sets. 203 | \end{problem} 204 | \begin{proof} 205 | Let $\mathfrak{q} \in \Spec(B)$; we claim $f^*(\V(\mathfrak{q})) = \V(f^*(\mathfrak{q}))$. By AM Exercise $1.21iii)$, we see $f^*(\V(\mathfrak{q})) \subseteq \overline{f^*(\V(\mathfrak{q}))} = \V(f^{-1}(\mathfrak{q})) = \V(f^*(\mathfrak{q}))$. Conversely, if $\mathfrak{p} \in \V(f^*(\mathfrak{q}))$, then $f^*(\mathfrak{q}) \subseteq \mathfrak{p}$, and so $f(f^*(\mathfrak{q})) \subseteq f(\mathfrak{p})$ is a chain of prime ideals in $f(A)$. Since $f(f^*(\mathfrak{q})) = f(f^{-1}(\mathfrak{q})) = \mathfrak{q} \cap f(A)$, we have that $\mathfrak{q}$ lies over $f(f^*(\mathfrak{q}))$, and so since $B$ is integral over $f(A)$, by Going Up there exists $\mathfrak{r} \in \Spec(B)$ such that $\mathfrak{r} \supseteq \mathfrak{q}$ and $\mathfrak{r} \cap f(A) = f(\mathfrak{p})$. Thus, $\mathfrak{p} = f^{-1}(f(\mathfrak{p})) = f^{-1}(\mathfrak{r} \cap f(A)) = f^{-1}(\mathfrak{r}) = f^*(\mathfrak{r})$, where $\mathfrak{r} \in \V(\mathfrak{q})$. This implies $\mathfrak{p} \in f^*(\V(\mathfrak{q}))$; thus, $f^*(\V(\mathfrak{q})) = \V(f^*(\mathfrak{q}))$, i.e., $f^*$ is closed. 206 | \end{proof} 207 | 208 | \begin{problem}[AM Exercise $5.10ii$] 209 | In the same situation as in Problem $3.1$, show that $f$ has the going-down property $\iff$ for any prime ideal $\mathfrak{q}$ of $B$, if $\mathfrak{p} = \mathfrak{q}^c$, then $f^*:\Spec(B_\mathfrak{q}) \to \Spec(A_\mathfrak{p})$ is surjective. 210 | \end{problem} 211 | \begin{proof} 212 | $(b) \Rightarrow (c)$. Suppose $\mathfrak{q} \in \Spec(B)$ and let $\mathfrak{p} = f^{-1}(\mathfrak{q}) = f^{-1}(\mathfrak{q} \cap f(A))$. Consider $\mathfrak{p}' \in \Spec(A) \setminus \V(\mathfrak{p})$; note that $\Spec(A) \setminus \V(\mathfrak{p}) \xleftrightarrow{1-1\:} \Spec(A_\mathfrak{p})$, and so showing that $\mathfrak{p}' \in \Im f^*$ suffices. Then, $\mathfrak{p}' \subseteq \mathfrak{p}$, and so we have $f(\mathfrak{p}') \subseteq f(\mathfrak{p})$ a chain of prime ideals in $f(A)$ and $\mathfrak{q} \cap f(A) = f(\mathfrak{p})$. By $(b)$, the Going Down property, there exists $\mathfrak{q}' \subseteq B$ such that $\mathfrak{q}' \cap f(A) = f(\mathfrak{p}')$. Finally, $f^{-1}(\mathfrak{q}' \cap f(A)) = f^*(\mathfrak{q}') = f^{-1}(f(\mathfrak{p}')) = \mathfrak{p}'$, and so $f^*$ is surjective. 213 | \par $(c) \Rightarrow (b)$. Suppose $\mathfrak{p}_1 \subseteq \mathfrak{p}_2$ is a chain of prime ideals in $f(A)$ and $\mathfrak{q}_2 \subseteq B$ lies over $\mathfrak{p}_2$. Note that $f^{-1}(\mathfrak{p}_2)$ is prime since it is a contraction of a prime ideal, and also $f^{-1}(\mathfrak{p}_2) = f^{-1}(\mathfrak{q}_2 \cap f(A)) = (\mathfrak{q}_2)^c$. By $(c)$, $f^* : \Spec(B_{\mathfrak{q}_2}) \to \Spec(A_{f^{-1}(\mathfrak{p}_2)})$ is surjective, and so $f^{-1}(\mathfrak{p}_1) \in \Spec(A_{f^{-1}(\mathfrak{p}_2)})$ has preimage $\mathfrak{q}_1 = (f^*)^{-1}(f^{-1}(\mathfrak{p}_1)) \in \Spec(B_{\mathfrak{q}_2})$. $\mathfrak{q}_1 \in \Spec(B_{\mathfrak{q}_2}) \implies \mathfrak{q}_1 \subseteq \mathfrak{q}_2$, and $f^{-1}(\mathfrak{p}_1) = f^*(\mathfrak{q}_1) = f^{-1}(\mathfrak{q}_1 \cap f(A)) \implies \mathfrak{p}_1 = \mathfrak{q}_1 \cap f(A)$, i.e., $\mathfrak{q}_1$ lies over $\mathfrak{p}_1$. By induction as in the original proof of Going Down, we are done. 214 | \end{proof} 215 | 216 | \begin{problem}[AM Exercise 5.11] 217 | Let $f: A \to B$ be a flat homomorphism of rings. Then $f$ has the going-down property. 218 | \end{problem} 219 | \begin{proof} 220 | By AM Exercise $3.18$, we know that if $\mathfrak{q} \in \Spec(B)$ and $\mathfrak{p} = \mathfrak{q}^c$, then $f^* : \Spec(B_\mathfrak{q} \to \Spec(A_\mathfrak{p}))$ is surjective. By AM Exercise $5.10ii)$ above, this implies $f$ has the going-down property. 221 | \end{proof} 222 | 223 | \begin{problem}[cf.~AM Ch 5 \# 12,13] 224 | Let $G$ be a finite group of automorphisms of a ring $A$, and let $A^G$ denote the subring of $G$-invariants, that is of all $x \in A$ such that $\sigma(x) = x$ for all $\sigma \in G$. 225 | \begin{enum} 226 | \item Prove that $A$ is integral over $A^G$. 227 | \item Fixing a prime ideal $P \subseteq A^G$, let $\mathcal{Q}$ be the set of primes in $A$ whose contraction is $P$. Show that $G$ acts transitively on $\mathcal{Q}$ (i.e., ``$G$ acts transitively on the fibers of the quotient $X = \Spec(A) \to \Spec(A^G) = X/G$''). 228 | \end{enum} 229 | \end{problem} 230 | \begin{proof}[Proof of $(a)$] 231 | %^$A^G$ is a subring since if $x,y \in A^G$, $xy = \sigma(x)\sigma(y) = \sigma(xy)$, $x\pm y = \sigma(x)\pm\sigma(y) = \sigma(x\pm y)$, and $1 \in A^G$ by the fact that $0 \ne x = \sigma(x) = \sigma(1x) = \sigma(1)x \implies \sigma(1) = 1$. 232 | Let $x \in A$ and consider 233 | \begin{equation*} 234 | f(t) = \prod_{\sigma \in G} (t - \sigma(x)) = (t-x)\prod_{\sigma \ne e}(t-\sigma(x)), 235 | \end{equation*} 236 | since $e \in G$. Thus, $f(x) = 0$, and since the highest order term of $f(t)$ is $t^{|G|}$, we see that $f$ is monic. We then show $f(t)$ has coefficients in $A^G$. We see that the coefficients of $f$ are elementary symmetric polynomials, i.e., of the form 237 | \begin{equation*} 238 | a_k = \sum_{i_1 < \cdots < i_k} \sigma_{i_1}(x)\cdots\sigma_{i_k}(x), 239 | \end{equation*} 240 | where we enumerate $G=\{\sigma_i\}$. $\tau(a_k) = a_k$ for any $\tau\in G$, for $\sigma$ just permutes terms in the summation; thus, $a_k \in A^G$ for all $k$, and $A$ is integral over $A^G$. 241 | \end{proof} 242 | \begin{proof}[Proof of $(b)$] 243 | Consider $Q_1,Q_2 \in \mathcal{Q}$. If $x \in Q_1$, then $\prod_{\sigma \in G}\sigma(x) \in Q_1 \cap A^G = P \subseteq Q_2$, for $e \in G \implies x \mid \prod\sigma(x)$ and $\tau(\prod\sigma(x)) = \prod(\sigma(x))$ for any $\tau \in G$ since $G$ is a group. It follows $\sigma(x) \in Q_2$ for some $\sigma$ since $Q_2$ is prime, i.e., $x \in \sigma^{-1}(Q_2)$. Thus $Q_1 \subseteq \bigcup_{\sigma \in G}\sigma^{-1}(Q_2)$. $\sigma^{-1}(Q_2) = Q_2^c$ and so they are prime; by prime avoidance $Q_1 \subseteq \sigma^{-1}(Q_2)$ for some $\sigma \in G$. But since $Q_1 \cap A^G = \sigma^{-1}(Q_2) \cap A^G = P$, by Incomparability we have $Q_1 = \sigma^{-1}(Q_2)$, i.e., $G$ acts transitively on $\mathcal{Q}$. 244 | \end{proof} 245 | 246 | \begin{problem}[cf.~AM Ch 5 \# 14,15] 247 | Suppose that $A$ is a normal domain with field of fractions $K$. Let $L$ be any extension field of $K$, and $B$ the integral closure of $A$ inside of $L$. 248 | \begin{enum} 249 | \item Recall that an automorphism of $L$ over $K$ is a ring isomorphism $\phi : L \to L$ such that $\phi(x) = x$ for all $x \in K$, and that the set of all such automorphisms $\Aut(L/K)$ form a group. Show that $\Aut(L/K)$ acts on $B$, i.e.~$\phi(B) = B$ for all $\phi \in \Aut(L/K)$. 250 | \item If $L$ is a finite Galois extension of $K$ with $G = \Aut(L/K)$, then show $A = B^G$ and the previous problem applies. What happens if instead the extension if purely inseparable? 251 | \end{enum} 252 | \end{problem} 253 | \begin{proof}[Proof of $(a)$] 254 | Suppose $b \in B$, and $f(x) = x^n + a_1x^{n-1} + \cdots + a_n$ with $a_i \in A$ is such that $f(b) = 0$. Since $\phi(f(x)) = \phi(x)^n + a_1\phi(x)^{n-1} + \cdots + a_n = f(\phi(x))$ by the fact that $\phi \in \Aut(L/K)$ holds $K \supseteq A$ fixed, we have $0 = f(b) = \phi(f(b)) = f(\phi(b))$, i.e., $\phi(B)$ is integral over $A$ and so $\phi(B) \subseteq B$ for all $\phi \in \Aut(L/K)$. In particular, we also have $\phi^{-1}(B) \subseteq B$ and so $B \subseteq \phi(B)$, which implies $\phi(B) = B$ for all $\phi \in \Aut(L/K)$. 255 | \end{proof} 256 | \begin{proof}[Proof of $(b)$] 257 | We have that $A \subseteq B \implies A = A^G \subseteq B^G$ by the fact that $A \subseteq K$, and so it suffices to show $B^G \subseteq A$. We have that $A \subseteq B^G \subseteq L^G = K$ by definition of $\Aut(L/K)$. But then, $B^G$ is integral over $A$, and $A$ normal implies it is integrally closed in $K$, which implies $B^G = A$. Thus, letting $G = \Aut(L/K)$ (finite since $[L:K] < \infty$) act on the ring $B$, we see that fixing a prime ideal $P \subseteq A = B^G$, $G$ acts transitively on the set of primes in $B$ whose contraction is $P$ by Problem $5(b)$. 258 | \par Now consider the case when the extension $L \supseteq K$ is purely inseparable; we want to show an analogue of Problem $5(b)$. So, suppose $P \subseteq A$ is prime, and $\mathcal{Q}$ is the set of primes in $B$ whose contraction is $P$. We claim that $Q = \{x \in B \mid x^{p^m} \in P~\text{for some}~m \ge 0\}$ for any $Q \in \mathcal{Q}$, where $p = \Char(K)$ prime since we have a purely inseparable extension. Now if $x^{p^m} \in P$ for some $m \ge 0$, then $x^{p^m} \in Q$ and so $x \in Q$. Conversely, if $x \in Q$ then $x^{p^m} \in K$ for some $m \ge 0$ since $L/K$ is purely inseparable. Then, $x^{p^m} \in Q \cap K = P$, and so $Q = \{x \in B \mid x^{p^m} \in P~\text{for some}~m \ge 0\}$. This implies that there is only one unique $Q \in \mathcal{Q}$, and so $G = \Aut(L/K)$ trivially acts transitively on $\mathcal{Q}$. Moreover, $\Spec(B) \leftrightarrow \Spec(A)$ is a bijection since $Q$ is uniquely defined by $P$, and vice versa. 259 | \end{proof} 260 | 261 | \begin{problem}[AM Exercise 5.28] 262 | Let $A$ be an integral domain, $K$ its field of fractions. Show that the following are equivalent: 263 | \begin{enum} 264 | \item $A$ is a valuation ring of $K$; 265 | \item If $\mathfrak{a},\mathfrak{b}$ are any two ideals of $A$, then either $\mathfrak{a} \subseteq \mathfrak{b}$ or $\mathfrak{b} \subseteq \mathfrak{a}$. 266 | \end{enum} 267 | Deduce that if $A$ is a valuation ring and $\mathfrak{p}$ is a prime ideal of $A$, then $A_{\mathfrak{p}}$ and $A/\mathfrak{p}$ are valuation rings of their fields of fractions. 268 | \end{problem} 269 | \begin{proof} 270 | $(1) \Rightarrow (2)$. Consider two ideals $\mathfrak{a},\mathfrak{b} \subseteq A$. Suppose, without loss of generality, that there exists $x \in \mathfrak{a} \setminus \mathfrak{b}$, and let $0 \ne y \in \mathfrak{b}$. Then, $x/y \notin A$, for otherwise $x \in \mathfrak{b}$ since $\mathfrak{b}$ is an ideal, and so $y/x \in A$, and so $y \in \mathfrak{a}$. Thus, $\mathfrak{b} \subseteq \mathfrak{a}$. 271 | \par $(2) \Rightarrow (1)$. Suppose there exist $a,b \in A$ such that $b \ne 0$ and $a/b \notin K$; in particular, this implies $a \ne 0$. Let $\mathfrak{a} = (a),\mathfrak{b} = (b)$. If $\mathfrak{a} \subseteq \mathfrak{b}$, then there exists $c \in A$ such that $a = bc$, i.e., $a/b = c \in A$, a contradiction. Thus, $\mathfrak{b} \subseteq \mathfrak{a}$, and so there exists $c \in A$ such that $b = ac$, i.e., $b/a = c \in A$, and so $A$ is a valuation ring of $K$. 272 | \par Now suppose $\mathfrak{p} \in \Spec(A)$. Any two ideals in $A_{\mathfrak{p}}$ (resp.~$A/\mathfrak{p}$) are of the form $\mathfrak{a}_{\mathfrak{p}},\mathfrak{b}_{\mathfrak{p}}$ (resp.~$\mathfrak{a}/\mathfrak{p},\mathfrak{b}/\mathfrak{p}$), where $\mathfrak{a},\mathfrak{b}$ are ideals of $A$. Since $A$ is a valuation ring, without loss of generality $\mathfrak{b} \subseteq \mathfrak{a}$, and so $\mathfrak{b}_{\mathfrak{p}} \subseteq \mathfrak{a}_{\mathfrak{p}}$ (resp.~$\mathfrak{b}/\mathfrak{p} \subseteq \mathfrak{a}/\mathfrak{p}$). Thus, $A_{\mathfrak{p}}$ (resp.~$A/\mathfrak{p}$) is a valuation ring of its field of fractions. 273 | \end{proof} 274 | 275 | \begin{problem}[AM Exercise 5.29] 276 | Let $A$ be a valuation ring of a field $K$. Show that every subring of $K$ which contains $A$ is a local ring of $A$. 277 | \end{problem} 278 | \begin{lemma*}[AM Exercise 5.27] 279 | Let $A,B$ be two local rings. $B$ is said to \emph{dominate} $A$ if $A$ is a subring of $B$ and the maximal ideal $\mathfrak{m}$ of $A$ is contained in the maximal ideal $\mathfrak{n}$ of $B$ (or, equivalently, $\mathfrak{m} = \mathfrak{n} \cap A$). Let $K$ be a field and let $\Sigma$ be the set of all local subrings of $K$. If $\Sigma$ is ordered by the relation of domination, show that $\Sigma$ has maximal elements and that $A \in \Sigma$ is maximal if and only if $A$ is a valuation ring of $K$. 280 | \end{lemma*} 281 | \begin{proof}[Proof of Lemma] 282 | Let $\{A_\alpha\}$ be a chain in $\Sigma$; let $A = \bigcup A_\alpha, \mathfrak{m} = \bigcup \mathfrak{m}_\alpha$. We claim $A$ is a local ring with maximal ideal $\mathfrak{m}$. $x \in A \setminus \mathfrak{m}$ implies $x \in A_\alpha \setminus \mathfrak{m}_\alpha$ for some $\alpha$. By MR Corollary $1.9$, we then see $x \in A_\alpha^\times$, and so $x \in A^\times$. By AM Proposition $1.6i$, then, $(A,\mathfrak{m})$ is local. By Zorn's lemma, $\Sigma$ therefore has a maximal element. 283 | \par $\Rightarrow$. Letting $f_\alpha : A_\alpha \hookrightarrow \overline{K}$ be the canonical inclusions, $\Sigma' = \{(A_\alpha,f_\alpha)\}$ has the same maximal elements as $\Sigma$, and moreover satisfies the construction on p.~$65$ in AM. Thus, by AM Theorem $5.21$, if $A \in \Sigma$ is maximal, it is a valuation ring of $K$. 284 | \par $\Leftarrow$. Suppose $(A,\mathfrak{m})$ is a valuation ring properly dominated by $(B,\mathfrak{n})$. Choose $x \in B \setminus A$ such that $x^{-1} \in A$, which exists since $A$ is a valuation ring. Then, $x^{-1} \in \mathfrak{m}$ since $x \notin A \implies x^{-1} \notin A^\times$, and then by MR Corollary $1.9$. But $x \in B$ implies $x^{-1} \notin \mathfrak{n}$ by MR Corollary $1.9$, and so $\mathfrak{m} \not\subseteq \mathfrak{n}$, a contradiction. 285 | \end{proof} 286 | \begin{proof} 287 | Let $A \subseteq B \subseteq K$; we want to show $B = A_{\mathfrak{p}}$ for some $\mathfrak{p} \subseteq A$. By AM Proposition $5.18i,ii$, we see $B$ is a valuation ring of $K$, and is moreover local. Let $\mathfrak{M}$ be the maximal ideal of $B$. Let $\mathfrak{p} = A \cap \mathfrak{M}$, which is prime since contractions of prime ideals of prime. 288 | \par We claim $B = A_{\mathfrak{p}}$. First, consider $x \in A \setminus \mathfrak{p}$. Then, $x \notin \mathfrak{M}$ by definition of $\mathfrak{p}$, and so $x \in B^\times$ by MR Corollary $1.9$. Thus, $a \in A \implies a/x \in B$, and so $A_{\mathfrak{p}} \subseteq B$. Now since $A_{\mathfrak{p}} \subseteq B$, we also have $\mathfrak{p}A_{\mathfrak{p}} \subseteq \mathfrak{M}$, where both ideals are maximal, i.e., $B$ dominates $A_{\mathfrak{p}}$. Since $A_{\mathfrak{p}}$ is a valuation ring by AM Exercise 5.28, we see that $A_{\mathfrak{p}}$ is maximal with respect to domination by AM Exercise $5.27$, i.e., $B = A_{\mathfrak{p}}$. 289 | \end{proof} 290 | 291 | \begin{problem}[AM Exercise 5.33] 292 | Let $\Gamma$ be a totally ordered abelian group. We shall show how to construct a field $K$ and a valuation $v$ of $K$ with $\Gamma$ as value group. Let $k$ be any field and let $A = k[\Gamma]$ be the group algebra of $\Gamma$ over $k$. By definition, $A$ is freely generated as a $k$-vector space by elements $x_\alpha~(\alpha \in \Gamma)$ such that $x_\alpha x_\beta = x_{\alpha+\beta}$. Show that $A$ is an integral domain. 293 | \par If $u = \lambda_1x_{\alpha_1} + \cdots + \lambda_nx_{\alpha_n}$ is any non-zero element of $A$, where the $\lambda_i$ are all $\ne 0$ and $\alpha_1 < \cdots < \alpha_n$, define $v_0(u)$ to be $\alpha_1$. Show that the mapping $v_0 : A - \{0\} \to \Gamma$ satisfies the conditions $(1)$ and $(2)$ of Exercise $31$. 294 | \par Let $K$ be the field of fractions of $A$. Show that $v_0$ can be uniquely extended to a valuation $v$ of $K$, and that the value group of $v$ is precisely $\Gamma$. 295 | \end{problem} 296 | \begin{proof} 297 | Suppose $uv = 0$ for nonzero $u,v \in A$. Then, $u = \lambda_1x_{\alpha_1} + \cdots + \lambda_mx_{\alpha_m}, v = \eta_1x_{\beta_1} + \cdots + \eta_nx_{\beta_n}$, where we assume without loss of generality that $\alpha_i,\beta_j$ are totally ordered by the order on $\Gamma$. Then, the least term in $uv$ with respect to this order is $\lambda_1\eta_1x_{\alpha_1+\beta_1}$, which is nonzero since $\lambda_1\eta_1 \ne 0$, a contradiction. Thus, $A$ is a domain. 298 | \par Consider $u,v$ as above. Then, $v_0(uv) = \alpha_1+\beta_1 = v_0(u)+v_0(v)$, and $v_0(u+v) \ge \min\{\alpha_1,\beta_1\} = \min\{v_0(u),v_0(v)\}$, and so $v_0$ satisfies $(1),(2)$ in AM Exercise 5.31. 299 | \par We want to extend $v_0$ to a valuation $v : K \setminus \{0\} \to \Gamma$. Such an extension must satisfy $(1)$ in AM Exercise $5.31$, i.e., $v(a/s) + v(s) = v(a)$. Thus, $v(a/s) = v(a) - v(s)$ uniquely determines $v$, and since $v_0$ maps onto $\Gamma$ and every element $-v(s) \in \Gamma$ by $\Gamma$'s additive group structure, we see the value group of $v$ is precisely $\Gamma$. 300 | \end{proof} 301 | 302 | 303 | \printbibliography 304 | \end{document} 305 | -------------------------------------------------------------------------------- /References.bib: -------------------------------------------------------------------------------- 1 | %% This BibTeX bibliography file was created using BibDesk. 2 | %% http://bibdesk.sourceforge.net/ 3 | 4 | %% Created for Takumi Murayama at 2015-05-09 21:12:41 -0400 5 | 6 | 7 | %% Saved with string encoding Unicode (UTF-8) 8 | 9 | 10 | 11 | @book{HS06, 12 | Address = {Cambridge}, 13 | Author = {Huneke, Craig and Swanson, Irena}, 14 | Date-Added = {2015-05-10 01:12:40 +0000}, 15 | Date-Modified = {2015-05-10 01:12:40 +0000}, 16 | Isbn = {978-0-521-68860-4; 0-521-68860-4}, 17 | Mrclass = {13B22 (13A18 13A30 13A35 13H15 14A05)}, 18 | Mrnumber = {2266432 (2008m:13013)}, 19 | Mrreviewer = {Liam O'Carroll}, 20 | Number = {336}, 21 | Publisher = {Cambridge University Press}, 22 | Series = {London Mathematical Society Lecture Note Series}, 23 | Title = {Integral closure of ideals, rings, and modules}, 24 | Year = {2006}, 25 | Bdsk-Url-1 = {http://www.ams.org.proxy.lib.umich.edu/mathscinet-getitem?mr=2266432}} 26 | 27 | @misc{stacks-project, 28 | Author = {{The Stacks Project Authors}}, 29 | Date-Added = {2015-05-10 01:10:38 +0000}, 30 | Date-Modified = {2015-05-10 01:10:38 +0000}, 31 | Howpublished = {\url{http://stacks.math.columbia.edu}}, 32 | Shorthand = {Stacks}, 33 | Title = {\itshape Stacks Project}, 34 | Year = {2014}} 35 | 36 | @book{Mac98, 37 | Address = {New York}, 38 | Author = {Mac Lane, Saunders}, 39 | Date-Added = {2015-05-10 00:09:30 +0000}, 40 | Date-Modified = {2015-05-10 00:09:30 +0000}, 41 | Edition = {Second ed}, 42 | Isbn = {0-387-98403-8}, 43 | Mrclass = {18-02}, 44 | 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F. and Macdonald, I. G.}, 337 | Date-Added = {2013-04-22 02:31:09 +0000}, 338 | Date-Modified = {2013-05-19 19:26:14 +0000}, 339 | Keywords = {Commutative Algebra}, 340 | Mrclass = {13.00}, 341 | Mrnumber = {0242802 (39 \#4129)}, 342 | Mrreviewer = {J. A. Johnson}, 343 | Publisher = {Addison-Wesley Publishing Co.}, 344 | Title = {Introduction to commutative algebra}, 345 | Year = {1969}, 346 | Bdsk-File-1 = 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\newcommand{\coh}{\mathsf{coh}\mathchar`-} 137 | \newcommand{\vect}{\mathsf{vect}\mathchar`-} 138 | \newcommand{\imm}[1][imm]{\hspace{0.75ex}\raisebox{0.58ex}{% 139 | \begin{tikzpicture}[commutative diagrams/every diagram] 140 | \draw[commutative diagrams/.cd, every arrow, every label,hook,{#1}] (0,0ex) -- (2.25ex,0ex); 141 | \end{tikzpicture}}\hspace{0.75ex}} 142 | \newcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{tikzcd}[column sep=small,ampersand replacement=\&] {#1} \rar[dashed] \& {#2} \end{tikzcd}\hspace{-0.7em}}} 143 | 144 | \usepackage{todonotes} 145 | 146 | \title{Hartshorne Ch.~I, \S1 Affine Varieties} 147 | \author{Takumi Murayama and Kyu Jun} 148 | 149 | \begin{document} 150 | \maketitle 151 | \setcounter{section}{1} 152 | \begin{problem}\mbox{}\label{exc:I.1.1} 153 | \begin{enuma} 154 | \item Let $Y$ be the plane curve $y = x^2$ (i.e., $Y$ is the zero set of 155 | the polynomial $f = y - x^2$). 156 | Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over 157 | $k$. 158 | \item Let $Z$ be the plane curve $xy = 1$. 159 | Show that $A(Z)$ is not isomorphic to a polynomial ring in one variable 160 | over $k$. 161 | \item Let $f$ be any irreducible quadratic polynomial in $k[x,y]$, and let 162 | $W$ be the conic defined by $f$. 163 | Show that $A(W)$ is isomorphic to $A(Y)$ or $A(Z)$. 164 | Which one is it when? 165 | \end{enuma} 166 | \end{problem} 167 | \begin{proof}[Proof of $(a)$] 168 | 169 | Mapping $k[x,y] \to k[t]$ by $x \mapsto t,y \mapsto t^2$, we have a 170 | surjective map with kernel $(y - x^2)$, and so $k[x,y]/(y-x^2) \cong k[t]$. 171 | Thus, $(y-x^2)$ is a prime ideal since $k[t]$ is a domain, and in particular 172 | it is radical; thus $I(Y) = (y-x^2)$, and $A(Y) = k[x,y]/(y-x^2) \cong k[t]$. 173 | \end{proof} 174 | \begin{proof}[Proof of $(b)$] 175 | $k[x,y]/(xy-1) \cong k[s,s^{-1}]$ by the map $x \mapsto s,y \mapsto s^{-1}$, 176 | which is a domain, hence $(xy-1)$ is prime, and in particular it is radical. 177 | Thus, $I(Z) = (xy-1)$, and $A(Z) \cong k[s,s^{-1}]$. 178 | This is not isomorphic to a polynomial ring, for any homomorphism 179 | $\varphi\colon k[s,s^{-1}] \to k[t]$ must have $\varphi(r) \in k$ for any 180 | $r \in k$ or $r \in \{s,s^{-1}\}$ since they are units, and so 181 | $\varphi(k[s,s^{-1}]) \subset k$. 182 | \end{proof} 183 | \begin{claim}[$c$] 184 | Let $\Char k \ne 2$, and let 185 | $f = a_{11}x^2 + a_{12}xy + a_{22}y^2 + b_1x + b_2y + c$, 186 | where is irreducible. Consider the matrix 187 | \begin{equation*} 188 | M = \begin{pmatrix} 189 | a_{11} & \dfrac{a_{12}}{2}\\ 190 | \dfrac{a_{12}}{2} & a_{22} 191 | \end{pmatrix} 192 | \end{equation*} 193 | Then, $A(W) \cong A(Y)$ if $\det M = 0$, and $A(W) \cong A(Z)$ if 194 | $\det M \ne 0$. 195 | \par If $\Char k =2$, then $A(W) \cong A(Y)$ if $a_{12} = 0$, and 196 | $A(W) \cong A(Z)$ if $a_{12} \ne 0$. 197 | \end{claim} 198 | \begin{proof}[Proof of $(c)$] 199 | By the structure theorem for symmetric bilinear forms 200 | \cite[Ch.~XV, Thm.~3.1]{Lan02}, we can assume after change of coordinates 201 | in $\AA^2$ that $a_{12} = 0$. 202 | Since (after possibly interchanging $x \leftrightarrow y$) $a_{11} \ne 0$, 203 | changing coordinates $x \mapsto x - b_1/2a_{11}$ cancels out the $x$ term in 204 | $f$. 205 | If $a_{22} \ne 0$, we can do the same process to $y$ to eliminate the $y$ 206 | term; otherwise, we can eliminate the $c$ term by replacing 207 | $y \mapsto y - c/b_2$. 208 | Thus, up to change in coordinates $f$ has one of the following two forms: 209 | \begin{equation*} 210 | a_{11}x^2 + a_{22}y^2 + c, \qquad a_{11}x^2 + b_2y. 211 | \end{equation*} 212 | We note in the first case that $c \ne 0$ for otherwise $f$ would be 213 | reducible (recalling that $k$ is algebraically closed), and that this 214 | corresponds to the case when $\det M \ne 0$. 215 | We note in the second case that $a_{11} \ne 0$ for $f$ was assumed to be 216 | quadratic, and $b_2 \ne 0$ for otherwise $f$ would be reducible; this 217 | corresponds to the case when $\det M = 0$. 218 | \par It remains to show the isomorphisms in either case. 219 | Note that since none of our coefficients above are zero, we can re-normalize 220 | our variables to get $x^2 - y^2 - 1$ and $x^2 - y$ (recalling again that $k$ 221 | is algebraically closed). 222 | In the latter case, we clearly have that $A(W) \cong A(Y)$, and so the former 223 | case remains. 224 | In the former case, substituting $x \mapsto (x+y)/2$ and $y \mapsto (x-y)/2$ 225 | gives that $f$ is of the form $xy - 1$, and so we have that $A(W) \cong A(Z)$. 226 | \par Finally, we show the claim for when $\Char k = 2$. First consider the 227 | case $a_{12} \ne 0$. Let $\alpha$ such that 228 | $\alpha^2 + a_{12}\alpha + a_{22} = 0$, which exists since $k$ is 229 | algebraically closed. 230 | Then, changing coordinates $x \mapsto x - \alpha y$, 231 | \begin{equation*} 232 | a_{11}x^2 + a_{12}xy + a_{22}y^2 \mapsto a_{11}x^2 + \alpha y^2 + a_{12}xy + a_{12}\alpha y^2 + a_{22}y^2 = a_{11}x^2 + a_{12}xy, 233 | \end{equation*} 234 | and so we can assume $a_{22} = 0$. 235 | Then, replacing $y \mapsto y + (a_{11}/a_{12})x$, we get 236 | \begin{equation*} 237 | a_{11}x^2 + a_{12}xy \mapsto a_{11}x^2 + a_{12}xy + a_{11}x^2 = a_{12}xy, 238 | \end{equation*} 239 | and so we can moreover assume $a_{11} = 0$. 240 | Finally, replacing $x \mapsto x - (b_2/a_{12})$ and 241 | $y \mapsto y - (b_1/a_{12})$, which we note are $k$-algebra isomorphisms, 242 | we get $f \mapsto a_{12}xy + c$ for some $c$. 243 | If $c = 0$ then $f$ was originally the union of two lines up to affine 244 | transformation, hence we have $c \ne 0$. 245 | So, $A(W)$ is isomorphic to $k[x,y]/(xy - 1) \cong A(Z)$ after rescaling 246 | of $x,y$. 247 | \par Now consider when $a_{12} = 0$; then, one of $a_{11},a_{22} \ne 0$. 248 | Suppose both are nonzero; then, changing coordinates 249 | $x \mapsto x + \sqrt{a_{22}/a_{11}}y$, we have 250 | \begin{equation*} 251 | a_{11}x^2 + a_{22}y^2 \mapsto a_{11}x^2 + a_{22}y^2 + a_{22}y^2 = a_{11}x^2, 252 | \end{equation*} 253 | and so we can assume $a_{22} = 0$, i.e., $f = a_{11}x^2 + b_1x + b_2y + c$. 254 | If $b_2 = 0$, then $f$ is a reducible polynomial defining one or two lines, 255 | a contradiction, so $b_2 \ne 0$. 256 | Then, changing coordinates $y \mapsto y + (b_1/b_2)x$, we have 257 | \begin{equation*} 258 | b_1x + b_2y \mapsto b_1x + b_2y + b_1x = b_2y, 259 | \end{equation*} 260 | and so we can assume $b_1 = 0$. 261 | Changing coordinates again such that $x \mapsto x + \sqrt{c}$, 262 | \begin{equation*} 263 | a_{11}x^2 + b_2y + c \mapsto a_{11}x^2 + c + b_2y + c = a_{11}x^2 + b_2y, 264 | \end{equation*} 265 | so renormalizing gives $f$ in the form $y - x^2$. 266 | Finally, this implies that $A(Z) = k[x,y]/(y-x^2) \cong A(Y)$. 267 | \end{proof} 268 | 269 | \begin{problem}[The Twisted Cubic Curve] 270 | Let $Y \subseteq \AA^3$ be the set $Y = \{(t,t^2,t^3) \mid t \in k\}$. 271 | Show that $Y$ is an affine variety of dimension $1$. 272 | Find generators for the ideal $I(Y)$. 273 | Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$. 274 | We say that $Y$ is given by the \emph{parametric representation} $x =t$, 275 | $y = t^2$, $z = t^3$. 276 | \end{problem} 277 | \begin{proof} 278 | We first show $\mathfrak{a} = (y - x^2,z - x^3)$ is prime and in particular, 279 | radical in $k[x,y,z]$. Define the $k$-algebra homomorphism 280 | $k[x,y,z] \to k[t]$ by $x \mapsto t,y\mapsto t^2,z \mapsto t^3$; clearly the 281 | kernel of this map contains $\mathfrak{a}$, and so this induces a map 282 | $\varphi\colon k[x,y,z]/\mathfrak{a} \to k[t]$. 283 | We claim $\varphi$ is injective. For suppose $f +\mathfrak{a} \mapsto 0$. 284 | By replacing $z$ with $xy$ and $y$ with $x^2$ in the expression for a 285 | particular representative $f$, $f + \mathfrak{a}$ is equivalent to a 286 | polynomial purely in $x\bmod\mathfrak{a}$. 287 | But any polynomial in $x$ maps to $0$ through $\varphi$ if and only if it is 288 | zero, hence $\varphi$ is injective. 289 | Since it was defined as the quotient of a surjective map, $\varphi$ is also 290 | an isomorphism of $k$-algebras, and so $\mathfrak{a}$ is prime and also 291 | radical since $k[t]$ is a domain. 292 | \par Now we claim $I(Y) = \mathfrak{a}$. 293 | $\supset$ is trivial since if $f \in \mathfrak{a}$, then $f(t,t^2,t^3) = 0$ 294 | for any $t \in k$ since the relations $y - x^2,z - x^3$ are all satisfied 295 | when $x=t,y=t^2,z=t^3$. 296 | $\subset$. We first show $Y \supset Z(\mathfrak{a})$. 297 | Suppose $(x,y,z)$ satisfy $y - x^2,z - x^3$. 298 | Then, $(x,y,z) = (x,x^2,x^3)$ by using the relations repeatedly, and so 299 | $(x,y,z) \in Y$. 300 | Now by Prop.~$1.2$, $Y \supset Z(\mathfrak{a})$ implies 301 | $I(Y) \subset I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}} = \mathfrak{a}$ 302 | since $\mathfrak{a}$ is radical from above. 303 | \par We check that $Y$ is closed; however, 304 | $\overline{Y} = Z(I(Y)) = Z(\mathfrak{a})$, and any point $(x,y,z)$ 305 | satisfying $y - x^2,z - x^3$ can be written $(x,x^2,x^3)$ as above, hence 306 | $Y \supset Z(\mathfrak{a})$, and so $Y = \overline{Y}$ is closed. 307 | \par Finally, by the first paragraph, 308 | $A(Y) = k[x,y,z]/I(Y) = k[x,y,z]/\mathfrak{a} \cong k[t]$, and so since 309 | $I(Y) = \mathfrak{a}$ is prime by the above with $Z(\mathfrak{a}) = Y$ and 310 | $k[t]$ is of dimension $1$, $Y$ is an affine variety of dimension $1$ by 311 | Cor.~1.4 and Prop.~1.7. 312 | \end{proof} 313 | 314 | \begin{problem} 315 | Let $Y$ be the algebraic set in $\AA^3$ defined by the two polynomials 316 | $x^2 - yz$ and $xz - x$. 317 | Show that $Y$ is a union of three irreducible components. 318 | Describe them and find their prime ideals. 319 | \end{problem} 320 | \begin{proof} 321 | We first know $Y = Z(x^2-yz,x(z-1))$, and so we claim 322 | \begin{equation*} 323 | Y = Z(x^2-yz,x(z-1)) = Z(yz,x) \cup Z(x^2-y,z-1). 324 | \end{equation*} 325 | $\subset$. If $P = (x,y,z) \in Z(x^2-yz,x(z-1))$, then either $x=0$ or 326 | $z-1=0$. 327 | In the former case, this implies $yz =0$ as well, hence $P \in Z(yz,x)$. 328 | In the latter case, this implies $x^2-y$ as well, hence $P \in Z(x^2-y,z-1)$. 329 | $\supset$. If $P \in Z(yz,x)$, then clearly $x^2-yz$ and $x(z-1)$ are 330 | satisfied; similarly, if $P \in Z(x^2-y,z-1)$, then $x^2-yz$ and $z-1$ are 331 | clearly satisfied. 332 | \par We now claim that $Z(yz,x) = Z(y,x) \cup Z(z,x)$. $\subset$ is clear 333 | since $x=0$ and one of $y = 0$ or $z=0$ must hold, so any point 334 | $P \in Z(yz,x)$ is in either $Z(y,x)$ or $Z(z,x)$. 335 | $\supset$ is trivial. 336 | \par We therefore have 337 | \begin{equation*} 338 | Y = Z(x,y) \cup Z(x,z) \cup Z(x^2-y,z-1). 339 | \end{equation*} 340 | We claim each component is irreducible. 341 | It suffices to show by Cor.~1.4 that each of the defining ideals is prime. 342 | $(x,y)$ is prime since $k[x,y,z]/(x,y) \cong k[z]$; similarly, $(x,z)$ is 343 | prime since $k[x,y,z]/(x,z) \cong k[y]$. 344 | $x^2-y,z-1$ is also prime since 345 | $k[x,y,z]/(x^2-y,z-1) \cong k[x,y]/(x^2-y) \cong k[t]$ as in Problem 346 | $\ref{exc:I.1.1}(a)$. 347 | \par We finally note that $Z(x,y)$ is the $z$-axis in $\AA^3$, $Z(x,z)$ is 348 | the $y$-axis in $\AA^3$, and $Z(x^2-y,z-1)$ is the parabola $x^2-y$ lying on 349 | the $z=1$ hyperplane. 350 | \end{proof} 351 | 352 | \begin{problem} 353 | If we identify $\AA^2$ with $\AA^1 \times \AA^1$ in the natural way, show 354 | that the Zariski topology on $\AA^2$ is not the product topology of the 355 | Zariski topologies on the two copies of $\AA^1$. 356 | \end{problem} 357 | \begin{proof} 358 | $Z(x-y) \subset \AA^2$ is closed. Now suppose $\AA^2$ had the product 359 | topology; then, by \cite[Exc.~17.13]{Mun00}, this would imply $\AA^1$ is 360 | Hausdorff, contradicting Ex.~1.1.1. 361 | \end{proof} 362 | 363 | \begin{problem} 364 | Show that a $k$-algebra $B$ is isomorphic to the affine coordinate ring of 365 | some algebraic set in $\AA^n$, for some $n$, if and only if $B$ is a 366 | finitely generated $k$-algebra with no nilpotent elements. 367 | \end{problem} 368 | \begin{proof} 369 | $\Leftarrow$. Suppose $B$ is a finitely generated $k$-algebra with no 370 | nilpotent elements. 371 | By definition of finitely-generated $k$-algebras, 372 | $B \cong k[x_1,\ldots,x_n]/I$ for some ideal 373 | $I \subset k[x_1,\ldots,x_n] \eqqcolon A$. 374 | Since $B$ has no nilpotent elements, 375 | $\sqrt{I} = \pi^{-1}(\mathfrak{N}_{A/I}) = I$ so 376 | $A(Z(I)) = A/I(Z(I)) = A/\sqrt{I} = A/I = B$ by Prop.~1.2. 377 | \par $\Rightarrow$. Suppose $B \cong A(X)$ for some algebraic set 378 | $X \subset \AA^n$. 379 | Then, $B \cong A/I(X)$, where $A$ is as before, hence $B$ is a finitely 380 | generated $k$-algebra. 381 | Now suppose $f + I(X) \in B$ were nilpotent, i.e., $(f + I(X))^r = 0$ for 382 | some $r$. 383 | Then, $f^r + I(X) = 0$, hence $f^r \in I(X)$ and so $f \in I(X)$ by 384 | Cor.~1.4 since $I(X)$ is radical, and so $f = 0$ in $B$. 385 | Thus, $B$ has no nilpotent elements. 386 | \end{proof} 387 | 388 | \begin{problem}\label{exc:1.6} 389 | Any nonempty open subset of an irreducible topological space is dense and 390 | irreducible. 391 | If $Y$ is a subset of a topological space $X$, which is irreducible in its 392 | induced topology, then the closure $\overline{Y}$ is also irreducible. 393 | \end{problem} 394 | \begin{remark} 395 | By definition, $X$ is irreducible if and only if in any decomposition 396 | $X = A \cup B$ for $A,B$ closed subsets of $X$, we have either $X = A$ or 397 | $X = B$. 398 | \end{remark} 399 | \begin{proof} 400 | Let $U \subset X$ be a nonempty open subset. 401 | Then, $X = (X \setminus U) \cup \overline{U}$, and since $X \setminus U$ is 402 | a proper subset of $X$, irreducibility of $X$ implies $X = \overline{U}$. 403 | \par Now let $U \subset X$ be an open subset of $X$, and let $A,B \subset X$ 404 | be closed such that $U = (A \cap U) \cup (B \cap U) = (A \cup B) \cap U$. 405 | By the previous paragraph, 406 | $X = \overline{U} = \overline{(A \cup B) \cap U} = A \cup B$, hence by 407 | irreducibility of $X$, without loss of generality $A \supset U$, and so 408 | $U = U \cap A$, i.e., $U$ is irreducible. 409 | \par Finally, suppose $\overline{Y} = A \cup B$ for $A,B$ closed in 410 | $\overline{Y}$. 411 | Then, $Y = (Y \cap A) \cup (Y \cap B)$, and since $Y$ is irreducible, without 412 | loss of generality $Y = Y \cap A$. 413 | Then, $\overline{Y} = \overline{Y \cap A} = A$, hence $\overline{Y}$ is 414 | irreducible. 415 | \end{proof} 416 | 417 | \begin{problem}\mbox{} 418 | \begin{enuma} 419 | \item Show that the following conditions are equivalent for a topological 420 | space $X$: $(i)$ $X$ is noetherian; $(ii)$ every nonempty family of 421 | closed subsets has a minimal element; $(iii)$ $X$ satisfies the ascending 422 | chain condition for open subsets; $(iv)$ every nonempty family of open 423 | subsets has a maximal element. 424 | \item A noetherian topological space is \emph{quasi-compact,} i.e., every 425 | open cover has a finite subcover. 426 | \item Any subset of a noetherian topological space is noetherian in its 427 | induced topology. 428 | \item A noetherian space which is also Hausdorff must be a finite set with 429 | the discrete topology. 430 | \end{enuma} 431 | \end{problem} 432 | \begin{proof}[Proof of $(a)$] 433 | $(i) \Rightarrow (ii)$. If we had a nonempty family of closed subsets with 434 | no minimal element, then we could inductively construct a non-terminating 435 | strictly decreasing sequence of closed subsets of $X$, contradicting that 436 | $X$ is noetherian. 437 | \par $(ii) \Rightarrow (i)$. Any descending chain of closed subsets 438 | $Y_1 \supset Y_2 \supset \cdots$ has a minimal element $Y_r$, i.e., 439 | $Y_r = Y_{r+1} = \cdots$, hence $X$ is noetherian. 440 | \par $(i) \Leftrightarrow (iii)$ and $(ii) \Leftrightarrow (iv)$ follow by 441 | taking complements. 442 | \end{proof} 443 | \begin{proof}[Proof of $(b)$] 444 | Let $\{X_i\}_{i \in I}$ be an open cover, and choose a well ordering on $I$. 445 | Then, consider the decreasing sequence of closed sets 446 | \begin{equation*} 447 | X \setminus X_1 \supseteq X \setminus (X_1 \cup X_2) \supseteq X \setminus (X_1 \cup X_2 \cup X_3) \supseteq \cdots. 448 | \end{equation*} 449 | Then, since $X$ is noetherian there exists a finite integer $r$ such that 450 | \begin{equation*} 451 | X \setminus (X_1 \cup \cdots \cup X_r) = \bigcap_{i \in I} X \setminus (X_1 \cup \cdots \cup X_i) = X \setminus \bigcup_{i \in I} X_i = \emptyset, 452 | \end{equation*} 453 | hence $X_1,\ldots,X_r$ is a finite subcover of $\{X_i\}_{i \in I}$. 454 | \end{proof} 455 | \begin{proof}[Proof of $(c)$] 456 | Let $A_1 \supset A_2 \supset \cdots$ be a descending chain of closed sets 457 | in $Y \subset X$; writing $A_i = Y \cap B_i$ for closed subsets 458 | $B_i \subset X$, we have that $B_1 \supset B_2 \supset \cdots$ stabilizes 459 | since $X$ is noetherian, so $A_1 \supset A_2 \supset \cdots$ does as well. 460 | \end{proof} 461 | \begin{proof}[Proof of $(d)$] 462 | Let $X$ be noetherian and Hausdorff. 463 | Hausdorff implies every finite point set is closed \cite[Thm.~17.8]{Mun00}, 464 | hence it suffices to show $X$ is finite. 465 | So suppose $X$ is infinite; let $\Sigma$ be the family of infinite closed 466 | subsets, which is nonempty since $X \in \Sigma$. 467 | $\Sigma$ has a minimal element $Z$ by property $(ii)$ from $(a)$. 468 | Let $p,q \in Z$; since $X$ is Hausdorff there exist $U,V$ disjoint that 469 | contain $p,q$ respectively. 470 | We then have $X = (X \setminus U) \cup (X \setminus V)$, and so 471 | $Z = (Z \cap (X \setminus U) \cup Z\cap(X \setminus V))$. 472 | Then each of $Z \cap (X \setminus U)$ and $Z \cap (X \setminus V)$ are 473 | closed and strictly contained in $Z$, hence are finite by minimality of $Z$. 474 | This implies $Z$ is also finite, a contradiction. 475 | \end{proof} 476 | 477 | \begin{problem}\label{exc:1.8} 478 | Let $Y$ be an affine variety of dimension $r$ in $\AA^n$. 479 | Let $H$ be a hypersurface in $\AA^n$, and assume $Y \not\subseteq H$. 480 | Then every irreducible component of $Y \cap H$ has dimension $r-1$. 481 | (See $(7.1)$ for a generalization.) 482 | \end{problem} 483 | \begin{proof} 484 | Since by Cor.~$1.4$, the irreducible components of $Y \cap H$ correspond to 485 | the minimal primes in $A(Y \cap H)$, it suffices to find the coheight of each 486 | minimal prime in $A(Y \cap H)$ by Thm.~1.8A$(b)$. 487 | Note that $Y \cap H = Z(I(Y) + (f))$ by the proof of Prop.~1.1, and so 488 | $I(Y \cap H) = \sqrt{I(Y) + (f)}$ by Prop.~1.2. 489 | So to find the coheight of each minimal prime in $A(Y \cap H)$, it suffices by 490 | \cite[Prop.~1.1]{AM69} to find the coheight of each minimal prime containing 491 | $\sqrt{I(Y) + (f)}$ in $k[x_1,\ldots,x_n]$. 492 | By \cite[Prop.~1.14]{AM69}, these primes are the same as the minimal primes 493 | containing $I(Y) + (f)$ in $k[x_1,\ldots,x_n]$, and by \cite[Prop.~1.1]{AM69} 494 | again, these are the same as the minimal primes containing $f$ in $A(Y)$. 495 | By the Hauptidealsatz (Thm.~1.11A), the height of each minimal prime 496 | containing $f$ in $A(Y)$ is $1$, hence the coheight of these minimal primes is 497 | $r-1$. 498 | Thus, by our reductions above we see that the dimension of each irreducible 499 | component is $r-1$. 500 | \end{proof} 501 | 502 | \begin{problem} 503 | Let $\mathfrak{a} \subseteq A = k[x_1,\ldots,x_n]$ be an ideal which can be 504 | generated by $r$ elements. 505 | Then every irreducible component of $Z(\mathfrak{a})$ has dimension $\ge n-r$. 506 | \end{problem} 507 | \begin{proof} 508 | We induce on the number of generators $r$; let $\{f_1,\ldots,f_r\}$ be 509 | a set of generators. 510 | The $r=1$ case is Prop.~1.13. 511 | So suppose the claim holds for $\mathfrak{b} = (f_1,\ldots,f_{r-1})$; let $H = Z(f_r)$. 512 | Let $Y_i$ be the irreducible components of $Y \coloneqq Z(\mathfrak{b})$; each has 513 | dimension $\ge n-r+1$ by inductive hypothesis. 514 | Then, $Z(\mathfrak{a}) = Y \cap H = \bigcup_i Y_i \cap H$ as on p.~2. 515 | If $Y_i \subset H$, then $Y_i \cap H = Y_i$ is an irreducible component of 516 | $Z(\mathfrak{a})$ and has dimension $\ge n-r+1$. 517 | If $Y_i \not\subset H$, then by Exercise \ref{exc:1.8}, the dimension of each 518 | irreducible component of $Y_i \cap H$ is $\ge n-r$. 519 | Since in either case the dimension of each irreducible component of 520 | $Z(\mathfrak{a})$ is $\ge n-r$, we are done. 521 | \end{proof} 522 | 523 | \begin{problem} \mbox{} 524 | \begin{enuma} 525 | \item If $Y$ is any subset of a topological space $X$ then $\dim Y \leq \dim X$. 526 | \item If $X$ is a topological space which is covered by a family of open 527 | subsets $\{U_{i}\}$, then $\dim X = \sup \dim U_i$. 528 | \item Give an example of a topological space $X$ and a dense open subset $U$ 529 | with $\dim U < \dim X$. 530 | \item If Y is a closed subset of an irreducible finite-dimensional topological space $X$, and if $\dim Y = \dim X$, then $Y= X$. 531 | \item Give an example of a noetherian topological space of infinite dimension. 532 | \end{enuma} 533 | \end{problem} 534 | \begin{proof}[Proof of $(a)$] 535 | Let $\dim Y = n$ and $Y_0 \subsetneq Y_1 \subsetneq \cdots \subsetneq Y_n \subset 536 | Y$ be a chain of distinct irreducible closed subsets of $Y$. Take the closure of 537 | each $Y_i$. By Exercise \ref{exc:1.6}, each $\overline{Y_i}$ is irreducible. 538 | Hence, we see that 539 | $\overline{Y_0} \subsetneq \overline{Y_1} \subsetneq \cdots \subsetneq 540 | \overline{Y_n}$ is an ascending chain of distinct irreducible closed subsets 541 | of $X$ ($\overline{Y_i} \neq \overline{Y_{i+1}}$ since if the two are equal, 542 | $Y_i = \overline{Y_i} \cap Y = \overline{Y_{i+1}} \cap Y = Y_{i+1}$, 543 | which is a contradiction). 544 | Hence $\dim X \geq n = \dim Y$. 545 | \end{proof} 546 | \begin{proof}[Proof of $(b)$] 547 | $\dim X \ge \sup \dim U_i$ by $(a)$. 548 | For the opposite inequality, we induce on dimension $n = \dim X$. 549 | If $n = 0$, the claim is trivial. 550 | Suppose $0 < n < \infty$. Then, there exists an ascending chain 551 | \begin{equation*} 552 | X_0 \subsetneq X_1 \subsetneq\cdots \subsetneq X_{n-1} \subsetneq X_n 553 | \subset X. 554 | \end{equation*} 555 | By inductive hypothesis, since $U_i \cap X_{n-1}$ cover $X_{n-1}$, there 556 | exists an ascending chain 557 | \begin{equation*} 558 | W_0 \subsetneq W_1 \subsetneq \cdots \subsetneq W_{n-2} \subsetneq W_{n-1} 559 | = U_i \cap X_{n-1} 560 | \end{equation*} 561 | for some $i$. Now $X_n \setminus X_{n-1}$ and $U_i \cap X_n$ are both open and 562 | dense in $X_n$ by Exercise \ref{exc:1.6}, hence they have nonempty 563 | intersection. 564 | This implies $U_i \cap X_{n-1} \subsetneq U_i \cap X_n$, hence $U_i \cap X_n$ 565 | has dimension $\ge \dim X_{n-1} + 1 = \dim X_n = \dim X$, and we are done. 566 | \par Now if $\dim X = \infty$, then for every $m$ there exists a closed 567 | irreducible subset $X_m$ of dimension $\ge m$. By the above, the open covers $U_i 568 | \cap X_m$ of $X_m$ satisfy $\dim X_m = \sup_i\dim (X_m \cap U_i)$, and so 569 | $\sup_m \dim X_m = \sup_{m,i}(U_i \cap X_m) = \infty$. 570 | %Let $\dim X = n$. 571 | %From (a), $\dim U_i \le \dim X$. Hence, $\sup \dim U_i \le \dim X$. 572 | %Now, consider an ascending chain $X_0 \subsetneq X_1 \subsetneq \cdots 573 | %\subsetneq X_n$ of distinct irreducible closed subsets of $X$. 574 | %Choose an open set $U_i$ such that $X_0 \cap U_i \neq \emptyset$. 575 | %We can choose such $U_i$ because $\{U_i\}$ covers $X$. 576 | %Then, $U_i \cap X_j \neq \emptyset$ because $X_j \supset X_0$. 577 | %Hence, $X_0 \cap U_i \subsetneq \cdots \subsetneq X_n \cap U_i$ is an 578 | %ascending chain of irreducible closed subsets (with respect to the subspace 579 | %topology) of $U_i$. Hence, $\sup \dim U_i \geq \dim X$. 580 | \end{proof} 581 | \begin{proof}[Proof of $(c)$] 582 | Let $X = {0,1}$, and $\tau = \{ \emptyset, X, \{0\}\}$ be the set of open sets of $X$. 583 | This is clearly a topology. 584 | Now, $\overline{\{0\}} = X$ so $U = \{0\}$ is dense in $X$. 585 | Note that $\{0\},\{1\},\{1,2\}$ are all irreducible sets as they cannot be 586 | written as the union of two proper closed subsets with respect to the subspace 587 | topology. 588 | The dimension of $U$ is $0$ as the only irreducible closed subset (with respect 589 | to the subspace topology) of $U$ is $U$ itself. However, as $\{1\}$ is 590 | irreducible, we have an ascending chain of closed irreducible subsets of 591 | $X$ with length $1$, which is $\{1\} \subsetneq \{0,1\} = X$. 592 | \end{proof} 593 | \begin{proof}[Proof of $(d)$] 594 | Let Y be a closed subset of an irreducible finite-dimensional topological space $X$ with $\dim Y = \dim X$. 595 | Assume that $Y \subsetneq X$. Let $Y_0 \subsetneq \cdots \subsetneq Y_n 596 | \subset Y$ be the ascending chain of irreducible closed sets. Then, $Y_i$ is 597 | an irreducible closed subset of $X$ as Y is closed. Because $X$ is also irreducible we have $Y_0 \subsetneq \cdots \subsetneq Y_n \subsetneq X$, so $\dim X \geq n+1 > \dim Y$, which is a contradiction. Hence $X = Y$. 598 | \end{proof} 599 | \begin{proof}[Proof of $(e)$] 600 | Let $X = \mathbb{N}$. Let $\mathscr{C} = \{ \emptyset, X, \{1\}, 601 | \{1,2\}, \{1,2,3\},\ldots\}$ be closed in $X$; this defines a topology on $X$. 602 | Let $Y_0 \supset Y_1 \supset \cdots$ be any sequence of closed subsets. 603 | $X$ is noetherian because there exists an integer $r$ such that $Y_r = Y_{r+1} 604 | = \cdots$ by the fact that $\mathbb{N}$ is well-ordered. 605 | \par Now, we show that the dimension of $X$ is infinite. 606 | We note that any closed set $\{1, \ldots,n\}$ is irreducible, so we have an 607 | infinite chain of closed irreducible subsets of $X$: $\{1\} \subsetneq \{1,2\} 608 | \subsetneq \cdots \subsetneq \{1,\ldots,n\} \subsetneq \cdots$. Hence $\dim X 609 | = \infty$. 610 | \end{proof} 611 | 612 | \begin{problem} 613 | Let $Y \subseteq \AA^3$ be the curve given parametrically by 614 | $x = t^3$, $y= t^4$, $z = t^5$. 615 | Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which cannot 616 | be generated by $2$ elements. 617 | We say $Y$ is \emph{not a local complete 618 | intersection}---cf.~\emph{(Ex.~$2.17$)}. 619 | \end{problem} 620 | \begin{proof} 621 | Let $\varphi\colon k[x,y,z] \to k[t^3,t^4,t^5]$ be the map $x \mapsto t^3$, $y 622 | \mapsto t^4$, $z \mapsto t^5$. 623 | By definition of $I(Y)$, $\ker\varphi = I(Y)$, and since this map is 624 | surjective, we have $A(Y) \cong k[t^3,t^4,t^5]$. 625 | Since $A(Y)$ is a domain, $I(Y)$ is prime. 626 | By Prop.~1.7 and Thm.~1.8A$(a)$, $\dim Y = \dim A(Y) = \trdeg_k \Frac(A(Y))$, 627 | and since $\Frac(A(Y)) \cong \Frac(k[t^3,t^4,t^5]) \cong k(t)$, 628 | $\trdeg_k \Frac(A(Y)) = 1$. 629 | Thus, $\dim Y = 1$, hence $Y$ is a curve, and also $\Ht I(Y) = 2$ by 630 | Thm.~1.8A$(b)$. 631 | \par We claim that $I(Y)$ is generated by a minimum of three elements. 632 | Suppose $f \in \ker\varphi$. 633 | Consider $k[x,y,z]$ as a graded ring with weights $3,4,5$ on $x,y,z$; then 634 | $\varphi$ is a graded ring homomorphism, so its kernel is a homogeneous ideal 635 | \cite[Ch.~II, \S11.3, Prop.~3]{Bou74}. 636 | We want find $(\ker\varphi)_{d}$. 637 | We have 638 | \begin{equation*} 639 | f = \sum_{\substack{(a,b,c)\\3a+4b+5c=d}} k_{abc}x^ay^bz^c \in 640 | (\ker\varphi)_d \iff \sum_{\substack{(a,b,c)\\3a+4b+5c=d}} k_{abc} = 0 641 | \end{equation*} 642 | since the sum on the right gives the coefficient of the $t^d$ term in 643 | $\varphi(f)$. 644 | We list the solutions for $3a+4b+5c=d$ for certain values of $d$: 645 | \begin{alignat*}{10} 646 | d &= 0&:& \quad (a,b,c) = (0,0,0)\\ 647 | d &= 1&:& \quad \text{\it impossible}\\ 648 | d &= 2&:& \quad \text{\it impossible}\\ 649 | d &= 3&:& \quad (a,b,c) = (1,0,0)\\ 650 | d &= 4&:& \quad (a,b,c) = (0,1,0)\\ 651 | d &= 5&:& \quad (a,b,c) = (0,0,1)\\ 652 | d &= 6&:& \quad (a,b,c) = (2,0,0)\\ 653 | d &= 7&:& \quad (a,b,c) = (1,1,0)\\ 654 | d &= 8&:& \quad (a,b,c) \in \{(1,0,1),(0,2,0)\}\\ 655 | d &= 9&:& \quad (a,b,c) \in \{(3,0,0),(0,1,1)\}\\ 656 | d &= 10&:& \quad (a,b,c) \in \{(2,1,0),(0,0,2)\} 657 | \end{alignat*} 658 | The cases $d=0,3,4,5,6,7$ imply that $(\ker \varphi)_d = 0$ for these values. 659 | The cases $d=8,9,10$ imply that 660 | \begin{align*} 661 | (\ker\varphi)_8 &= k\langle xz-y^2 \rangle, & (\ker\varphi)_9 &= k\langle 662 | x^3-yz \rangle, & (\ker\varphi)_{10} &= k\langle x^2y-z^2 \rangle, 663 | \end{align*} 664 | where $k\langle g \rangle$ denotes the span of $g$ as a $k$-vector subspace of 665 | $k[x,y,z]_d$. 666 | Now, all three of $xz-y^2,x^3-yz,x^2y - z^2$ must appear in a generating set for 667 | $\ker\varphi$, for there are no lower degree elements in $\ker\varphi$ that 668 | can generate them, and they cannot generate each other by looking at their 669 | degree. 670 | Thus, $I(Y) = \ker\varphi$ cannot be generated by $2$ elements, and $Y$ is not 671 | a local complete intersection. 672 | \end{proof} 673 | 674 | \begin{problem} 675 | Give an example of an irreducible polynomial $f \in \mathbb{R}[x,y]$, 676 | whose zero set $Z(f)$ in $\AA_{\mathbb{R}}^2$ is not irreducible 677 | \emph{(cf.~$1.4.2$)}. 678 | \end{problem} 679 | \begin{proof} 680 | Let $f(x,y) = y^2 + x^2(x-1)^2$. This is irreducible in $\mathbb{R}[x,y]$ 681 | since as a polynomial in $y$ over $\mathbb{R}[x]$, it is the sum of two 682 | nonzero squares, hence has no roots in $\mathbb{R}[x]$. Now, $Z(f) = 683 | \{(0,0),(1,0)\}$, which is not irreducible since each point is closed. 684 | \end{proof} 685 | 686 | \printbibliography 687 | \end{document} 688 | -------------------------------------------------------------------------------- /Hartshorne/II Schemes/6 Divisors/II.6 Divisors.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt,letterpaper]{article} 2 | \usepackage{geometry} 3 | \geometry{letterpaper} 4 | \usepackage{microtype} 5 | \usepackage{amsmath,amssymb,amsthm,mathrsfs} 6 | \usepackage{mathtools} 7 | \usepackage{ifpdf} 8 | \ifpdf 9 | \setlength{\pdfpagewidth}{8.5in} 10 | \setlength{\pdfpageheight}{11in} 11 | \else 12 | \fi 13 | \usepackage{hyperref} 14 | 15 | \usepackage{tikz} 16 | \usepackage{tikz-cd} 17 | \usetikzlibrary{decorations.markings} 18 | \tikzset{ 19 | open/.style = {decoration = {markings, mark = at position 0.5 with { \node[transform shape] {\tikz\draw[fill=white] (0,0) circle (.3ex);}; } }, 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\DeclareFieldFormat{multipostnote}{#1} 38 | 39 | \renewcommand{\theenumi}{$(\alph{enumi})$} 40 | \renewcommand{\labelenumi}{\theenumi} 41 | 42 | \newcounter{enumacounter} 43 | \newenvironment{enuma} 44 | {\begin{list}{$(\alph{enumacounter})$}{\usecounter{enumacounter} \parsep=0em \itemsep=0em \leftmargin=2.75em \labelwidth=1.5em \topsep=0em}} 45 | {\end{list}} 46 | \newcounter{enumdcounter} 47 | \newenvironment{enumd} 48 | {\begin{list}{$(\arabic{enumdcounter})$}{\usecounter{enumdcounter} \parsep=0em \itemsep=0em \leftmargin=1.75em \labelwidth=1.5em \topsep=0em}} 49 | {\end{list}} 50 | \newtheorem*{theorem}{Theorem} 51 | \newtheorem*{universalproperty}{Universal Property} 52 | \newtheorem{problem}{Problem}[section] 53 | \newtheorem{subproblem}{Problem}[problem] 54 | \newtheorem{lemma}{Lemma}%[section] 55 | \newtheorem{proposition}{Proposition} 56 | \newtheorem{property}{Property}[problem] 57 | \newtheorem*{lemma*}{Lemma} 58 | \theoremstyle{definition} 59 | \newtheorem*{definition}{Definition} 60 | \newtheorem*{claim}{Claim} 61 | \theoremstyle{remark} 62 | \newtheorem*{remark}{Remark} 63 | 64 | \numberwithin{equation}{section} 65 | \numberwithin{figure}{problem} 66 | \renewcommand{\theequation}{\arabic{section}.\arabic{equation}} 67 | 68 | \DeclareMathOperator{\Ann}{Ann} 69 | \DeclareMathOperator{\Ass}{Ass} 70 | \DeclareMathOperator{\Supp}{Supp} 71 | \DeclareMathOperator{\WeakAss}{\widetilde{Ass}} 72 | \let\Im\relax 73 | \DeclareMathOperator{\Im}{im} 74 | \DeclareMathOperator{\Spec}{Spec} 75 | \DeclareMathOperator{\SPEC}{\mathbf{Spec}} 76 | \DeclareMathOperator{\Sp}{sp} 77 | \DeclareMathOperator{\maxSpec}{maxSpec} 78 | \DeclareMathOperator{\Hom}{Hom} 79 | \DeclareMathOperator{\Soc}{Soc} 80 | \DeclareMathOperator{\Ht}{ht} 81 | \DeclareMathOperator{\A}{\mathcal{A}} 82 | \DeclareMathOperator{\V}{\mathbf{V}} 83 | \DeclareMathOperator{\Aut}{Aut} 84 | \DeclareMathOperator{\Char}{char} 85 | \DeclareMathOperator{\Frac}{Frac} 86 | 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\newcommand{\uqgr}{\underline{\mathsf{qgr}}\mathchar`-} 135 | \newcommand{\qcoh}{\mathsf{qcoh}\mathchar`-} 136 | \newcommand{\Alg}{\mathsf{Alg}\mathchar`-} 137 | \newcommand{\coh}{\mathsf{coh}\mathchar`-} 138 | \newcommand{\vect}{\mathsf{vect}\mathchar`-} 139 | \newcommand{\imm}[1][imm]{\hspace{0.75ex}\raisebox{0.58ex}{% 140 | \begin{tikzpicture}[commutative diagrams/every diagram] 141 | \draw[commutative diagrams/.cd, every arrow, every label,hook,{#1}] (0,0ex) -- (2.25ex,0ex); 142 | \end{tikzpicture}}\hspace{0.75ex}} 143 | \newcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{tikzcd}[column sep=small,ampersand replacement=\&] {#1} \rar[dashed] \& {#2} \end{tikzcd}\hspace{-0.7em}}} 144 | 145 | \usepackage{todonotes} 146 | %\usepackage[notref,notcite]{showkeys} 147 | 148 | \title{Hartshorne Ch.~II, \S6 Divisors} 149 | \author{Takumi Murayama} 150 | 151 | \begin{document} 152 | \maketitle 153 | \setcounter{section}{6} 154 | \begin{problem} 155 | Let $X$ be a scheme satisfying $(*)$. Then $X \times \mathbf{P}^n$ also satisfies $(*)$, and $\Cl(X \times \mathbf{P}^n) \cong (\Cl X) \times \mathbf{Z}$. 156 | \end{problem} 157 | \begin{lemma}\label{irredlem} 158 | $X$ is irreducible if and only if there exists an open cover $U_i$ where each $U_i$ is irreducible, and $U_i \cap U_j \ne \emptyset$ for all $i,j$. 159 | \end{lemma} 160 | \begin{proof}[Proof of Lemma \ref{irredlem}] 161 | $X$ is reducible $\Leftrightarrow X = A \cup B \Leftrightarrow\emptyset = (X\setminus A) \cap (X\setminus B)$, and so $X$ is irreducible if and only if every open subset is dense. 162 | \par $\Rightarrow$ Clearly $U_i \cap U_j \ne \emptyset$ for all $i,j$. If $U \subset U_i$, $\overline{U}$ is of the form $A \cap U_i$ for $A$ closed in $X$. But since $U$ is open in $X$, $A = X$ and so $\overline{U} = U_i$. 163 | \par $\Leftarrow$ Let $U \subset X$. Suppose $U \cap U_i \ne \emptyset$ but $U \cap U_j = \emptyset$; then, $U \cap (U_i \cap U_j) = \emptyset$, contradicting that $U$ is dense in $U_i$. So $U$ intersects all $U_i$, and since a set containing $U$ is closed if and only if it is closed in every element of the open cover $U_i$, and since $U$ is dense in each $U_i$, the closure must contain all $U_i$'s, hence is equal to $X$. 164 | \end{proof} 165 | \begin{proof} 166 | Let $X$ have a finite open cover $U_i = \Spec A_i$ for $A_i$ noetherian domains. Then, $X \times \mathbf{P}^n$ has cover $\Spec R_{ij} \coloneqq \Spec A_i[x_0/x_j,\ldots,x_n/x_j]$ by construction in Thm.~3.3 and Prop.~$2.5(b)$, and so is noetherian since the $R_{ij}$ are noetherian by the Hilbert basis theorem \cite[Thm.~7.5]{AM69}. To show integrality, it suffices to show $X \times \mathbf{P}^n$ is reduced and irreducible by Prop.~3.1. $X \times \mathbf{P}^n$ is reduced since any local ring $\OO_\mathfrak{p}$ is a localization of $R_{ij}$, hence has zero nilradical by \cite[Cor.~3.12]{AM69}. $X \times \mathbf{P}^n$ is irreducible since it has a cover $R_{ij}$ of irreducible open sets (these rings are domains since $A_i$ is a domain \cite[Prop.~7.2.4]{DF04}) that pairwise intersect (their projections in $X$, $\mathbf{P}^n$ pairwise intersect). To show $X \times \mathbf{P}^n$ is separated, $X \times \mathbf{P}^n \to \Spec \mathbf{Z} \times \mathbf{P}^n = \mathbf{P}^n$ is a base change of a separated morphism hence is separated by Cor.~$4.6(c)$, and since $\mathbf{P}^n \to \Spec \mathbf{Z}$ is separated by Thm.~4.9, the composition of these two morphisms $X \times \mathbf{P}^n \to \Spec \mathbf{Z}$ is separated by Cor.~$4.6(b)$. 167 | \par To show $X \times \mathbf{P}^n$ is reduced in codimension one, we note that every local ring $\OO_\mathfrak{p}$ of dimension one is a localization of an $R_{ij}$ at a prime $\mathfrak{p}$ of height one. There are two types of height one primes $\mathfrak{p}$: Type 1 is such that $\Ht \mathfrak{p} \cap A_i = 1$, and Type 2 is such that $\mathfrak{p} \cap A_i = 0$. Note these correspond to Types 1 and 2 in the proof of Prop.~6.6. 168 | \par Suppose $\mathfrak{p}$ is of Type $1$, with $(p) = \mathfrak{p} \cap A_i$ by \cite[Prop.~9.2]{AM69}. We claim $\mathfrak{p} \cap \mathbf{Z}[x_0/x_j,\ldots,x_n/x_j] \subset \mathbf{Z}$. For, suppose not, and it contains non-constant $f \in \mathbf{Z}[x_0/x_j,\ldots,x_n/x_j]$. Then, $f \in \mathfrak{p}$, and so $\mathfrak{p} \supset (p,f) \supsetneq (p)$, i.e., $\Ht \mathfrak{p} > 1$, a contradiction. Thus, $\OO_{\mathfrak{p}} = \OO_{X,\mathfrak{p} \cap A_i}(x_0/x_j,\ldots,x_n/x_j)$, which is a DVR since $\OO_{X,\mathfrak{p} \cap A_i}$ is by extending the valuation on $\OO_{X,\mathfrak{p} \cap A_i}$ by $0$ for all $x_k/x_j$, hence is a regular local ring by \cite[Prop.~9.2]{AM69}. 169 | \par Suppose $\mathfrak{p}$ is of Type 2; then, letting $K = \Frac(A_i)$, $\OO_{\mathfrak{p}}$ is a localization of $S \coloneqq K[x_0/x_j,\ldots,x_n/x_j]$ at a height one prime $\mathfrak{q}$. But $S$ is a UFD, hence every height one prime is principal by I, Prop.~1.12A, and so the maximal ideal of $S$ is principal, hence $\OO_{\mathfrak{p}}$ is a regular local ring by \cite[Prop.~9.2]{AM69}. Thus we have shown $X \times \mathbf{P}^n$ satisfies $(*)$. 170 | \par It now remains to show the isomorphism $\Cl(X \times \mathbf{P}^n) \cong (\Cl X) \times \mathbf{Z}$. First, letting $Z = X \times V(x_0)$, which is of codimension one since $V(x_0)$ is of codimension one in $\mathbf{P}^n$ and is irreducible by Lemma \ref{irredlem} since it is covered by $\Spec A_i[x_1/x_j,\ldots,x_n/x_j]$ for $j\ne0$ which are irreducible and pairwise intersect (by looking at their projections in $X$, $V(x_0)$), we have $X \times \mathbf{P}^n \setminus Z = X \times D_+(x_0) \cong X \times \mathbf{A}^n$, and so we have the exact sequence 171 | \begin{equation*} 172 | \mathbf{Z} \overset{\iota}{\longrightarrow} \Cl(X \times \mathbf{P}^n) \overset{j}{\longrightarrow} \Cl(X) \longrightarrow 0 173 | \end{equation*} 174 | where $j$ is the composition of $\Cl(X \times \mathbf{P}^n) \to \Cl(X \times \mathbf{A}^n)$ and the isomorphism $\Cl(X \times \mathbf{A}^n) \isoto \Cl(X)$ which exists by applying Prop.~6.6 and the noting that $\mathbf{A}^n = \Spec \mathbf{Z}[x_1,\ldots,x_n] \cong \Spec (\mathbf{Z}[x] \otimes \cdots \otimes \mathbf{Z}[x]) = \mathbf{A}^1 \times \cdots \times \mathbf{A}^1$ by Thm.~3.3. We first show the map $\iota\colon d \mapsto d \cdot Z$ is injective. So suppose $d \cdot Z$ is principal for some $d \in \mathbf{Z}$; then, letting $K = K(X) = \Frac(A_i)$ as above, there exists $f \in K(X \times \mathbf{P}^n)^\times$ such that $v_Z(f) = d$ and $v_Y(f) = 0$ for all other prime divisors $Y$. This means $f$ is equal (up to unit factor in $\OO_{\eta(Z)}$, where $\eta(Z)$ is the generic point at $Z$) to 175 | \begin{equation*} 176 | f(x_0,\ldots,x_n) = \frac{x_0^dg(x_0,\ldots,x_n)}{h(x_0,\ldots,x_n)}, 177 | \end{equation*} 178 | where $g,h$ are homogeneous polynomials in $K[x_0,\ldots,x_n]$ such that $x_0 \nmid g,h$ and such that they share no common factors, and moreover $d + \deg g = \deg h$. $g,h \in K$, since any irreducible factor of either defines a prime divisor on which $f$ would have nonzero valuation. But this contradicts that $d + \deg g = \deg h$. This means that we have the short exact sequence 179 | \begin{equation*} 180 | 0 \longrightarrow \mathbf{Z} \overset{\iota}{\longrightarrow} \Cl(X \times \mathbf{P}^n) \overset{j}{\longrightarrow} \Cl(X) \longrightarrow 0 181 | \end{equation*} 182 | and to show the claim, it suffices to show this sequence splits; by the splitting lemma \cite[Prop.-Def.~2.10]{Rei95}, it suffices to show that $j$ has a section. Let $s \colon \Cl(X) \to \Cl(X \times \mathbf{P}^n)$ be defined by $\sum n_iY_i \mapsto \sum n_i p_1^{-1}(Y_i)$. Then, composing with $j$ gives $\sum n_i p_1^{-1}(Y_i) \mapsto \sum n_i (p_1^{-1}(Y_i) \cap (X \times \mathbf{A}^n)) \mapsto \sum n_iY_i$ since $p_1^{-1}(Y_i) \cap (X \times \mathbf{A}^n)$ is the preimage of $Y_i$ through the projection $X \times \mathbf{A}^n \to X$. Thus, the sequence splits and we have $\Cl(X \times \mathbf{P}^n) \cong (\Cl X) \times \mathbf{Z}$. 183 | \end{proof} 184 | 185 | \begin{problem} 186 | \emph{Varieties in Projective Space.} Let $k$ be an algebraically closed field, and let $X$ be a closed subvariety of $\mathbf{P}^n_k$ which is nonsingular in codimension one (hence satisfies $(*)$). For any divisor $D = \sum n_iY_i$ on $X$, we define the \emph{degree} of $D$ to be $\sum n_i\deg Y_i$, where $\deg Y_i$ is the degree of $Y_i$, considered as a projective variety itself \emph{(I,\S7)}. 187 | \begin{enuma} 188 | \item Let $V$ be an irreducible hypersurface in $\mathbf{P}^n$ which does not contain $X$, and let $Y_i$ be the irreducible components of $V \cap X$. They all have codimension $1$ by \emph{(I, Ex.~1.8)}. For each $i$, let $f_i$ be a local equation for $V$ on some open set $U_i$ of $\mathbf{P}^n$ for which $Y_i \cap U_i \ne \emptyset$, and let $n_i = v_{Y_i}(\bar{f_i})$, where $\bar{f_i}$ is the restriction of $f_i$ to $U_i \cap X$. Then we define the \emph{divisor} $V.X$ to be $\sum n_iY_i$. Extend by linearity, and show that this gives a well-defined homomorphism from the subgroup of $\Div \mathbf{P}^n$ consisting of divisors, none of whose components contain $X$, to $\Div X$. 189 | \item If $D$ is a principal divisor on $\mathbf{P}^n$, for which $D.X$ is defined as in $(a)$, show that $D.X$ is principal on $X$. Thus we get a homomorphism $\Cl\mathbf{P}^n \to \Cl X$. 190 | \item Show that the integer $n_i$ defined in $(a)$ is the same as the intersection multiplicity $i(X,V;Y_i)$ defined in \emph{(I,\S7)}. Then use the generalized B\'ezout theorem \emph{(I,7.7)} to show that for any divisor $D$ on $\mathbf{P}^n$, none of whose components contain $X$, 191 | \begin{equation*} 192 | \deg(D.X) = (\deg D)\cdot(\deg X). 193 | \end{equation*} 194 | \item If $D$ is a principal divisor on $X$, show that there is a rational function $f$ on $\mathbf{P}^n$ such that $D = (f).X$. Conclude that $\deg D = 0$. Thus the degree function defines a homomorphism $\deg\colon \Cl X \to \mathbf{Z}$ (This gives another proof of $(6.10)$, since any complete nonsingular curve is projective.) Finally, there is a commutative diagram 195 | \begin{equation*} 196 | \begin{tikzcd} 197 | \Cl \mathbf{P}^n \rar \dar{\deg}[swap]{\cong} & \Cl X \dar{\deg}\\ 198 | \mathbf{Z} \rar{\cdot(\deg X)} & \mathbf{Z}\mathrlap{,} 199 | \end{tikzcd} 200 | \end{equation*} 201 | and in particular, we see that the map $\Cl\mathbf{P}^n \to \Cl X$ is injective. 202 | \end{enuma} 203 | \end{problem} 204 | \begin{proof}[Proof of $(a)$] 205 | First, the numbers $n_i$ are well-defined since if we have different local equations $\overline{f_i}$ for $Y_i$, they must match on intersections up to units in $\OO_X(U_i \cap U_j)$, but the generic point of $Y_i$ will be contained in all $U_i \subset \mathbf{P}^n$ for which $Y_i \cap U_i \ne \emptyset$, hence the valuation of $\overline{f_i}$ will not depend on choice of $\overline{f_i}$. The map $V \mapsto V.X$ for $V \not\supset X$ is then clearly a well-defined homomorphism by extending linearly. 206 | \end{proof} 207 | \begin{proof}[Proof of $(b)$] 208 | Since $X$ is a closed subvariety of $\mathbf{P}^n_k = \Proj k[x_0,\ldots,x_n] \eqqcolon \Proj S$ satisfying $(*)$, it is of the form $V(I)$ for homogeneous prime ideal $I \subset S$. Now let $D = (f)$ for $f \in S_{(0)}^\times$; using the fact that $S$ is a UFD, $f$ is (up to unit in $S_{(I)}$) of the form 209 | \begin{equation}\label{princdiveq} 210 | f(x_0,\ldots,x_n) = \frac{g_1^{d_1} \cdots g_r^{d_r}}{h_1^{e_1} \cdots h_s^{e_s}}, 211 | \end{equation} 212 | where $g_p,h_q$ are homogeneous and irreducible polynomials in $k[x_0,\ldots,x_n]$, there are no shared common factors between the $g_p$ and $h_q$, and we have 213 | \begin{equation*} 214 | \sum_{p=1}^r d_p\deg g_p = \sum_{q=1}^s e_q \deg h_q. 215 | \end{equation*} 216 | Note this means that 217 | \begin{equation*} 218 | D = \sum_{p=1}^r d_p V(g_p) - \sum_{q=1}^s e_q V(h_q). 219 | \end{equation*} 220 | Now since $D.X$ is defined as in $(a)$, we know none of the prime divisors $V(g_p)$ or $V(h_q)$ contain $X$; thus, none of the $g_p$ or $h_q$ are in $I$. Then, the image $\bar{f}$ of $f$ in $K(X)^\times$ is still of the form \eqref{princdiveq}, except replacing $g_p$, $h_q$ with their images in $k[x_0,\ldots,x_n]/I$. Let $\bar{g}_p = \prod g'_{p\alpha}$ and $\bar{h}_q = \prod h'_{q\beta}$ be irreducible decompositions in $k[x_0,\ldots,x_n]/I$. Then, $D.X = (\bar{f})$ since 221 | \begin{equation*} 222 | D.X = \sum_{p=1}^r d_p \sum_\alpha V(g'_{p\alpha}) - \sum_{q=1}^s e_q \sum_\beta V(h'_{q\beta}), 223 | \end{equation*} 224 | and so $D.X$ is principal on $X$. We therefore get a homomorphism $\Cl_X \mathbf{P}^n \to \Cl X$ by the universal property of quotients, where $\Cl_X \mathbf{P}^n$ is the quotient of the subgroup of $\Div \mathbf{P}^n$ consisting of divisors none of whose components contain $X$ by the subgroup consisting of principal divisors in that subgroup. 225 | \par To show this extends to a homomorphism $\Cl \mathbf{P}^n \to \Cl X$, it suffices to define a map $\Cl \mathbf{P}^n \to \Cl_X \mathbf{P}^n$. Let $D \in \Cl \mathbf{P}^n$. Picking a hyperplane $H$ containing a point $p \notin X$, $D \sim \deg D \cdot H$ by Prop.~$6.4(a)$, and so the map mapping $H$ to itself in $\Cl_X \mathbf{P}^n$ defines a homomorphism $\Cl \mathbf{P}^n \to \Cl_X \mathbf{P}^n$ by Prop.~$6.4(c)$ and extending linearly. 226 | \end{proof} 227 | \begin{proof}[Proof of $(c)$] 228 | Let $v_{Y_i}(\bar{f}_i) = \ell$, i.e., if $Y_i \cap U_i = V(g_i) \subset X \cap U_i$, then $\bar{f}_i = g_i^\ell$ up to a unit in $\OO_{\eta(Y_i)} = \OO_{(g_i)}$; this last equality follows since we can first localize to a $U_i$ and then further to the local ring at $(g_i)$. Let $(g_i)_{(g_i)} = \mathfrak{m}$ denote the maximal ideal of this local ring. Then, we have the filtration 229 | \begin{equation}\label{filtr} 230 | 0 \subset \mathfrak{m}^\ell\OO_{(g_i)}/(I_X + I_V) \subset \mathfrak{m}^{\ell-1}\OO_{(g_i)}/(I_X + I_V) \subset \cdots \subset \OO_{(g_i)}/(I_X + I_V) 231 | \end{equation} 232 | of length $\ell$, where the ideals we quotient out by are extended in $\OO_{(g_i)}$. Then, the quotient of adjacent terms is $\mathfrak{m}^{\ell-1}/\mathfrak{m}^\ell \cong \kappa(g_i)$ by \cite[Prop.~$2.1i$]{AM69}, which are simple as $\OO_{(g_i)}$-modules. We claim $\ell = i(X,V;Y_i)$. By I, Prop.~7.4 $S/(I_X + I_V)$ has a filtration $0 = M^0 \subset M^1 \subset \cdots \subset M^r = M$, where $M^j/M^{j-1} \cong (S/\mathfrak{p}_j)(l_j)$, for $l_j \in \mathbf{Z}$ and $\mathfrak{p}_j \subset S$ a homogeneous prime ideal; moreover, if $\mathfrak{p}$ is a minimal prime of $M$, then the number of times $\mathfrak{p}$ occurs in the set $\{\mathfrak{p}_j\}$ is equal to the length of $M_\mathfrak{p}$ over $S_\mathfrak{p}$, i.e., $\mu_{\mathfrak{p}}(M)$ times. Letting $\mathfrak{p}_i$ be the ideal defining $Y_i$, we see that $\mathfrak{p}_i$ is a minimal prime of $M$, and thus appears $i(X,V;Y_i) = \mu_{\mathfrak{p}_i}(M)$ times in the set $\{\mathfrak{p}_j\}$ by the above. But restricting to $U_i$ and then to the ideal $\mathfrak{p}_i$ (now defined by $g_i$) gives a filtration of length $i(X,V;Y_i)$. Thus, $\ell = v_{Y_i}(\bar{f}_i) \le i(X,V;Y_i)$. But $i(X,V;Y_i) \le \ell$ since the $\mathfrak{m}^{\ell-1}/\mathfrak{m}^\ell$ are simple and then by \cite[Prop.~6.7]{AM69}. 233 | \par Finally, for any divisor $D$ on $\mathbf{P}^n$, none of whose components contain $X$, by I, Thm.~7.7 we have 234 | \begin{equation*} 235 | (\deg D) \cdot (\deg X) = \sum_i i(X,D;Y_i) \cdot \deg Y_i = \sum_i n_i \deg Y_i = \deg(D.X).\qedhere 236 | \end{equation*} 237 | \end{proof} 238 | \begin{proof}[Proof of $(d)$] 239 | Suppose $D = (g)$ on $X$, and so $g = p/q \in K(X)^\times = (S/I_X)_{(0)}^\times$ up to unit in $\OO_{\eta(D)}$, where $p,q \in S/I_X$ are of the same degree and share no common factors. Then, we can lift $p,q$ to $S$, giving a lift of $g$ in $K(\mathbf{P}^n) = S_{(0)}$. This lift $f$ is such that $D = (f).X$, for restricting the equation $f$ back to $K(X)$ gives $g$. By $(c)$, $\deg D = \deg ((f).X) = \deg (f) \cdot \deg X = 0$ by Prop.~$6.4(b)$. Thus, we have a homomorphism $\deg\colon \Cl X \to \mathbf{Z}$ by the universal property of the quotient from the degree homomorphism $\Div X \to \mathbf{Z}$ we have already from the problem statement. 240 | \par Finally, we have the commutative diagram given where the top horizontal map is given by $D \mapsto dH \mapsto dH.X$ for $d = \deg D$ by the equation in $(c)$. $\Cl\mathbf{P}^n \to \Cl X$ is injective since $\deg X \ne 0$ in the definition of the bottom horizontal map, and so the composition in either direction around the square is injective. 241 | \end{proof} 242 | 243 | \begin{problem} 244 | \emph{Cones.} In this exercise we compare the class group of a projective variety $V$ to the class group of its cone \emph{(I, Ex.~2.10)}. So let $V$ be a projective variety in $\mathbf{P}^n$, which is of dimension $\ge 1$ and nonsingular in codimension $1$. Let $X = C(V)$ be the affine cone over $V$ in $\mathbf{A}^{n+1}$, and let $\bar{X}$ be its projective closure in $\mathbf{P}^{n+1}$. Let $P \in X$ be the vertex of the cone. 245 | \begin{enuma} 246 | \item Let $\pi\colon\bar{X} - P \to V$ be the projection map. Show that $V$ can be covered by open subsets $U_i$ such that $\pi^{-1}(U_i) \cong U_i \times \mathbf{A}^1$ for each $i$, and then show as in $(6.6)$ that $\pi^*\colon\Cl V \to \Cl(\bar{X} - P)$ is an isomorphism. Since $\Cl \bar{X} \cong \Cl(\bar{X} - P)$, we have also $\Cl V \cong \Cl \bar{X}$. 247 | \item We have $V \subseteq \bar{X}$ as the hyperplane section at infinity. Show that the class of the divisor $V$ in $\Cl \bar{X}$ is equal to $\pi^*(\text{class of}~V.H)$ where $H$ is any hyperplane of $\mathbf{P}^n$ not containing $V$. Thus conclude using $(6.5)$ that there is an exact sequence 248 | \begin{equation*} 249 | 0 \to \mathbf{Z} \to \Cl V \to \Cl X \to 0, 250 | \end{equation*} 251 | where the first arrow sends $1 \mapsto V.H$, and the second is $\pi^*$ followed by the restriction to $X - P$ and inclusion in $X$. (The injectivity of the first arrow follows from the previous exercise.) 252 | \item Let $S(V)$ be the homogeneous coordinate ring of $V$ (which is also the affine coordinate ring of $X$). Show that $S(V)$ is a unique factorization domain if and only if $(1)$ $V$ is projectively normal \emph{(Ex.~5.14)}, and $(2)$ $\Cl V \cong \mathbf{Z}$ and is generated by the class of $V.H$. 253 | \item Let $\OO_P$ be the local ring of $P$ on $X$. Show that the natural restriction map induces an isomorphism $\Cl X \to \Cl(\Spec \OO_P)$. 254 | \end{enuma} 255 | \end{problem} 256 | \begin{proof}[Proof of $(a)$] 257 | Let $S$ be the homogeneous coordinate ring of $V$ as in Problem $5.14$; it is generated by the images of $x_i \in k[x_0,\ldots,x_n]$ modulo the homogeneous ideal defining $V$, and so $V$ is covered by $U_i \coloneqq D_+(x_i)$. Then, $\bar{X} = \Proj S[t]$. Consider the maps $S_{(x_i)} \hookrightarrow S[t]_{(x_i)}$; these define a map $\psi_i \colon D_+(x_i) \to U_i \subset V$ where $D_+ \subset \bar{X}$ by Prop.~$2.3(b)$ and using Prop.~$2.5(b)$. These maps clearly glue together by the universal property for localization \cite[Prop.~3.1]{AM69} since the maps on $U_i \cap U_j$ arise from the dashed map in the diagram 258 | \begin{equation*} 259 | \begin{tikzcd}[row sep=scriptsize] 260 | S_{(x_i)} \rar\dar & S[t]_{(x_i)}\dar\\ 261 | S_{(x_ix_j)}\rar[dashed] & S[t]_{(x_ix_j)}\\ 262 | S_{(x_j)} \rar\uar & S[t]_{(x_j)}\uar 263 | \end{tikzcd} 264 | \end{equation*} 265 | to give a map $\pi\colon\bar{X} - P \to V$ since $\bar{X} \setminus \bigcup D_+(x_i) = \bigcap V(x_i) = V(x_0,\ldots,x_n) = P$. By construction, $D_+(x_i) \subset \pi^{-1}(U_i)$. In the other direction, suppose $\pi(Q) \in U_i$ but $Q \notin D_+(x_i)$, i.e., $Q \ni x_i$; then $Q \in D_+(x_j)$ for some $j$, and so $\pi(Q) = \varphi_j^{-1}(Q) \ni x_i/x_j$, and so $\pi(Q) \notin U_i$, a contradiction. So $\pi^{-1}(U_i) = D_+(x_i) = \Spec S[t]_{(x_i)}$. We claim the map obtained by letting 266 | \begin{equation*} 267 | S[t]_{(x_i)} \to S_{(x_i)}[t], \qquad \frac{ft^e}{x_i^{d+e}} \mapsto \frac{f}{x_i^d} t^e 268 | \end{equation*} 269 | is an isomorphism. It is clearly well-defined since other representatives 270 | \begin{equation*} 271 | \frac{x_i^cft^e}{x_i^{c+d+e}} \mapsto \frac{x_i^cf}{x_i^{c+d}} t^e = \frac{f}{x_i^d} t. 272 | \end{equation*} 273 | It is clearly unital, respects addition by definition since we extended linearly, and is multiplicative since 274 | \begin{equation*} 275 | \frac{ft^e}{x^{d+e}} \cdot \frac{f't^{e'}}{x^{d'+e'}} = \frac{ff't^{e+e'}}{x^{d+d'+e+e'}} \mapsto \frac{ff'}{x^{d+d'}}t^{e+e'} = \frac{f}{x^d}t^e \cdot \frac{f'}{x^{d'}}t^{e'}, 276 | \end{equation*} 277 | hence is a morphism. It is clearly surjective since any element of $S_{(x_i)}[t]$ can be written as the sum of elements of the form on the right, and is injective since a polynomial in $t$ is zero if and only if all of its coefficients are zero, which implies $f=0$. Thus, applying Problem $2.4$, we have $\pi^{-1}(U_i) \cong U_i \times \mathbf{A}^1$ as claimed, since $U_i \times \mathbf{A}^1 = \Spec S_{(x_i)} \times \Spec \mathbf{Z}[t] \cong \Spec S_{(x_i)} \otimes \mathbf{Z}[t] \cong \Spec S_{(x_i)}[t]$ by Thm.~3.3. Note in particular this implies $U_i \times \mathbf{A}^1$ are affine open sets in $\bar{X} - P$. 278 | \par Now to show $\Cl V \cong \Cl(\bar{X} - P)$, we define $\pi^*\colon\Cl V \to \Cl(\bar{X} - P)$ as in Prop.~6.6 to be $\sum n_iY_i \mapsto \sum n_i\pi^{-1}(Y_i)$ on divisors. We note that if $K = K(V)$, then $K(\bar{X} - P) = \Frac(S_{(x_i)}[t]) = \Frac(S_{(x_i)})(t) = K(t)$ using the fact that the $U_i \times \mathbf{A}^1$ are affine opens in $\bar{X} - P$, and the isomorphism above. Note also that the same argument implies $K(\bar{X} - P) = K(\bar{X}) = S[t]_{(0)}$ since $U_i \times \mathbf{A}^1$ are also open affines in $S_{(0)}$. So, if $f \in K^\times$, then $\pi^*((f))$ is the divisor of $f$ considered as an element of $K(t)$. Thus we have a homomorphism $\Cl V \to \Cl(\bar{X} - P)$. 279 | \par To show $\pi^*$ is an isomorphism, we first note that as in the proof of Prop.~6.6, codimension one points in $\bar{X} - P$ have image in $V$ of either codimension zero (type 1) or one (type 2). To show that $\pi^*$ is injective, suppose $\pi^*D = (f)$ for some $f \in K(t)^\times$. $\pi^*D$ then involves prime divisors of type 1, and so $f \in K$. Otherwise we could write $f = g/h \in S[t]_{(0)}$ with $g,h \in S[t]$ relatively prime (up to unit factor in $\OO_{\eta(\pi^*D)}$), and of $g,h$ are not both in $S$, then $(f)$ will involve some prime divisor of type 2 on $\bar{X} - P$. Now if $f \in K$, then $D = (f)$, so $\pi^*$ is injective. To show that $\pi^*$ is surjective, it suffices as in the proof of Prop.~6.6 that any prime divisor of type 2 on $\bar{X} - P$ is linearly equivalent to a linear combination of prime divisors of type $1$. Let $\eta$ be the generic point for this divisor; it is then contained in some $U_i \times \mathbf{A}^1$ by the above, hence $\pi(\eta) \in U_i$. Localizing at the generic point of $V$, this gives a prime divisor in $\Spec K[t]$, corresponding to a prime ideal $\mathfrak{p} \subset K[t]$. This is principal by I, Prop.~1.13 since $K[t]$ is a UFD, so let $f$ be a generator. Then $f \in K(t)$, and $(f)$ consists of $Z$ plus perhaps something purely of type 1: it can't involve any other prime divisors of type 2 since otherwise $Z$ would not be irreducible. Thus $Z$ is linearly equivalent to a divisor purely of type 1. 280 | \par Finally, since Prop.~$6.5(b)$ implies $\Cl \bar{X} \cong \Cl(\bar{X} - P)$, we have $\Cl V \cong \Cl \bar{X}$. 281 | \end{proof} 282 | \begin{proof}[Proof of $(b)$] 283 | $H_\ell$ is given by a linear equation $\ell$ in $\mathbf{P}^n$ and so the same equation in $\ell$ gives $\pi^*(H)$ in $\mathbf{P}^{n+1}$. The hyperplane at infinity $H_\infty$ is given by $t$ in $\mathbf{P}^{n+1}$ such that $H_\infty \cap \bar{X} = V$. Thus, $\bar{\ell} \in S$ and $\bar{t} \in S[t]$ define $V.H$ and $V$ respectively, and $\pi^*(V.H)$ is defined by $\bar{\ell}$ as an element of $S[t]$. Thus, letting $f = \bar{\ell} - \bar{t}$, we have that $\pi^*(V.H) - V = (f)$. 284 | \par Now using Prop.~$6.5(c)$ with $Z = V \subset \bar{X}$, we claim we have the exact sequences 285 | \begin{equation*} 286 | \begin{tikzcd} 287 | 0 \rar & \mathbf{Z} \rar\dar[equal] & \Cl \bar{X} \rar\dar{\pi^*} & \Cl X \rar\dar[equal] & 0\\ 288 | 0 \rar & \mathbf{Z} \rar & \Cl V \rar & \Cl X \rar & 0 289 | \end{tikzcd} 290 | \end{equation*} 291 | In the first row, the first map $\mathbf{Z} \to \Cl \bar{X}$ is the composition $\mathbf{Z} \isoto \Cl \mathbf{P}^{n-1} \to \Cl \bar{X}$ mapping $1 \mapsto H \mapsto V.H$, which is injective by Problem $6.2(d)$. Thus, the bottom row is exact also since $\pi^*$ is an isomorphism by $(a)$. 292 | \end{proof} 293 | \begin{proof}[Proof of $(c)$] 294 | Consider the exact sequence in $(b)$ for $X = \Spec S(V)$ the affine cone over $V$. $S(V)$ is a unique factorization domain if and only if $S(V)$ is integrally closed (i.e., $V$ is projectively normal by Problem $5.14$) and $\Cl S(V) = 0$ by Prop.~6.2. But then, $\Cl S(V) = 0 \iff \Cl V \cong \mathbf{Z}$ by $(b)$, and so we are done. 295 | \end{proof} 296 | \begin{proof}[Proof of $(d)$] 297 | $X = \Spec S$, hence $\OO_P = S_P$. The localization map $S \to \OO_P$ then induces a map of spectra $\Spec \OO_P \to \Spec S$ by Prop.~$2.3(b)$. Recall that $\Spec \OO_P$ (as a set) consists of all prime ideals of $S$ that are contained in $P$ \cite[Cor.~3.13]{AM69}, and that the map $\Spec \OO_P \to \Spec S$ by construction is exactly this one-to-one correspondence. 298 | \par We claim that the group homomorphism obtained by extending linearly the restriction of this correspondence to height one primes induces an isomorphism $\rho\colon \Cl \Spec \OO_P \isoto \Cl X$. Note that any principal divisor $(f) \in \Cl \Spec \OO_P$ with $f \in \Frac(\OO_P)^\times$ in fact defines a divisor on $\Cl X$, since $K(X) = \Frac(\OO_P)$ by Problem $3.6$. Thus, we have a homomorphism $\rho$ by the universal property of the quotient. 299 | \par Now, we want to show $\rho$ is injective and surjective. It is clearly injective since if $D \in \Cl \Spec \OO_P$ is such that $\rho(D) = (f)$ for some $f \in \Frac(S)^\times$, then the equality $K(X) = \Frac(\OO_P)$ from above implies that in fact $D = (f)$. To show surjectivity, it suffices to show that any divisor in $\Cl X$ is linearly equivalent to one defined by a prime $\mathfrak{q}$ contained in the prime ideal $\mathfrak{p}$ associated to $P$. 300 | \end{proof} 301 | 302 | \begin{problem} 303 | Let $k$ be a field of characteristic $\ne 2$. Let $f \in k[x_1,\ldots,x_n]$ be a \emph{square-free} nonconstant polynomial, i.e., in the unique factorization of $f$ into irreducible polynomials, there are no repeated factors. Let $A = k[x_1,\ldots,x_n,z](z^2-f)$. Show that $A$ is an integrally closed ring. 304 | \end{problem} 305 | \begin{proof} 306 | We first compute the quotient field $K$ of $A$. Clearly $K \supset k(x_1,\ldots,x_n)[z]/(z^2-f)$; we claim these are equal. For, suppose we have an element $1/(g+zh) \in K$; then, we can write 307 | \begin{equation*} 308 | \frac{1}{g+zh}\frac{g-zh}{g-zh} = \frac{g-zh}{g^2-fh^2}, 309 | \end{equation*} 310 | where we use $z^2 - f$; hence, every element in $K$ can be written in the form $g' + zh'$, where $g',h' \in k(x_1,\ldots,x_n)$. Hence $K = k(x_1,\ldots,x_n)[z]/(z^2-f)$, and this is a degree $2$ extension, thus Galois with automorphism $z \mapsto -z$. Let $\alpha = g + hz \in K$, where $g,h \in k(x_1,\ldots,x_n)$. Then, $\alpha$ has minimal polynomial $X^2 - 2gX + (g^2-h^2f)$, for 311 | \begin{equation*} 312 | (g+hz)^2 - 2g(g+hz) + (g^2 - h^2f) = g^2 + 2ghz + h^2z^2 - 2g^2 - 2ghz + g^2 - h^2f = 0. 313 | \end{equation*} 314 | Thus, $\alpha$ is integral over $k[x_1,\ldots,x_n]$ if and only if $2g,g^2-h^2f \in k[x_1,\ldots,x_n]$ if and only if $g, h^2f \in k[x_1,\ldots,x_n]$. Now if $\alpha$ is integral, and $h$ had nontrivial denominator, then $h^2 \notin k[x_1,\ldots,x_n]$ since $f$ is square-free; hence $h \in k[x_1,\ldots,x_n]$ so $\alpha \in A$. Conversely, if $\alpha \in A$, then $g,h^2f \in k[x_1,\ldots,x_n]$, hence $\alpha$ is integral over $k[x_1,\ldots,x_n]$. Thus $A$ is the integral closure of $k[x_1,\ldots,x_n]$ in $K$. 315 | \end{proof} 316 | 317 | \begin{problem} 318 | \emph{Quadric Hypersurfaces.} Let $\Char k \ne 2$, and let $X$ be the affine quadric hypersurface $\Spec k[x_0,\ldots,x_n]/(x_0^2 + x_1^2 + \cdots + x_r^2)$---cf.~\emph{(I,Ex.~5.12)}. 319 | \begin{enuma} 320 | \item Show that $X$ is normal if $r \ge 2$ (use \emph{(Ex.~6.4)}). 321 | \item Show by a suitable linear change of coordinates that the equation of $X$ could be written as $x_0x_1 = x_2^2 + \cdots + x_r^2$. Now imitate the method of $(6.5.2)$ to show that: 322 | \begin{enumd} 323 | \item If $r = 2$, then $\Cl X \cong \mathbf{Z}/2\mathbf{Z}$; 324 | \item If $r = 3$, then $\Cl X \cong \mathbf{Z}$ (use $(6.6.1)$ and \emph{(Ex.~6.3)} above); 325 | \item If $r \ge 4$ then $\Cl X = 0$. 326 | \end{enumd} 327 | \item Now let $Q$ be the projective quadric hypersurface in $\mathbf{P}^n$ defined by the same equation. Show that: 328 | \begin{enumd} 329 | \item If $r=2$, $\Cl Q \cong \mathbf{Z}$, and the class of a hyperplane section $Q.H$ is twice the generator; 330 | \item If $r = 3$, $\Cl Q \cong \mathbf{Z} \oplus \mathbf{Z}$; 331 | \item If $r \ge 4$, $\Cl Q \cong \mathbf{Z}$, generated by $Q.H$. 332 | \end{enumd} 333 | \item Prove Klein's theorem, which says that if $r \ge 4$, and if $Y$ is an irreducible subvariety of codimension $1$ on $Q$, then there is an irreducible hypersufrace $V \subseteq \mathbf{P}^n$ such that $V \cap Q = Y$, with multiplicity one. In other words, $Y$ is a complete intersection. 334 | \end{enuma} 335 | \end{problem} 336 | \begin{proof}[Proof of $(a)$] 337 | If $r \ge 2$, then $k[x_0,\ldots,x_n]/(x_0^2 + x_1^2 + \cdots + x_r^2)$ is of the form in Problem $6.4$, hence each $\OO_P$ is integrally closed by \cite[Prop.~5.13]{AM69}. 338 | \end{proof} 339 | \begin{proof}[Proof of $(b)$] 340 | Assume there is a squareroot of $-1$, which we denote by $i$. Then, let 341 | \begin{equation*} 342 | x_0 \mapsto \frac{y_0+y_1}{2}, \quad x_1 \mapsto \frac{y_0-y_1}{2i}, 343 | \end{equation*} 344 | and so 345 | \begin{equation*} 346 | x_0^2 + x_1^2 = 347 | \end{equation*} 348 | \end{proof} 349 | 350 | \printbibliography 351 | \end{document} 352 | -------------------------------------------------------------------------------- /Atiyah-Macdonald/3 Rings and Modules of Fractions/3 Rings and Modules of Fractions.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt,letterpaper]{article} 2 | \usepackage{geometry} 3 | \geometry{letterpaper} 4 | \usepackage{amsmath,amssymb,amsthm,mathrsfs} 5 | \usepackage{mathtools} 6 | \usepackage{ifpdf} 7 | \ifpdf 8 | \setlength{\pdfpagewidth}{8.5in} 9 | \setlength{\pdfpageheight}{11in} 10 | \else 11 | \fi 12 | \usepackage{hyperref} 13 | 14 | \usepackage{tikz} 15 | \usepackage{tikz-cd} 16 | \usetikzlibrary{decorations.markings} 17 | \tikzset{ 18 | open/.style = 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\end{tikzpicture}}\hspace{0.75ex}} 151 | \newcommand{\dashto}[2]{\smash{\hspace{-0.7em}\begin{tikzcd}[column sep=small,ampersand replacement=\&] {#1} \rar[dashed] \& {#2} \end{tikzcd}\hspace{-0.7em}}} 152 | 153 | %\usepackage{todonotes} 154 | %\usepackage[notref,notcite]{showkeys} 155 | 156 | \title{Atiyah-Macdonald Ch.~3 Rings and Modules of Fractions} 157 | \author{Takumi Murayama and Kyu Jun} 158 | 159 | \begin{document} 160 | \maketitle 161 | \setcounter{section}{3} 162 | \begin{problem}\label{exc:3.1} 163 | Let $S$ be a multiplicatively closed subset of a ring $A$, and let $M$ be a finitely generated $A$-module. Prove that $S^{-1}M = 0$ if and only if there exists $s \in S$ such that $sM = 0$. 164 | \end{problem} 165 | \begin{proof} 166 | Assume that $S^{-1}M = 0$, and let $\{x_1, \ldots, x_n\}$ be a set of 167 | generators of $M$, as an $A$-module. Then as $S^{-1}M = 0$, for each $i \in 168 | \{1, \ldots, n\}$, there exists $s_i \in S$ such that $s_i \cdot x_i=0$. Let $s= \prod_{i} s_i$. Then, for any $\sum_{i}a_ix_i \in M$, 169 | \begin{equation*} 170 | s \cdot \sum_{i}a_ix_i = \sum_{i}a_isx_i = 0, 171 | \end{equation*} 172 | hence $sM = 0$. 173 | \par Conversely, if there exists $s \in S$ such that $sM = 0$, then 174 | $S^{-1}M = 0$ since for each generator $x_i$, $sx_i = 0$. 175 | \end{proof} 176 | 177 | \begin{problem} 178 | Let $\mathfrak{a}$ be an ideal of a ring $A$, and let $S = 1 + \mathfrak{a}$. Show that $S^{-1}\mathfrak{a}$ is contained in the Jacobson radical of $S^{-1}A$. 179 | \par Use this result and Nakayama's lemma to give a proof of $(2.5)$ which does not depend on determinants. 180 | \end{problem} 181 | \begin{proof}[Proof of first claim] 182 | Suppose $x \in S^{-1}\mathfrak{a}$; recall that for $x \in 183 | \mathfrak{R}_{S^{-1}A}$ it suffices to show $1-xy$ is a unit in $S^{-1}A$ for 184 | all $y \in S^{-1}A$ by Prop.~$1.9$. Then, $x = m/(1+a)$ for $m,a \in \mathfrak{a}$. Let $y = n/(1+a')$ for $a' \in \mathfrak{a},n \in A$, then 185 | \begin{equation*} 186 | 1 - xy = 1 - \frac{m}{1+a} \cdot \frac{n}{1+a'} = 1 - \frac{mn}{1+a''} = \frac{1 + a''-mn}{1+a''} = \frac{1 + a'''}{1+a''}, 187 | \end{equation*} 188 | for some $a'',a''' \in \mathfrak{a}$, since $S$ is multiplicative and $a''-mn \in \mathfrak{a}$. Thus, we see that $(1+a'')/(1+a''') \in S^{-1}A$ is the inverse of $1-xy$, and so $x \in \mathfrak{R}_{S^{-1}A}$. 189 | \end{proof} 190 | \begin{corollary}[Cor.~$2.5$] 191 | Let $M$ be a finitely generated $A$-module and let $\mathfrak{a}$ be an ideal of $A$ such that $\mathfrak{a}M = M$. Then there exists $x \equiv 1\pmod{\mathfrak{a}}$ such that $xM = 0$. 192 | \end{corollary} 193 | \begin{proof}[Proof of Corollary] 194 | Since $M = \mathfrak{a}M$, we first have that, for any $m \in M$, $m = am' \in 195 | \mathfrak{a}M$ for some $m' \in M$. Then, we have $S^{-1}M = 196 | (S^{-1}\mathfrak{a})(S^{-1}M)$, by having $m/s = (a/1)(m'/s)$ as above. Since 197 | $S^{-1}\mathfrak{a} \subset \mathfrak{R}_{S^{-1}A}$ by the first part of the 198 | problem, we apply Nakayama to get that $S^{-1}M = 0$. By Exercise 199 | \ref{exc:3.1}, we see that there is then $x \in S^{-1}$ such that $xM = 0$, and $x \equiv 1 \pmod{\mathfrak{a}}$ by definition of $S$. 200 | \end{proof} 201 | 202 | \begin{problem} 203 | Let $A$ be a ring, let $S$ and $T$ be two multiplicatively closed subsets of 204 | $A$, and let $U$ be the image of $T$ in $S^{-1}A$. 205 | Show that the rings $(ST)^{-1}A$ and $U^{-1}(S^{-1}A)$ are isomorphic. 206 | \end{problem} 207 | \begin{proof} 208 | Let $g\colon A \to U^{-1}(S^{-1}A) \eqqcolon B$ be the canonical ring 209 | homomorphism $a \mapsto a/1$. 210 | To show $(ST)^{-1}A \cong U^{-1}(S^{-1}A)$, it suffices to show the 211 | properties in Cor.~3.2 hold. 212 | \par Suppose $st \in ST$ where $s\in S,t\in T$. 213 | Then, $g(st) = st/1 = (s/1)(t/1)$ is a unit in $B$ since 214 | $t/1 \in U$ and $s/1$ is a unit in $S^{-1}A$, hence $3.2i)$ holds. 215 | \par Suppose $g(a) = 0$. Then, $a/1 \equiv 0$ and so $(t/1)(a/1) = at/1 = 0$ in 216 | $S^{-1}A$ for some $t/1 \in U$, hence $ast = 0$ for some $st \in ST$, 217 | hence $3.2ii)$ holds. 218 | \par Now suppose we have a unit $(a/s)/(t/1)$ in $B$. We claim $(a/s)/(t/1) = 219 | g(a)g(st)^{-1}$. For, $g(a)g(st)^{-1} = (a/1)(st/1)^{-1} = (a/1)(1/st) = 220 | (a/1)/(st/1)$, which is equivalent to $(a/s)/(t/1)$ since $(a/1)(t/1) = 221 | (a/s)(st/1)$ in $S^{-1}A$. 222 | Thus, $3.2iii)$ holds, and so $(ST)^{-1}A \cong U^{-1}(S^{-1}A)$ by Cor.~3.2. 223 | \end{proof} 224 | 225 | \begin{problem} 226 | Let $f\colon A \to B$ be a homomorphism of rings and let $S$ be a multiplicatively closed subset of $A$. Let $T= f(S)$. Show that $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$ modules. 227 | \end{problem} 228 | \begin{proof} 229 | Let $\phi$ be a natural $S^{-1}A$-module homomorphism from $S^{-1}B$ to $T^{-1}B$ given by $b/s \mapsto b/f(s)$. As $f$ is a homomorphism of rings, $\phi$ is a homomorphism. Let $\psi: T^{-1}B\to S^{-1}B$ such that $\psi(b/f(s)) = b/s$ for each $b \in B, f(s) = T = f(s)$. It is immediate that $\phi$ and $\psi$ are inverses of each other. Hence, $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$ modules. 230 | \end{proof} 231 | 232 | \begin{problem}[AM Exercise 3.5] 233 | Let $A$ be a ring. Suppose that, for each prime ideal $\mathfrak{p}$, the local ring $A_\mathfrak{p}$ has no nilpotent element $\ne 0$. Show that $A$ has no nilpotent element $\ne 0$. If each $A_\mathfrak{p}$ is an integral domain, is $A$ necessarily an integral domain? 234 | \end{problem} 235 | \begin{proof} 236 | Suppose $A$ has a nilpotent element, i.e., $\mathfrak{N} \ne 0$. Then, by AM Corollary $3.12$, the nilradical of $A_\mathfrak{p}$ is $\mathfrak{N}_\mathfrak{p}$. But $\mathfrak{N} \ni x \ne 0 \implies x/1 \in \mathfrak{N}_\mathfrak{p}$, which is a contradiction. 237 | \par Now consider $A = k \times k$; clearly this is not an integral domain, for $(a,0)\cdot(0,b) = 0$. Note $\mathfrak{p} = k \times 0$ and $\mathfrak{q} = 0 \times k$ are the only non-trivial prime ideals of $A$. Now consider $\mathfrak{p}_\mathfrak{p}$; for any $a/b \in \mathfrak{p}_\mathfrak{p}$, we claim $a/b \equiv 0/1 = 0$. For, $a(0,1) = 0$ and $(0,1) \in A \setminus \mathfrak{p}$ and by definition of the equivalence relation. But recall from AM Example $1$ from p.~$38$ that this ideal $0$ is maximal in $A_\mathfrak{p}$, and so $A_\mathfrak{p}$ is a field. Since the same argument applies for $A_\mathfrak{q}$, we see that each $A_\mathfrak{p}$ being an integral domain does not imply $A$ is an integral domain. 238 | \par One another simple example would be $A = \mathbb{Z}/6\mathbb{Z} = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. Prime ideals of $A$ are copies of $\mathbb{Z}/(2)$ and $ \mathbb{Z}/(3)$ (i.e. $\mathfrak{p}_1 = \{0,3\}$ and $ \mathfrak{p}_2 = \{0,2,4\}$) ,and $A_{\mathfrak{p}} = \mathbb{Z}/(3)$ or $\mathbb{Z}/(2)$. Hence, $A_\mathfrak{p}$ is a domain for all of its prime ideals $\mathfrak{p}$, but $A$ itself is obviously not a domain as $2 \cdot 3 = 0$. 239 | \par In general Let $R$ be a ring with two nonzero prime ideals $\mathfrak{p}$ and $\mathfrak{q}$ such that each is not included in the other, and that $\Spec R = \{\mathfrak{p}, \mathfrak{q}\}.$ Now let $A = R/\mathfrak{p}\mathfrak{q}$. Then $A$ is not a domain because the image of any two element $p \in \mathfrak{p}/\mathfrak{q}$ and $q \in \mathfrak{q}/\mathfrak{p}$ are zero divisors of $A$. However, $A_{\mathfrak{p}}$ and $A_{\mathfrak{q}}$ are domains. 240 | \end{proof} 241 | 242 | \begin{problem}[AM Exercise 3.7] 243 | A multiplicatively closed subset $S$ of a ring $A$ is said to be \emph{saturated} if 244 | \begin{equation*} 245 | xy \in S \iff x \in S~\text{and}~y \in S. 246 | \end{equation*} 247 | Prove that 248 | \begin{enumi} 249 | \item $S$ is saturated $\iff A - S$ is a union of prime ideals. 250 | \item If $S$ is any multiplicatively closed subset of $A$, there is a unique smallest saturated multiplicatively closed subset $\overline{S}$ containing $S$, and that $\overline{S}$ is the complement in $A$ of the union of the prime ideals which do not meet $S$. ($\overline{S}$ is called the \emph{saturation} of $S$.) 251 | \end{enumi} 252 | If $S = 1 + \mathfrak{a}$, where $\mathfrak{a}$ is an ideal of $A$, find $\overline{S}$. 253 | \end{problem} 254 | \begin{proof}[Proof $i)$] 255 | Suppose $A \setminus S = \bigcup \mathfrak{p}_i$ for some prime $\{\mathfrak{p}_i\}$. Then, if $xy \in A \setminus S$, we have that $xy \in \mathfrak{p}_i$ for some $i$, and so $x \in \mathfrak{p}_i \subseteq A \setminus S \lor y\in \mathfrak{p}_i \subseteq A \setminus S$, i.e., $x \notin S \lor y \notin S$. Conversely, if $x \notin S \lor y \notin S$, then $x \in \mathfrak{p}_i \lor y\in \mathfrak{p}_i$ for some $i$. But then, since $\mathfrak{p}_i$ is an ideal, $xy \in \mathfrak{p}_i \subseteq A \setminus S$, i.e., $xy \notin S$. 256 | \par Now suppose $S$ is saturated; we want to show that all $x \notin S$ are in some prime ideal disjoint from $S$. We see that $(x) \cap S = \emptyset$, for $S$ is saturated, and so is a proper ideal. Moreover, $(x)^e \subseteq S^{-1}A$ is not $(1)$, for $x/1$ is not a unit in $S^{-1}A$. Then, there exists a maximal ideal $\mathfrak{m} \supseteq (x)^e$ by AM Corollary $1.4$. We see by AM Proposition $3.11iv)$ implies $\mathfrak{m}^c \subseteq A$ is a prime ideal that does not intersect $S$, and $x \in \mathfrak{m}^c$. 257 | \end{proof} 258 | \begin{proof}[Proof of $ii)$] 259 | Let $\mathcal{S}$ be the set of saturated multiplicatively closed subsets of $A$ containing $S$, which is non-empty since it contains $A$. We claim $\overline{S} = \bigcap_{S \in \mathcal{S}} S$ works. Clearly it is unique. Since $A \setminus \overline{S} = A \setminus \bigcap_{T \in \mathcal{S}} S = \bigcup_{T \in \mathcal{S}} A \setminus T = \bigcup_{T \in \mathcal{S}} \bigcup \mathfrak{p}_i$, where the last union is over the prime ideals in the complement of each $S$ shown to exist in part $i)$, and each $\mathfrak{p}_i$ does not intersect $\overline{S}$. Moreover, if there is any $\mathfrak{p}$ that does not intersect $\overline{S}$, $A \setminus \mathfrak{p} \in \mathcal{S}$, and so $\mathfrak{p} \subseteq A \setminus \overline{S}$. 260 | \end{proof} 261 | \begin{proof}[Solution for last statement] 262 | First note $\mathfrak{p} \cap S \ne \emptyset \iff \exists a \in \mathfrak{a}$ such that $1 + a \in \mathfrak{p} \iff 1 \in \mathfrak{p} + \mathfrak{a}$. By part $ii)$, then, $A \setminus \overline{S} = \bigcup \mathfrak{p}$, where the union is taken over $\mathfrak{p}$ such that $1 \notin \mathfrak{p} + \mathfrak{a}$. 263 | \par We claim this equals the union $\bigcup \mathfrak{m}$ taken over maximal ideals $\mathfrak{m} \supseteq \mathfrak{a}$. Clearly, $1 \notin \mathfrak{m} + \mathfrak{a}$ since $\mathfrak{m}$ is proper yet contains $\mathfrak{a}$. Likewise, if $\mathfrak{p}$ is such that $1 \notin \mathfrak{p} + \mathfrak{a}$, we can find a maximal ideal $\mathfrak{m} \supseteq \mathfrak{p} + \mathfrak{a}$ by AM Corollary $1.4$, and we see $1 \notin \mathfrak{m} + \mathfrak{a}$. Thus, $\overline{S} = A \setminus \bigcup_{\mathfrak{m} \supseteq \mathfrak{a}} \mathfrak{m}$. 264 | \end{proof} 265 | 266 | \begin{problem}[AM Exercise 3.8] 267 | Let $S,T$ be multiplicatively closed subsets of $A$, such that $S \subseteq T$. Let $\phi : S^{-1}A \to T^{-1}A$ be the homomorphism which maps each $a/s \in S^{-1}A$ to $a/s$ considered as an element of $T^{-1}A$. Show that the following statements are equivalent: 268 | \begin{enumi} 269 | \item $\phi$ is bijective. 270 | \item For each $t \in T$, $t/1$ is a unit in $S^{-1}A$. 271 | \item For each $t \in T$ there exists $x \in A$ such that $xt \in S$. 272 | \item $T$ is contained in the saturation of $S$ (Exercise $7$). 273 | \item Every prime ideal which meets $T$ also meets $S$. 274 | \end{enumi} 275 | \end{problem} 276 | \begin{proof} 277 | $i) \Rightarrow ii)$. $(t/1)(1/t) = 1$ in $T^{-1}A$. $\phi$ has an inverse $\phi^{-1}$ so $1 = \phi^{-1}(1) = \phi^{-1}(t/1)\phi^{-1}(1/t) = t/1\phi^{-1}(1/t)$ in $S^{-1}A$, i.e., $t/1$ is a unit. 278 | \par $ii) \Rightarrow iii)$. Let $a/b = (t/1)^{-1} \in S^{-1}A$; then, $(a/b)(t/1) = 1$, i.e., $c(at-b) = 0$ for some $c \in S$. Then, $act = bc \in S$, and $ac \in A$; thus, $x = ac$ works. 279 | \par $iii) \Rightarrow iv)$. $t \in T \implies \exists x \in A$ such that $xt \in S \subseteq \overline{S} \implies t \in \overline{S}$. 280 | \par $iv) \Rightarrow v)$. Suppose $\mathfrak{p} \cap S \ne \emptyset$; then, $\mathfrak{p} \cap \overline{S} = \emptyset$ by AM Exercise $3.7ii)$, and so $\mathfrak{p} \cap T = \emptyset$ by $iv)$. This proves the contrapositive. 281 | \par $v) \Rightarrow iii)$. Suppose $iii)$ does not hold. Then, $(t) \cap S = \emptyset$ for some $t \in T$; note that $t$ is not a unit otherwise $(t) = A$. We then see that $(t)^e$ is contained in some maximal ideal $\mathfrak{m}$ of $S^{-1}A$ by AM Corollary $1.4$, and so $\mathfrak{m}^c \subseteq A$ is a prime ideal that does not intersect $S$. But then, $\mathfrak{m}^c \cap S = \emptyset$ while $t \in \mathfrak{m}^c \cap T$, a contradiction. 282 | \par $iii) \Rightarrow ii)$. Consider $t/1 \in S^{-1}A$. Since there exists $x \in A$ such that $xt \in S$, we see that $t/1 = xt/x$, and $x/xt \in S^{-1}A$ is the inverse of $t/1$. 283 | \par $ii) \Rightarrow i)$. Suppose $\phi(a/s) = \phi(a'/s')$ in $T^{-1}A$, i.e., there exists $t \in T$ such that $t(as'-a's) = 0$. Choosing $x \in A$ such that $xt \in S$, which we have since we have already shown $ii) \Rightarrow iii)$, we see that $xt(as'-a's) = 0$, and so $a/s = a'/s'$ in $S^{-1}A$, i.e., $\phi$ is injective. 284 | \par Now let $t \in T$ and let $a/s \in S^{-1}A$ such that $(t/1)(a/s) = 1$, i.e., $x(at-s) = 0$ for some $x \in S$. But since $S \subseteq T$, wesee that this implies $1/t = a/s$ in $T^{-1}A$. This implies $\phi$ is surjective since all elements of the form $1/t$ are in the image of $\phi$. 285 | \end{proof} 286 | 287 | \begin{problem}[AM Exercise 3.9] 288 | The set $S_0$ of all non-zero-divisors in $A$ is a saturated multiplicatively closed subset of $A$. Hence the set $D$ of zero-divisors in $A$ is a union of prime ideals (see Chapter $1$, Exercise $14$). Show that every minimal prime ideal of $A$ is contained in $D$. %Use Exercise 6 289 | \par The ring $S_0^{-1}A$ is called the \emph{total ring of fractions of $A$}. Prove that 290 | \begin{enumi} 291 | \item $S_0$ is the largest multiplicatively closed subset of $A$ for which the homomorphism $A \to S_0^{-1}A$ is injective. 292 | \item Every element in $S_0^{-1}A$ is either a zero-divisor or a unit. 293 | \item Every ring in which every non-unit is a zero-divisor is equal to its total ring of fractions (i.e., $A \to S_0^{-1}A$ is bijective). 294 | \end{enumi} 295 | \end{problem} 296 | \begin{proof}[Proof of first claim] 297 | We first show $S_0$ is saturated. Suppose $xy \in S_0$ but $x \notin S_0 \lor y \notin S_0$. Then, there is $z \ne 0$ such that $xz = 0 \lor yz = 0$, and so $xyz = 0$, so then $xy \notin S_0$. Now if $xy \notin S_0$, then there exists $z \ne 0$ such that $xyz = 0$. If $yz = 0$ then $y \notin S_0$, and if $yz \ne 0$ then $x \notin S_0$. $D = A \setminus S_0$ is then a union of prime ideals by AM Exercise $3.7i)$. 298 | \par We now show that every minimal prime ideal of $A$ is contained in $D$. So suppose not; then, there exists prime and minimal $\mathfrak{p}$ which contains some $x$ which is not a zero-divisor. Consider $A \setminus \mathfrak{p}$, which is a multiplicatively closed subset of $A$. Then, $\{x^iy \mid y \in A \setminus \mathfrak{p}, i \in \mathbb{N}\} \supsetneq A\setminus\mathfrak{p}$, and so we see that $A \setminus \mathfrak{p}$ is not maximal, which implies that $\mathfrak{p}$ is not a minimal prime ideal of $A$ by AM Exercise $3.6$, a contradiction. 299 | \end{proof} 300 | \begin{proof}[Proof of $i)$] 301 | Let $\varphi : A \to S_0^{-1}A$, and $S$ be the largest multiplicatively closed subset of $A$ for which $\varphi$ is injective. Suppose $a/1 = 0/1$ in $S_0^{-1}A$. Then, $sa = 0$ for some $s \in S_0$. But $s \in S_0 \implies a = 0$, and so $\varphi$ is injective. Thus, $S_0 \subseteq S$. 302 | \par Now we show the reverse inclusion by assuming $\varphi$ is injective. $a/1 = 0/1$ in $S^{-1}A$ implies that $sa = 0$ for some $s \in S$. But then, $a=0$ by the injectivity of $\varphi$, and so $s$ is a non-zero-divisor. Thus, $S \subseteq S_0$. 303 | \end{proof} 304 | \begin{proof}[Proof of $ii)$] 305 | $a/s \in S_0^{-1}A \implies a \in S_0 \lor a \notin S_0$. $a \in S_0 \implies (a/s)^{-1} = s/a \in S_0^{-1}A$, i.e., $a/s$ is a unit. $a \notin S_0 \implies \exists x \in A$ such that $ax = 0 \implies (a/s)(x/1) = 0$, i.e., $a/s$ is a zero-divisor. 306 | \end{proof} 307 | \begin{proof}[Proof of $iii)$] 308 | $\varphi : A \hookrightarrow S_0^{-1}A$, and so it suffices to show $\varphi$ is surjective. $S_0^{-1}A \ni a/x = ax^{-1}/xx^{-1} = ax^{-1}/1 = \varphi(ax^{-1})$, where $x^{-1}$ exists since $S_0$ consists of units. 309 | \end{proof} 310 | 311 | \begin{problem}[AM Exercise 3.10] 312 | Let $A$ be a ring. 313 | \begin{enumi} 314 | \item If $A$ is absolutely flat (i.e. every A-module is flat) and $S$ is any multiplicatively closed subset of $A$, then $S^{-1}A$ is absolutely flat. 315 | \item $A$ is absolutely flat $\iff$ $A_\mathfrak{m}$ is a field for each maximal ideal $\mathfrak{m}$. 316 | \end{enumi} 317 | \end{problem} 318 | \begin{proof}[Proof of $i$] We begin by citing a previous result: 319 | \begin{proposition} (AM Exercise 2.27) $A$ is absolutely flat $\iff$ Every principal ideal of $A$ is idempotent. 320 | \end{proposition} 321 | 322 | Assume $A$ is absolutely flat. Then from the proposition above, for every element $a \in A$ $(a)^2 = (a^2) = (a)$. That is, there exists an element $u \in A^{\times}$ such that $a^2 = ua$. 323 | 324 | Now let $\frac{a}{s} \in S^{-1}A$. Then there exists $\alpha, \beta \in A^{\times}$ such that $a^2 = \alpha a$, $s^2 = \beta s$. Then, 325 | $$\Big{(}\frac{a}{s}\Big{)}^2 = \Big{(}\frac{a^2}{s^2}\Big{)} = \Big{(}\frac{\alpha \beta^{-1} a}{s}\Big{)} \subseteq \Big{(}\frac{a}{s}\Big{)}$$ 326 | 327 | We note that $$\frac{\alpha^{-1}s^2}{s} \cdot \frac{\alpha \beta^{-1} a}{s} = \frac{a}{s}$$ Hence,$\Big{(}\frac{a}{s}\Big{)}^2 \supseteq \Big{(}\frac{a}{s}\Big{)}$. Thus, Every principal ideal in $S^{-1}A$ is an idempotent, and again from the proposition, $S^{-1}A$ is absolutely flat. 328 | 329 | \end{proof} 330 | 331 | \begin{proof}[Proof of $ii$] 332 | \end{proof} 333 | 334 | \begin{problem}[AM Exercise 3.11] 335 | Let $A$ be a ring. Prove that TFAE: 336 | \begin{enumi} 337 | \item $A/\mathfrak{N}$ is absolutely flat ($\mathfrak{N}$ being the nilradical of $A$) 338 | \item Every prime ideal of $A$ is maximal. 339 | \item $\Spec (A)$ is a $T_1$-space (i.e. every subset consisting of a single point is closed). 340 | \item $\Spec(A)$ is Hausdorff 341 | \end{enumi} 342 | 343 | \par If these conditions are satisfied, show that $\Spec (A)$ is compact and totally disconnected (i.e. the only connected subsets of $\Spec(A)$ are those consisting of a single point). 344 | \end{problem} 345 | 346 | \begin{proof}[Proof of equivalence] 347 | We will show $iv) \Rightarrow iii) \Rightarrow ii) \Rightarrow i) \Rightarrow iv)$ 348 | 349 | \par $iv) \Rightarrow iii)$: Trivial 350 | \par $iii) \Rightarrow ii)$: Assume that $\Spec(A)$ is a $T_1$-space. Let $\mathfrak{p} \in \Spec(A)$. Then, $V(\mathfrak{p}) = \overline{\{\mathfrak{p}\}} = \{\mathfrak{p}\}$ because $\Spec(A)$ is $T_1$. By definition of $V(\mathfrak{p})$, this means that $\mathfrak{p}$ is maximal. (AM Exercise 1.18). Hence, $iii)$ implies $ii)$. 351 | \par $ii) \Rightarrow i)$: Assume that every prime ideal of $A$ is maximal. Choose an element $x \in A$ and any $\mathfrak{p} \in \Spec(A)$. Then, either $x \in \mathfrak{p}$ or $x \notin \mathfrak{p} \Rightarrow x+ \mathfrak{p} = (1)$ because $\mathfrak{p}$ is maximal. Then, $x -1 \in \mathfrak{p}$. So for all $x \in A$ and $\mathfrak{p} \in \Spec(A)$, $x(x-1) \in \mathfrak{p}$. Therefore, any $\overline{x} \in A/\mathfrak{N}$ is idempotent. Using the proposition from the previous problem, we conclude that $A/\mathfrak{N}$ is absolutely flat. 352 | 353 | \par $i) \Rightarrow iv)$. From AM Exercise 1.21 $iv)$, $\Spec(A)$ and $\Spec(A/\mathfrak{N})$ are naturally homeomorphic. Hence, we are done if we show that if a ring $A$ is absolutely flat, $\Spec(A)$ is Hausdorff. 354 | 355 | If $A$ is absolutely flat, then for any $x \in A$, there exists $u \in A^{\times}$ such that $x^2 = ux$ (again, from the proposition). Then $x(x-u) = 0 \in \mathfrak{p}$ for any prime ideal $\mathfrak{p}$. Thus either $x \in \mathfrak{p}$ or $x-u \in \mathfrak{p}$. If both $x$ and $x-u$ is in $\mathfrak{p}$, then $x-(x-u) = u \in \mathfrak{p}$ so $\mathfrak{p} = (1)$, as $u$ is a unit. Hence, $x$ and $x-u$ cannot be in a same prime ideal, so $\Spec(A) = V(x) \sqcup V(x-u) = (V(x-u))^c \sqcup (V(x))^c = X_{x-u} \sqcup X_x$. Hence, $\Spec(A)$ is Hausdorff. 356 | 357 | \end{proof} 358 | \begin{proof}[Proof of the last claim] 359 | $\Spec A$ is always quasi-compact from AM Exercise 17 $v)$. As $\Spec A$ is Hausdorff by the assumption, it is compact. 360 | \par Assume that $C \subset \Spec(A)$ is a subset of $\Spec(A)$ with more than one element. Let $\mathfrak{p} \neq \mathfrak{q} \in C$. Without loss of generality assume $\mathfrak{p} \neq \emptyset$. Choose $x \in \mathfrak{p}/\mathfrak{q}$. Then we see from the last proof that $\mathfrak{p} \in V(x) = X_{x-u}$, and $\mathfrak{q} \in X_{x}$ where $X_{x-u} \sqcup X_x = \Spec(A)$. Then $C = (C \cap X_x) \sqcup (C \cap X_{x-u})$, so $C$ is disconnected. Hence, the only subsets of $\Spec(A)$ are those consisting of a single point. 361 | 362 | \end{proof} 363 | 364 | \begin{problem}[AM Exercise 3.12] 365 | Let $A$ be an integral domain and $M$ an $A$-module. An element $x \in M$ is a \emph{torsion element} of $M$ if $\Ann(x) \ne 0$, that is if $x$ is killed by some non-zero element of $A$. Show that the torsion elements of $M$ form a submodule of $M$. This submodule is called the \emph{torsion submodule} of $M$ and is denoted by $T(M)$. If $T(M) = 0$, the module $M$ is said to be torsion-free. Show that 366 | \begin{enumi} 367 | \item If $M$ is any $A$-module, then $M/T(M)$ is torsion-free. 368 | \item If $f : M \to N$ is a module homomorphism, then $f(T(M)) \subseteq T(N)$. 369 | \item If $0 \to M' \to M \to M''$ is an exact sequence, then the sequence $0 \to T(M') \to T(M) \to T(M'')$ is exact. 370 | \item If $M$ is any $A$-module, then $T(M)$ is the kernel of the mapping $x \mapsto 1 \otimes x$ of $M$ into $K \otimes_A M$, where $K$ is the field of fractions of $A$. 371 | \end{enumi} 372 | \end{problem} 373 | \begin{proof}[Proof of first claim] 374 | Let $x,y \in T(M)$ be distinct; we have $ax = 0, a'y = 0$ for some non-zero $a,a' \in \Ann(x)$. Then, $aa'(x-y) = aa'x - aa'y = 0$ where $aa' \ne 0$ since $A$ is an integral domain. Thus, $T(M)$ is an abelian group. 375 | \par $x \in T(M) \implies ax = 0$ for some non-zero $a \in \Ann(x)$. Then, $\Ann(\alpha x) \ni a$ for all $\alpha \in A$ since $\alpha xy = 0$. Thus, $T(M)$ is closed under external multiplication, and $T(M) \subseteq M$ is a submodule. 376 | \end{proof} 377 | \begin{proof}[Proof of $i)$] 378 | Suppose $\overline{x} \in T(M/T(M))$. Then, there is non-zero $a \in A$ such that $0 = a\overline{x} = \overline{ax}$, and so $ax \in T(M)$. There exists non-zero $a' \in A$ such that $a'ax = 0$, but $a'a \ne 0$ since $A$ is an integral domain, and so $x \in T(M)$, i.e., $\overline{x} = 0$. 379 | \end{proof} 380 | \begin{proof}[Proof of $ii)$] 381 | Suppose $x \in T(M)$, then there exists non-zero $a \in A$ such that $ax = 0$. We then see $0 = f(ax) = af(x)$, and so $a \in \Ann(f(x)) \implies f(x) \in T(N)$. 382 | \end{proof} 383 | \begin{proof}[Proof of $iii)$] 384 | We have the exact sequence 385 | \begin{equation*} 386 | 0 \to M' \xrightarrow{f} M \xrightarrow{g} M''. 387 | \end{equation*} 388 | We claim that 389 | \begin{equation*} 390 | 0 \to T(M') \xrightarrow{\overline{f}} T(M) \xrightarrow{\overline{g}} T(M'') 391 | \end{equation*} 392 | is exact, where $\overline{f}$ is the restriction of $f$ on $T(M')$, and likewise for $\overline{g}$. Note that the image of each map is contained in the next module of the sequence by $ii)$. At $T(M')$, we see that $f$ is injective and so $\overline{f}$ is injective since it is simply a restriction, and so the sequence is exact at $T(M')$. At $T(M)$, we claim $\Im(\overline{f}) = \ker(\overline{g})$. Suppose $m \in \ker(\overline{g}) \subseteq \ker(g) = \Im(f)$. Then, choose $m' \in M'$ such that $f(m') = m$; since $m \in T(M)$ there exists non-zero $a \in A$ such that $am = 0$. Then, $0 = am = af(m') = f(am')$. By the injectivity of $f$, we see $am' = 0$, and so $m' \in T(M')$, i.e., $\ker({g}) \subseteq \Im(\overline{f})$. Next we see $m \in \Im(\overline{f}) \subseteq \Im(f) = \ker(g)$, and so $\overline{g}(m) = 0$. Thus, $\Im(\overline{f}) \subseteq \ker(\overline{g})$, and our sequence is exact at $T(M)$. 393 | \end{proof} 394 | \begin{proof}[Proof of $iv)$] 395 | Let $\varphi: x \mapsto 1 \otimes x$. We see that $T(M) \subseteq \ker(\varphi)$ since $x \in T(M) \implies \exists a \ne 0 \in A$ such that $ax = 0 \implies x \mapsto 1 \otimes x = a/a \otimes x = 1/a \otimes ax = 1/a \otimes 0 = 0$. 396 | \par To show $\ker(\varphi) \subseteq T(M)$, we first show that 397 | \begin{equation*} 398 | K = \varinjlim_{\mathclap{a \in A \setminus \{0\}}} A[1/a]. 399 | \end{equation*} 400 | We see that for $A[1/a], A[1/a']$, $A[1/a] + A[1/a'] \subseteq A[1/(aa')]$, and so by AM Exercise $2.17$, we have that $K$ is the direct limit we claimed. Tensoring with $M$ gives 401 | \begin{equation*} 402 | K \otimes_A M = \varinjlim_{\mathclap{a \in A \setminus \{0\}}} (A[1/a] \otimes_A M), 403 | \end{equation*} 404 | by AM Exercise $2.20$. 405 | \par Now if $1 \otimes x = 0$ in $K \otimes_A M$, we see that $1 \otimes x = 0$ in some $A[1/a] \otimes_A M$ for some $a \in A$ by AM Exercise $2.15$. Since $1 = a/a$, we have $1 \otimes x = (a/a) \otimes x = (1/a) \otimes ax = 0$. Moreover, $A[1/a] \otimes_A M \hookrightarrow M$ is injective by having $b/a \otimes x \mapsto bx$, which only maps to zero if $b = 0$ by the fact that $A$ is an integral domain. Thus, $ax = 0$, and so $x \in T(M)$. 406 | \end{proof} 407 | 408 | \begin{problem}[AM Exercise 3.13] 409 | Let $S$ be a multiplicatively closed subset of an integral domain $A$. In the notation of Exercise $12$, show that $T(S^{-1}M) = S^{-1}(TM)$. Deduce that the following are equivalent: 410 | \begin{enumi} 411 | \item $M$ is torsion-free. 412 | \item $M_\mathfrak{p}$ is torsion-free for all prime ideals $\mathfrak{p}$. 413 | \item $M_\mathfrak{m}$ is torsion-free for all maximal ideals $\mathfrak{m}$. 414 | \end{enumi} 415 | \end{problem} 416 | \begin{proof}[Proof that $T(S^{-1}M) = S^{-1}(TM)$] 417 | Suppose $0 \notin S$, for this would make everything equal $0$, making the claim trivial. Let $m/s \in T(S^{-1}M)$; then, there exists non-zero $a/s' \in S^{-1}A$ such that $(m/s)(a/s') = am/(ss') = 0$, i.e., there exists $s''' \in S$ such that $s'''am = 0$. But since $A$ is an integral domain, we have $s''a \ne 0$ since $s'',a \ne 0$. Thus, $m \in T(M)$, and so $m/s \in S^{-1}(TM)$. 418 | \par Now let $m \in T(M)$, i.e., there exists non-zero $a \in A$ such that $am = 0$. Then, $a/1 \ne 0/1$ since $0 \notin S$, and therefore, $(m/s)(a/1) = 0$ for arbitrary $s$, i.e., $S^{-1}(TM) \subseteq T(S^{-1}M)$. 419 | \end{proof} 420 | \begin{proof}[Proof of equivalence] 421 | $i) \Rightarrow ii)$. By the previous proof, $T(M_\mathfrak{p}) = (TM)_\mathfrak{p} = 0_\mathfrak{p} = 0$. 422 | \par $ii) \Rightarrow iii)$. True since maximal $\implies$ prime. 423 | \par $iii) \Rightarrow i)$. Suppose $T(M) \ne \emptyset$. Then, there is some $x \in T(M)$ and so $ax = 0$ for some non-zero $a \in A$. Let $\mathfrak{m}$ be the maximal ideal that contains $\Ann(x)$, which exists via AM Corollary $1.4$. Since $A$ is an integral domain, $a/1 \ne 0$ in $M_\mathfrak{m}$. Then, $a(x/1) = 0$ in $M_\mathfrak{m}$ which is torsion-free, and so $x/1 = 0$ in $M_\mathfrak{m}$. This implies $a'x = 0$ for some $a' \in A \setminus \mathfrak{m}$, which contradicts our choice of $\mathfrak{m} \supseteq \Ann(x)$. 424 | \end{proof} 425 | 426 | \begin{problem} [AM Exercise 3.16] Let B be a flat $A$-algebra. Then TFAE: 427 | \begin{enumi} 428 | \item $\mathfrak{a}^{ec} = \mathfrak{a}$ for all ideals $\mathfrak{a}$ of $A$. 429 | \item $\Spec (B) \to \Spec(A)$ is surjective 430 | \item For every maximal ideal $\mathfrak{m}$ of A we have $\mathfrak{m}^e \neq (1)$. 431 | \item If $M$ is any non-zero $A$-module, them $M_B \neq 0$. 432 | \item For every A-module $M$, the mapping $x \mapsto 1 \otimes x$ of M into $M_B$ is injective. 433 | \end{enumi} 434 | \par $B$ is said to be faithfully flat over A 435 | \end{problem}1 436 | \begin{proof} 437 | \par $i) \Rightarrow ii)$: From AM prop 3.16, $\mathfrak{p} \in \Spec(A)$ is the contraction of a prime ideal of $B$ iff $\mathfrak{p}^{ec} = \mathfrak{p}$. From the assumption, $\mathfrak{a}^{ec} = \mathfrak{a}$ for all ideals $\mathfrak{a}$, so $\mathfrak{p} \in \Spec(A)$ is the contraction of a prime ideal of $B$. Hence, $\Spec (B) \to \Spec(A)$ is surjective. 438 | \par $ii) \Rightarrow iii)$: assume $\mathfrak{m}^e = (1)$. There exists a prime ideal $\mathfrak{q}$ of $B$ such that $\mathfrak{q}^c = \mathfrak{m}$. Then we see that $(1) = \mathfrak{m}^e = \mathfrak{q}^{ce} \supseteq \mathfrak{q}$, which contradicts to the assumption that $ \mathfrak{q}$ is a prime ideal. 439 | \par $iii) \Rightarrow iv)$: Let $0 \neq x \in M$, and let $M' = Ax$. We see that $0 \to M \to M'$ is exact. Since $B$ is flat over $A$, $0 \to M \otimes B \to M' \otimes B$ is exact. Hence, if $M'_B = M' \otimes B$ is nonzero, $M_B$ is also nonzero. 440 | $M' \simeq A/\mathfrak{a}$ for some ideal $\mathfrak{a} \neq (1)$ of A, so $M'_B \simeq B \otimes A/\mathfrak{a} = B/\mathfrak{a}^e \otimes A \simeq B/\mathfrak{a}^e$. 441 | Then, There is a maximal ideal $\mathfrak{m}$ that contains $\mathfrak{a}$ (Corollary 1.4), and $\mathfrak{a}^e \subseteq \mathfrak{m}^e \subsetneq (1)$ from the assumption. Hence, $M'_B \neq 0$. 442 | \par $iv) \Rightarrow v)$: Let $M'$ be the kernel of $M \to M_B$. Then $0 \to M' \to M \to M_B $ is exact. Because $B$ if flat, $0 \to M'_B \to M_B \to (M_B)_B$ is exact. Using AM Exercise 2.13, $M_B \to (M_B)_B$ is injective, so $M'_B =0 \Rightarrow M' = 0$. Hence, the mapping $x \mapsto 1 \otimes x$ of $M$ into $M_B$ is injective. 443 | \par $v) \Rightarrow i)$ Choose an ideal $\mathfrak{a}$ of $A$. Let $M = A/\mathfrak{a}$. Then we see that the mapping $x \mapsto 1 \otimes x$ of $M = A/\mathfrak{a}$ into $M_B = A/\mathfrak{a} \otimes B = A \otimes B/\mathfrak{a}^e \simeq B/\mathfrak{a}^e$ is injective. Hence, $\mathfrak{a}^{ec} \subseteq \mathfrak{a}$. $\mathfrak{a}^{ec} \supseteq \mathfrak{a}$ from AM proposition 1.17, so $\mathfrak{a}^{ec} = \mathfrak{a}$. 444 | 445 | \end{proof} 446 | 447 | \begin{problem} [AM Exercise 3.17] Let $A \xrightarrow{f} B \xrightarrow{g} C$ be ring homomorphisms. If $g \circ f$ is flat and $g$ is faithfully flat, then $f$ is flat. 448 | \end{problem} 449 | 450 | \begin{proof} 451 | $C$ is a flat $A$-module, and a faithfully flat $B$-module. 452 | \par Let M, N be $A$-modules, $\phi: M \to N$ be an injective map. Then, as $C$ is flat $A$-module, $\phi \otimes_A 1: M \otimes_A C \to N \otimes_A C$ is injective (AM proposition 2.19). As $C \simeq C \otimes_B B$, this means that $\phi \otimes_A (1 \otimes_B 1): M \otimes_A (B \otimes_B C) \to N \otimes_A( B \otimes_B C)$ is injective. As tensor product commutes, $(\phi \otimes_A 1) \otimes_B 1: (M \otimes_A B) \otimes_B \to (N \otimes_A B) \otimes_B C$ is injective. As $C$ is faithfully flat $B$-module, $x \mapsto 1 \otimes_B x$ is injective, so $\phi \otimes_A 1: M \otimes_A B \to N \otimes_A B$ is injective. Thus, B is a flat $A$-module, again by the proposition 2.19 453 | 454 | \end{proof} 455 | 456 | \begin{problem}[AM Exercise 3.21]\mbox{} 457 | \begin{enumi} 458 | \item Let $A$ be a ring, $S$ a multiplicatively closed subset of $A$, and $\phi : A \to S^{-1}A$ the canonical homomorphism. Show that $\phi^* : \Spec(S^{-1}A) \to \Spec(A)$ is a homeomorphism of $\Spec(S^{-1}A)$ onto its image in $X = \Spec(A)$. Let this image be denoted by $S^{-1}X$. 459 | \par In particular, if $f \in A$, the image of $\Spec(A_f)$ in $X$ is the basic open set $X_f$ (Chapter $1$, Exercise $17$). 460 | \item Let $f : A \to B$ be a ring homomorphism. Let $X = \Spec(A)$ and $Y = \Spec(B)$, and let $f^* : Y \to X$ be the mapping associated with $f$. Identifying $\Spec(S^{-1}A)$ with its canonical image $S^{-1}X$ in $X$, and $\Spec(S^{-1}B) (=\Spec(f(S)^{-1}B))$ with its canonical image $S^{-1}Y$ in $Y$, show that $S^{-1}f^* : \Spec(S^{-1}B) \to \Spec(S^{-1}A)$ is the restriction of $f^*$ to $S^{-1}Y$, and that $S^{-1}Y = f^{*-1}(S^{-1}X)$. 461 | \item Let $\mathfrak{a}$ be an ideal of $A$ and let $\mathfrak{b} = \mathfrak{a}^e$ be its extension in $B$. Let $\overline{f} : A/\mathfrak{a} \to B/\mathfrak{b}$ be the homomorphism induced by $f$. If $\Spec(A/\mathfrak{a})$ is identified with its canonical image $V(\mathfrak{a})$ in $X$, and $\Spec(B/\mathfrak{b})$ with its image $V(\mathfrak{b})$ in $Y$, show that $\overline{f}^*$ is the restriction of $f^*$ to $V(\mathfrak{b})$. 462 | \item Let $\mathfrak{p}$ be a prime ideal of $A$. Take $S = A - \mathfrak{p}$ in $ii)$ and then reduce mod $S^{-1}\mathfrak{p}$ as in $iii)$. Deduce that the subspace $f^{*-1}(\mathfrak{p})$ of $Y$ is naturally homeomorphic to $\Spec(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p}) = \Spec(k(\mathfrak{p})\otimes_A B)$, where $k(\mathfrak{p})$ is the residue field of the local ring $A_\mathfrak{p}$. 463 | \par $\Spec(k(\mathfrak{p}) \otimes_A B)$ is called the \emph{fiber} of $f^*$ over $\mathfrak{p}$. 464 | \end{enumi} 465 | \end{problem} 466 | \begin{proof}[Proof of $i)$] 467 | $\phi^*$ is continuous by AM Exercise $1.21i)$. Since every prime ideal of $S^{-1}A$ is an extended ideal by AM Proposition $3.11i)$, we see that $\phi^*$ is injective by AM Exercise $3.20ii)$, and therefore bijective onto its image. We consider $S^{-1}X = \Im(\phi^*)$; this is the set of prime ideals that do not meet $S$ by Proposition $3.11iv)$. It therefore remains to show that $\phi^*$ is closed. Since any arbitrary ideal of $S^{-1}A$ is an extended ideal by AM Proposition $3.11i)$, we only have to consider basis elements of the form $\mathbb{V}(\mathfrak{a}^e) \subseteq \Spec(S^{-1}A)$ for $\mathfrak{a} \in A$. We claim $\phi^*(\mathbb{V}(\mathfrak{a}^e)) = S^{-1}X \cap \mathbb{V}(\mathfrak{a}^{ec})$ (note the latter is closed in $S^{-1}X$ since $\mathbb{V}(\mathfrak{a}^{ec})$ is closed in $X$). If $\mathfrak{p} \in \phi^*(\mathbb{V}(\mathfrak{a}^e))$, then $\mathfrak{p} \cap S = \emptyset$ and $\mathfrak{a}^e \subseteq \mathfrak{p}^e \implies \mathfrak{a}^{ec} \subseteq \mathfrak{p}^{ec} = \mathfrak{p} \implies \mathfrak{p} \in S^{-1}X \cap \mathbb{V}(\mathfrak{a}^{ec})$. On the other hand if $\mathfrak{p} \in S^{-1}X \cap \mathbb{V}(\mathfrak{a}^{ec})$, then $\mathfrak{p} \cap S = \emptyset$ and $\mathfrak{a}^{ec} \subseteq \mathfrak{p} \implies \mathfrak{a}^e \subseteq \mathfrak{p}^e$. Thus $\phi$ is a homeomorphism onto its image. 468 | \par In particular, if $f \in A$, $\phi^*(\Spec(A_f)) = X_f$ by having $S = \braket{f}$. 469 | \end{proof} 470 | \begin{proof}[Proof of $ii)$] 471 | We first claim the diagram, with $\varphi_A : A \hookrightarrow S^{-1}A, \varphi_B : B \hookrightarrow S^{-1}B$ the natural embedding maps, 472 | \begin{center} 473 | \begin{tikzcd} 474 | A \arrow{r}{f}\arrow{d}[swap]{\varphi_A} & B \arrow{d}{\varphi_B}\\ 475 | S^{-1}A \arrow{r}{S^{-1}f} & S^{-1}B 476 | \end{tikzcd} 477 | \end{center} 478 | commutes. But this is clear since $S^{-1}B \simeq f(S)^{-1}B$ by AM Exercise $3.4$, and since $S^{-1}f(a/s) = f(a)/f(s)$. Moreover, calculating explicitly, if $a \in A$, $(\varphi_B \circ f) (a) = \varphi_B (f(a)) = f(a)/1$, while $(S^{-1}f \circ \varphi_A)(a) = S^{-1}f (a/1) = f(a)/1$. Since AM Exercise $1.21$ implies $f^* \circ \varphi_B^* = (\varphi_B \circ f)^* = (S^{-1}f \circ \varphi_A)^* = \varphi_A^* \circ (S^{-1}f)^*$, we have the commutative diagram 479 | \begin{center} 480 | \begin{tikzcd} 481 | \Spec(S^{-1}B) = S^{-1}Y \arrow{r}{(S^{-1}f)^*}\arrow{d}[swap]{\varphi_B^*} & \Spec(S^{-1}A) = S^{-1}X \arrow{d}{\varphi_A^*}\\ 482 | \Spec(B) = Y \arrow{r}{f^*} & \Spec(A) = X 483 | \end{tikzcd} 484 | \end{center} 485 | where the identification is by $i)$, which shows the compatibility of $(S^{-1}f)^*$ and $f^*$. 486 | \par We now show $S^{-1}Y = f^{*-1}(S^{-1}X)$. The diagram above shows that $f^*(S^{-1}Y) \subseteq S^{-1}X$, and so $S^{-1}Y \subseteq f^{*-1}(S^{-1}X)$. On the other hand, suppose $\mathfrak{p} \in f^{*-1}(S^{-1}X)$. Then, $\mathfrak{p}^c = f^*(\mathfrak{p}) \in S^{-1}X$, so $\mathfrak{p}^c \cap S = \emptyset$ by AM Proposition $3.11iv)$. To show $\mathfrak{p} \in S^{-1}Y$, it suffices to show $\mathfrak{p} \cap f(S) = \emptyset$. So, suppose $x \in \mathfrak{p} \cap f(S)$; then, $x = f(s)$ for some $s \in S \cap \mathfrak{p}^c = \emptyset$, which is a contradiction. Thus, $S^{-1}Y \supseteq f^{*-1}(S^{-1}X)$, and $S^{-1}Y = f^{*-1}(S^{-1}X)$. 487 | \end{proof} 488 | \begin{proof}[Proof of $iii)$] 489 | Letting $\pi_A : A \twoheadrightarrow A/\mathfrak{a},\pi_B : B \twoheadrightarrow B/\mathfrak{b}$ be the natural quotient maps, we have the commutative diagram 490 | \begin{center} 491 | \begin{tikzcd} 492 | A \arrow{r}{f}\arrow{d}[swap]{\pi_A} & B \arrow{d}{\pi_B}\\ 493 | A/\mathfrak{a} \arrow{r}{\overline{f}} & B/\mathfrak{b} 494 | \end{tikzcd} 495 | \end{center} 496 | and by the same argument as in $iv)$, we get 497 | \begin{center} 498 | \begin{tikzcd} 499 | \Spec(B/\mathfrak{b}) = \mathbb{V}(\mathfrak{b}) \arrow{r}{\overline{f}^*}\arrow{d}[swap]{\pi_B^*} & \Spec(A/\mathfrak{a}) = \mathbb{V}(\mathfrak{a}) \arrow{d}{\pi_A^*}\\ 500 | \Spec(B) = Y \arrow{r}{f^*} & \Spec(A) = X 501 | \end{tikzcd} 502 | \end{center} 503 | The identification comes from AM Exercise $1.21iv)$, which says $\pi_B^*$ is a homeomorphism $\Spec(B/\mathfrak{b}) \to \mathbb{V}(\ker(\pi_B)) = \mathbb{V}(\mathfrak{b})$ and $\pi_A^*$ is a homeomorphism $\Spec(A/\mathfrak{a}) \to \mathbb{V}(\ker(\pi_A)) = \mathbb{V}(\mathfrak{a})$. Thus, $\overline{f}^*$ and $f^*$ are compatible. 504 | %\par Much like in $ii)$, we also show $\mathbb{V}(\mathfrak{b}) = f^{*-1}(\mathbb{V}(\mathfrak{a}))$. The diagram above shows that $f^*(\mathbb{V}(\mathfrak{b})) \subseteq \mathbb{V}(\mathfrak{a})$, and so $\mathbb{V}(\mathfrak{b}) \subseteq f^{*-1}(\mathbb{V}(\mathfrak{a}))$. On the other hand, suppose $\mathfrak{p} \in f^{*-1}(\mathbb{V}(\mathfrak{a}))$. Then, $\mathfrak{p}^c = f^{*-1}(\mathfrak{p}) \in \mathbb{V}(\mathfrak{a})$, and so $\mathfrak{p}^c \supseteq \mathfrak{a}$. Taking the extension of both, we get $\mathfrak{p} = \mathfrak{p}^{ce} \supseteq \mathfrak{a}^e = \mathfrak{b}$, and so $\mathfrak{p}^c \in \mathbb{V}(\mathfrak{b})$. Thus, $f^{*-1}(\mathbb{V}(\mathfrak{a})) \subseteq \mathbb{V}(\mathfrak{b})$, and $\mathbb{V}(\mathfrak{b}) = f^{*-1}(\mathbb{V}(\mathfrak{a}))$. 505 | \end{proof} 506 | \begin{proof}[Proof of $iv)$] 507 | Following the steps given, we get the commutative diagram 508 | \begin{center} 509 | \begin{tikzcd} 510 | \Spec(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p}) \arrow{r}{\overline{f_\mathfrak{p}}^*}\arrow{d}[swap]{\pi_B^*} & \Spec(A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p})\arrow{d}{\pi_A^*}\\ 511 | \Spec(B_\mathfrak{p}) \arrow{r}{f_\mathfrak{p}^*}\arrow{d}[swap]{\varphi_B^*} & \Spec(A_\mathfrak{p}) \arrow{d}{\varphi_A^*}\\ 512 | \Spec(B) \arrow{r}{f^*} & \Spec(A) 513 | \end{tikzcd} 514 | \end{center} 515 | By $iii)$, $\Spec(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p})$ is homeomorphic to $\mathbb{V}(\mathfrak{p}B_\mathfrak{p})$. By $ii)$, $\mathbb{V}(\mathfrak{p}B_\mathfrak{p})$ is homeomorphic to $\varphi_B^*(\mathbb{V}(\mathfrak{p}B_\mathfrak{p}))$. We now claim that $\varphi_B^*(\mathbb{V}(\mathfrak{p}B_\mathfrak{p})) = f^{*-1}(\mathfrak{p})$. Suppose $\mathfrak{q} \in f^{*-1}(\mathfrak{p})$, which gives $\mathfrak{p} \in \Im(\varphi_A^*) \implies \mathfrak{q} \in \Im(\varphi_B^*)$. Then, since $\mathfrak{p} = f^{-1}(\mathfrak{q})$, i.e., $f(\mathfrak{p}) \subseteq \mathfrak{q}$, i.e., $\mathfrak{p}B \subseteq \mathfrak{q}$. Thus, $\mathfrak{q}_\mathfrak{p}$ is prime in $B_\mathfrak{p}$ and contains $\mathfrak{p}B_\mathfrak{p}$, i.e., $f^{*-1}(\mathfrak{p}) \subseteq \varphi_B^*(\mathbb{V}(\mathfrak{p}B_\mathfrak{p}))$. In the other direction, suppose $\mathfrak{q} \in \varphi_B^*(\mathbb{V}(\mathfrak{p}B_\mathfrak{p}))$. Then, $\mathfrak{p}B_\mathfrak{p} \subseteq \mathfrak{q}_\mathfrak{p}$ so $\mathfrak{p}B \subseteq \mathfrak{q}_\mathfrak{p}^c = \mathfrak{q}$, i.e., $f(\mathfrak{p}) \subseteq \mathfrak{q}$. So, $\mathfrak{p} \subseteq f^{-1}(\mathfrak{q})$. On the other hand, $f^{-1}(\mathfrak{q}) \subseteq \mathfrak{p}$ since $\mathfrak{q} \cap f(A \setminus \mathfrak{p}) = \emptyset$ by choice of $\mathfrak{q}$. Thus, $\mathfrak{p} = f^{-1}(\mathfrak{q})$, and so $\mathfrak{q} \in f^{*-1}(\mathfrak{p})$, i.e., $\varphi_B^*(\mathbb{V}(\mathfrak{p}B_\mathfrak{p})) = f^{*-1}(\mathfrak{p})$. 516 | \par We now want to show the isomorphism given. We have $B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p} \simeq B_\mathfrak{p}/(\mathfrak{p}B)_\mathfrak{p} \simeq (B/\mathfrak{p}B)_\mathfrak{p}$ by AM Proposition $3.11v)$. Then, $(B/\mathfrak{p}B)_\mathfrak{p} \simeq A_\mathfrak{p} \otimes_A B/\mathfrak{p}B$ by Proposition $3.5$. AM Exercise $1.2$ gives $A_\mathfrak{p} \otimes_A B/\mathfrak{p}B \simeq A_\mathfrak{p} \otimes_A (A/\mathfrak{p} \otimes_A B)$. The associativity of the tensor product from AM Proposition $2.14ii)$ then gives $A_\mathfrak{p} \otimes_A (A/\mathfrak{p} \otimes_A B) \simeq (A/\mathfrak{p} \otimes_A A_\mathfrak{p}) \otimes_A B$. Applying AM Exercise $1.2$ again yields $(A/\mathfrak{p} \otimes_A A_\mathfrak{p}) \otimes_A B \simeq A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} \otimes_A B$. But then $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} = A_\mathfrak{p}/\mathfrak{p}_\mathfrak{p} = k(\mathfrak{p})$, since $A_\mathfrak{p}$ is local and therefore $\mathfrak{p}_\mathfrak{p}$ can only be the maximal ideal, and so we get the isomorphism $B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p} \simeq k(\mathfrak{p}) \otimes_A B$. We note that this also preserves ring structure since this is just the map $b/f(x) + \mathfrak{p}B_\mathfrak{p} \leftrightarrow (1/x + \mathfrak{p}_\mathfrak{p}) \otimes b$. 517 | \end{proof} 518 | 519 | \printbibliography 520 | \end{document} 521 | --------------------------------------------------------------------------------