├── Cover-page.pdf ├── ELasker.jpg ├── Jean-Louis_Koszul.jpg ├── JiaoShouShengYa.jpg ├── LICENSE ├── Makefile ├── Nakayama.png ├── OZariski.jpg ├── README.md ├── WanFeng.jpg ├── YAlg3-1.tex ├── YAlg3-2.tex ├── YAlg3-3.tex ├── YAlg3-4.tex ├── YAlg3-5.tex ├── YAlg3-6.tex ├── YAlg3-7.tex ├── YAlg3-8.tex ├── YAlg3-intro.tex ├── YAlg3.bib ├── YAlg3.pdf ├── YAlg3.tex ├── by-nc.eps ├── myarrows.sty └── mycommand.sty /Cover-page.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/Cover-page.pdf -------------------------------------------------------------------------------- /ELasker.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/ELasker.jpg -------------------------------------------------------------------------------- /Jean-Louis_Koszul.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/Jean-Louis_Koszul.jpg -------------------------------------------------------------------------------- /JiaoShouShengYa.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/JiaoShouShengYa.jpg -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | Attribution 4.0 International 2 | 3 | ======================================================================= 4 | 5 | Creative Commons Corporation ("Creative Commons") is not a law firm and 6 | does not provide legal services or legal advice. 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For 392 | the avoidance of doubt, this paragraph does not form part of the 393 | public licenses. 394 | 395 | Creative Commons may be contacted at creativecommons.org. 396 | -------------------------------------------------------------------------------- /Makefile: -------------------------------------------------------------------------------- 1 | MAINFILE=YAlg3 2 | COVER=Cover-page 3 | LATEXMK=latexmk 4 | 5 | default: 6 | # Generating $(MAINFILE).pdf 7 | $(LATEXMK) -pdf -pdflatex="xelatex -synctex=1 -shell-escape -interaction=nonstopmode %O %S" $(MAINFILE) 8 | 9 | nosync: 10 | # Generating $(MAINFILE).pdf 11 | $(LATEXMK) -pdf -pdflatex="xelatex -shell-escape -interaction=nonstopmode %O %S" $(MAINFILE) 12 | 13 | clean: 14 | # Cleaning... 15 | @rm -f *.aux *.log *.idx *.ind *.ilg *.thm *.toc *.blg *.bbl *.bcf *.out 16 | @rm -f *.fls *.fdb_latexmk *.run.xml *.synctex.gz *.xdv *~ *.lof *.lot 17 | @rm -f .metadonnees* 18 | 19 | tarball: 20 | @rm -f ../Yanqi-Algebra-3.tar.gz 21 | @tar --exclude .git -zcvf ../Yanqi-Algebra-3.tar.gz . 22 | 23 | zip: 24 | @rm -f ../Yanqi-Algebra-3.zip 25 | @zip -r ../Yanqi-Algebra-3.zip . 26 | 27 | .PHONY: clean 28 | -------------------------------------------------------------------------------- /Nakayama.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/Nakayama.png -------------------------------------------------------------------------------- /OZariski.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/OZariski.jpg -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | This are the XeLaTeX sources for the my lecture notes on commutative algebra, entitled somewhat pompously as **Yanqi Lake Lectures on Algebra, Part III**. These notes have been used at 2 | * [University of Chinese Academy of Sciences](http://www.ucas.ac.cn) (Yanqi Lake campus), whence the title; 3 | * [Peking University](http://www.pku.edu.cn). 4 | 5 | These notes are slightly outdated, poorly organized, and the mathematical contents have not been thoroughly checked yet. Please use them at your own risk. The author does not intend to publish these notes. 6 | 7 | # How to compile 8 | 9 | ## System requirements 10 | The source codes are to be compiled using XeLaTeX. The reader is assumed to work under the UN*X + bash environment. 11 | 12 | The recipes below can be tweaked to work under Windows, but this is not recommended. The simplest solution is to go open-source. 13 | 14 | We only need the standard packages and fonts, such as 15 | - [TeX Live](https://tug.org/texlive), including the programs latexmk, xindy and biber. 16 | - Standard fonts included in TeX Live. 17 | - The [Noto Serif CJK](https://github.com/googlei18n/noto-cjk) fonts. 18 | - The [League Spartan](https://www.theleagueofmoveabletype.com/league-spartan) font. 19 | 20 | The aforementioned OpenType fonts should be installed system-wide to be accessible by XeLaTeX. 21 | 22 | For some strange reason, I used and installed the fonts TeX Gyre Heros Cn and TeX Gyre Pagella. In case of error messages related to these fonts, please look for the OTF files (in the directories in your computer which store TeX-related fonts) whose names begin with **texgyreheroscn** and **texgyrepagella**, then install them manually in your system. 23 | 24 | Make sure that all the relevant packages/programs are installed. For reference, the author made the compilation using Arch-based Linux distributions with TeX Live 2018; the packages **biber** and **texlive-science** are required. 25 | 26 | ## Clone the files 27 | Assume that [Git](https://git-scm.com/) has been installed on your computer. As a preparation for the compilation process, we will clone the files into `~/Yanqi-Algebra-3` in our home directory. In command line, type 28 | ``` 29 | cd ~ 30 | git clone https://github.com/wenweili/Yanqi-Algebra-3 31 | ``` 32 | 33 | All the source files are encoded in UTF-8, which is the de facto standard for storing multilingual texts (although the document is largely written in English). If you encounter problems in opening the source files under Windows, try to re-configure your editor or convert the encoding manually. 34 | 35 | ## Compile the TeX source 36 | 37 | Move to the directory 38 | ``` 39 | cd ~/Yanqi-Algebra-3 40 | ``` 41 | Then, either type 42 | ``` 43 | latexmk -pdf -pdflatex="xelatex -shell-escape -interaction=nonstopmode %O %S" YAlg3 44 | ``` 45 | under bash, or more simply 46 | ``` 47 | make 48 | ``` 49 | 50 | The resulting PDF file should appear as **YAlg3.pdf** in the same directory. Note that the main file is **YAlg3.tex**. 51 | 52 | To clean up everything in our directory except the PDF file, type 53 | ``` 54 | make clean 55 | ``` 56 | 57 | # The source codes 58 | These notes are based on the standard *book* document class from LaTeX. Some other macros are outsourced to **mycommands.sty** and **myarrows.sty**. 59 | 60 | # The cover page 61 | The cover page is in the file **Cover-page.pdf**, which will automatically be included in the resulting main PDF file after compilation. It is made from the open source software [Scribus](https://www.scribus.net/); the source file in *.sla* format is not included here. 62 | 63 | # Feedback 64 | In case of problems of compilation, please kindly report to the author. Make sure that all the system requirements above are met, and provide detailed error messages. Other suggestions are also welcome. 65 | 66 | # License 67 | Except possibly the photos and the logo of UCAS, the entire codebase is under [CC BY-NC 4.0](https://creativecommons.org/licenses/by-nc/4.0/). 68 | 69 | # Star History 70 | 71 | [![Star History Chart](https://api.star-history.com/svg?repos=wenweili/Yanqi-Algebra-3&type=Date)](https://star-history.com/#wenweili/Yanqi-Algebra-3&Date) 72 | -------------------------------------------------------------------------------- /WanFeng.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/wenweili/Yanqi-Algebra-3/ecc87c193f9f2d49c2c68b78a7977bb9802893d4/WanFeng.jpg -------------------------------------------------------------------------------- /YAlg3-1.tex: -------------------------------------------------------------------------------- 1 | % To be compiled by XeLaTeX, preferably under TeX Live. 2 | % LaTeX source for ``Yanqi Lake Lectures on Algebra'' Part III. 3 | % Copyright 2019 李文威 (Wen-Wei Li). 4 | % Permission is granted to copy, distribute and/or modify this 5 | % document under the terms of the Creative Commons 6 | % Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) 7 | % https://creativecommons.org/licenses/by-nc/4.0/ 8 | 9 | % To be included 10 | \chapter{Warming up} 11 | 12 | The reader might be familiar with most of the materials in this lecture. Our goal is to fix notation and present the basic structural results on Noetherian and Artinian rings or modules, including the celebrated Nakayama's Lemma which will be used repeatedly. 13 | 14 | \section{Review on ring theory} 15 | Let $R$ be a ring, supposed to be commutative with unit $1 \neq 0$ as customary. Recall that an ideal $I \subsetneq R$ is called 16 | \begin{compactitem} 17 | \item \emph{prime}, if $ab \in I \iff (a \in I) \vee (b \in I)$; 18 | \item \emph{maximal}, if $I$ is maximal among the proper ideals of $R$ with respect to inclusion. 19 | \end{compactitem} 20 | Recall the following standard facts 21 | \begin{itemize} 22 | \item $I$ is prime if and only if $R/I$ is an integral domain, i.e. has no zero divisors except $0$; 23 | \item $I$ is maximal if and only if $R/I$ is a field; in particular, maximal ideals are prime; 24 | \item every proper ideal $I$ is contained in a maximal ideal (an application of Zorn's Lemma).\index{Zorn's Lemma} 25 | \end{itemize} 26 | 27 | \begin{definition}[Local rings] \index{local ring} 28 | The ring $R$ is called \emph{local} if it has a unique maximal ideal, \emph{semi-local} if it has only finitely many maximal ideals. 29 | 30 | Let $\mathfrak{m}$ be the maximal ideal of a local ring $R$. We call $R/\mathfrak{m}$ the \emph{residue field} of $R$. A local homomorphism between local rings $\varphi: R_1 \to R_2$ is a ring homomorphism such that $\varphi(\mathfrak{m}_1) \subset \mathfrak{m}_2$. Consequently, local homomorphisms induce embeddings on the level of residue fields. 31 | \end{definition} 32 | Sometimes we denote a local ring by the pair $(R, \mathfrak{m})$. 33 | 34 | \begin{remark} 35 | Let $R$ be a local ring with maximal ideal $\mathfrak{m}$, then $R^\times = R \smallsetminus \mathfrak{m}$. The is easily seen as follows. An element $x \in R$ is invertible if and only if $Rx = R$. Note that $Rx=R$ is equivalent to that $x$ is not contained in any maximal ideal, and the only maximal ideal is $\mathfrak{m}$. 36 | \end{remark} 37 | 38 | Throughout these lectures, we shall write \index{Spec@$\Spec(R)$} \index{MaxSpec@$\MaxSpec(R)$} 39 | \begin{align*} 40 | \Spec(R) & := \{ \text{prime ideals of } R \}, \\ 41 | \MaxSpec(R) & := \{ \text{maximal ideals of } R \}. 42 | \end{align*} 43 | They are called the \emph{spectrum} and the \emph{maximal spectrum} of $R$, respectively. The upshot is that $\Spec(R)$ comes with a natural topology. 44 | 45 | \begin{definition}[Zariski topology] \index{Zariski topology}\index{$V(\mathfrak{a})$} 46 | For any ideal $\mathfrak{a} \subset R$, set $V(\mathfrak{a}) := \{ \mathfrak{p} \in \Spec(R): \mathfrak{p} \supset \mathfrak{a} \}$. Then there is a topology on $\Spec(R)$, called the \emph{Zariski topology}, whose closed subset are precisely $V(\mathfrak{a})$, for various ideals $\mathfrak{a}$. 47 | \end{definition} 48 | Indeed, we only have to prove the family of subsets $\{ V(\mathfrak{a}) : \mathfrak{a} \subset R \}$ is closed under finite union and arbitrary intersections. It boils down to the easy observation that $V(\mathfrak{a}) \cup V(\mathfrak{b}) = V(\mathfrak{a}\mathfrak{b})$ (check this!) and $\bigcap_{\mathfrak{a} \in \mathcal{A}} V(\mathfrak{a}) = V\left( \sum_{\mathfrak{a} \in \mathcal{A}} \mathfrak{a} \right)$, where $\mathcal{A}$ is any family of ideals. 49 | 50 | Given a ring homomorphism $\varphi: R_1 \to R_2$, if $I \subset R_2$ is an ideal, then $\varphi^{-1}(I) \subset R_1$ is also an ideal. 51 | \begin{proposition} 52 | Given $\varphi$ as above, it induces a continuous map 53 | \begin{align*} 54 | \varphi^\sharp: \Spec(R_2) & \longrightarrow \Spec(R_1) \\ 55 | \mathfrak{p} & \longmapsto \varphi^{-1}(\mathfrak{p}) 56 | \end{align*} 57 | with respect to the Zariski topologies on spectra. 58 | \end{proposition} 59 | \begin{proof} 60 | Clearly, $ab \in \varphi^{-1}(\mathfrak{p})$ is equivalent to $\varphi(a)\varphi(b) \in \mathfrak{p}$, which is in turn equivalent to $(\varphi(a) \in \mathfrak{p}) \vee (\varphi(b) \in \mathfrak{p})$ when $\mathfrak{p}$ is prime. 61 | 62 | To show the continuity of $\varphi^\sharp$, observe that for any ideal $\mathfrak{a} \subset R_1$ and $\mathfrak{p} \in \Spec(R_2)$, we have $\varphi^{-1}(\mathfrak{p}) \supset \mathfrak{a}$ if and only if $\mathfrak{p} \supset \varphi(\mathfrak{a})$, i.e. $\mathfrak{p} \in V(\varphi(\mathfrak{a}) R_2)$. Hence the preimage of closed subsets are still closed. 63 | \end{proof} 64 | 65 | More operations on spectra: 66 | \begin{itemize} 67 | \item Take $R_1$ to be a subring of $R_2$ and $\varphi$ be the inclusion map, the map above becomes $\mathfrak{p} \mapsto \mathfrak{p} \cap R_1$. 68 | \item Take $\varphi: R \twoheadrightarrow R/I$ to be a quotient homomorphism, then $\varphi^{-1}$ is the usual bijection from $\Spec(R/I)$ onto $V(I)$. 69 | \item In general, $\varphi^{-1}$ does not induce $\MaxSpec(R_2) \to \MaxSpec(R_1)$, as illustrated in the case $\varphi: \Z \hookrightarrow \Q$. 70 | \end{itemize} 71 | 72 | At this stage, we can prove a handy result concerning prime ideals. 73 | \begin{proposition}[Prime avoidance]\label{prop:prime-avoidance} \index{prime avoidance} 74 | Let $I$ and $\mathfrak{p}_1, \ldots, \mathfrak{p}_n$ be ideals of $R$ such that $I \subset \bigcup_{i=1}^n \mathfrak{p}_i$. Suppose that 75 | \begin{compactitem} 76 | \item either $R$ contains an infinite field, or 77 | \item at most two of the ideals $\mathfrak{p}_1, \ldots, \mathfrak{p}_n$ are non-prime, 78 | \end{compactitem} 79 | then there exists $1 \leq i \leq n$ such that $I \subset \mathfrak{p}_i$. 80 | \end{proposition} 81 | \begin{proof} 82 | If $R$ contains an infinite field $F$, the ideals are automatically $F$-vector subspaces of $R$. Since $I = \bigcup_{i=1}^r I \cap \mathfrak{p}_i$ whereas an $F$-vector space cannot be covered by finitely many proper subspaces, there must exist some $i$ with $I \cap \mathfrak{p}_i = I$. 83 | 84 | Under the second assumption, let us argue by induction on $n$ that $\forall i \; I \not\subset \mathfrak{p}_i$ implies $I \not\subset \bigcup_{i=1}^n \mathfrak{p}_i$. The case $n=1$ is trivial. When $n \geq 2$, by induction we may choose, for each $i$, an element $x_i \in I \smallsetminus \bigcup_{j \neq i} \mathfrak{p}_j$. Suppose on the contrary that $I \subset \bigcup_{j=1}^n \mathfrak{p}_j$, then we would have $x_i \in \mathfrak{p}_i$, for all $i=1, \ldots, n$. 85 | 86 | When $n=2$ we have $x_1 + x_2 \notin \mathfrak{p}_1 \cup \mathfrak{p}_2$ and $x_1 + x_2 \in I$, a contradiction. When $n > 2$, we may assume $\mathfrak{p}_1$ is prime, therefore 87 | \[ x_1 + \prod_{j=2}^n x_j \notin \bigcup_{i=1}^n \mathfrak{p}_i, \] 88 | again a contradiction. 89 | \end{proof} 90 | 91 | \begin{exercise} 92 | The following construction from \cite[Exercise 3.17]{Eis95} shows that the assumptions of Proposition \ref{prop:prime-avoidance} cannot be weakened. Take $R = (\Z/2\Z)[X, Y] / (X,Y)^2$, which has a basis $\{1, X, Y\}$ (modulo $(X,Y)^2$) as a $\Z/2\Z$-vector space. Show that the image $\mathfrak{m}$ of $(X, Y)$ in $R$ is the unique prime ideal, and can be expressed as a union of three ideals properly contained in $\mathfrak{m}$. 93 | \end{exercise} 94 | 95 | \section{Localization of rings and modules} \index{localization}\index{$M[S^{-1}]$} 96 | Let $S$ be a \emph{multiplicative subset} of $R$, which means that 97 | \begin{inparaenum}[(a)]\index{multiplicative subset} 98 | \item $1 \in S$, 99 | \item $S$ is closed under multiplication, and 100 | \item $0 \notin S$. 101 | \end{inparaenum} 102 | The \emph{localization} of $R$ with respect to $S$ is the ring $R[S^{-1}]$ formed by classes $[r,s]$ with $r \in R$, $s \in S$, modulo the equivalence relation 103 | \[ [r,s] = [r',s'] \iff \exists t \in S, \; (rs' - r's)t = 0. \] 104 | 105 | You should regard $[r,s]$ as a token for $r/s$; the ring structure of $R[S^{-1}]$ is therefore evident. In brief, localization amounts to formally inverting the elements of $S$, whence the notation $R[S^{-1}]$. Note that condition (c) guarantees $R[S^{-1}] \neq \{0\}$. 106 | 107 | \begin{exercise} 108 | Given $R$ and $S$, show that $r \mapsto r/1$ yields a natural homomorphism $R \to R[S^{-1}]$ and show that its kernel equals $\{r: \exists s \in S, \; sr=0 \}$. 109 | \end{exercise} 110 | 111 | The universal property of $R \to R[S^{-1}]$ can be stated using commutative diagrams as follows. 112 | \[ 113 | \forall \left\{ \begin{array}{l} 114 | \varphi: R \to R': \; \text{ring homomorphism} \\ 115 | \text{s.t. } \varphi(S) \subset (R')^\times 116 | \end{array}\right. ,\quad 117 | \begin{tikzcd} 118 | R \arrow[r] \arrow[rd, "\varphi"'] & R[S^{-1}] \arrow[d, dashed, "\exists!"] \\ 119 | & R' 120 | \end{tikzcd} 121 | \] 122 | 123 | Consequently, if $S \subset R^\times$ then $R \simeq R[S^{-1}]$ canonically. Furthermore, the homomorphism $R \to R[S^{-1}]$ induces a bijection 124 | \begin{equation}\label{eqn:localization-Spec} \begin{tikzcd}[row sep=tiny, column sep=small] 125 | \Spec(R[S^{-1}]) \arrow[leftrightarrow, r, "1:1"] & \left\{ \mathfrak{p} \in \Spec(R): \mathfrak{p} \cap S = \emptyset \right\} \arrow[phantom, r, "\subset" description] & \Spec(R) \\ 126 | \mathfrak{p} R[S^{-1}] = \{ r/s: r \in \mathfrak{p}, \; s \in S \} & \mathfrak{p} \arrow[mapsto, l] & \\ 127 | \mathfrak{q} \arrow[mapsto, r] & \text{its preimage} . 128 | \end{tikzcd}\end{equation} 129 | 130 | \begin{exercise} 131 | Check the properties above. 132 | \end{exercise} 133 | 134 | Let us review some important instances of the localization procedure. 135 | \begin{enumerate} 136 | \item Take $S$ to be the subset of non zero-divisors of $R$. This is easily seen to be a multiplicative subset (check it!) and $K(R) := R[S^{-1}]$ is called the \emph{total fraction ring} of $R$. The reader is invited to check that $R \to K(R)$ is the ``biggest localization'' such that the natural homomorphism $R \to R[S^{-1}]$ is injective. Hint: state this in terms of universal properties.\index{total fraction ring} 137 | 138 | When $R$ is an integral domain, we shall take $S := R \smallsetminus \{0\}$; in this case the total fraction ring $\text{Frac}(R) := K(R)$ is the well-known \emph{field of fractions} of $R$.\index{field of fractions} 139 | 140 | \item Take any $\mathfrak{p} \in \Spec(R)$ and $S := R \smallsetminus \mathfrak{p}$. From the definition of prime ideals, one infers that $S$ is a multiplicative subset of $R$. The corresponding localization is denoted by $R \to R_{\mathfrak{p}} := R[S^{-1}]$. We see from \eqref{eqn:localization-Spec} that 141 | \[ \MaxSpec(R_{\mathfrak{p}}) = \{ \mathfrak{p} R_{\mathfrak{p}} \}; \] 142 | in particular, $R_{\mathfrak{p}}$ is a local ring with maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$. This is the standard way to produce local rings; we say that $R_{\mathfrak{p}}$ is the localization of $R$ at the prime $\mathfrak{p}$. 143 | 144 | \item Suppose $f \in R$ is not nilpotent, that is, $f^n \neq 0$ for every $n$. Take $S := \{ f^n : n \geq 0 \}$. The corresponding localization is denoted by the self-explanatory notation $R \to R[f^{-1}]$. 145 | \end{enumerate} 146 | 147 | \begin{exercise} 148 | Describe the following localizations explicitly. 149 | \begin{enumerate}[(a)] 150 | \item $R = \Z$, and we localize at the prime ideal $(p)$ where $p$ is a prime number. 151 | \item $R = \CC[X_1, \ldots, X_n]$ and we localize at the maximal ideal generated by $X_1, \ldots, X_n$. 152 | \item $R = \CC\llbracket X \rrbracket$ (the ring of formal power series) and $S := \{X^n : n \geq 0 \}$. 153 | \end{enumerate} 154 | \end{exercise} 155 | 156 | \begin{exercise} 157 | Prove that $R[X]/(fX-1) \rightiso R[f^{-1}]$ by mapping $X \mapsto f^{-1}$. Hint: use the universal property. 158 | \end{exercise} 159 | 160 | Always let $S \subset R$ be a multiplicative subset. The localization $M[S^{-1}]$ of an $R$-module $M$ can be defined in the manner above, namely as the set of equivalence classes $[m,s]$ with $m \in M$ and $s \in S$, such that 161 | \[ [m,s] = [m',s'] \iff \exists t \in S, \; t(s'm - sm')=0. \] 162 | As in the case of rings, we shall write $m/s$ instead of $[m,s]$. It is an $R[S^{-1}]$-module, equipped with a natural homomorphism $M \to M[S^{-1}]$ of $R$-modules. This yields a functor: for any homomorphism $f: M \to N$ we have a natural $f[S^{-1}]: M[S^{-1}] \to N[S^{-1}]$, mapping $m/s$ to $f(m)/s$; furthermore $f[S^{-1}] \circ g[S^{-1}] = (f \circ g)[S^{-1}]$ whenever composition makes sense. 163 | 164 | A slicker interpretation is to use the natural isomorphism $M[S^{-1}] \rightiso R[S^{-1}] \dotimes{R} M$ which maps $m/s$ to $(1/s) \otimes m$. Hereafter, we shall identify $M[S^{-1}]$ and $R[S^{-1}] \dotimes{R} M$ without further comments. 165 | 166 | In the same vein, we may define $M[f^{-1}]$ and $M_{\mathfrak{p}}$, for non-nilpotent $f \in R$ and $\mathfrak{p} \in \Spec(R)$ respectively. Localization ``commutes'' with several standard operation on modules, which we sketch below. The details are left to the reader. 167 | \begin{itemize} 168 | \item For $R$-modules $M$, $N$, we have a natural isomorphism of $R[S^{-1}]$-modules 169 | \begin{gather*} 170 | (M \dotimes{R} N)[S^{-1}] \simeq M[S^{-1}] \dotimes{R[S^{-1}]} N[S^{-1}], \\ 171 | (M \oplus N)[S^{-1}] \simeq M[S^{-1}] \oplus N[S^{-1}]. 172 | \end{gather*} 173 | Same for arbitrary direct sums. This is easily seen by viewing $M[S^{-1}]$ as $M \otimes_R R[S^{-1}]$. 174 | \item Note that $\Hom_R(M, N)$ is also an $R$-module: simply set $(rf)(m) = r \cdot f(m)$ for any $f \in \Hom_R(M, N)$. There is a natural homomorphism 175 | \begin{gather}\label{eqn:localization-Hom} 176 | \Hom_R(M, N)[S^{-1}] \to \Hom_{R[S^{-1}]} \left( M[S^{-1}], N[S^{-1}] \right) 177 | \end{gather} 178 | sending $s^{-1} \otimes \varphi$ to $s^{-1}\varphi: m/t \mapsto \varphi(m)/st$. It is clearly an isomorphism for $M = R$, thus also for $M = R^a$ where $a \in \Z_{\geq 0}$, but this is not the case in general, as there is no uniform bound for the ``denominators'' for any given $\psi: M[S^{-1}] \to N[S^{-1}]$ --- some finiteness condition is needed. Let us assume $M$ to be \emph{finitely presented}\index{finitely presented}, i.e. there is an exact sequence 179 | \[ R^a \to R^b \to M \to 0, \quad a,b \in \Z_{\geq 0}. \] 180 | In this case \eqref{eqn:localization-Hom} is an isomorphism, as easily seen from the commutative diagram with exact rows: 181 | \[ \begin{tikzcd}[column sep=tiny] 182 | 0 \arrow[r] & \Hom(M,N)[S^{-1}] \arrow[r] \arrow[d] & \Hom(R^b, N)[S^{-1}] \arrow[r] \arrow[d, "\simeq"] & \Hom(R^a, N)[S^{-1}] \arrow[d, "\simeq"] \\ 183 | 0 \arrow[r] & \Hom(M[S^{-1}], N[S^{-1}]) \arrow[r] & \Hom(R[S^{-1}]^b, N[S^{-1}]) \arrow[r] & \Hom(R[S^{-1}]^a, N[S^{-1}]) 184 | \end{tikzcd}\] 185 | Here we used the fact that localization preserves exactness: see the Proposition \ref{prop:localization-exactness} below. 186 | \item Let $\varphi: R \to R'$ be a ring homomorphism and $S \subset R$ be a multiplicative subset, so that $S' := \varphi(S) \subset R'$ is also multiplicative. View $R'$ as an $R$-module, then we have 187 | \[\begin{tikzcd}[row sep=tiny] 188 | \underbracket{R'[(S')^{-1}]}_{\text{as ring}} \arrow[r, "\sim"] & \underbracket{R'[S^{-1}]}_{\text{as module}} \arrow[equal, r] & R[S^{-1}] \dotimes{R} R' \\ 189 | r'/\varphi(s) \arrow[mapsto, rr] & & (1/s) \otimes r' \\ 190 | r'\varphi(r)/\varphi(s) & & (r/s) \otimes r' \arrow[mapsto, ll] 191 | \end{tikzcd}\] 192 | \item As a special case, take $\varphi$ to be a quotient homomorphism $R \to R/I$, we get a natural isomorphism 193 | \[ (R/I)[(S')^{-1}] \simeq R[S^{-1}] \dotimes{R} (R/I) = \frac{R[S^{-1}]}{I[S^{-1}]} \] 194 | from the right exactness of $\otimes$. 195 | \end{itemize} 196 | 197 | We prove an easy yet fundamental property of localizations, namely they are exact functors. 198 | \begin{proposition}[Exactness of localization]\label{prop:localization-exactness} 199 | Let $S$ be any multiplicative subset of $R$. If 200 | \[ \cdots \to M_i \xrightarrow{f_i} M_{i+1} \to \cdots \] 201 | is an exact sequence of $R$-modules, then 202 | \[ \cdots \to M_i[S^{-1}] \xrightarrow{f_i[S^{-1}]} M_{i+1}[S^{-1}] \to \cdots \] 203 | is also exact. 204 | \end{proposition} 205 | \begin{proof} 206 | By homological common sense, we are reduced to the exact sequences 207 | \begin{inparaenum}[(i)] 208 | \item $0 \to M' \to M \to M''$ (i.e. left exactness), 209 | \item $M' \to M \to M'' \to 0$ (i.e. right exactness). 210 | \end{inparaenum} 211 | The case (ii) is known for tensor products in general. 212 | 213 | As for (i), note that for every homomorphism $g: M \to M''$, 214 | \begin{align*} 215 | \Ker\left(g[S^{-1}]\right) & = \left\{ \frac{y}{s} \in M[S^{-1}]: \exists t \in S, \; t g(y)=0 \right\} \xlongequal{\because y/s = ty/ts} \left\{ \frac{y}{s} : y \in \Ker(g) \right\} \\ 216 | & = \Image\left[ \Ker(g)[S^{-1}] \to M[S^{-1}] \right]. 217 | \end{align*} 218 | Thus it remains to show that if $f: M' \hookrightarrow M$, then $f[S^{-1}]: M'[S^{-1}] \to M[S^{-1}]$ is injective as well. If $x/s \mapsto f(x)/s = 0$ under $f[S^{-1}]$, there exists $t \in S$ such that $tf(x) = f(tx) = 0$ in $M$, therefore $tx=0$ in $M'$, but the latter condition implies $x/s = 0$ in $M'[S^{-1}]$. 219 | \end{proof} 220 | 221 | \begin{lemma} 222 | Let $M$ be an $R$-module. The localizations $M \to M_{\mathfrak{m}}$ for various maximal ideals $\mathfrak{m}$ assemble into an injection 223 | \[ M \hookrightarrow \prod_{\mathfrak{m} \in \MaxSpec(R)} M_{\mathfrak{m}}. \] 224 | \end{lemma} 225 | \begin{proof} 226 | Let $m \in M$ be such that $m \mapsto 0 \in M_{\mathfrak{m}}$ for all $\mathfrak{m}$. This means that for all $\mathfrak{m}$ there exists $s \in R \smallsetminus \mathfrak{m}$ such that $sm=0$. Hence the annihilator ideal $\text{ann}_R(m) := \{r \in R: rm=0 \}$ is not contained in any maximal ideal, thus $\text{ann}_R(m) = R$. 227 | \end{proof} 228 | There is an analogue for rings. Observe that when $R$ is an integral domain, all the localizations $R[S^{-1}]$ can be regarded as subrings of the field of fractions $\text{Frac}(R)$. 229 | \begin{lemma} 230 | Let $R$ be an integral domain, then $R = \bigcap_{\mathfrak{m} \in \MaxSpec(R)} R_{\mathfrak{m}}$ as subrings of $\mathrm{Frac}(R)$. 231 | \end{lemma} 232 | \begin{proof} 233 | Only the inclusion $\supset$ requires proof. Let $x \in \text{Frac}(R)$ and define $D := \{r \in R: rx \in R \}$ (the ideal of denominators). Suppose $x \in R_{\mathfrak{m}}$ for all maximal $\mathfrak{m}$, then $D \not\subset \mathfrak{m}$ for all maximal $\mathfrak{m}$. The same reasoning as above leads to $D=R$. 234 | \end{proof} 235 | It will be important to gain finer control on the ideals $\mathfrak{m}$ in the assertion above, say by using some prime ideals ``lower'' than the maximal ones. We will return to this issue later. 236 | 237 | \section{Radicals and Nakayama's lemma} 238 | We begin with two important notions of radicals. The first version is defined as follows. Given an ideal $I$ of $R$, the \emph{nilpotent radical} $\sqrt{I}$ is defined to be 239 | \[ \sqrt{I} := \{ r \in R: \exists n, \; r^n \in I \}. \] 240 | It is readily seen to be an ideal from the binomial identity $(a+b)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} a^k b^{2n-k}$. It should also be clear that $\sqrt{I} \subset R$ equals the preimage of $\sqrt{0} \subset R/I$. \index{nilpotent radical} 241 | 242 | \begin{exercise} 243 | Show that $\sqrt{\sqrt{I}} = I$ for all ideal $I$. 244 | \end{exercise} 245 | 246 | \begin{proposition} 247 | Let $I$ be a proper ideal of $R$. We have 248 | \[ \sqrt{I} = \bigcap_{\substack{\mathfrak{p} \in \Spec(R) \\ \mathfrak{p} \supset I }} \mathfrak{p}. \] 249 | \end{proposition} 250 | \begin{proof} 251 | By replacing $R$ by $R/I$, this is easily reduced to the case $I = \{0\}$. If $r$ is nilpotent and $\mathfrak{p} \in \Spec(R)$, then $r^n = 0 \in \mathfrak{p}$ implies $r \in \mathfrak{p}$. Conversely, suppose that $r$ is not nilpotent. There exists a prime ideal in $R[r^{-1}]$, which comes from some $\mathfrak{p} \in \Spec(R)$ with $r \in R \smallsetminus \mathfrak{p}$ by \eqref{eqn:localization-Spec}. Hence $r \notin \mathfrak{p}$. 252 | \end{proof} 253 | \begin{remark}\index{reduced ring} 254 | A ring $R$ is called \emph{reduced} if $\sqrt{0} = \{0\}$. In any case, $R_\text{red} := R \big/ \sqrt{0}$ is a reduced ring. Furthermore, any reduced quotient of $R$ factors through $R_\text{red}$. 255 | \end{remark} 256 | 257 | The second radical is probably familiar to the readers. The \emph{Jacobson radical} $\text{rad}(R)$ of the ring $R$ is the intersection of all maximal ideals. The previous proposition implies $\text{rad}(R) \supset \sqrt{(0)}$. Note that 258 | \[ a \in \text{rad}(R) \implies (1+a) \in R^\times . \] 259 | Indeed, $1+a$ cannot be contained in any maximal ideal $\mathfrak{m}$, for otherwise $1 = (1+a)-a \in \mathfrak{m}$, which is absurd. \index{Jacobson radical} 260 | 261 | To prove the celebrated Nakayama's Lemma, let us recall an easy variant of the Cayley--Hamilton theorem from linear algebra. 262 | \begin{lemma}\label{prop:Cayley-Hamilton}\index{Cayley--Hamilton theorem} 263 | Suppose that $I \subset R$ is an ideal, $M$ is an $R$-module with generators $x_1, \ldots, x_n$ and $\varphi \in \End_R(M)$ satisfies $\varphi(M) \subset IM$, then there exists a polynomial $P(X) = X^n + a_{n-1} X^{n-1} + \cdots + a_0 \in R[X]$ with $a_i \in I$, such that $P(\varphi) = \varphi^n + a_{n-1} \varphi^{n-1} + \cdots + a_0 = 0$. 264 | \end{lemma} 265 | \begin{proof} 266 | Write $\varphi(x_i) = \sum_{j=1}^n a_{ij} x_j$ where $a_{ij} \in I$. Set $A := (a_{ij})_{1 \leq i,j \leq n} \in \text{Mat}_n(R)$. Regard $M$ as an $R[X]$-module by letting $X$ act as $\varphi$. Then we have the matrix equation over $R[X]$ 267 | \[ (X \cdot \identity_{n \times n} - A) \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix} \] 268 | Multiplying by the cofactor matrix $(X \cdot \identity_M - A)^\vee$ on the left, we see that $P(X) := \det(X \cdot \identity_{n \times n} - A) \in R[X]$ acts as $0$ on each $x_i$, thus on the whole $M$. This is the required polynomial. 269 | \end{proof} 270 | 271 | \begin{theorem}[Nakayama's Lemma]\label{prop:NAK}\index{Nakayama's lemma} 272 | Suppose that $M$ is a finitely generated $R$-module and $I$ is an ideal of $R$ such that $IM = M$. Then there exists $a \in I$ such that $(1+a)M=0$. If $I \subset \text{rad}(R)$, then we have $M = \{0\}$ under these assumptions. 273 | \end{theorem} 274 | \begin{proof} 275 | Write $M = Rx_1 + \cdots + Rx_n$. Plug $\varphi = \identity_M$ into Lemma \ref{prop:Cayley-Hamilton} to deduce that $P(\identity_M) = 1 + \underbracket{a_{n-1} + \cdots + a_0}_{=: a \in I}$ acts as $0$ on $M$. This proves the first part. Assume furthermore that $I \subset \text{rad}(R)$, then $1+a \in R^\times$ so that $M = (1+a)M = 0$. 276 | \end{proof} 277 | 278 | \begin{figure}[h] 279 | \centering \includegraphics[height=180pt]{Nakayama.png} \\ \vspace{1em} 280 | \begin{minipage}{0.7\textwidth} 281 | \small Nakayama's Lemma is named after Tadashi Nakayama (1912---1964). Picture borrowed from \cite{obi-NAK}. 282 | \end{minipage} 283 | \end{figure} 284 | 285 | \begin{corollary}\label{prop:NAK-generation} 286 | Let $M$ be a finitely generated $R$-module, and let $I \subset \mathrm{rad}(R)$ be an ideal of $R$. If the images of $x_1, \ldots, x_n \in M$ in $M/IM$ form a set of generators, then $x_1, \ldots, x_n$ generate $M$. 287 | \end{corollary} 288 | \begin{proof} 289 | Apply Theorem \ref{prop:NAK} to $N := M/(Rx_1 + \cdots + Rx_n)$; our assumption $M = IM + Rx_1 + \cdots + Rx_n$ entails that $IN=N$, thus $N=0$. 290 | \end{proof} 291 | 292 | We record another amusing consequence of Theorem \ref{prop:NAK}. 293 | \begin{proposition} 294 | Let $M$ be a finitely generated $R$-module and $\psi \in \End_R(M)$. If $\psi$ is surjective then $\psi$ is an automorphism. 295 | \end{proposition} 296 | \begin{proof} 297 | Introduce a variable $Y$. Make $M$ into an $R[Y]$-module by letting $Y$ act as $\psi$. Put $I := (Y)$ so that $IM=M$. Theorem \ref{prop:NAK} yields some $Q(Y) \in R[Y]$ satisfying $(1 - Q(Y)Y) M = 0$, that is, $Q(\psi)\psi = \identity_M$. 298 | \end{proof} 299 | 300 | \section{Noetherian and Artinian rings} 301 | An $R$-module $M$ is called \emph{Noetherian} (resp. \emph{Artinian}) if every ascending (resp. descending) chain of submodules eventually stabilizes. Recall that in a short exact sequence $0 \to M' \to M \to M'' \to 0$, we have $M$ is Noetherian (resp. Artinian) if and only if $M'$ and $M''$ are. Being both Noetherian and Artinian is equivalent to being a module of \emph{finite length}\index{length}, that is, a module admitting composition series. \index{Artinian}\index{Noetherian} 302 | 303 | If we take $M := R$ on which $R$ acts by multiplication, then the submodules are precisely the ideals of $R$. We say that $R$ is a Noetherian (resp. Artinian) ring if $R$ as an $R$-module is Noetherian (resp. Artinian); this translates into the corresponding chain conditions on the ideals. Both chain conditions are preserved under passing to quotients and localizations. Finitely generated modules over a Noetherian ring are Noetherian. The following result ought to be known to the readers. 304 | \begin{theorem}[Hilbert's Basis Theorem] \index{Hilbert's basis theorem} 305 | If $R$ is Noetherian, then so is the polynomial algebra $R[X_1, \ldots, X_n]$ for any $n \in \Z_{\geq 1}$. 306 | \end{theorem} 307 | Joint with the foregoing remarks, we infer that finitely generated algebras over Noetherian rings are still Noetherian. 308 | 309 | On the other hand, being Artinian is a rather stringent condition on rings. 310 | \begin{theorem}\label{prop:Artinian-length} 311 | A ring $R$ is Artinian if and only if $R$ is of finite length as an $R$-module. Such rings are semi-local. 312 | \end{theorem} 313 | \begin{proof} 314 | As noticed before, having finite length implies that $R$ is Noetherian as well as Artinian. Assume conversely that $R$ is an Artinian ring. First we claim that $\MaxSpec(R)$ is finite, i.e. $R$ is semi-local. If there were an infinite sequence of distinct maximal ideals $\mathfrak{m}_1, \mathfrak{m}_2, \ldots$, we would have an infinite chain 315 | \[ \mathfrak{m}_1 \supset \mathfrak{m}_1 \mathfrak{m}_2 \supset \mathfrak{m}_1 \mathfrak{m}_2 \mathfrak{m}_3 \supset \cdots. \] 316 | This chain is strictly descending, since $\mathfrak{m}_1 \cdots \mathfrak{m}_i = \mathfrak{m}_1 \cdots \mathfrak{m}_{i+1}$ would imply $\mathfrak{m}_{i+1} \supset \mathfrak{m}_1 \cdots \mathfrak{m}_i$, hence $\mathfrak{m}_{i+1} \supset \mathfrak{m}_j$ for some $1 \leq j \leq i$ because maximal ideals are prime. From the Artinian property we conclude that there are only finitely many maximal ideals $\mathfrak{m}_1, \ldots, \mathfrak{m}_n$ of $R$. 317 | 318 | Set $\mathfrak{a} := \mathfrak{m}_1 \cdots \mathfrak{m}_n$. Since $R$ is Artinian we must have $\mathfrak{a}^k = \mathfrak{a}^{k+1}$ for some $k > 0$. We claim that $\mathfrak{a}^k = 0$. 319 | 320 | Put $\mathfrak{b} := \{r \in R: r\mathfrak{a}^k = 0 \}$, we have to show $\mathfrak{b}=R$. If not, let $\mathfrak{b}'$ be a minimal ideal lying strictly over $\mathfrak{b}$. Thus $\mathfrak{b}' = Rx + \mathfrak{b}$ for any $x \in \mathfrak{b}' \smallsetminus \mathfrak{b}$. We must have $\mathfrak{a}x + \mathfrak{b} \subsetneq \mathfrak{b}'$, for otherwise $M := \mathfrak{b}'/\mathfrak{b}$ is finitely generated (say by $x$) and satisfies $M = \mathfrak{a}M$, then Theorem \ref{prop:NAK} plus $\mathfrak{a} \subset \text{rad}(R)$ would imply $M = \{0\}$, which is absurd. By minimality we have $\mathfrak{b} = \mathfrak{a}x + \mathfrak{b}$. It follows that $\mathfrak{a}x \subset \mathfrak{b}$, i.e. $\mathfrak{a}^{k+1} x = 0$; from $\mathfrak{a}^k = \mathfrak{a}^{k+1}$ we infer $x \in \mathfrak{b}$. Contradiction. 321 | 322 | All in all, we obtain a descending chain of ideals 323 | \begin{align*} 324 | R & \supset \mathfrak{m}_1 \supset \mathfrak{m}_1 \mathfrak{m}_2 \supset \cdots \supset \mathfrak{m}_1 \cdots \mathfrak{m}_n = \mathfrak{a} \\ 325 | & \supset \mathfrak{a} \mathfrak{m}_1 \supset \mathfrak{a} \mathfrak{m}_1 \mathfrak{m}_2 \supset \cdots \supset \mathfrak{a} \mathfrak{m}_1 \cdots \mathfrak{m}_n = \mathfrak{a}^2 \\ 326 | & \supset \cdots \supset \mathfrak{a}^k = \{0\}. 327 | \end{align*} 328 | Each subquotient thereof, which is \textit{a priori} an $R$-module, is actually an $R/\mathfrak{m}_i$-vector space for some $1 \leq i \leq n$. Such a vector space must also satisfy the descending chain condition on vector subspaces, otherwise pulling-back will contradict the Artinian assumption on $R$. Artinian vector spaces must be finite-dimensional. It follows that all these subquotients are of finite length, hence so is $R$ itself. 329 | \end{proof} 330 | 331 | \begin{corollary}\label{prop:Artinian-dim-0} 332 | A ring $R$ is Artinian if and only if it is Noetherian and every prime ideal of $R$ is maximal. 333 | \end{corollary} 334 | \begin{proof} 335 | If $R$ is Artinian, then $R$ is of finite length, hence is Noetherian as well. For every prime ideal $\mathfrak{p}$, we have $\mathfrak{p} \supset \{0\} = (\mathfrak{m}_1 \cdots \mathfrak{m}_n)^k$ in the notations of the proof above, therefore $\mathfrak{p} \supset \mathfrak{m}_i$ for some $i$, so $\mathfrak{p} = \mathfrak{m}_i$ is maximal. 336 | 337 | Conversely, if $R$ is Noetherian but of infinite length, then the nonempty set of ideals 338 | \[ \mathcal{S} := \left\{ \text{ideals } I \subsetneq R: R/I \text{ has infinite length} \right\} \] 339 | contains a maximal element $\mathfrak{p}$. We contend that $\mathfrak{p}$ is prime. If $xy \in \mathfrak{p}$ with $x,y \notin \mathfrak{p}$, then $R/(\mathfrak{p}+Rx)$ has finite length by the choice of $\mathfrak{p}$; on the other hand, $\mathfrak{a} := \{ r \in R: rx \in \mathfrak{p}\} \supsetneq \mathfrak{p}$ (as $y \in \mathfrak{a}$), hence $R/\mathfrak{a}$ has finite length as well. From the short exact sequence $0 \to R/\mathfrak{a} \xrightarrow{x} R/\mathfrak{p} \to R/(\mathfrak{p}+Rx) \to 0$ we see $R/\mathfrak{p}$ has finite length, contradiction. 340 | 341 | If we assume moreover that every prime ideal is maximal, then for the $\mathfrak{p}$ chosen above, $R/\mathfrak{p}$ will be a field, thus of finite length. This is impossible. 342 | \end{proof} 343 | 344 | Rings whose prime ideals are all maximal are said to have \emph{dimension zero}, in the sense of Krull dimensions; we shall return to this point in \S\ref{sec:Krull-dimension}. 345 | 346 | \section{What is commutative algebra?} 347 | In broad terms, \emph{commutative algebra} is the study of commutative rings. Despite its intrinsic beauty, we prefer to motivate from an external point of view. See also \cite{Eis95}. 348 | 349 | \begin{asparaenum}[(A)] 350 | \item \textbf{Algebraic geometry}\index{variety}. To simplify matters, we consider affine algebraic varieties over an algebraically closed field $\Bbbk$. Roughly speaking, such a variety is the zero locus $\mathcal{X} = \{ f_1 = \cdots = f_r = 0 \}$ in $\mathbb{A}^n = \Bbbk^n$ of $f_i \in \Bbbk[X_1, \ldots, X_n]$. The choice of equations is of course non-unique: what matters is the ideal $I(\mathcal{X}) := \{ f \in \Bbbk[X_1, \ldots, X_n] : f|_{\mathcal{X}} = 0 \}$. Conversely, every ideal $\mathfrak{a}$ defines a subset $V(\mathfrak{a}) := \{x \in \Bbbk^n : \forall f \in \mathfrak{a}, \; f(x)=0 \}$. As consequences of Hilbert's Nullstellensatz\index{Nullstellensatz}, which we will discuss later, we have 351 | \[ V \circ I(\mathcal{X}) = \mathcal{X}, \qquad I \circ V(\mathfrak{a}) = \sqrt{\mathfrak{a}}. \] 352 | One can deduce from this that the (closed) subvarieties of $\mathbb{A}^n$ are in bijection with ideals $\mathfrak{a}$ satisfying $\sqrt{\mathfrak{a}} = \mathfrak{a}$. 353 | 354 | Furthermore, $\Bbbk[\mathcal{X}] := \Bbbk[X_1, \ldots, X_n]/I(\mathcal{X})$ may be regarded as the ring of ``regular functions'' (i.e. functions definable by means of polynomials) on $\mathcal{X}$, and $\MaxSpec(\Bbbk[\mathcal{X}])$ is in bijection with the points of $\mathcal{X}$: to $x = (x_1, \ldots, x_n) \in \mathcal{X}$ we attach 355 | \[ \mathfrak{m}_x = (X_1 - x_1, \ldots, X_n - x_n) \supset I(\mathcal{X}). \] 356 | By passing to the ring $\Bbbk[\mathcal{X}]$, we somehow obtain a description of $\mathcal{X}$ that is independent of embeddings into affine spaces. Moreover, $\mathcal{X}$ inherits the Zariski topology from that on $\Spec(\Bbbk[\mathcal{X}])$. 357 | 358 | Naively speaking, the geometric properties of $\mathcal{X}$ transcribe in ring-theoretic terms to the reduced Noetherian $\Bbbk$-algebra $\Bbbk[\mathcal{X}]$. For example, assume that $f \in \Bbbk[\mathcal{X}]$ is not nilpotent, then the formation of $\Bbbk[\mathcal{X}][f^{-1}]$ corresponds to taking the Zariski-open subset $\mathcal{X}_f = \{x \in \mathcal{X}: f(x) \neq 0 \}$ of $\mathcal{X}$. This may be explained as follows: 359 | \[ \mathcal{X}_f \simeq \left\{ (x_1, \ldots, x_n, y) : f_1(x_1, \ldots, x_n) = \cdots = f_r(x_1, \ldots, x_n) = 0, \; f(x_1, \ldots, x_n)y = 1 \right\} \] 360 | which is also an affine algebraic variety in $\Bbbk^{n+1}$, and one may verify that $\Bbbk[\mathcal{X}_f] = \Bbbk[\mathcal{X}][f^{-1}]$. 361 | 362 | \item \textbf{Invariant theory}\index{invariant theory}. Let $G$ be a group acting on a finite-dimensional $\Bbbk$-vector space $V$ from the right, and let $\Bbbk[V]$ be the $\Bbbk$-algebra of polynomials on $V$. Thus $\Bbbk[V]$ carries a left $G$-action by $gf(v) = f(vg)$. For ``reasonable'' groups $G$, say finite or $\Bbbk$-algebraic ones, the classical invariant theory seeks to describe the subalgebra $\Bbbk[V]^G$ of invariants\footnote{More generally, we are also interested in the algebra of invariant differential operators with polynomial coefficients.} in terms of \emph{generators} and \emph{relations}. 363 | 364 | In particular, one has to know when is the algebra $\Bbbk[V]^G$ finitely generated. This is actually the source of many results in commutative algebra, such as the Basis Theorem and Nullstellensatz of Hilbert. For example, let the symmetric group $G = \mathfrak{S}_n$ act on $V = \Bbbk^n$ in the standard manner, then our question is completely answered by the following classical result: $\Bbbk[V]^G$ equals the polynomial algebra $\Bbbk[e_1, \ldots, e_n]$, where $e_i$ stands for the $i$-th elementary symmetric function in $n$ variables. 365 | 366 | The same questions may be posed for any affine algebraic variety $V$. From the geometric point of view, if $\Bbbk[V]^G$ is finitely generated, it will consist of regular functions of some kind of quotient variety $V /\!/ G$. The study of quotients in this sense naturally leads to \emph{geometric invariant theory}, for which we refer to \cite{AG4} for details. 367 | \end{asparaenum} 368 | -------------------------------------------------------------------------------- /YAlg3-2.tex: -------------------------------------------------------------------------------- 1 | % To be compiled by XeLaTeX, preferably under TeX Live. 2 | % LaTeX source for ``Yanqi Lake Lectures on Algebra'' Part III. 3 | % Copyright 2019 李文威 (Wen-Wei Li). 4 | % Permission is granted to copy, distribute and/or modify this 5 | % document under the terms of the Creative Commons 6 | % Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) 7 | % https://creativecommons.org/licenses/by-nc/4.0/ 8 | 9 | % To be included 10 | \chapter{Primary decompositions} 11 | 12 | We shall follow \cite[\S 3]{Eis95} and \cite[\S 8]{Mat80} closely. 13 | 14 | \section{The support of a module} 15 | Fix a ring $R$. For any $R$-module $M$ and $x \in M$, we define the \emph{annihilator} $\text{ann}_R(x) := \{ r \in R: rx=0 \}$; it is an ideal of $R$. Also define $\text{ann}_R(M) := \{r \in R: rM=0 \} = \bigcap_{x \in M} \text{ann}_R(x)$. \index{$\text{ann}_R(M), \text{ann}_R(x)$} 16 | 17 | \begin{definition}\index{support}\index{SuppM@$\Supp(M)$} 18 | The \emph{support} of an $R$-module $M$ is 19 | \[ \Supp(M) := \left\{ \mathfrak{p} \in \Spec(R): M_{\mathfrak{p}} \neq 0 \right\}. \] 20 | \end{definition} 21 | Let us unwind the definition: $\mathfrak{p} \notin \Supp(M)$ means that every $x \in M$ maps to $0 \in M_{\mathfrak{p}}$, equivalently $\text{ann}_R(x) \not\subset \mathfrak{p}$. To test whether $\mathfrak{p} \notin \Supp(M)$, we only need to check the foregoing condition for $x$ ranging over a generating set of $M$. 22 | 23 | \begin{proposition} 24 | If $M$ is finitely generated then $\Supp(M) = V(\mathrm{ann}_R(M))$; in particular it is Zariski-closed in $\Spec(R)$. 25 | \end{proposition} 26 | \begin{proof} 27 | Suppose $M = Rx_1 + \cdots + Rx_n$. Set $I_i := \text{ann}_R(x_i)$ and observe that $\bigcap_{i=1}^n I_i = \text{ann}_R(M)$. Then $\mathfrak{p} \in \Supp(M)$ if and only if $\mathfrak{p} \supset I_i$ for some $i$, that is, $\mathfrak{p} \in \bigcup_{i=1}^n V(I_i)$. To conclude, note that $V(I) \cup V(J) = V(IJ) = V(I \cap J)$ for any ideals $I, J \subset R$ (an easy exercise). 28 | \end{proof} 29 | 30 | Finite generation is needed in the result above. Consider the $\Z$-module $M := \bigoplus_{a \geq 1} \Z/p^a \Z$. Its support is $\{p\}$ but $\mathrm{ann}(M) = \{0\}$. 31 | 32 | \begin{proposition} 33 | For an exact sequence $0 \to M' \to M \to M'' \to 0$ we have $\Supp(M) = \Supp(M') \cup \Supp(M'')$. For arbitrary direct sums we have $\Supp(\bigoplus_i M_i) = \bigcup_i \Supp(M_i)$. 34 | \end{proposition} 35 | \begin{proof} 36 | Again, we use the exactness of localization for the first assertion. For $\mathfrak{p} \in \Spec(R)$ we have an exact $0 \to M'_{\mathfrak{p}} \to M_{\mathfrak{p}} \to M''_{\mathfrak{p}} \to 0$, hence $M_{\mathfrak{p}} \neq 0$ if and only if $\mathfrak{p} \in \Supp(M') \cup \Supp(M'')$. The second assertion is obvious. 37 | \end{proof} 38 | 39 | From an $R$-module $M$, one can build a ``field of modules'' over $\Spec(R)$ by assigning to each $\mathfrak{p}$ the $R_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$, and $\Supp(M)$ is precisely the subset of $M_{\mathfrak{p}}$ over which the field is non-vanishing. This is how the support arises in algebraic geometry. A more precise description will require the notion of quasi-coherent sheaves on schemes. 40 | 41 | \begin{exercise} 42 | Let $M$, $N$ be finitely generated $R$-modules. Show that $\Supp(M \dotimes{R} N) = \Supp(M) \cap \Supp(N)$. Hint: since localization commutes with $\otimes$, it suffices to prove that $M \dotimes{R} N \neq \{0\}$ when $M, N$ are both nonzero finitely generated modules over a local ring $R$. Nakayama's Lemma implies that $M \dotimes{R} \Bbbk$ and $\Bbbk \dotimes{R} N$ are both nonzero where $\Bbbk$ is the residue field of $R$. Now 43 | \[ (M \dotimes{R} \Bbbk) \dotimes{\Bbbk} (\Bbbk \dotimes{R} N) \simeq M \dotimes{R} (\Bbbk \dotimes{\Bbbk} \Bbbk) \dotimes{R} N \simeq (M \dotimes{R} N) \dotimes{R} \Bbbk \] 44 | is nonzero. 45 | \end{exercise} 46 | 47 | \section{Associated primes} 48 | Throughout this section, $R$ will be a Noetherian ring. 49 | 50 | \begin{definition}\index{associated prime}\index{$\text{Ass}(M)$} 51 | A prime ideal $\mathfrak{p}$ is said to be an \emph{associated prime} of an $R$-module $M$ if $\text{ann}_R(x) = \mathfrak{p}$ for some $x \in M$; equivalently, $R/\mathfrak{p}$ embeds into $M$. Denote the set of associated primes of $M$ by $\text{Ass}(M)$. 52 | \end{definition} 53 | 54 | \begin{example} 55 | For the $\Z$-module $M := \Z/n\Z$ with $n \in \Z_{> 1}$, one easily checks that $\text{Ass}(M)$ is the set of prime factors of $n$. 56 | \end{example} 57 | 58 | \begin{lemma}\label{prop:Ass-maximal} 59 | Consider the set $\mathcal{S} := \{ \mathrm{ann}_R(x) : x \in M, \; x \neq 0 \}$ of ideals, partially ordered by inclusion. Every maximal element in $\mathcal{S}$ is prime. 60 | \end{lemma} 61 | \begin{proof} 62 | Let $\mathfrak{p} = \text{ann}_R(x)$ be a maximal element of $\mathcal{S}$ and suppose $ab \in \mathfrak{p}$. If $b \notin \mathfrak{p}$, then 63 | \[ bx \neq 0, \quad abx = 0, \quad R \neq \text{ann}_R(bx) \supset \text{ann}_R(x) = \mathfrak{p}. \] 64 | Hence $a \in \text{ann}_R(bx) = \mathfrak{p}$ by the maximality of $\mathfrak{p}$ in $\mathcal{S}$. 65 | \end{proof} 66 | 67 | \begin{definition}\index{zero divisor} 68 | Call $r \in R$ a \emph{zero divisor} on $M$ if $rx=0$ for some $x \in M \smallsetminus \{0\}$. 69 | \end{definition} 70 | 71 | \begin{theorem}\label{prop:Ass-properties} 72 | Let $M$ be an $R$-module. 73 | \begin{enumerate}[(i)] 74 | \item We have $M = \{0\}$ if and only if $\mathrm{Ass}(M) = \emptyset$. 75 | \item The union of all $\mathfrak{p} \in \mathrm{Ass}(M)$ equals the set of zero divisors on $M$. 76 | \item For any multiplicative subset $S \subset R$, we have 77 | \[ \mathrm{Ass}(M[S^{-1}]) = \left\{ \mathfrak{p}[S^{-1}] : \mathfrak{p} \in \mathrm{Ass}(M), \; \mathfrak{p} \cap S = \emptyset \right\} . \] 78 | \item If $0 \to M' \to M \to M''$ is exact, then $\mathrm{Ass}(M') \subset \mathrm{Ass}(M) \subset \mathrm{Ass}(M') \cup \mathrm{Ass}(M'')$. 79 | \end{enumerate} 80 | \end{theorem} 81 | \begin{proof} 82 | \begin{asparaenum}[(i)] 83 | \item Clearly $\text{Ass}(\{0\}) = \emptyset$. If $M \neq 0$, the set $\mathcal{S}$ in Lemma \ref{prop:Ass-maximal} is then nonempty, hence contains a maximal element $\mathfrak{p}$ because $R$ is Noetherian; this yields $\mathfrak{p} \in \text{Ass}(M)$. 84 | 85 | \item Elements of any $\mathfrak{p} \in \text{Ass}(M)$ are all zero divisors by the very definition of associated primes. Conversely, if $r \in \text{ann}_R(x)$ for some $x \in M \smallsetminus \{0\}$, there must exist some maximal element $\mathfrak{p}$ of $\mathcal{S}$ with $\mathfrak{p} \supset \text{ann}_R(x)$ as $R$ is Noetherian; so $\mathfrak{p}$ is the required associated prime containing $r$. 86 | 87 | \item If $\mathfrak{p} \in \Spec(R)$, $\mathfrak{p} \cap S = \emptyset$ and there is some $R/\mathfrak{p} \hookrightarrow M$, then 88 | \[ \{0\} \neq R[S^{-1}]/\mathfrak{p}[S^{-1}] \simeq (R/\mathfrak{p})[S^{-1}] \hookrightarrow M[S^{-1}] \] 89 | by the exactness of localizations, hence $\mathfrak{p}[S^{-1}] \in \text{Ass}(M[S^{-1}])$. Conversely, every element of $\text{Ass}(M[S^{-1}])$ has the form $\mathfrak{p}[S^{-1}]$ for some $\mathfrak{p} \in \Spec(R)$ disjoint from $S$. Also recall that $\mathfrak{p}$ equals the preimage of $\mathfrak{p}[S^{-1}]$ under $R \to R[S^{-1}]$. There exist $x \in M$ and $s \in S$ such that $\mathfrak{p}[S^{-1}] = \text{ann}_{R[S^{-1}]}(x/s) = \text{ann}_{R[S^{-1}]}(x)$. Ideals in a Noetherian ring being finitely generated, we infer that $\exists t \in S$ with $\mathfrak{p} \subset \text{ann}_R(tx)$. It remains to show $\mathfrak{p} = \text{ann}_R(tx)$. If $rtx=0$ for some $r \in R$, then $r$ maps into $r/1 \in \mathfrak{p}[S^{-1}] = \text{ann}_{R[S^{-1}]}(x)$; thus $r \in \mathfrak{p}$. 90 | 91 | \item It suffices to treat the second $\subset$. Suppose that $\mathfrak{p} \in \text{Ass}(M)$ and $R/\mathfrak{p} \simeq N$ for some submodule $N \subset M$. Identify $M'$ with $\Ker(M \to M'')$. If $M' \cap N = \{0\}$ then $R/\mathfrak{p} \simeq N \hookrightarrow M''$, so $\mathfrak{p} \in \text{Ass}(M'')$. If there exists $x \in M' \cap N \subset N$ with $x \neq 0$, then we have $\text{ann}_R(x) = \mathfrak{p}$ since $N \simeq R/\mathfrak{p}$ and $\mathfrak{p}$ is prime; in this case $\mathfrak{p} \in \text{Ass}(M')$. 92 | \end{asparaenum} 93 | \end{proof} 94 | 95 | We remark that $0 \to M' \to M \to M'' \to 0$ being exact does not imply $\text{Ass}(M) = \text{Ass}(M') \cup \text{Ass}(M'')$. To see this, consider $R=\Z$ and $0 \to \Z \xrightarrow{p} \Z \to \Z/p\Z \to 0$ for some prime number $p$. 96 | 97 | \begin{exercise} 98 | Show that $\text{Ass}(M_1 \oplus M_2) = \text{Ass}(M_1) \cup \text{Ass}(M_2)$. 99 | \end{exercise} 100 | 101 | \begin{theorem}\label{prop:Supp-Ass} 102 | For every $R$-module $M$ we have $\Supp(M) = \bigcup_{\mathfrak{p} \in \mathrm{Ass}(M)} V(\mathfrak{p})$, in particular $\mathrm{Ass}(M) \subset \Supp(M)$. Furthermore, every minimal element of $\Supp(M)$ with respect to inclusion is actually a minimal element of $\mathrm{Ass}(M)$. 103 | \end{theorem} 104 | \begin{proof} 105 | For any prime $\mathfrak{q}$ we have $M_{\mathfrak{q}} \neq 0 \iff \text{Ass}(M_{\mathfrak{q}}) \neq \emptyset$, and the latter condition holds precisely when there exists $\mathfrak{p} \in \text{Ass}(M)$ with $\mathfrak{p} \cap (R \smallsetminus \mathfrak{q}) = \emptyset$, i.e. $\mathfrak{q} \supset \mathfrak{p}$. This proves the first assertion. The second assertion is a direct consequence. 106 | \end{proof} 107 | 108 | Observe that if $\mathfrak{p} \in \Supp(M)$, then $\mathfrak{q} \supset \mathfrak{p} \implies \mathfrak{q} \in \Supp(M)$: the reason is that 109 | \begin{gather}\label{eqn:localization-in-stages} 110 | M_{\mathfrak{p}} = (M_{\mathfrak{q}})_{\mathfrak{p}R_{\mathfrak{q}}} 111 | \end{gather} 112 | thus the occurrence of non-minimal elements in $\Supp(M)$ is unsurprising. In contrast, the non-minimal elements in $\text{Ass}(M)$ are somehow mysterious. These non-minimal associated primes are called \emph{embedded primes}. \index{embedded prime} 113 | 114 | \begin{exercise} 115 | Prove the formula \eqref{eqn:localization-in-stages} of ``localization in stages''. 116 | \end{exercise} 117 | 118 | Hereafter we impose finite generation on $M$. This implies $M$ is Noetherian. 119 | \begin{theorem} 120 | Let $M$ be a finitely generated $R$-module. There exists a chain $M = M_n \supset M_{n-1} \supset \cdots \supset M_0 = \{0\}$ of submodules such that for every $0 < i \leq n$, the subquotient $M_i/M_{i-1}$ is isomorphic to $R/\mathfrak{p}_i$ for some prime ideal $\mathfrak{p}_i$. 121 | 122 | Furthermore we have $\mathrm{Ass}(M) \subset \{\mathfrak{p}_1, \ldots, \mathfrak{p}_n \}$; in particular $\mathrm{Ass}(M)$ is a finite set. 123 | \end{theorem} 124 | \begin{proof} 125 | Assume $M \supsetneq M_0 := \{0\}$. There exists $\mathfrak{p}_1 \in \text{Ass}(M)$ together with a submodule $M_1 \subset M$ isomorphic to $R/\mathfrak{p}_1$. Furthermore $\text{Ass}(M_1) = \text{Ass}(R/\mathfrak{p}_1) = \{ \mathfrak{p}_1\}$ as easily seen. Hence the Theorem \ref{prop:Ass-properties} entails $\text{Ass}(M) \subset \{\mathfrak{p}_1\} \cup \text{Ass}(M/M_1)$. 126 | 127 | If $M_1 = M$ we are done. Otherwise we start over with $M/M_1$, finding $M_1 \subset M_2 \subset M$ with $M_2/M_1 \simeq R/\mathfrak{p}_2$ where $\mathfrak{p}_2 \in \text{Ass}_R(M/M_1)$, and so forth. This procedure terminates in finite steps since $M$ is Noetherian. 128 | \end{proof} 129 | 130 | \section{Primary and coprimary modules} 131 | The classical framework of primary decompositions concerns ideals, but it is advantageous to allow modules here. As before, the ring $R$ is Noetherian. 132 | 133 | \begin{definition}\index{coprimary module}\index{primary module} 134 | An $R$-module $M$ is called \emph{coprimary} if $\text{Ass}(M)$ is a singleton. A submodule $N \subsetneq M$ is called a $\mathfrak{p}$-\emph{primary} submodule if $M/N$ is coprimary with associated prime $\mathfrak{p} \in \Spec(R)$. 135 | \end{definition} 136 | 137 | \begin{proposition}\label{prop:coprimary-locnil} 138 | The following are equivalent for a nonzero $R$-module $M$. 139 | \begin{enumerate}[(i)] 140 | \item $M$ is coprimary; 141 | \item for every zero divisor $r \in R$ for $M$ and every $x \in M$, there exists $n \geq 1$ such that $r^n x = 0$. 142 | \end{enumerate} 143 | \end{proposition} 144 | The condition (ii) is usually called the local-nilpotency of $r$ on $M$. When $M = R/I$ for some ideal $I$, it translates into: all zero divisors of the ring $R/I$ are nilpotent. 145 | \begin{proof} 146 | (i) $\implies$ (ii): Suppose $\text{Ass}(M) = \{ \mathfrak{p}\}$ and $x \in M \smallsetminus \{0\}$. From $\emptyset \neq \text{Ass}(Rx) \subset \text{Ass}(M)$ we infer that $\text{Ass}(Rx) = \{\mathfrak{p}\}$, thus by Theorem \ref{prop:Supp-Ass} we see $V(\mathfrak{p}) = \Supp(Rx) = V(\text{ann}(Rx))$. Hence $\mathfrak{p} = \sqrt{\text{ann}(Rx)}$. This implies (ii) by the definition of $\sqrt{\hspace{5pt}}$. 147 | 148 | (ii) $\implies$ (i): It is routine to check that 149 | \[ \mathfrak{p} := \left\{ r \in R: \forall x \in M \; \exists n \geq 1, \; r^n x = 0 \right\} \] 150 | is an ideal of $R$. For every $\mathfrak{q} \in \text{Ass}(M)$ there exists $x \in M$ with $\text{ann}_R(x) = \mathfrak{q}$. Every $r \in \mathfrak{p}$ has some power falling in $\mathfrak{q}$, thus $\mathfrak{p} \subset \mathfrak{q}$. Conversely, (ii) and Theorem \ref{prop:Ass-properties} imply $\mathfrak{q} = \text{ann}_R(x) \subset \mathfrak{p}$. From $\mathfrak{q}=\mathfrak{p}$ we conclude $M$ is coprimary with the unique associated prime $\mathfrak{p}$. 151 | \end{proof} 152 | 153 | \begin{exercise}[Classical definition of primary ideals]\label{ex:primary-ideal} 154 | Let $I$ be a proper ideal of $R$. Show that $R/I$ is coprimary if and only if the following holds: 155 | \[ \forall a,b \in R, \; (ab \in I) \wedge (a \notin I) \implies \exists n \geq 1, \; b^n \in I. \] 156 | In this case we also say $I$ is a \emph{primary ideal} of $R$. Show that $\{\sqrt{I}\} = \text{Ass}(R/I)$ if $I$ is a primary ideal. Hint: apply Proposition \ref{prop:coprimary-locnil}. 157 | \end{exercise} 158 | 159 | \begin{exercise}\label{ex:maximal-primary} 160 | Let $\mathfrak{m}$ be a maximal ideal of $R$. Show that every ideal $I \subsetneq R$ containing some power of $\mathfrak{m}$ is primary, and $\text{Ass}(R/I) = \{\mathfrak{m}\}$. Hint: show that $\mathfrak{m}$ is the only prime ideal containing $I = \text{ann}_R(R/I)$. 161 | \end{exercise} 162 | 163 | \begin{lemma}\label{prop:intersection-primary} 164 | Let $\mathfrak{p} \in \Spec(R)$ and $N_1, N_2 \subset M$ are $\mathfrak{p}$-primary submodules. Then $N_1 \cap N_2$ is a $\mathfrak{p}$-primary submodule of $M$. 165 | \end{lemma} 166 | \begin{proof} 167 | We have $M/N_1 \cap N_2 \hookrightarrow M/N_1 \oplus M/N_2$. Since $N_1 \cap N_2 \neq M$, we have 168 | \[ \emptyset \neq \text{Ass}(M/N_1 \cap N_2) \subset \text{Ass}(M/N_1) \cup \text{Ass}(M/N_2) = \{\mathfrak{p}\} \] 169 | by Theorem \ref{prop:Ass-properties}. 170 | \end{proof} 171 | 172 | \section{Primary decomposition: the main theorem} 173 | We still assume $R$ Noetherian and fix a finitely generated $R$-module $M$. 174 | 175 | \begin{theorem}[Lasker--Noether]\label{prop:primary-decomp}\index{primary decomposition} 176 | Let $N \subsetneq M$ be an $R$-submodule. Then we can express $N$ as 177 | \[ N = M_1 \cap \cdots \cap M_n, \] 178 | for some $n \geq 1$ and primary $R$-submodules $M_i$, say with $\mathrm{Ass}(M/M_i) = \{\mathfrak{p}_i\}$ for $i = 1, \ldots, n$. Such a decomposition is called a \emph{primary decomposition} of $N$. We say it is \emph{irredundant} if none of the $M_i$ can be dropped, and \emph{minimal} if there is no such decomposition with fewer terms. 179 | \begin{enumerate}[(i)] 180 | \item We have $\mathrm{Ass}(M/N) \subset \{ \mathfrak{p}_1, \ldots, \mathfrak{p}_n \}$, equality holds when the decomposition is irredundant. 181 | \item If the decomposition is minimal, then for every $\mathfrak{p} \in \mathrm{Ass}(M/N)$ there exists a unique $1 \leq i \leq n$ with $\mathfrak{p} = \mathfrak{p}_i$; consequently $n = |\mathrm{Ass}(M/N)|$. 182 | \item Consider a primary decomposition of $N$. Let $S \subset R$ be any multiplicative subset, and assume without loss of generality that $\mathfrak{p}_1, \ldots, \mathfrak{p}_m$ are the primes among $\{\mathfrak{p}_1, \ldots, \mathfrak{p}_n \}$ which are disjoint from $S$, then $m \geq 1 \iff N[S^{-1}] \subsetneq M[S^{-1}]$ and 183 | \[ N[S^{-1}] = M_1[S^{-1}] \cap \cdots \cap M_m[S^{-1}] \] 184 | is a primary decomposition of the $R[S^{-1}]$-submodule $N[S^{-1}] \subsetneq M[S^{-1}]$; this decomposition of $N[S^{-1}]$ is minimal if the one for $N$ is. 185 | \end{enumerate} 186 | \end{theorem} 187 | Note that the ``irredundant'' condition in \cite[(8.D)]{Mat80} corresponds to minimality here. The rings whose ideals all have primary decompositions are called \emph{Laskerian rings}, thus part (i) of the Theorem says Noetherian implies Laskerian, but there exist non-Noetherian examples. 188 | 189 | \begin{proof} 190 | Establish the existence of primary decompositions first. Replacing $M$ by $M/N$, we may assume $N = \{0\}$ from the outset. We claim that 191 | \begin{gather}\label{eqn:Lasker-Noether-aux} 192 | \forall \mathfrak{p} \in \text{Ass}(M), \; \exists Q(\mathfrak{p}) \subset M, \; 193 | \left\{ \begin{array}{l} 194 | Q(\mathfrak{p}) \text{ is } \mathfrak{p}\text{-primary}, \\ 195 | \text{Ass}(Q(\mathfrak{p})) = \text{Ass}(M) \smallsetminus \{ \mathfrak{p} \}. 196 | \end{array}\right. 197 | \end{gather} 198 | Granting this, $Q := \bigcap_{\mathfrak{p} \in \text{Ass}(M)} Q(\mathfrak{p})$ yields the required decomposition since $\text{Ass}(M)$ is finite and $\text{Ass}(Q) = \emptyset$. 199 | 200 | Establish \eqref{eqn:Lasker-Noether-aux} as follows. Put $\Psi := \{ \mathfrak{p} \}$. By Zorn's Lemma\index{Zorn's Lemma} we get a maximal element $Q(\mathfrak{q})$ from the set 201 | \[ \emptyset \neq \left\{ Q \subset M: \text{submodule}, \; \text{Ass}(Q) \subset \text{Ass}(M) \smallsetminus \Psi \right\} \] 202 | which is partially ordered by inclusion (details omitted, and you may also use the Noetherian property of $M$). Since 203 | \[ \text{Ass}(M) \subset \text{Ass}(M/Q(\mathfrak{p})) \cup \text{Ass}(Q(\mathfrak{p})), \] 204 | it suffices to show $\text{Ass}(M/Q(\mathfrak{p})) \subset \Psi$. Let $\mathfrak{q} \in \text{Ass}(M/Q(\mathfrak{p}))$ so that there exists $M \supset Q' \supset Q(\mathfrak{p})$ with $Q'/Q(\mathfrak{p}) \simeq R/\mathfrak{q}$. Since $\text{Ass}(Q') \subset \text{Ass}(Q(\mathfrak{p})) \cup \{ \mathfrak{q} \}$, maximality forces $\mathfrak{q} \in \Psi$ (otherwise $\text{Ass}(Q') \subset \text{Ass}(M) \smallsetminus \Psi$), whence \eqref{eqn:Lasker-Noether-aux}. Now we turn to the properties (i) --- (iii). 205 | 206 | \begin{asparaenum}[(i)] 207 | \item The obvious embedding 208 | \[ M/N \hookrightarrow \bigoplus_{i=1}^n M/M_i \] 209 | together with Theorem \ref{prop:Ass-properties} yield $\text{Ass}(M/N) \subset \bigcup_{i=1}^n \text{Ass}(M/M_i) = \{ \mathfrak{p}_1, \ldots, \mathfrak{p}_n \}$. 210 | 211 | Now assume the given primary decomposition is irredundant, we have 212 | \begin{align*} 213 | \{0\} & \neq \frac{M_2 \cap \cdots \cap M_n}{N} = \frac{M_2 \cap \cdots \cap M_n}{M_1 \cap (M_2 \cap \cdots \cap M_n)} \\ 214 | & \simeq \frac{M_1 + M_2 \cap \cdots \cap M_n}{M_1} \hookrightarrow M/M_1. 215 | \end{align*} 216 | Thus $\text{Ass}(M/N)$ contains $\text{Ass}((M_2 \cap \cdots \cap M_n)/N) = \{\mathfrak{p}_1\}$. Same for $\mathfrak{p}_2, \ldots, \mathfrak{p}_n$. 217 | \item Suppose $N = M_1 \cap \cdots \cap M_n$ is an irredundant primary decomposition, so that $\text{Ass}(M/N) = \{ \mathfrak{p}_1, \ldots, \mathfrak{p}_n \}$. If $\mathfrak{p}_i = \mathfrak{p}_j$ for some $1 \leq i \neq j \leq n$, Lemma \ref{prop:intersection-primary} will imply that $M_i \cap M_j$ is primary, leading to a shorter primary decomposition. This is impossible when the primary decomposition is minimal. 218 | \item Suppose $S \cap \mathfrak{p}_i = \emptyset$ (equivalently, $i \leq m$). By Theorem \ref{prop:Ass-properties} and the exactness of localization, $M_i[S^{-1}] \subset M[S^{-1}]$ will be $\mathfrak{p}_i[S^{-1}]$-primary. On the other hand $S \cap \mathfrak{p}_i \neq \emptyset$ implies $\text{Ass}((M/M_i)[S^{-1}]) = \emptyset$ by Theorem \ref{prop:Ass-properties}, thus $M[S^{-1}]/M_i[S^{-1}] = 0$. Since localization respects intersections, we obtain 219 | \[ N[S^{-1}] = \bigcap_{i=1}^m M_i[S^{-1}]. \] 220 | In particular $N[S^{-1}]$ is proper if and only if $m \geq 1$. 221 | 222 | It remains to show $\text{Ass}(M[S^{-1}]/N[S^{-1}])$ has $m$ elements if the original primary decomposition is minimal. Indeed, that set is just $\{ \mathfrak{p}[S^{-1}] : \mathfrak{p} \in \text{Ass}(M/N), \; \mathfrak{p} \cap S = \emptyset \}$ by Theorem \ref{prop:Ass-properties}, which equals $\{ \mathfrak{p}_1[S^{-1}], \ldots, \mathfrak{p}_m[S^{-1}] \}$ (distinct) by (ii). 223 | \end{asparaenum} 224 | \end{proof} 225 | 226 | A natural question arises: to what extent are minimal primary decompositions unique? For those $M_i$ whose associated primes are minimal in $\text{Ass}(M/N)$, the answer turns out to be positive. 227 | 228 | \begin{corollary}\label{prop:p-primary-component} 229 | Let $N = M_1 \cap \cdots \cap M_n$ be a minimal primary decomposition of $N \subsetneq M$. Suppose that $M_1$ is $\mathfrak{p}$-primary where $\mathfrak{p} := \mathfrak{p}_1$ is a minimal element in $\mathrm{Ass}(M/N)$, then $M_1$ equals the preimage of $N_{\mathfrak{p}}$ under $M \to M_{\mathfrak{p}}$. Call it the $\mathfrak{p}$-\emph{primary component} of $N$. 230 | \end{corollary} 231 | \begin{proof} 232 | Recall the proof of Theorem \ref{prop:primary-decomp}, especially the part (iii); here we localize with respect to $S := R \smallsetminus \mathfrak{p}$. The minimality assumption entails 233 | \[ N_{\mathfrak{p}} = M_{1, \mathfrak{p}} \subset M_{\mathfrak{p}}. \] 234 | It remains to show that the preimage of $M_{1, \mathfrak{p}}$ under $M \to M_{\mathfrak{p}}$ equals $M_1$, in other words the injectivity of the natural map $M/M_1 \to (M/M_1)_{\mathfrak{p}} = M_{\mathfrak{p}}/M_{1, \mathfrak{p}}$. Indeed, $\bar{x} \in M/M_1$ maps to $0$ if and only if there exists $s \notin \mathfrak{p}$ with $s\bar{x}=0$, but Theorem \ref{prop:Ass-properties} implies that the zero divisors of $M/M_1$ must lie in $\mathfrak{p}$. 235 | \end{proof} 236 | 237 | It follows that the non-uniqueness of minimal primary decompositions can only arise from embedded primes in $\text{Ass}(M/N)$. 238 | 239 | \section{Examples and remarks} 240 | Primary decompositions are most often applied in the case $M = R$ and $N = I$ is a proper ideal. The goal is to express $I$ as an intersection of primary ideals. To begin with, let us take $R = \Z$. Observations: 241 | \begin{itemize} 242 | \item The primary ideals of $R$ take the form $(p)^n$, where $n \geq 1$ and $p$ is a prime number or zero. This may be deduced from Exercise \ref{ex:primary-ideal} or directly from definitions. 243 | \item The irredundant primary decompositions of $\Z/n\Z$, for $n > 1$, corresponds to the factorization of $n$ into prime-powers. There are no embedded primes in this case; the irredundant primary decomposition is unique and automatically minimal. 244 | \end{itemize} 245 | 246 | \begin{exercise} 247 | Justify the foregoing assertions. 248 | \end{exercise} 249 | 250 | \begin{figure}[h] 251 | \centering \includegraphics[height=170pt]{ELasker.jpg} \\ \vspace{1em} 252 | \begin{minipage}{0.7\textwidth} 253 | \small Emanuel Lasker (1868--1941) first obtained the primary decomposition for finitely generated $\Bbbk$-algebras and the algebras of convergent power series in 1905. His method involves techniques from \emph{elimination theory}. His result is then generalized and rewritten by Emmy Noether in 1921, in which the ascending chain condition plays a pivotal role. Lasker is best known for being the World Chess Champion from 1894 to 1921. (Picture taken from \href{https://commons.wikimedia.org/w/index.php?curid=5676713}{Wikimedia Commons}) 254 | \end{minipage} 255 | \end{figure} 256 | 257 | Therefore one can regard primary decompositions as a generalization of factorization of integers, now performed on the level of ideals. The most important case is $R = \Bbbk[X_1, \ldots, X_n]$ (fix some field $\Bbbk$), as it is naturally connected to classical problems in algebraic geometry. Let us consider a simple yet non-trivial example from \cite[\S 3]{Eis95}. 258 | 259 | \begin{example} 260 | Take $R=\Bbbk[X,Y]$ and $I = (X^2, XY)$. The reader is invited to check that 261 | \[ I = (X) \cap (X^2, XY, Y^2) = (X) \cap (X^2, Y). \] 262 | Claim: this gives two minimal primary decompositions of $I$. The ideal $(X)$ is prime, hence primary. In fact, $(X^2, XY, Y^2) = (X,Y)^2$ and $(X^2, Y)$ are both primary ideals associated to the maximal ideal $(X, Y)$. This follows either by direct arguments or by Exercise \ref{ex:maximal-primary}, noting that $(X,Y)^2 = (X^2, XY, Y^2)$ is contained in $(X^2, Y)$. The embedded prime $(X,Y)$ is seen to be responsible non-uniqueness of primary decompositions. 263 | 264 | To see the geometry behind, recall that $V(\mathfrak{a}) \cup V(\mathfrak{b}) = V(\mathfrak{a}\mathfrak{b}) = V(\mathfrak{a} \cap \mathfrak{b})$ for any ideals $\mathfrak{a}, \mathfrak{b}$, thus expressing $I$ as an intersection means breaking the corresponding geometric object into a union of simpler pieces. Also recall that for an ideal $I \subset \Bbbk[X,Y]$, the points in $\bigcap_{f \in I} \{f=0\}$ are in bijection with the maximal ideals lying over $I$, at least for $\Bbbk$ algebraically closed (Nullstellensatz). Thus we may interpret these primary decompositions as equalities among ``geometric objects'' embedded in $\Bbbk^2$: 265 | \[ \left\{ X^2=0, XY=0 \right\} = \begin{cases} \left\{ X = 0 \right\} \cup \left\{X^2=XY=Y^2=0 \right\} \\ \left\{ X = 0 \right\} \cup \left\{ X^2=Y=0 \right\}. \end{cases} \] 266 | \begin{enumerate} 267 | \item The geometric object defined by $X=0$ inside $\Bbbk^2$ is certainly the $Y$-axis: the regular functions living on this space form the $\Bbbk$-algebra $\Bbbk[X,Y]/(X) = \Bbbk[Y]$. 268 | \item The geometric object defined by $X^2=XY=Y^2=0$ looks ``physically'' like the origin $(0,0)$, but the $\Bbbk$-algebra of ``regular functions'' (in an extended sense) living on it equals $\Bbbk[X,Y]/(X^2,XY,Y^2)$: by restricting a polynomial function $f(X,Y)$ to this ``thickened point'', we see not only $f(0,0)$ but also $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$. In other words, we shall view $X^2=XY=Y^2=0$ as the first-order infinitesimal neighborhood of $(0,0) \in \Bbbk^2$. 269 | \item In a similar vein, $X^2=Y=0$ physically defines $(0,0)$, but by restricting $f$ to that thickened point, we retrieve $f(0,0)$ as well as $\frac{\partial f}{\partial x}(0,0)$. Therefore we obtain the first-order infinitesimal neighborhood of $0$ inside the $X$-axis. 270 | \end{enumerate} 271 | 272 | Both decomposition says that we obtain the $Y$-axis together with first-order infinitesimal information at the origin $(0,0)$. This is also a nice illustration of the use of nilpotent elements in scheme theory. 273 | \end{example} 274 | 275 | \begin{example}[Symbolic powers] 276 | Let $\mathfrak{p}$ be a prime ideal in a Noetherian ring $R$ and fix $n \geq 1$. Observe that $\mathfrak{p}$ is the unique minimal element in $\Supp(R/\mathfrak{p}^n) = V(\mathfrak{p}^n)$. By Theorem \ref{prop:Supp-Ass} $\mathfrak{p} \in \text{Ass}(R/\mathfrak{p}^n)$, so it makes sense to denote by $\mathfrak{p}^{(n)}$ the $\mathfrak{p}$-primary component (Corollary \ref{prop:p-primary-component}) of $\mathfrak{p}^n$, called the $n$-th \emph{symbolic power} of $\mathfrak{p}$. In general $\mathfrak{p}^{(n)} \supsetneq \mathfrak{p}^n$. For a nice geometric interpretation of symbolic powers due to Nagata and Zariski, we refer to \cite[\S 3.9]{Eis95}. 277 | \end{example} 278 | 279 | Getting primary decomposition of ideals in polynomial algebras is a non-trivial task. Thanks to the pioneers in computational commutative algebra, this can now achieved on your own computer, eg. by the open-source \href{http://www.sagemath.org}{SageMath} system.\index{SageMath} 280 | -------------------------------------------------------------------------------- /YAlg3-4.tex: -------------------------------------------------------------------------------- 1 | % To be compiled by XeLaTeX, preferably under TeX Live. 2 | % LaTeX source for ``Yanqi Lake Lectures on Algebra'' Part III. 3 | % Copyright 2019 李文威 (Wen-Wei Li). 4 | % Permission is granted to copy, distribute and/or modify this 5 | % document under the terms of the Creative Commons 6 | % Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) 7 | % https://creativecommons.org/licenses/by-nc/4.0/ 8 | 9 | % To be included 10 | \chapter{Going-up, going-down, gradings and filtrations} 11 | 12 | This lecture will be less self-contained than the other ones. 13 | 14 | \section{Going-up and going-down} 15 | In geometry, it is often crucial to know properties of a morphism between algebraically defined geometric objects. Rephrased in terms of commutative algebra, our goal is to understand the image of $\varphi^\sharp: \Spec(B) \to \Spec(A)$ where $\varphi: A \to B$ is a given ring homomorphism. 16 | 17 | \begin{itemize} 18 | \item We say the \emph{going-up} property holds for $\varphi$ if for every $\mathfrak{p} \subset \mathfrak{p}'$ in $\Spec(A)$ and $\mathfrak{q} \in \Spec(B)$ with $\varphi^\sharp(\mathfrak{q}) = \mathfrak{p}$ (and we say $\mathfrak{q}$ \emph{lies over} $\mathfrak{p}$...) there exists $\mathfrak{q}' \supset \mathfrak{q}$ lying over $\mathfrak{p}'$. \index{going-up} 19 | \item We say the \emph{going-down} property holds for $\varphi$ if for every $\mathfrak{p} \subset \mathfrak{p}'$ in $\Spec(A)$ and $\mathfrak{q}' \in \Spec(B)$ lying over $\mathfrak{p}'$, there exists $\mathfrak{q} \subset \mathfrak{q}'$ lying over $\mathfrak{p}$. \index{going-down} 20 | \end{itemize} 21 | Pictorially: 22 | \begin{center}\begin{tikzpicture} 23 | \node (B) at (2.5, 2) {$B$}; 24 | \node (P) at (0,1) {$\mathfrak{q}$}; 25 | \node (P') at (1,2) {$\mathfrak{q}'$}; 26 | \node (A) at (2.5, 0) {$A$}; 27 | \node (p) at (0,-1) {$\mathfrak{p}$}; 28 | \node (p') at (1,0) {$\mathfrak{p}'$}; 29 | \draw (P') -- (P) -- (p) -- (p') -- (P'); 30 | \end{tikzpicture}\end{center} 31 | 32 | \begin{lemma}[Existence of minimal over-primes]\label{prop:minimal-prime}\index{minimal prime ideal} 33 | Let $\mathfrak{a}$ be a proper ideal in a ring $R$. There exists a prime ideal $\mathfrak{p}$ which is minimal among all primes containing $\mathfrak{a}$. If $\mathfrak{P}$ is an ideal containing $\mathfrak{a}$, one can choose $\mathfrak{p} \subset \mathfrak{P}$. 34 | \end{lemma} 35 | \begin{proof} 36 | One easily reduces to the case $\mathfrak{a} = \{0\}$. We want to use Zorn's Lemma\index{Zorn's Lemma} to find a minimal prime. It boils down to show that any chain $(\mathfrak{p}_i)_{i \in I}$ of prime ideals ($I$: totally ordered set with $j > i \implies \mathfrak{p}_i \supset \mathfrak{p}_j$) has a lower bound; we assume $\mathfrak{p}_i \subset \mathfrak{P}$ when $\mathfrak{P}$ is prescribed. It suffices to show $\mathfrak{p} := \bigcap_{i \in I} \mathfrak{p}_i$ is prime: if $xy \in \mathfrak{p}$ but there exists $i$ with $x \notin \mathfrak{p}_i$, then $x \notin \mathfrak{p}_j$ whenever $j \geq i$; in this case $j \geq i \implies y \in \mathfrak{p}_j$. This entails $y \in \mathfrak{p}$. 37 | \end{proof} 38 | 39 | \begin{lemma} 40 | The going-down property for $\varphi$ is equivalent to the following: for every $\mathfrak{p} \in \Spec(A)$ with $\varphi(\mathfrak{p})B \neq B$ and any minimal over-prime $\mathfrak{q}$ of $\varphi(\mathfrak{p})B$, we have $\varphi^\sharp(\mathfrak{q}) = \mathfrak{p}$. 41 | \end{lemma} 42 | \begin{proof} 43 | Assuming going-down for $\varphi$, let $\mathfrak{p}, \mathfrak{q}$ be as above. Evidently $\varphi^\sharp(\mathfrak{q}) \supset \varphi^{-1}(\varphi(\mathfrak{p})B) \supset \mathfrak{p}$. If we have $\supsetneq$, then going-down guarantees the existence of $\mathfrak{q}^\flat \subsetneq \mathfrak{q}$ lying over $\mathfrak{p}$. Thus $\varphi^{-1}(\mathfrak{q}^\flat) = \mathfrak{p}$ implies $\mathfrak{q}^\flat \supset \varphi(\mathfrak{p})B$, contradicting the minimality of $\mathfrak{q}$. 44 | 45 | To show the converse, consider $\mathfrak{p} \subset \mathfrak{p}'$ with $\mathfrak{q}'$ lying over $\mathfrak{p}'$. We have $\varphi(\mathfrak{p})B \subset \varphi(\mathfrak{p}')B \subset \mathfrak{q}' \neq B$. Take $\mathfrak{q}$ to be a minimal over-prime of $\varphi(\mathfrak{p})B$ (which exists by Lemma \ref{prop:minimal-prime}) to verify the going-down property. 46 | \end{proof} 47 | 48 | \begin{theorem}\label{prop:going-down-flat} 49 | Going-down holds for flat $\varphi: A \to B$. 50 | \end{theorem} 51 | \begin{proof} 52 | Consider $\mathfrak{p} \subset \mathfrak{p}'$ and $\mathfrak{q}'$ lying over $\mathfrak{p}'$ in the setting of going-down. First, $B_{\mathfrak{q}'}$ is flat over $A_{\mathfrak{p}'}$ by Proposition \ref{prop:flatness-localized}. Secondly, $A_{\mathfrak{p}'} \to B_{\mathfrak{q}'}$ is faithfully flat since it is local by Theorem \ref{prop:faithfully-flat-criterion} (iii), therefore induces a surjection on spectra. Take any prime of $B_{\mathfrak{q}'}$ mapping to $\mathfrak{p}A_{\mathfrak{p}} \in \Spec(A_{\mathfrak{p}'})$ and to $\mathfrak{q} \in \Spec(B)$. In view of the commutative diagrams 53 | \[\begin{tikzcd} 54 | B \arrow[r] & B_{\mathfrak{q}'} \\ 55 | A \arrow[u, "\varphi"] \arrow[r] & A_{\mathfrak{p}'} \arrow[u, "\varphi_{\mathfrak{p}'}"'] 56 | \end{tikzcd} \qquad \begin{tikzcd} 57 | \Spec(B) \arrow[d, "\varphi^\sharp"'] & \Spec(B_{\mathfrak{q}'}) \arrow[d, "\varphi_{\mathfrak{p}'}^\sharp"] \arrow[hookrightarrow, l] \\ 58 | \Spec(A) & \Spec(A_{\mathfrak{p}'}) \arrow[hookrightarrow, l] 59 | \end{tikzcd}\] 60 | we see $\mathfrak{q}$ is the required prime in going-down. 61 | \end{proof} 62 | 63 | \begin{theorem}[Krull--Cohen--Seidenberg]\label{prop:Cohen-Seidenberg} 64 | Suppose the ring $B$ is integral over its subring $A$. The following holds. 65 | \begin{enumerate}[(i)] 66 | \item The map $\Spec(B) \to \Spec(A)$ given by $\mathfrak{q} \mapsto \mathfrak{q} \cap A$ is surjective. 67 | \item There are no inclusion relations in any fiber of $\Spec(B) \to \Spec(A)$. 68 | \item Going-up holds for $A \hookrightarrow B$. 69 | \item If $A$ is a local ring and $\mathfrak{p} \in \MaxSpec(A)$, then the fiber $\{ \mathfrak{q} : \mathfrak{q} \cap A = \mathfrak{p} \}$ equals $\MaxSpec(B)$. 70 | \item Assume $A,B$ are domains and $A$ is normal. Then going-down holds for $A \hookrightarrow B$. 71 | \item Assume furthermore that $B$ is the integral closure of $A$ in a normal field extension $L \supset K := \mathrm{Frac}(A)$, then $\Gamma := \mathrm{Aut}(L/K)$ acts transitively on each fiber of $\Spec(B) \to \Spec 72 | (A)$. 73 | \end{enumerate} 74 | \end{theorem} 75 | The last assertion should be familiar to readers with a background in algebraic number theory. 76 | \begin{proof} 77 | (iv): Let $\mathfrak{q} \in \MaxSpec(B)$ and $\mathfrak{p}_0 := \mathfrak{q} \cap A$. We know $B/\mathfrak{q}$ is a field, integral over its subring $A/\mathfrak{p}_0$. We claim that $A/\mathfrak{p}_0$ is also a field, therefore $\mathfrak{p}_0 = \mathfrak{p}$. Let $x \in A/\mathfrak{p}_0 \smallsetminus \{0\}$. Its inverse in $B/\mathfrak{q}$ satisfies an integral relation $(\frac{1}{x})^n + a_{n-1} (\frac{1}{x})^{n-1} + \cdots + a_0 = 0$ over $A/\mathfrak{p}_0$. Multiplying both sides by $x^{n-1}$ yields $\frac{1}{x} \in (A/\mathfrak{p}_0)[x]$. 78 | 79 | Conversely, we have to show any $\mathfrak{q} \in \Spec(B)$ with $\mathfrak{q} \cap A = \mathfrak{p}$ is maximal. Again, there is an integral extension of domains $A/\mathfrak{p} \hookrightarrow B/\mathfrak{q}$. Consider $y \in B/\mathfrak{q}$ satisfying $y^n + a_{n-1} y^{n-1} + \cdots + a_0 = 0$, with the smallest possible $n$. If $y \neq 0$ then $a_0 \neq 0$. Since $A/\mathfrak{p}$ is a field, the recipe to produce $y^{-1} \in (A/\mathfrak{p})[y]$ is well-known. 80 | 81 | (i), (ii): Fix $\mathfrak{p}$ and consider the inclusion $A_{\mathfrak{p}} \hookrightarrow B_{\mathfrak{p}} = B \dotimes{A} A_{\mathfrak{p}}$ which is still integral (note that $A \smallsetminus \mathfrak{p}$ is a multiplicative subset of $B$, and $B_{\mathfrak{p}}$ is nonzero). We are reduced to the case $A$ is local with maximal ideal $\mathfrak{p}$. By (iv) the fiber of $\mathfrak{p}$ in $\Spec(B)$ is nothing but $\MaxSpec(B)$. This establishes (i) and (ii) since there are no inclusions among maximal ideals. 82 | 83 | (iii): Consider $\mathfrak{p} \subset \mathfrak{p}'$ and $\mathfrak{q}$ over $\mathfrak{p}$ in the setting of going-up. Then (i) is applicable to $A/\mathfrak{p} \hookrightarrow B/\mathfrak{q}$ and yields the required $\mathfrak{q}' \in \Spec(B/\mathfrak{q}) \hookrightarrow \Spec(B)$. 84 | 85 | (vi): Observe that every $\sigma \in \Gamma$ induces an $A$-automorphism of $B$. Let $K' := L^\Gamma$. By (infinite) Galois theory we know $L/K'$ is Galois and $K'/K$ is purely inseparable. Let $A'$ be the integral closure of $A$ in $K'$. First observe that 86 | \[ \Spec(A') \longrightarrow \Spec(A), \quad \mathfrak{p}' \mapsto \mathfrak{p} = \mathfrak{p}' \cap A \] 87 | is a bijection. Indeed, $K' \neq K$ only when $p := \text{char}(K) > 0$, in which case the inverse is given by $\mathfrak{p}' = \left\{t \in A': t^{p^m} \in \mathfrak{p}, \; m \gg 0 \right\}$. Thus we assume henceforth that $L/K$ is Galois. 88 | 89 | Deal with the case $[L:K] < \infty$ first. Consider $\mathfrak{q}, \mathfrak{q}' \in \Spec(B)$ in the fiber over $\mathfrak{p}$. Suppose on the contrary that $\Gamma \mathfrak{q}$ does not meet $\mathfrak{q}'$, then by (ii) we have $\mathfrak{q}' \not\subset \sigma(\mathfrak{q})$ for each $\sigma \in \Gamma$. By the prime avoidance (Proposition \ref{prop:prime-avoidance}), there exists $x \in \mathfrak{q}' \smallsetminus \bigcup_\sigma \sigma(\mathfrak{q})$, since $\Gamma$ is finite. Now define the norm $y := N_{L/K}(x) \in K$, which is some positive power (namely $[L:K]_i$) of $\prod_{\sigma \in \Gamma} \sigma(x)$, hence belongs to $B$. Notice that 90 | \begin{compactitem} 91 | \item $A$ normal implies $y \in A$; 92 | \item $x \notin \sigma^{-1}(\mathfrak{q})$ for all $\sigma \in \Gamma$ implies $y \notin \mathfrak{p}$; 93 | \item however $y \in \mathfrak{q}' \cap A = \mathfrak{p}$ since $x \in \mathfrak{q}'$. Contradiction. 94 | \end{compactitem} 95 | 96 | Now suppose $[L:K]$ is infinite and $\mathfrak{q}, \mathfrak{q}'$ in the fiber over $\mathfrak{p}$. We need to use the Krull topology on $\Gamma$. For every finite, Galois subextension $E/K$ of $L/K$, define the set 97 | \[ \mathcal{T}(E) := \left\{ \sigma \in \Gamma: \sigma(\mathfrak{q} \cap E) = \mathfrak{q}' \cap E \right\}. \] 98 | By the finite case we know $\mathcal{T}(E) \neq \emptyset$. Furthermore, 99 | \begin{compactitem} 100 | \item $\mathcal{T}(E)$ is closed in $\Gamma$ (it is the preimage of some subset of $\text{Gal}(E/K)$); 101 | \item $E' \subset E \implies \mathcal{T}(E') \supset \mathcal{T}(E)$; 102 | \item the intersection of all $\mathcal{T}(E)$ is nonempty (by the compactness of $\Gamma$ and the previous step). 103 | \end{compactitem} 104 | Taking $\sigma \in \bigcap_E \mathcal{T}(E)$ gives $\sigma(\mathfrak{q}) = \mathfrak{q}'$. 105 | 106 | (v): Define $L_0 := \text{Frac}(B)$ and $K := \text{Frac}(A)$. Then $L_0/K$ is algebraic and we may take a normal closure $L$ of $L_0$ over $K$. Let $C$ be the integral closure of $A$ (thus of $B$) in $L$. Consider the setting $\mathfrak{p} \subset \mathfrak{p}'$ and $\mathfrak{q} \cap A = \mathfrak{p}$ of going-down. Take any $\mathfrak{r} \in \Spec(C)$ mapping to $\mathfrak{p}$. By (iii) for $A \to C$ we obtain $\mathfrak{r}_1 \in \Spec(C)$ such that $\mathfrak{r}_1 \mapsto \mathfrak{p}'$ and $\mathfrak{r}_1 \supset \mathfrak{r}$. Next, take $\mathfrak{r}_2 \in \Spec(C)$ mapping to $\mathfrak{q}'$; by (vi) there exists $\sigma \in \text{Aut}(L/K)$ with $\sigma(\mathfrak{r}_1) = \mathfrak{r}_2$. One can check that $\mathfrak{q} := \sigma(\mathfrak{r}) \cap B$ is the required prime ideal. Explained pictorially: 107 | \begin{center}\begin{tikzpicture} 108 | \node (R) at (-2, 2) {$\mathfrak{r}$}; 109 | \node (R1) at (-0.5, 2) {$\mathfrak{r}_1$}; 110 | \node (R2) at (2, 2) {$\mathfrak{r}_2$}; 111 | \node[fill=gray!80] (SR) at (0.5, 2) {}; 112 | \node[fill=gray!40] (Q) at (-0.25, 1) {}; 113 | \node (Q') at (1.5, 1) {$\mathfrak{q}'$}; 114 | \node (P) at (-1, 0) {$\mathfrak{p}$}; 115 | \node (P') at (1, 0) {$\mathfrak{p}'$}; 116 | 117 | \node at (3, 2) {$C$}; \node at (3, 1) {$B$}; \node at (3, 0) {$A$}; 118 | 119 | \draw (P) -- (R) -- (R1) -- (P') -- (P); 120 | \draw[dashed] (P) -- (SR) -- (R2) -- (Q')-- (P'); 121 | \draw[dashed] (Q) -- (Q'); 122 | 123 | \draw (-1, 2.5) edge[->, ultra thick, bend left=40] node[above, midway] {$\sigma$} (1, 2.5); 124 | \end{tikzpicture}\end{center} 125 | we first construct $\mathfrak{r}$ and then $\mathfrak{r}_1$ by going-up, then ``tilt'' it via some $\sigma$ to match $\mathfrak{r}_1$ with some chosen $\mathfrak{r}_2$ above $\mathfrak{q}'$, so that $\sigma(\mathfrak{r}) \cap B$ produces the required going-down: 126 | \end{proof} 127 | 128 | \begin{exercise}\label{exo:integral-closed} 129 | Let $A \subset B$ be an integral extension of rings. Show that $\Spec(B) \to \Spec(A)$ is a closed map (Cf.\ Proposition \ref{prop:going-up-closed}.) Hint: Let $\mathfrak{b} \subset B$ be a proper ideal, then $A/\mathfrak{b} \cap A \hookrightarrow B/\mathfrak{b}$ is still integral. Reduce the problem to showing that $V(\{0_B\}) = \Spec(B)$ has closed image in $\Spec(A)$. 130 | \end{exercise} 131 | 132 | \section{Subsets in the spectrum} 133 | \begin{proposition}\label{prop:going-up-closed} 134 | Let $\varphi: A \to B$ be a ring homomorphism with going-up property and suppose $B$ is Noetherian. Then $\varphi^\sharp: \Spec(B) \to \Spec(A)$ is a closed map: it maps closed subsets to closed subsets. 135 | \end{proposition} 136 | \begin{proof} 137 | Consider a closed subset $V(\mathfrak{b})$ of $\Spec(B)$. First, every $\mathfrak{q} \in \Spec(B)$ with $\mathfrak{q} \supset \mathfrak{b}$ lies over a minimal over-prime of $\mathfrak{b}$, by Lemma \ref{prop:minimal-prime}. Secondly, $B$ is Noetherian implies $\text{Ass}(B/\mathfrak{b})$ is finite; in particular there are only finitely many minimal over-primes $\mathfrak{q}_1, \ldots, \mathfrak{q}_n$ of $\mathfrak{b}$. 138 | 139 | Set $\mathfrak{p}_i := \varphi^\sharp(\mathfrak{q}_i)$ for all $i$. By going-up, $V(\mathfrak{p}_i)$ is contained in $\varphi^\sharp(V(\mathfrak{b}))$. On the other hand, every $\mathfrak{p} = \varphi^\sharp(\mathfrak{q})$ with $\mathfrak{q} \in V(\mathfrak{b})$ lies over some $\mathfrak{p}_i = \varphi^\sharp(\mathfrak{q}_i)$ by the foregoing discussion. This shows $\varphi^\sharp(V(\mathfrak{b})) = \bigcup_{i=1}^n V(\mathfrak{p}_i)$ is closed. 140 | \end{proof} 141 | 142 | \begin{corollary} 143 | Suppose $B$ is Noetherian and integral over a subring $A$. Then $\Spec(B) \to \Spec(A)$ is a closed surjection with finite fibers. 144 | \end{corollary} 145 | \begin{proof} 146 | Apply Theorem \ref{prop:Cohen-Seidenberg} with Proposition \ref{prop:going-up-closed} to show that $\Spec(B) \to \Spec(A)$ is closed and surjective. 147 | 148 | To show the finiteness of the fiber over $\mathfrak{p} \in \Spec(A)$, note that the preimage of $V(\mathfrak{p})$ in $\Spec(B)$ equals $V(\mathfrak{p}B)$. Since there are no inclusions in the fiber over $\mathfrak{p}$, every element in that fiber must be a minimal over-prime of $\mathfrak{p}B$. We have seen in the proof of Proposition \ref{prop:going-up-closed} that there are only finitely many such minimal-over primes. 149 | \end{proof} 150 | 151 | Note that the ``closed surjection'' part applies to any integral extension of rings. See Exercise \ref{exo:integral-closed}. 152 | 153 | In order to obtain further results of this type, we have to introduce more notions. Let $R$ be a ring. 154 | \begin{definition} 155 | For $\mathfrak{p}, \mathfrak{p}' \in \Spec(R)$ satisfying $\mathfrak{p} \subset \mathfrak{p}'$, we say $\mathfrak{p}$ is a \emph{generalization} of $\mathfrak{p}'$, and $\mathfrak{p}'$ is a \emph{specialization} of $\mathfrak{p}$. 156 | \end{definition} 157 | To make geometric meaning from it, being ``specialized'' signifies that there are ``more equations'' in $\mathfrak{p}'$, therefore it corresponds a smaller embedded geometric object. For example, in $R=\CC[X,Y]$ the prime ideal $(X,Y)$ is a specialization of $(X)$, as the origin $X=Y=0$ belongs to the line $X=0$. 158 | 159 | \begin{lemma} 160 | With respect to the Zariski topology, $\mathfrak{p}$ is a generalization of $\mathfrak{p}'$ if and only if $\mathfrak{p}' \in \overline{\{\mathfrak{p}\}}$. 161 | \end{lemma} 162 | \begin{proof} 163 | The condition $\mathfrak{p}' \in \overline{\{\mathfrak{p}\}}$ means that for every ideal $\mathfrak{a}$, if $\mathfrak{p} \supset \mathfrak{a}$ then $\mathfrak{p}' \supset \mathfrak{a}$. Taking $\mathfrak{a} = \mathfrak{p}$ yields $\mathfrak{p}' \supset \mathfrak{p}$, and the converse is even easier. 164 | \end{proof} 165 | A subset is called stable under specialization (resp. generalization) if the specialization (resp. generalization) of any member still belongs to that set. The following is straightforward. 166 | 167 | \begin{lemma}\label{prop:stability-vs-closeness-0} 168 | Any closed subset is stable under specialization; any open subset is stable under generalization. 169 | \end{lemma} 170 | 171 | \begin{definition}\index{constructible subset} 172 | Suppose $R$ is Noetherian. A subset is called \emph{locally closed} if it is the intersection of an open with a closed subset, called \emph{constructible} if it is a finite union of locally closed subsets. % A possibly infinite intersection (resp. union) of constructible subsets is called \emph{pro-constructible} (resp. \emph{ind-constructible}). 173 | 174 | Closed subsets of the form $V(\mathfrak{p})$, where $\mathfrak{p} \in \Spec(R)$, are called \emph{irreducible}; in this case we call $\mathfrak{p}$ the \emph{generic point} of $Z$, which is uniquely characterized as the point which generalizes every member of $Z$. 175 | \end{definition} 176 | \begin{itemize} 177 | \item The foregoing definition is standard only for $R$ Noetherian. The general definition in EGA differs. 178 | \item These notions can be applied to any topological space $X$. In practice one usually suppose $X$ to be 179 | \begin{compactitem} 180 | \item \emph{Noetherian}: the closed subsets satisfy descending chain condition, 181 | \item \emph{sober}: every irreducible has a generic point, 182 | \end{compactitem} 183 | in order to get interesting results. This explains our Noetherian assumption. 184 | \item If $\Bbbk$ is algebraically closed, $R = \Bbbk[X_1, \ldots, X_n]/\mathfrak{a}$ corresponds to an affine algebraic variety $\mathcal{X} \subset \Bbbk^n$ and we work with $\MaxSpec(R)$ instead of $\Spec(R)$, then a subset $E \subset \mathcal{X}$ being locally closed means that it can be defined by a formula using the usual language of algebraic operations over $\CC$, with coordinate variables $X_1, \ldots, X_n$ and the symbols $=, \neq$, but \emph{without using the quantifiers $\exists, \forall$}. For example, the formula 185 | \[ (\neg X = 0) \vee (X = 0 \wedge Y=0) \] 186 | defines the constructible subset $E := \{(x,y) \in \Bbbk^2 : x \neq 0 \} \cup \{(0,0) \}$, which is neither closed nor open. Note that $E$ is the image of the polynomial map $\Bbbk^2 \to \Bbbk^2$ given by $(x,y) \mapsto (x,xy)$. 187 | \end{itemize} 188 | 189 | \begin{exercise} 190 | Show that the set of constructible subsets is stable under finite $\cup$, finite $\cap$ and taking complements. 191 | \end{exercise} 192 | 193 | \begin{exercise} 194 | Show that irreducible closed subsets $Z$ admit the following topological characterization: if $X = A \cup B$ with $A,B$ closed, then either $X=A$ or $X=B$. 195 | \end{exercise} 196 | 197 | Suppose $R$ is Noetherian. Given any closed subset $Z \subset \Spec(R)$, we may write $Z = Z_1 \cup \cdots \cup Z_n$ with each $Z_i$ irreducible. One way to do this is to use the primary decomposition for $\mathfrak{a}$, where we assume $Z = V(\mathfrak{a})$; then $Z_1, \ldots, Z_n$ will correspond to the minimal elements in $\text{Ass}(R/\mathfrak{a})$. One can show by purely topological means that such an irreducible decomposition is unique if we require $i \neq j \implies Z_i \not\subset Z_j$. See \cite[I.1.5]{Har77}. 198 | 199 | \begin{lemma}\label{prop:char-constructible} 200 | Let $E$ be a subset of $\Spec(R)$ where $R$ is a Noetherian ring. The following are equivalent: 201 | \begin{enumerate}[(i)] 202 | \item $E$ is constructible; 203 | \item for every irreducible subset $Z$ of $\Spec(R)$, either $Z \cap E$ is not dense in $Z$ or $Z \cap E$ contains a nonempty open subset of $Z$. 204 | \end{enumerate} 205 | \end{lemma} 206 | \begin{proof} 207 | Omitted. See \cite[(6.C)]{Mat80}. 208 | \end{proof} 209 | 210 | \begin{theorem}[C.\ Chevalley]\label{prop:Chevalley}\index{Chevalley's theorem} 211 | Let $\varphi: A \to B$ be a ring homomorphism such that $A$ is Noetherian and $B$ is a finitely generated $A$-algebra. Then $\varphi^\sharp$ maps constructible subsets to constructible subsets. 212 | \end{theorem} 213 | \begin{proof} 214 | Omitted. See \cite[(6.E)]{Mat80}. 215 | \end{proof} 216 | 217 | Now we can give a partial converse to Lemma \ref{prop:stability-vs-closeness-0}, albeit not in the strongest form. 218 | \begin{proposition} 219 | Suppose $R$ is Noetherian and $E$ is a constructible subset of $\Spec(R)$. If $E$ is stable under specialization (resp. generalization), then $E$ is closed (resp. open) in $\Spec(R)$. 220 | \end{proposition} 221 | \begin{proof} 222 | It suffices to treat the specialization-stable case by taking complements. Write the Zariski-closure $\bar{E}$ as a finite union of irreducibles components $Z$, without inclusion relations. For each irreducible component $Z$, notice that $Z \cap E$ is dense in $Z$ for all $Z$, since otherwise $\bar{E} = \overline{\bigcup_Z (Z \cap E)} = \bigcup_Z \overline{Z \cap E}$ would lead to another irreducible decomposition. Thus $Z \cap E$ contains a nonempty open of $Z$ by Lemma \ref{prop:char-constructible}. This open subset of $E$ must contain the generic point of $Z$. As $E$ is stable under specialization, we obtain $Z \subset E$. This being true for all $Z$, we deduce that $E = \bar{E}$. 223 | \end{proof} 224 | 225 | \begin{proposition} 226 | Consider a ring homomorphism $\varphi: A \to B$ satisfying going-down. Suppose $A$ is Noetherian and $B$ is a finitely generated $A$-algebra, then $\varphi^\sharp: \Spec(B) \to \Spec(A)$ is an open map. 227 | \end{proposition} 228 | \begin{proof} 229 | Let $U = \Spec(B) \smallsetminus V(\mathfrak{a})$ be an open subset. Going-down implies that $\varphi^\sharp(U)$ is stable under generalization. It suffices to show $\varphi^\sharp(U)$ is constructible, and this is the content of Chevalley's Theorem \ref{prop:Chevalley}. 230 | \end{proof} 231 | 232 | \section{Graded rings and modules} 233 | Let $(\Gamma, +)$ be a commutative monoid. In most cases we consider $\Gamma = \Z_{\geq 0}$. 234 | 235 | \begin{definition}\index{graded} 236 | A $\Gamma$-graded ring is a ring $R$ whose underlying additive group is endowed with a decomposition $R = \bigoplus_{\gamma \in \Gamma} R_\gamma$, such that $R_\gamma R_\eta \subset R_{\gamma + \eta}$ for all $\gamma, \eta \in \Gamma$. 237 | 238 | For $R$ as above, a $\Gamma$-graded $R$-module is an $R$-module $M$ whose underlying additive group decomposes as $M = \bigoplus_{\gamma \in \Gamma} M_\gamma$, such that $R_\gamma M_\eta \subset M_{\gamma + \eta}$ for all $\gamma, \eta \in \Gamma$; in particular, $R$ itself is a $\Gamma$-graded $R$-module. If $x \in M_\gamma \smallsetminus \{0\}$, we say $x$ is homogeneous of degree $\gamma$.\index{homogeneous} 239 | \end{definition} 240 | We will often omit $\Gamma$ when there is no worry of confusion. Note that if $0$ is allowed to be homogeneous, as people sometimes do, it will be homogeneous of any degree. 241 | 242 | \begin{exercise} 243 | Show that in a graded ring $R$ we always have $1 \in R_0$, provides that $(\Gamma, +)$ satisfies the cancellation law: $\gamma+\eta=\eta \iff \gamma=0$. Hint: let $e_0$ be the component of $1_R$ in degree 0, argue that $x e_0 = x = e_0 x$ for all homogeneous $x \in R$. The condition $1 \in R_0$ is sometimes built into the definition of graded rings. 244 | \end{exercise} 245 | 246 | \begin{definition} 247 | A graded submodule of a graded $R$-module $M$ is a submodule $N$ with $N = \bigoplus_\gamma (N \cap M_\gamma)$, which gives rise to a natural grading $N_\gamma := N \cap M_\gamma$ on $N$. 248 | \end{definition} 249 | For graded $N \subset M$, the quotient $R$-module $M/N = \bigoplus_\gamma M_\gamma/N_\gamma$ is again graded. As a special case, we have the notion of graded ideals of $R$ (also known as \emph{homogeneous ideals}), and the quotient ring $R/\mathfrak{a}$ with respect to graded $\mathfrak{a}$ inherits the evident grading. 250 | 251 | Let $N$ be an $R$-submodule of $M$ and suppose $M$ is graded. The following are easily seen to be equivalent: 252 | \begin{enumerate}[(i)] 253 | \item $N \subset M$ is graded; 254 | \item $N$ is generated by homogeneous elements; 255 | \item if $x = \sum_\gamma x_\gamma \in N$ with each $x_\gamma$ homogeneous of degree $\gamma$, then $\forall \; x_\gamma \in N$. 256 | \end{enumerate} 257 | 258 | \begin{example} 259 | Let $A$ be a ring and $R := A[X_1, \ldots, X_n]$. Then $R$ is naturally $\Z_{\geq 0}$-graded by degrees: for each $d \in \Z_{\geq 0}$, let $R_d$ be the set of homogeneous polynomials of total degree $d$. Ideals generated by homogeneous polynomials are precisely the graded ideals. The importance of this grading comes from projective algebraic geometry. 260 | 261 | On the other hand, $R$ can also be graded by monomials by taking $\Gamma = \Z_{\geq 0} \times \cdots \times \Z_{\geq 0}$ ($n$ copies), and we set $R_{(d_1, \ldots, d_n)} = A \cdot X_1^{d_1} \cdots X_n^{d_n}$ 262 | \end{example} 263 | 264 | Many constructions in commutative algebra can be generalized to the graded case. Let us illustrate what one can do in an important case, the primary decomposition (cf.\ \cite[\S 3.5 and Exercise 3.5]{Eis95}). In the $\Z$-graded context, it says that for a finitely generated graded module $M$ over a Noetherian graded ring, the associated primes of $M$ are all graded ideals, and one can write $\{0\} = N_1 \cap \cdots \cap N_m$ where $N_i \subset M$ are graded submodules with $\text{Ass}(M/N_i) = \{\mathfrak{p}_i\}$, $\mathfrak{p}_i \in \text{Ass}(M)$, etc. Most of the arguments in the ungraded case carry over verbatim, and the only new technique is the following 265 | 266 | \begin{lemma} 267 | Let $M$ be a $\Z$-graded module over a $\Z$-graded ring $R$. If $x \in M$ and $\mathfrak{p} := \mathrm{ann}(x)$ is a prime ideal, then 268 | \begin{enumerate}[(i)] 269 | \item $\mathfrak{p}$ is a homogeneous ideal, and 270 | \item $\mathfrak{p} = \mathrm{ann}(y)$ for some homogeneous element $y \in M$. 271 | \end{enumerate} 272 | \end{lemma} 273 | \begin{proof} 274 | We begin with (i). Let $t \in \mathfrak{p}$ and $x \in M$ be such that $\mathfrak{p} = \text{ann}(M)$. Write 275 | \[ t = \sum_{\gamma \in \mathcal{A}} t_\gamma, \quad x = \sum_{\eta \in \mathcal{B}} x_\eta, \] 276 | where $t_\gamma \in R_\gamma \smallsetminus \{0\}$ and $x_\eta \in M_\eta \smallsetminus \{0\}$ for all $\gamma, \eta$. Denote by $\gamma_0$ and $\eta_0$ the minimal elements in $\mathcal{A}$ and $\mathcal{B}$, respectively. Homogenity amounts to $t_\gamma \in \mathfrak{p}$ for each $\gamma \in \mathcal{A}$, and this will be done by induction on $|\mathcal{B}|$. By a further induction on $|\mathcal{A}|$, for fixed $x$ and $\mathfrak{p}$, this can be reduced to showing $t_{\gamma_0} \in \mathfrak{p}$. 277 | 278 | First of all, considerations of degrees and $tx=0$ lead to $t_{\gamma_0} x_{\eta_0} = 0$. If $x = x_{\eta_0}$ (i.e.\ $|\mathcal{B}| = 1$), we obtain $t_{\gamma_0} \in \text{ann}(x) = \mathfrak{p}$ as required. In general: 279 | \begin{itemize} 280 | \item Suppose that $\mathfrak{p} = \text{ann}(t_{\gamma_0} x)$; since the decomposition 281 | \[ t_{\gamma_0} x = \sum_{\substack{\eta \in \mathcal{B} \\ \eta \neq \eta_0}} t_{\gamma_0} x_\eta \] 282 | involves fewer homogeneous terms, $\mathfrak{p}$ is then homogeneous by our induction hypothesis on $|\mathcal{B}|$. 283 | \item Suppose there exists $s \in R \smallsetminus \mathfrak{p}$ such that $s(t_{\gamma_0} x) = 0$, then $st_{\gamma_0} \in \mathfrak{p}$, hence $t_{\gamma_0} \in \mathfrak{p}$. This concludes the homogeneity (i). 284 | \end{itemize} 285 | 286 | From the homogeneity $\mathfrak{p}$ we infer that $\mathfrak{p} \subset \text{ann}(x_\eta)$ for each $\eta \in \mathcal{B}$. Now that 287 | \[ \mathfrak{p} = \text{ann}(x) \supset \prod_{\eta \in \mathcal{B}} \text{ann}(x_\eta), \] 288 | we have $\mathfrak{p} \supset \text{ann}(x_\eta)$ for some $\eta$, hence $\mathfrak{p} = \text{ann}(x_\eta)$. Take $y = x_\eta$ to obtain (ii). 289 | \end{proof} 290 | 291 | \section{Filtrations} 292 | Now turn to filtrations. We only deal with decreasing filtrations indexed by $\Z_{\geq 0}$. 293 | \begin{definition}\index{filtration}\index{gr@$\gr_F(M)$} 294 | A \emph{filtration} on a ring $R$ is a descending sequence 295 | \[ R = F^0 R \supset F^1 R \supset F^2 R \supset \cdots \] 296 | of ideals such that $F^i R \cdot F^j R \subset F^{i+j} R$. Define the associated $\Z_{\geq 0}$-graded ring 297 | \[ \gr_F(R) := \bigoplus_{n \geq 0} \underbracket{F^n R \big/ F^{n+1} R}_{=: \gr_F^n R} \] 298 | whose multiplication is defined as follows: if $x \in F^n R / F^{n+1} R$ and $y \in F^m R/F^{m+1} R$, choose liftings $\tilde{x} \in F^n R$ and $\tilde{y} \in F^m R$ and define 299 | \[ xy := \text{the image of }\; \tilde{x}\tilde{y}\; \text{ in }\; F^{n+m} R / F^{n+m+1} R; \] 300 | this is readily seen to be well-defined. The multiplication of arbitrarily many homogeneous elements can be obtained in the same recipe. The datum $(R, F^\bullet R)$ is called a \emph{filtered ring}, 301 | 302 | Given a filtered ring $R$, a \emph{filtered $R$-module} $M$ is an $R$-module $M$ equipped with a descending sequence\footnote{In view of later applications, the filtration on a module is indexed by $\Z$ instead of $\Z_{\geq 0}$.} of submodules 303 | \[ \cdots \supset F^i M \supset F^{i+1} M \supset \cdots, \quad i \in \Z \] 304 | such that $F^i R \cdot F^j M \subset F^{i+j} M$. Define the associated graded module as the $\Z$-graded $\gr_F R$-module 305 | \[ \gr_F(M) := \bigoplus_{n \in \Z} \underbracket{F^n M \big/ F^{n+1} M}_{=: \gr_F^i M} \] 306 | whose scalar multiplication is defined using liftings as above. 307 | \end{definition} 308 | 309 | The subscript $F$ in $\gr$ will often be omitted. To guarantee that $\gr_F(R) \neq \{0\}$, we usually impose the harmless condition 310 | \[ R = F^0 R \supsetneq F^1 R. \] 311 | 312 | \begin{example}\index{filtration!$\mathfrak{a}$-adic} 313 | Let $\mathfrak{a}$ be a proper ideal of $R$, then $F^i R := \mathfrak{a}^i$ defines a filtration on $R$, called the \emph{$\mathfrak{a}$-adic filtration}. 314 | \end{example} 315 | 316 | \begin{definition}\label{def:a-stable} 317 | Equip $R$ with the $\mathfrak{a}$-adic filtration. A filtered $R$-module $M$ is called $\mathfrak{a}$-\emph{stable} if $\mathfrak{a} \cdot F^i M = F^{i+1} M$ for $i \gg 0$. 318 | \end{definition} 319 | Recall that $\mathfrak{a} \cdot F^i M \subset F^{i+1} M$ holds for all $M$, which is a part of our assumption. 320 | 321 | As an easy example, set $F^i M := \mathfrak{a}^i M$ for $i \geq 1$, and $F^{\leq 0} M := M$; this is the $\mathfrak{a}$-adic filtration on $M$, which is obviously $\mathfrak{a}$-stable. 322 | 323 | \begin{proposition}\label{prop:gr-Noetherian} 324 | Suppose $\mathfrak{a}$ is a proper ideal of $R$. If $R$ is Noetherian, so is $\gr(R)$ with respect to the $\mathfrak{a}$-adic filtration. In fact $\gr(R)$ is finitely generated over $\gr^0(R)$. 325 | \end{proposition} 326 | \begin{proof} 327 | Let $x_1, \ldots, x_n$ be generators of $\mathfrak{a}$, with images $\bar{x}_i \in \mathfrak{a}/\mathfrak{a}^2$. Then $\gr(R)$ is generated by $\bar{x}_1, \ldots \bar{x}_n$ over $R/\mathfrak{a} = \gr^0 R$, which is also Noetherian. Now apply Hilbert's Basissatz. 328 | \end{proof} 329 | 330 | \begin{proposition}\label{prop:gr-fg} 331 | Let $\mathfrak{a}$ be a proper ideal of $R$ and $M$ a finitely generated $R$-module. Suppose $M$ is endowed with an $\mathfrak{a}$-stable filtration such that $F^i M$ is finitely generated for each $i$, and $F^{\leq 0} M = M$. Then $\gr(M)$ is a finitely generated $\gr(R)$-module. 332 | \end{proposition} 333 | \begin{proof} 334 | Take $n$ such that $\mathfrak{a} \cdot F^i M = F^{i+1} M$ for all $i \geq n$. Then in $\gr(M) = \bigoplus_i \gr^i M$ we have 335 | \[ (\mathfrak{a}/\mathfrak{a}^2) \cdot \gr^i M = \gr^{i+1} M, \quad i \geq n. \] 336 | Therefore it suffices to take generators from $\gr^0 M, \ldots, \gr^n M$, each of whom is finitely generated over $R/\mathfrak{a} = \gr^0 R$. 337 | \end{proof} 338 | 339 | \section{Theorems of Artin--Rees and Krull}\label{sec:Artin-Rees} 340 | Conserve the conventions in the previous section on filtrations, etc. 341 | 342 | \begin{definition}[Morphisms between filtered objects] 343 | Let $(A, F^\bullet A)$ and $(B, F^\bullet B)$ be filtered rings. A \emph{morphism} between them means a ring homomorphism $\varphi: A \to B$ satisfying $\varphi(F^i A) \subset F^i B$ for all $i$. Similarly, suppose $A$ is filtered and let $M, N$ be filtered $A$-modules. A morphism $M \to N$ means a homomorphism $\psi: M \to N$ of $R$-modules satisfying $\psi(F^i M) \subset F^i N$ for all $i$. 344 | \end{definition} 345 | This makes the filtered rings and the filtered modules over a filtered ring into categories. Obviously, morphisms $\varphi$ between filtered objects induce graded morphisms $\gr \varphi$ between the associated graded objects. Therefore we obtain a functor from the category of filtered rings or modules into their graded avatars. 346 | 347 | \begin{remark} 348 | Suppose $\varphi: M \to N$ is a morphism between filtered $A$-modules. The quotient $M/\Ker(\varphi)$ inherits a filtration from $M$, whereas the submodule $\Image(\varphi)$ inherits one from $N$. When the natural isomorphism $M/\Ker(\varphi) \to \Image(\varphi)$ is an isomorphism between filtered modules, or equivalently 349 | \[ \forall i \in \Z, \; \varphi(F^i M) = \varphi(M) \cap F^i N, \] 350 | we say $\varphi$ is a \emph{strict morphism}. 351 | \end{remark} 352 | 353 | It is often useful to relate properties of a filtered module or morphism to its graded counterpart. Propositions \ref{prop:gr-Noetherian} and \ref{prop:gr-fg} are such examples. Here is an example for the other direction. We say that a filtration on $M$ is \emph{exhaustive} if $\bigcup_i F^i M = M$, \emph{separating} (or \emph{Hausdorff}) if $\bigcap_i F^i M = \{0\}$ 354 | 355 | \begin{proposition}\label{prop:gr-injective} 356 | Suppose that $\varphi: M \to N$ is a morphism between filtered $R$-modules. If $M$ is exhaustive and separating, and $\gr\varphi$ is injective, then $\varphi$ is also injective. 357 | \end{proposition} 358 | \begin{proof} 359 | Let $x \in \Ker(\varphi)$. There exists $n$ such that $x \in F^n M$. Regard $x + F^{n+1} M$ as an element of $\gr^n M$. Then $\gr(\varphi)(x + F^{n+1} M) = \varphi(x) + F^{n+1} N = 0$, so $x \in F^{n+1} M$. Iterating this argument, we have $x \in \bigcap_{k \geq n} F^k M = \{0\}$. 360 | \end{proof} 361 | See Lemma \ref{prop:gr-surjective} for the case of surjections. 362 | 363 | In what follows, $\mathfrak{a}$ always denotes a proper ideal of a ring $R$. 364 | 365 | \begin{definition}[Blow-up algebras and modules]\index{blow-up algebra} 366 | Introduce an indeterminate $X$ and define the $\Z_{\geq 0}$-graded $R$-algebra 367 | \[ \text{Bl}_{\mathfrak{a}} R := \bigoplus_{n \geq 0} \mathfrak{a}^n X^n \; \subset R[X]. \] 368 | If an $R$-module $M$ is endowed with a filtration $(F^i M)_{i \geq 0}$ compatible with $\mathfrak{a}$, we define 369 | \[ \text{Bl}(M) := \bigoplus_{n \geq 0} (F^n M) \otimes X^n \subset M \dotimes{R} R[X]. \] 370 | Clearly, $\text{Bl}(M)$ is a graded $\text{Bl}_{\mathfrak{a}} R$-module. 371 | \end{definition} 372 | 373 | \begin{remark} 374 | The indeterminate $X$ is somehow a placeholder. Enlarge $\text{Bl}_{\mathfrak{a}} R$ to $\tilde{B}$ by setting the negative graded pieces to be $R \cdot X^{< 0}$, and set $T = X^{-1}$. We see that $\tilde{B}$ is actually an $R[T]$-algebra, sometimes called the \emph{Rees algebra} of $\mathfrak{a}$. Under the specialization $T=0$ we get \index{Rees algebra} 375 | \[ \frac{ \tilde{B}}{T \tilde{B}} \simeq \bigoplus_{n \geq 0} \frac{\mathfrak{a}^n}{\mathfrak{a}^{n+1}} \cdot T^{-n} \simeq \gr(R). \] 376 | On the other hand, inverting $T$ yields 377 | \[ \tilde{B}[T^{-1}] = \bigoplus_{n \in \Z} R \cdot T^{-n} = R[T^{\pm 1}]. \] 378 | This reflects a well-known deformation construction in geometry; $\gr(R)$ is actually the graded $R$-algebra corresponding to the \emph{normal cone} defined by $\mathfrak{a} \subset R$. We refer to \cite[\S 5.1]{Fu98} for details. 379 | \end{remark} 380 | 381 | \begin{lemma}\label{prop:Bl-fg} 382 | Consider a ring $R$ with proper ideal $\mathfrak{a}$, together with a filtered $R$-module $M$, assume furthermore that each $F^i M$ is finitely generated over $R$. The following are equivalent: 383 | \begin{enumerate}[(i)] 384 | \item $\mathrm{Bl}(M)$ is finitely generated over $\mathrm{Bl}_{\mathfrak{a}} R$; 385 | \item the filtration on $M$ is $\mathfrak{a}$-stable. (Definition \ref{def:a-stable}) 386 | \end{enumerate} 387 | \end{lemma} 388 | \begin{proof} 389 | (i) $\implies$ (ii): Choose homogeneous generators $x_1, \ldots, x_n \in \text{Bl}(M)$ with degrees $d_1, \ldots, d_n$ respectively. It is then routine to see that 390 | \[ i \geq \max\{d_1, \ldots, d_n \} \implies (F^{i+1} M) X^{i+1} = \mathfrak{a} X \cdot (F^i M) X^i, \] 391 | that is, $\mathfrak{a} \cdot F^i M = F^{i+1} M$ for these $i$. 392 | 393 | (ii) $\implies$ (i): Suppose $\mathfrak{a} \cdot F^i M = F^{i+1} M$ for $i \geq d$, then $\text{Bl}(M)$ is generated by $\bigoplus_{j \leq d} (F^j M) X^j$, and each $F^j M$ is finitely generated over $R = (\text{Bl}_{\mathfrak{a}} R)_0$. 394 | \end{proof} 395 | 396 | \begin{theorem}[Artin--Rees]\label{prop:Artin-Rees} 397 | Let $R$ be a Noetherian ring endowed with $\mathfrak{a}$-adic filtration. Let $M$ be a finitely generated $R$-module and $N \subset M$ an $R$-submodule. Then the filtration on $N$ induced by the $\mathfrak{a}$-adic filtration of $M$, namely $F^i N := \mathfrak{a}^i M \cap N$, is $\mathfrak{a}$-stable. 398 | \end{theorem} 399 | \begin{proof} 400 | Define $\text{Bl}(N)$ using the induced filtration $F^i N := \mathfrak{a}^i M \cap N$, which is a submodule of the finitely generated $\text{Bl}_{\mathfrak{a}} R$-module $\text{Bl}(M)$ (Lemma \ref{prop:Bl-fg}). If $\mathfrak{a} = (a_1, \ldots, a_m)$ than $\text{Bl}_{\mathfrak{a}} R = R[a_1 X, \ldots, a_m X] \subset R[X]$, hence Noetherian by Hilbert's Basissatz. We deduce that $\text{Bl}(N)$ is finitely generated over $\text{Bl}_{\mathfrak{a}} R$. In turn, this implies $F^\bullet N$ is an $\mathfrak{a}$-stable filtration on $N$ by Lemma \ref{prop:Bl-fg}. 401 | \end{proof} 402 | 403 | \begin{theorem}\label{prop:intersection-thm} 404 | For $R, \mathfrak{a}, M$ as in the previous theorem, we set $N := \bigcap_{n \geq 0} \mathfrak{a}^n M$. Then $\mathfrak{a}N = N$. 405 | \end{theorem} 406 | \begin{proof} 407 | Since the induced filtration on $N$ is $\mathfrak{a}$-stable by Theorem \ref{prop:Artin-Rees}, for $n \gg 0$ we have 408 | \[ N = \mathfrak{a}^n M \cap N = \mathfrak{a} \cdot (\mathfrak{a}^{n-1} M \cap N) = \mathfrak{a} N. \] 409 | The assertion follows. 410 | \end{proof} 411 | 412 | \begin{corollary}[Krull]\label{prop:Krull-intersection-rad} 413 | If $\mathfrak{a} \subset \mathrm{rad}(R)$, then $\bigcap_{n \geq 0} \mathfrak{a}^n M = \{0\}$ for any finitely generated $R$-module $M$. In particular $\bigcap_{n \geq 0} \mathfrak{a}^n = \{0\}$ whenever $\mathfrak{a} \subset \mathrm{rad}(R)$. 414 | \end{corollary} 415 | \begin{proof} 416 | Theorem \ref{prop:intersection-thm} together with Nakayama's lemma imply $N = \{0\}$. 417 | \end{proof} 418 | 419 | \begin{corollary}[Krull's Intersection Theorem]\label{prop:Krull-intersection-domain}\index{Krull's intersection theorem} 420 | Let $R$ be a Noetherian domain and $\mathfrak{a}$ a proper ideal. Then $\bigcap_{n \geq 0} \mathfrak{a}^n = \{0\}$. 421 | \end{corollary} 422 | \begin{proof} 423 | Define $N := \bigcap_{n \geq 0} \mathfrak{a}^n \subset R$. By Theorem \ref{prop:intersection-thm} we have $\mathfrak{a}N = N$, thus there exists $r \in \mathfrak{a}$ with $1+r \in \text{ann}(N)$ by Nakayama's Lemma (Theorem \ref{prop:NAK}). As $\mathfrak{a}$ is proper, $1+r$ cannot be zero. Since $R$ is a domain containing $N$, the only possibility is $N=\{0\}$ as asserted. 424 | \end{proof} -------------------------------------------------------------------------------- /YAlg3-6.tex: -------------------------------------------------------------------------------- 1 | % To be compiled by XeLaTeX, preferably under TeX Live. 2 | % LaTeX source for ``Yanqi Lake Lectures on Algebra'' Part III. 3 | % Copyright 2019 李文威 (Wen-Wei Li). 4 | % Permission is granted to copy, distribute and/or modify this 5 | % document under the terms of the Creative Commons 6 | % Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) 7 | % https://creativecommons.org/licenses/by-nc/4.0/ 8 | 9 | % To be included 10 | \chapter{Dimension of finitely generated algebras} 11 | 12 | The main reference is \cite[\S 13]{Mat80}. 13 | 14 | \section{Dimensions in fibers} 15 | Consider a homomorphism $\varphi: A \to B$, which induces $\varphi^\sharp: \Spec(B) \to \Spec(A)$ on prime spectra. Given $\mathfrak{p} \in \Spec(A)$, we are interested in the fiber $(\varphi^\sharp)^{-1}(\mathfrak{p})$; the prime ideals $\mathfrak{q}$ therein are described by $\varphi^{-1}(\mathfrak{q}) = \mathfrak{p}$, or equivalently: 16 | \[ \varphi^{-1}(\mathfrak{q}) \cap (A \smallsetminus \mathfrak{p}) = \emptyset, \quad \mathfrak{q} \supset \varphi(\mathfrak{p}). \] 17 | Adopt the convention that a zero ring has $\Spec = \emptyset$. The first condition says that $\mathfrak{q}$ comes from $\Spec(B_{\mathfrak{p}})$, where $B_{\mathfrak{p}}$ is the localization with respect to $\varphi(A \smallsetminus \mathfrak{p})$ (possibly zero). The second condition then says that $\mathfrak{q}B_{\mathfrak{p}}$ lies over the image of $\mathfrak{p}A_{\mathfrak{p}}$. Set 18 | \[ \kappa(\mathfrak{p}) := A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}. \] 19 | The fiber $(\varphi^\sharp)^{-1}(\mathfrak{p})$ is then identified with the spectrum of 20 | \[ B \dotimes{A} \kappa(\mathfrak{p}) = \left( B \dotimes{A} A_{\mathfrak{p}} \right) \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p}) = B_{\mathfrak{p}} \dotimes{A_{\mathfrak{p}}} (A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} ), \] 21 | which is empty if and only if $B \dotimes{A} \kappa(\mathfrak{p})$ is zero. This also equips $(\varphi^\sharp)^{-1}(\mathfrak{p})$ with an extra structure: it is the spectrum of an explicit quotient ring of $B_{\mathfrak{p}}$. 22 | 23 | Observe that for all $\mathfrak{q} \in (\varphi^\sharp)^{-1}(\mathfrak{p})$, the localization of $B_{\mathfrak{p}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p})$ at the image of $\mathfrak{q}$ is canonically isomorphic to $B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p})$. 24 | 25 | \begin{proposition}\label{prop:fiber-ineq} 26 | Assume $A, B$ to be Noetherian. Let $\mathfrak{q} \in \Spec(B)$ and $\mathfrak{p} := \varphi^\sharp(\mathfrak{q})$. We have 27 | \begin{enumerate}[(i)] 28 | \item $\dim(B_{\mathfrak{q}}) \leq \dim(A_{\mathfrak{p}}) + \dim B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p})$ (note that the existence of $\mathfrak{q}$ ensures that $B_{\mathfrak{p}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p}) \neq \{0\}$, hence $B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p}) \neq \{0\}$); 29 | \item equality holds if going-down holds for $\varphi$; 30 | \item if going-down holds and $\varphi^\sharp$ is surjective, then $\dim(B) \geq \dim(A)$, and for all ideal $\mathfrak{a} \subsetneq A$ we have $\varphi(\mathfrak{a}) B \neq B$ and $\mathrm{ht}(\mathfrak{a}) = \mathrm{ht}(\varphi(\mathfrak{a})B)$. 31 | \end{enumerate} 32 | \end{proposition} 33 | 34 | It is crucial to notice that $\dim B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p}) = \text{ht}(\mathfrak{q}B_{\mathfrak{q}}/\varphi(\mathfrak{p}) B_{\mathfrak{q}}) = \text{ht}(\mathfrak{q}/\varphi(\mathfrak{p})B)$. 35 | 36 | This bounds the source dimension of a morphism by the target dimension plus the fiber dimension, localized both at $\mathfrak{p}$ and $\mathfrak{q} \in (\varphi^\sharp)^{-1}(\mathfrak{p})$. In order to have an equality, a certain submersion-like condition on $\varphi^\sharp$ is evidently required; this explains the going-down condition. Cf. \cite[(6.H)]{Mat80}. 37 | 38 | \begin{proof} 39 | We have an induced local homomorphism $A_{\mathfrak{p}} \to B_{\mathfrak{q}}$ since $\mathfrak{q} \mapsto \mathfrak{p}$. Since (i) and (ii) depend only on this induced homomorphism, we may assume from the outset that $A$, $B$ are local with maximal ideals $\mathfrak{p}, \mathfrak{q}$, and $\varphi$ is a local homomorphism. Let $d := \dim A$ and take a parameter ideal $I = (t_1, \ldots, t_d)$ of $A$, so that $\mathfrak{p}^k \subset I \subset \mathfrak{p}$ for some $k$. It follows that $\sqrt{\varphi(\mathfrak{p}) B} = \sqrt{\varphi(I)B}$, therefore $\dim(B \dotimes{A} \kappa(\mathfrak{p})) = \dim(B/\varphi(\mathfrak{p})B) = \dim (B/\varphi(I)B)$; denote this number as $e$. Take a $s_1, \ldots, s_e \in \mathfrak{q}$ whose images generate a parameter ideal for $B/\varphi(I)B$, and put $J := (\varphi(t_1), \ldots, \varphi(t_d), s_1, \ldots, s_e)$. Then $B/J$ is Artinian, therefore $\dim B \leq d+e$ establishes (i). 40 | 41 | As for (ii), we conserve the same hypotheses and take a prime chain $\mathfrak{q} = \mathfrak{q}_0 \supsetneq \cdots \supsetneq \mathfrak{q}_e$ with $\mathfrak{q}_e \supset \varphi(\mathfrak{p})B$ in $B$, as well as a prime chain $\mathfrak{p} = \mathfrak{p}_0 \supsetneq \cdots \supsetneq \mathfrak{p}_d$ in $A$. Note that $\varphi^{-1}(\mathfrak{q}_e) = \mathfrak{p}$ since $e = \dim B/\varphi(\mathfrak{p})B$. By applying going-down repeatedly to the chain $\mathfrak{p}_i$, we obtain a prime chain in $B$ 42 | \[ \mathfrak{q}_e \supsetneq \cdots \supsetneq \mathfrak{q}_{e+d}, \quad \varphi^{-1}(\mathfrak{q}_{e+i}) = \mathfrak{p}_i. \] 43 | Concatenation with $\mathfrak{q}_0 \supsetneq \cdots$ gives $\dim(B) = d + e$. This shows (ii). 44 | 45 | The first assertion of (iii) results from (ii). To show the remaining one, let us show $\varphi(\mathfrak{a})B \neq B$: if $\varphi^\sharp(\mathfrak{q}) = \mathfrak{p} \supset \mathfrak{a}$, then $\mathfrak{q} \supset \varphi(\mathfrak{a}) B$. Next, take a minimal over-prime $\mathfrak{q} \supset \varphi(\mathfrak{a})B$ with $\text{ht}(\mathfrak{q}) = \text{ht}(\varphi(\mathfrak{a})B)$. With $\mathfrak{p} := \varphi^{-1}(\mathfrak{q}) \supset \mathfrak{a}$, we must have $\dim B_{\mathfrak{q}} \otimes \kappa(\mathfrak{p}) = \text{ht}(\mathfrak{q}/\varphi(\mathfrak{p})B) = 0$ by the minimality of $\mathfrak{q}$. An application of (ii) yields 46 | \begin{gather*} 47 | \text{ht}(\varphi(\mathfrak{a})B) = \text{ht}(\mathfrak{q}) = \text{ht}(\mathfrak{p}) \geq \text{ht}(\mathfrak{a}). 48 | \end{gather*} 49 | To obtain $\leq$, choose $\mathfrak{p} \supset \mathfrak{a}$ with $\text{ht}(\mathfrak{p}) = \text{ht}(\mathfrak{a})$ and take $\mathfrak{q} \in \Spec(B)$ with $\mathfrak{p} = \varphi^{-1}(\mathfrak{q})$; this implies $\mathfrak{q} \supset \varphi(\mathfrak{p})B \supset \varphi(\mathfrak{a})B$. Upon shrinking $\mathfrak{q}$, we may even assume $\mathfrak{q}$ is minimal over $\varphi(\mathfrak{p})B$, i.e. $\text{ht}(\mathfrak{q}/ \varphi(\mathfrak{p})B) = 0$. Using (ii), this entails $\text{ht}(\mathfrak{a}) = \text{ht}(\mathfrak{p}) = \text{ht}(\mathfrak{q}) \geq \text{ht}(\varphi(\mathfrak{a})B)$. 50 | \end{proof} 51 | 52 | \vspace{1em} 53 | \begin{center} 54 | \begin{tikzpicture} 55 | \draw (-2,1) -- (2,1); 56 | \draw[line width=6pt, white] (0, 2) -- (0, -1.5); 57 | \draw[line width=2pt, black] (0,2) -- (0, 0.7); 58 | \draw[line width=1.5pt, dashed] (0, 0.65) -- (0, 0); 59 | \draw[line width=2pt, black] (0,0) -- (0, -1.2); 60 | \draw[line width=5pt, white] (-3,0) -- (3,0); 61 | \draw (-3,0) -- (3,0); 62 | \draw (-3,0) -- (-2,1); 63 | \draw (3,0) -- (2,1); 64 | \fill[opacity=0.2, shade, color=black!50!white] (-2,1) -- (2,1) -- (3,0) -- (-3,0) -- (-2,1); 65 | \node[anchor=west, text width=4cm, align=left] at (4.5, 0.7) {\small The minimal primes in a ring $B$ may have different heights when the scheme $\Spec(B)$ is not equi-dimensional.}; 66 | \end{tikzpicture} 67 | \end{center} 68 | \vspace{1em} 69 | 70 | Going-down holds for flat $\varphi$ by Theorem \ref{prop:going-down-flat}, therefore the dimension equality 71 | \[ \dim(B_{\mathfrak{q}}) = \dim(A_{\mathfrak{p}}) + \dim B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa(\mathfrak{p}) \] 72 | holds for flat ring homomorphisms. 73 | 74 | \begin{remark} 75 | In general, if $B$ is a finitely generated algebra over Noetherian $A$ such that $\Spec(B) \to \Spec(A)$ is a closed map, the fiber dimension $\mathfrak{p} \mapsto \dim B/\mathfrak{p}B$ is an \emph{upper semi-continuous function} on the target space $\Spec(A)$. More concretely, the fiber dimension is non-decreasing under specialization of $\mathfrak{p}$. Cf. \cite[\S 14.3]{Eis95} or \cite[(13.E)]{Mat80}. Try to understand this phenomenon intuitively. 76 | \end{remark} 77 | 78 | \begin{proposition}\label{prop:dim-integral} 79 | Suppose $B$ is integral over a subring $A$. 80 | \begin{enumerate}[(i)] 81 | \item We have $\dim A = \dim B$. 82 | \item If we assume moreover that $A, B$ are both Noetherian, then $\mathrm{ht}(\mathfrak{q}) \leq \mathrm{ht}(\mathfrak{q} \cap A)$ for every $\mathfrak{q} \in \Spec(B)$. 83 | \item Furthermore, if going-down also holds for $A \hookrightarrow B$, we have $\mathrm{ht}(J) = \mathrm{ht}(J \cap A)$ for every ideal $J \subsetneq B$. 84 | \end{enumerate} 85 | \end{proposition} 86 | \begin{proof} 87 | Going-up holds and $\Spec(B) \twoheadrightarrow \Spec(A)$ in the situation of (i) by Theorem \ref{prop:Cohen-Seidenberg}, hence $\dim B \geq \dim A$ by lifting prime chains. To prove $\leq$, observe that $\mathfrak{q} \subsetneq \mathfrak{q}'$ implies $\mathfrak{q} \cap A \subsetneq \mathfrak{q}' \cap A$ since there are no inclusion relations in the fibers of $\Spec(B) \to \Spec(A)$. 88 | 89 | As for (ii), note that $\text{ht}(\mathfrak{q}) \leq \text{ht}(\mathfrak{p}) + \text{ht}(\mathfrak{q}/\mathfrak{p}B)$ where $\mathfrak{p} := \mathfrak{q} \cap A$; as the are no inclusion relations in fibers, the last term must be $0$. 90 | 91 | Now assume going-down and consider (iii). Take $\mathfrak{q} \in \Spec(B)$ with $\text{ht}(\mathfrak{q}) = \text{ht}(J)$. Put $\mathfrak{p} := \mathfrak{q} \cap A \supset J　\cap A$. Again, since there are no inclusions in the fiber over $\mathfrak{p}$ of $\Spec(B) \twoheadrightarrow \Spec(A)$, we have $\dim(B_{\mathfrak{q}}/ \mathfrak{p}B_{\mathfrak{q}}) = 0$. Proposition \ref{prop:fiber-ineq} (ii) implies $\text{ht}(\mathfrak{q})=\text{ht}(\mathfrak{p})$, therefore $\text{ht}(J) \geq \text{ht}(J \cap A)$. 92 | 93 | On the other hand, for any $\mathfrak{p} \supset J \cap A$ with $\text{ht}(\mathfrak{p}) = \text{ht}(J \cap A)$, since $A/J \cap A \hookrightarrow B/J$ is integral, there exists $\mathfrak{q} \supset J$ with $\mathfrak{q} \cap A = \mathfrak{p}$. Together with Proposition \ref{prop:fiber-ineq} (i) and $\mathrm{ht}(\mathfrak{q}/\mathfrak{p}B) = 0$, this implies $\text{ht}(J) \leq \text{ht}(\mathfrak{q}) \leq \text{ht}(\mathfrak{p}) = \text{ht}(J \cap A)$ by (ii). 94 | \end{proof} 95 | 96 | \section{Calculation for polynomial algebras} 97 | Let us apply the results from the previous section to elucidate the Krull dimension of polynomial algebras. 98 | 99 | \begin{theorem}\label{prop:dim-polynomial-alg} 100 | Let $A$ be a Noetherian ring, we have $\dim A[X_1, \ldots, X_n] = \dim A + n$ for any $n \geq 0$. In particular, $\dim A[X_1, \ldots, X_n] = n$ if $A$ is Artinian (eg.\ a field). 101 | \end{theorem} 102 | \begin{proof} 103 | Evidently we may assume $n=1$. We shall apply Proposition \ref{prop:fiber-ineq} to $A \hookrightarrow B = A[X]$. Take any $\mathfrak{p} \in \Spec(A)$ and let $\mathfrak{q}$ be a maximal element in $\{\mathfrak{q}' \in \Spec(B): \mathfrak{q} \cap A = \mathfrak{p} \}$. Put $\kappa := \kappa(\mathfrak{p})$. It suffices to show that $B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa$ has dimension one, since $B$ is free hence flat over $A$, and Proposition \ref{prop:fiber-ineq} will imply 104 | \[ \dim B_{\mathfrak{q}} = \dim A_{\mathfrak{p}} + 1 \] 105 | and taking supremum over $\mathfrak{p} \in \Spec(A)$ gives the result. 106 | 107 | Indeed, put $B_{\mathfrak{p}} := B[(A \smallsetminus \mathfrak{p})^{-1}] = A_{\mathfrak{p}}[X]$ and $\mathfrak{q}' := \mathfrak{q}[(A \smallsetminus \mathfrak{p})^{-1}] \in \Spec(B_{\mathfrak{p}})$. As $\mathfrak{q}' \supset \mathfrak{p}A_{\mathfrak{p}}$, we have $\overline{\mathfrak{q}'} := \mathfrak{q}' \supset \mathfrak{p}A_{\mathfrak{p}} \in \Spec(\kappa[X])$, and $\overline{\mathfrak{q}'}$ is maximal in the fiber over $\{0\}$ of $\Spec(\kappa[X]) \to \Spec(\kappa)$, i.e.\ in $\MaxSpec(\kappa[X])$. Localization in stages yields 108 | \[ B_{\mathfrak{q}} \dotimes{A_{\mathfrak{p}}} \kappa \simeq \frac{B_{\mathfrak{q}}}{\mathfrak{p} B_{\mathfrak{q}}} \simeq \left( \frac{B_{\mathfrak{p}}}{ \mathfrak{p} A_{\mathfrak{p}} B_{\mathfrak{p}}} \right)_{\mathfrak{q}' } \simeq \kappa[X]_{\overline{\mathfrak{q}'}}. \] 109 | As $\kappa[X]$ is a principal ideal domain which is not a field, every maximal ideal thereof has height one. Hence $\dim \kappa[X]_{\overline{\mathfrak{q'}}} = \mathrm{ht}(\mathfrak{q}') = 1$. 110 | \end{proof} 111 | 112 | \begin{corollary}\label{prop:poly-alg-ht} 113 | Let $\Bbbk$ be a field, then for every $0 \leq i \leq n$ we have $\mathrm{ht}(X_1, \ldots, X_i) = i$ in $\Bbbk[X_1, \ldots, X_n]$. 114 | \end{corollary} 115 | \begin{proof} 116 | The prime chain $\{0\} \subset (X_1) \subsetneq \cdots \subsetneq (X_1, \ldots, X_n)$ has length $n = \dim \Bbbk[X_1, \ldots, X_n]$. Thus for each $0 \leq i \leq n$, the chain $\{0\} \subset (X_1) \subsetneq \cdots \subsetneq (X_1, \ldots, X_i)$ has maximal length among all prime chains starting with $(X_1, \ldots, X_i)$. 117 | \end{proof} 118 | 119 | Combined with Theorem \ref{prop:dim-polynomial-alg}, we see that for $\Bbbk$ a field, $R := \Bbbk[X_1, \ldots, X_n]$ and $\mathfrak{p} := (X_1, \ldots, X_i)$, the equality 120 | \[ \text{ht}(\mathfrak{p}) + \dim R/\mathfrak{p} = \dim R. \] 121 | holds. This will be generalized to finitely generated domains over $\Bbbk$. 122 | 123 | We record another simple consequence for later use. 124 | \begin{corollary}\label{prop:fg-dimension-bound} 125 | Let $\Bbbk$ be a field. Any $\Bbbk$-algebra $A$ with $n$ generators has finite dimension $\leq n$. 126 | \end{corollary} 127 | \begin{proof} 128 | Writing $A = \Bbbk[X_1, \ldots, X_n]/I$ for some ideal $I$, we have $\dim A \leq \Bbbk[X_1, \ldots, X_n]$ since every prime chain in $A$ lifts to $\Bbbk[X_1, \ldots, X_n]$. Now apply Theorem \ref{prop:dim-polynomial-alg}. 129 | \end{proof} 130 | 131 | \section{Noether normalization and its consequences} 132 | Fix a field $\Bbbk$. A few preparatory results are in order. 133 | 134 | \begin{lemma}\label{prop:normalization-Nagata} 135 | Suppose $\Bbbk$ is a field and $t \in \Bbbk[X_1, \ldots X_e] \smallsetminus \Bbbk$. There exist $t_1, \ldots, t_{e-1} \in \Bbbk[X_1, \ldots, X_e]$ such that $\Bbbk[X_1, \ldots, X_e]$ is finitely generated as a module over the $\Bbbk$-subalgebra $S := \Bbbk[t_1, \ldots, t_{e-1}, t]$. 136 | \end{lemma} 137 | \begin{proof} 138 | We seek $t_i$ of the form $X_i - X_e^{k^i}$ where $k$ is a large integer. Then $t$ can be uniquely expressed as a polynomial of $t_1, \ldots, t_{e-1}, X_e$. We claim that upon modifying $t$ by $\Bbbk^\times$, which is clearly harmless, one can choose $k$ such that $t$ is monic as an element of $\Bbbk[t_1, \ldots, t_{e-1}][X_e]$, say of some degree $\delta$. If this is the case, 139 | \[ t = X_e^\delta + \sum_{0 \leq j < \delta} \left(\text{polynomial in } t_1, \ldots, t_{e-1}\right) X_e^j \] 140 | says that $X_e$ is integral over $S$, and then $\Bbbk[X_1, \ldots, X_e]$ is generated as an $S$-module by $1, X_e, \ldots, X_e^{\delta-1}$. 141 | 142 | To choose $k$, one stares at the expansion 143 | \[ X_1^{a_1} \cdots X_e^{a_e} = \left(t_1 + X_e^{k^1}\right)^{a_1} \cdots \left(t_{e-1} + X_e^{k^{e-1}}\right)^{a_{e-1}} \cdot X_e^{a_e} = X_e^{a_e + a_1 k^1 + \cdots + a_{e-1} k^{e-1}} + \text{mixed terms}. \] 144 | If $k > \max\{a_1, \ldots, a_e \}$, the exponent of $X_e$ is simply the base $k$ expression with digits $a_e, a_1, \ldots, a_{e-1}$. Now write $t$ as a linear combination of monomials $X_1^{a_1} \cdots X_e^{a_e}$ and expand them in terms of $t_1, \ldots, t_{e-1}, X_e$. From the observation above, different $(a_1, \ldots, a_e)$ contributes a different exponent of $X_e$ whenever $k \gg 0$. Adjusting $t$ by $\Bbbk^\times$, we get the asserted property. 145 | \end{proof} 146 | 147 | \begin{lemma}\label{prop:normalization-aux} 148 | Let $\mathfrak{a}$ be a nonzero ideal of a domain $R$, then $\dim(R/\mathfrak{a}) + 1 \leq \dim R$. 149 | \end{lemma} 150 | \begin{proof} 151 | Any prime chain $\mathfrak{p}_0 \supsetneq \cdots \supsetneq \mathfrak{p}_n$ in $R$ with $\mathfrak{p}_n \supset \mathfrak{a}$ can be extended to a prime chain length $n+1$, namely by adjoining $\mathfrak{p}_{n+1} := \{0\}$. 152 | \end{proof} 153 | 154 | \begin{theorem}[E.\ Noether, M.\ Nagata]\label{prop:Noether-normalization}\index{Noether normalization} 155 | Let $B$ be a finitely generated $\Bbbk$-algebra of dimension $n$. Consider a chain of proper ideals $I_1 \subsetneq \cdots \subsetneq I_m$, with $\dim(B/I_j) = d_j$ and $d_1 > \cdots > d_m \geq 0$. Then there exist a $\Bbbk$-subalgebra $A \subset B$ together with an isomorphism $A \simeq \Bbbk[X_1, \ldots, X_n]$, satisfying 156 | \begin{compactitem} 157 | \item $B$ is a finitely generated $A$-module, in particular $B$ is integral over $A$; 158 | \item $I_j \cap A \simeq (X_{d_j+1}, \ldots, X_n)$ under the isomorphism above, for each $1 \leq j \leq m$. 159 | \end{compactitem} 160 | \end{theorem} 161 | 162 | Note that the assumption $I_j \subsetneq I_{j+1}$ is merely for convenience. Allowing $I_j = I_{j+1}$ and $d_j = d_{j+1}$ for some $j$ is surely possible.. 163 | 164 | \begin{proof} 165 | Write $B = \Bbbk[Y_1, \ldots, Y_r]/J$, where $r \geq n$ (Corollary \ref{prop:fg-dimension-bound}), and denote the preimage of $I_j$ in $\Bbbk[Y_1, \ldots, Y_r]$ by $\tilde{I}_j$. The first step is to reduce to the case $J=\{0\}$. To see this, we adjoin $\tilde{I}_0 = \{0\}$ (with $d_0 = n$) into the ideal chain; it may happen that $\tilde{I}_0 = \tilde{I}_1$, but that's harmless. Suppose we can find $\tilde{A} \subset \Bbbk[Y_1, \ldots, Y_r]$, $\tilde{A} \simeq \Bbbk[X_1, \ldots, X_r]$ with the required properties relative to $\tilde{I}_\bullet$. Taking quotient by $J$, we obtain the corresponding properties for $I_\bullet$. Indeed, the passage from $\tilde{A}$ to $A := \tilde{A}/(\tilde{A} \cap J)$ truncates the variables $X_{n + 1}, \ldots, X_r$, whereas 166 | \[ I_j \cap A = \frac{\tilde{I}_j \cap (\tilde{A} + J)}{J} = \frac{(\tilde{I}_j \cap \tilde{A}) + J}{J} \simeq \frac{\tilde{I}_j \cap \tilde{A}}{J \cap \tilde{A}}; \] 167 | try to convince yourself of the middle equality. 168 | 169 | Secondly, having reduced to the case $B = \Bbbk[Y_1, \ldots, Y_r]$ (thus $r = n$), it suffices to pick $x_1, \ldots, x_n \in B$ such that 170 | \begin{enumerate}[(a)] 171 | \item $B$ is finitely generated as a module over $A := \Bbbk[x_1, \ldots, x_n]$, and 172 | \item $I_j \cap A \supset (x_{d_j+1}, \ldots, x_n)$ for all $j$. 173 | \end{enumerate} 174 | Indeed, (a) implies $\mathrm{Frac}(A)$ and $\mathrm{Frac}(B)$ have the same transcendence degree $n$ over $\Bbbk$, therefore $x_1, \ldots, x_n$ must be algebraically independent over $\Bbbk$. On the other hand, $\dim(B/I_j) = \dim(A/I_j \cap A)$ by (a) and Proposition \ref{prop:dim-integral}; but if $I_j \cap A \supsetneq (x_{d_j+1}, \ldots, x_n)$ then $A/I_j \cap A$ is a proper quotient of the domain $\Bbbk[x_1, \ldots, x_{d_j}]$, therefore would have dimension $< d_j$ by Lemma \ref{prop:normalization-aux}. 175 | 176 | We shall construct $x_1, \ldots, x_n$ step by step. Suppose $0 \leq e \leq n$ and that we have produced elements $x'_1, \ldots, x'_e$ and $x_{e+1}, \ldots, x_n$ in $B$ satisfying 177 | \begin{enumerate}[(i)] 178 | \item $B$ is finitely generated as a module over $\Bbbk[x'_1, \ldots, x'_e, x_{e+1}, \ldots, x_n] =: S_e$; 179 | \item $I_j \cap S_e \supset (x_{e+1}, \ldots, x_n)$ when $d_j \leq e$; 180 | \item $I_j \cap S_e \supset (x_{d_j + 1}, \ldots, x_n)$ when $d_j \geq e$ (with $j = 1, \ldots, m$). 181 | \end{enumerate} 182 | 183 | For the initial case $e=n$, simply take $x'_i := Y_i$. Our aim is $e=0$. Let us explain the induction step from $e \geq 1$ to $e-1$. The prior argument based on transcendence degrees implies the algebraic independence among 184 | \[ x'_1, \ldots, x'_e, x_{e+1}, \ldots, x_n. \] 185 | If $e \leq d_j$ for all $j$, we are done. Otherwise set $j := \min\{j': e > d_{j'} \}$, we contend that 186 | \[ I_j \cap \Bbbk[x'_1, \ldots, x'_e] \neq \{0\}. \] 187 | If not, we would have $I_j \cap S_e = (x_{e+1}, \ldots, x_n)$ since $I_j \cap S_e \supset (x_{e+1}, \ldots, x_n)$ by (ii). We have $\dim(S_e/I_j \cap S_e) = \dim(B/I_j) = d_j$ by integrality, whilst $\dim(S_e/(x_{e+1}, \ldots, x_n)) = e$. Contradiction. 188 | 189 | Now take $x_e \in I_j \cap \Bbbk[x'_1, \ldots, x'_e] \smallsetminus \{0\}$. Note that $x_e \notin \Bbbk$ as $I_j \neq B$. By Lemma \ref{prop:normalization-Nagata}, we may choose $x''_1, \ldots, x''_{e-1} \in \Bbbk[x'_1, \ldots, x'_e]$ such that $\Bbbk[x'_1, \ldots, x'_e]$ is finitely generated over $\Bbbk[x''_1, \ldots, x''_{e-1}, x_e]$ as a module. It remains to verify that the new sequence 190 | \[ x''_1, \ldots, x''_{e-1}, x_e, \ldots, x_n \] 191 | satisfies (i)---(iii) above with $e-1$ replacing $e$. 192 | 193 | First, $S_e$ is a finitely generated module over its subalgebra $\Bbbk[x''_1, \ldots, x_n]$ by construction, hence so is $B$ and (i) follows. Next, let $1 \leq j' \leq m$. If $d_{j'} > e-1$, the procedure above does not affect $x_{d_j + 1}, \ldots, x_n$, so they belong to $I_{j'} \cap \Bbbk[x''_1, \ldots, x_n]$. If $d_{j'} < e$, setting $j := \min\{j'' : e > d_{j''} \}$ we have $j' \geq j$ and 194 | \[ I_{j'} \cap \Bbbk[x''_1, \ldots, x_n] \supset I_j \cap \Bbbk[x''_1, \ldots, x_n] \supset (x_{e+1}, \ldots, x_n) + (x_e). \] 195 | All in all, we obtain (ii) and (iii). 196 | \end{proof} 197 | 198 | \begin{corollary}[Dimension formula]\label{prop:fg-dim-formula}\index{dimension formula} 199 | Let $B$ be a finitely generated algebra over a field $\Bbbk$. Suppose $B$ is a domain, then for all $\mathfrak{q} \in \Spec(B)$ we have 200 | \[ \dim(B/\mathfrak{q}) + \mathrm{ht}(\mathfrak{q}) = \dim B. \] 201 | \end{corollary} 202 | \begin{proof} 203 | Choose a subalgebra $A$ of $B$ as in Theorem \ref{prop:Noether-normalization} and put $\mathfrak{p} = \mathfrak{q} \cap A$. Since $B$ is a domain and $A$ is normal, the Cohen--Seidenberg Theorem \ref{prop:Cohen-Seidenberg} asserts the going-down property for $A \hookrightarrow B$. Proposition \ref{prop:dim-integral} implies that $\dim A = \dim B$, $\dim A/\mathfrak{p} = \dim B/\mathfrak{q}$ and $\text{ht}(\mathfrak{p}) = \text{ht}(\mathfrak{q})$, so we are again reduced to the case $B = \Bbbk[X_1, \ldots, X_n]$ and $\mathfrak{q} = (X_{d+1}, \ldots, X_n)$. This is known by Corollary \ref{prop:poly-alg-ht}. 204 | \end{proof} 205 | 206 | The dimension formula allows us to compute $\dim B$ by choosing any prime $\mathfrak{q}$, a prime chain of longest length below $\mathfrak{q}$ (i.e. in $B_{\mathfrak{q}}$) and another one above $\mathfrak{q}$ (i.e. in $B/\mathfrak{q}$); their concatenation will then be a prime chain in $B$ with maximal length $\dim B$. By applying the dimension formula to $\mathfrak{q}, \mathfrak{q}' \subset B$ and to $\mathfrak{q}'/\mathfrak{q} \subset B/\mathfrak{q}$, it follows that $\text{ht}(\mathfrak{q}') = \text{ht}(\mathfrak{q}) + \text{ht}(\mathfrak{q}'/\mathfrak{q})$ for all $\mathfrak{q}' \supsetneq \mathfrak{q}$ in $B$. 207 | 208 | For a general Noetherian domain $B$, we call a prime chain \emph{maximal} if it is not properly contained in any prime chain. A priori, a maximal prime chain does not necessarily have length equal to $\dim B$. If 209 | \[ \forall \mathfrak{q}' \supset \mathfrak{q}: \text{primes}, \quad \text{ht}(\mathfrak{q}') = \text{ht}(\mathfrak{q}) + \text{ht}(\mathfrak{q}'/\mathfrak{q}), \] 210 | we say $B$ is a \emph{catenary} domain\index{catenary}. This means that $\text{ht}(\mathfrak{q}'/\mathfrak{q}) = \dim B_{\mathfrak{q}'}/\mathfrak{q}B_{\mathfrak{q}'}$ is the common length of all maximal prime chains between $\mathfrak{q}'$ and $\mathfrak{q}$. In particular, maximal prime chains in $B_{\mathfrak{q}'}$ are automatically longest. As shown above, finitely generated domains over a field are catenary. For a finer analysis of catenary and universally catenary rings, we refer to \cite[\S 14]{Mat80}. 211 | 212 | \begin{corollary} 213 | Suppose $B$ is a domain finitely generated over a field $\Bbbk$. Set $L := \mathrm{Frac}(B)$. Then $\dim B = \mathrm{tr.deg}_\Bbbk(L)$ and it is the common length of maximal prime chains. 214 | \end{corollary} 215 | \begin{proof} 216 | Choose a subalgebra $A$ of $B$ as in Theorem \ref{prop:Noether-normalization}. Since the field $\text{Frac}(B)$ is a finite extension of $\text{Frac}(A)$ and $\dim A = \dim B$ by Proposition \ref{prop:dim-integral}, the first assertion reduces immediately to the case $B = \Bbbk[X_1, \ldots, X_n]$, which is obvious. As to the second assertion, consider a maximal prime chain $\mathfrak{q}_0 \supsetneq \cdots \supsetneq \mathfrak{q}_n = \{0\}$ in $B$. Maximality implies $\text{ht}(\mathfrak{q}_i/\mathfrak{q}_{i+1}) = 1$ for all $i$ and $\dim B/\mathfrak{q}_0 = 0$. Applying Corollary \ref{prop:fg-dim-formula} repeatedly, we see 217 | \begin{multline*} 218 | \dim B = \dim B/\mathfrak{q}_n = \dim B/\mathfrak{q}_{n-1} + \text{ht}(\mathfrak{q}_{n-1}/\mathfrak{q}_n) = \\ 219 | \dim B/\mathfrak{q}_{n-2} + \text{ht}(\mathfrak{q}_{n-1}/\mathfrak{q}_{n-2}) + \text{ht}(\mathfrak{q}_{n-1}/\mathfrak{q}_n) = \cdots = \dim B/\mathfrak{q}_0 + \sum_{i=0}^{n-1} \text{ht}(\mathfrak{q}_i/\mathfrak{q}_{i+1}) 220 | \end{multline*} 221 | which equals $n$. 222 | \end{proof} 223 | 224 | \begin{remark} 225 | Recall that finitely generated domains over an algebraically closed field $\Bbbk$ are objects ``opposite'' to the irreducible affine $\Bbbk$-varieties. It is instructive to make a comparison with the analytic theory when $\Bbbk=\CC$. Let $\mathcal{X}$ be a compact connected complex manifold of (complex) dimension $n$. Denote by $\mathcal{M}(\mathcal{X})$ the field of meromorphic functions on $\mathcal{X}$. Siegel proved that $\text{tr.deg}_{\CC}(\mathcal{M}(\mathcal{X})) \leq n$. When $\mathcal{X}$ is a projective algebraic $\CC$-variety, equality holds and we have $\mathcal{M}(\mathcal{X}) = \text{Frac}(A)$ if $\Spec(A)$ is any open dense affine subscheme in $\mathcal{X}$. In general, the abundance of meromorphic functions on $\mathcal{X}$ is a subtle issue, cf. the case of Riemann surfaces ($n=1$). Compact connected complex manifolds with $\text{tr.deg}_{\CC}(\mathcal{M}(\mathcal{X})) = \dim_{\CC} \mathcal{X}$ are called \emph{Moishezon manifolds}.\index{Moishezon manifolds} 226 | 227 | Non-algebraic Moishezon manifolds do exist, and Moishezon proved that a Moishezon manifold is projective, hence algebraic, if and only if it is Kähler. Following M.\ Artin and D.\ Knutson, one can enlarge the category of $\CC$-schemes into that of \emph{algebraic spaces} over $\CC$, and there is an analytification functor $\mathcal{X} \mapsto \mathcal{X}_{\text{an}}$ that sends algebraic spaces of finite type over $\CC$ to complex analytic varieties. M. Artin \cite[\S 7]{Ar70} showed that the analytification establishes an equivalence between the category of smooth proper algebraic spaces of finite type over $\CC$ and that of Moishezon manifolds. Therefore, such manifolds still retain an algebraic flavor: they are quotients of certain $\CC$-schemes by étale equivalence relations. 228 | \end{remark} 229 | -------------------------------------------------------------------------------- /YAlg3-7.tex: -------------------------------------------------------------------------------- 1 | % To be compiled by XeLaTeX, preferably under TeX Live. 2 | % LaTeX source for ``Yanqi Lake Lectures on Algebra'' Part III. 3 | % Copyright 2019 李文威 (Wen-Wei Li). 4 | % Permission is granted to copy, distribute and/or modify this 5 | % document under the terms of the Creative Commons 6 | % Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) 7 | % https://creativecommons.org/licenses/by-nc/4.0/ 8 | 9 | % To be included 10 | \chapter{Serre's criterion for normality and depth} 11 | 12 | References: \cite[\S 11]{Eis95} and \cite[{X.1}]{Bour98}. Except in the last section, we will try to avoid the use of depth as in \cite{Eis95}. 13 | 14 | \section{Review of discrete valuation rings} 15 | Let $R$ be an integral domain. 16 | 17 | \begin{definition}\index{discrete valuation ring} 18 | A \emph{discrete valuation} on a field $K$ is a surjective map $v: K^\times \to \Z$ such that 19 | \begin{itemize} 20 | \item $v(xy) = v(x) + v(y)$, and 21 | \item $v(x+y) \geq \min\{v(x), v(y) \}$ 22 | \end{itemize} 23 | for all $x,y \in K^\times$, where we set $v(0) = +\infty$ for convenience. We say a domain $R$ with $K := \mathrm{Frac}(R)$ is a \emph{discrete valuation ring}, abbreviated as DVR, if there exists a discrete valuation $v$ such that 24 | \[ R = v^{-1}([0, +\infty]). \] 25 | We say $t \in R$ is a \emph{uniformizer} if $v(t)=1$.\index{uniformizer} 26 | \end{definition} 27 | It follows immediately that $R^\times = v^{-1}(0)$. Uniformizers are unique up to $R^\times$. Note that $v(K^\times) = \Z$ implies that $R$ cannot be a field. 28 | 29 | \begin{example} 30 | The ring $\Z_p$ of $p$-adic integers ($p$: prime number) together with the usual $p$-adic valuations are standard examples of DVR. The algebra of formal power series $\Bbbk\llbracket X\rrbracket$ are also DVR: the valuation of $\sum_n a_n X^n$ is the smallest $n$ such that $a_n \neq 0$. 31 | 32 | More generally, in the geometric context, discrete valuations can be defined by looking at the vanishing order of rational/meromorphic functions along subvarieties of codimension one with suitable regularities. 33 | \end{example} 34 | 35 | \begin{lemma}\label{prop:dvr-regular} 36 | Let $R$ be a discrete valuation ring with valuation $v$ and uniformizer $t$. Every ideal $\mathfrak{a} \neq \{0\}$ of $R$ has the form $(t^r)$ for a unique $r \geq 0$. In particular, $R$ is a local principal ideal domain which is not a field, hence is of dimension $1$. 37 | \end{lemma} 38 | \begin{proof} 39 | Take $r := \min\{v(x) : x \in \mathfrak{a} \}$. 40 | \end{proof} 41 | 42 | In the exercises below, we assume $R$ is a discrete valuation ring with valuation $v$. 43 | \begin{exercise} 44 | Show that $t$ is a uniformizer if and only if it generates the maximal ideal of $R$. 45 | \end{exercise} 46 | \begin{exercise} 47 | Reconstruct $v$ from the ring-theoretic structure of $R$. 48 | \end{exercise} 49 | 50 | Recall that a regular local ring $R$ with $\dim R = 1$ is a Noetherian local ring whose maximal ideal $\mathfrak{m}$ is principal and nonzero; elements generating $\mathfrak{m}$ are called the regular parameters for $R$. 51 | \begin{proposition}\label{prop:reg-local-dvr} 52 | Suppose $t$ is a regular parameter in a regular local ring $R$ of dimension one, then $R$ is a domain, and every element $x \in R \smallsetminus \{0\}$ can be uniquely written as $x = t^r u$ with $r \geq 0$ and $u \in R^\times$. This makes $R$ into a discrete valuation ring by setting $v(x) = r$, for which $t$ is a uniformizer. 53 | 54 | Therefore $R$ is a discrete valuation ring. Conversely, every discrete valuation ring is regular local of dimension $1$. 55 | \end{proposition} 56 | \begin{proof} 57 | From Theorem \ref{prop:regular-local-domain} we know regular local rings are Noetherian domains. By applying Krull's Intersection Theorem (Corollary \ref{prop:Krull-intersection-domain}) to the powers of $(t)$, we see that $r := \sup\{k \geq 0: x \in (t)^k \}$ is finite. Write $x = t^r u$. Since $R^\times = R \smallsetminus \mathfrak{m}$, we see $u \in R^\times$. As to uniqueness, suppose $t^r u = t^s w$ with $r \geq s$, then $t^{r-s} = u^{-1}w \in R^\times$ implies $r=s$, hence $u=w$ as $R$ is a domain. As every element of $\text{Frac}(R)^\times$ is uniquely expressed as $t^r u$ with $r \in \Z$, one readily checks that $v(t^r u) = r$ satisfies all the requirements of discrete valuation. 58 | 59 | The converse direction has been addressed in Lemma \ref{prop:dvr-regular}. 60 | \end{proof} 61 | 62 | To recap, in dimension one we have 63 | \[ \text{regular local ring} \iff \text{discrete valuation ring}. \] 64 | This will be related to normality later on. 65 | 66 | \begin{exercise} 67 | Explain that the regular local rings of dimension $0$ are just fields. 68 | \end{exercise} 69 | 70 | \section{Auxiliary results on the total fraction ring} 71 | Let $R$ be a ring, henceforth assumed Noetherian. If there exist a non zero-divisor $t \in R$ and $\mathfrak{p} \in \text{Ass}(R/(t))$, we say $\mathfrak{p}$ is \emph{associated to a non zero-divisor}. 72 | 73 | \begin{lemma}\label{prop:zero-test-ass} 74 | Let $M$ be a finitely generated $R$-module. An element $x \in M$ is zero if and only if its image in $M_{\mathfrak{p}}$ is zero for every maximal element $\mathfrak{p}$ in $\mathrm{Ass}(M)$. 75 | \end{lemma} 76 | \begin{proof} 77 | Suppose $x \neq 0$. Since $M$ is Noetherian, among ideals of the form $\text{ann}(y)$ there is a maximal one containing $\text{ann}(x)$, and we have seen in Lemma \ref{prop:Ass-maximal} that such an ideal $\mathfrak{p}$ belongs to $\text{Ass}(M)$. Since $\text{ann}(x) \subset \mathfrak{p}$, we have $x/1 \in M_{\mathfrak{p}} \smallsetminus \{0\}$. 78 | \end{proof} 79 | 80 | Call a ring \emph{reduced}\index{reduced ring} if it has no nilpotent element except zero. 81 | \begin{lemma} 82 | Suppose $R$ is reduced, then $\mathrm{Ass}(R)$ consists of minimal primes. 83 | \end{lemma} 84 | \begin{proof} 85 | As $R$ is reduced, $\{0\} = \sqrt{0_R}$ is the intersection of minimal prime ideals $\mathfrak{p}_1, \mathfrak{p}_2, \ldots$ (all lying in $\text{Ass}(R)$, hence finite in number). By the theory of primary decompositions, one infers that $\text{Ass}(R) = \{\mathfrak{p}_1, \ldots \}$. 86 | \end{proof} 87 | 88 | For the next result, we denote by $T$ the set of non zero-divisors of $R$. Recall that the \emph{total fraction ring} $K(R)$ is $R[T^{-1}]$; this is the largest localization such that $R \to K(R)$ is injective, and $K(R) = \text{Frac}(R)$ when $R$ is a domain. The map $\mathfrak{p} \mapsto \mathfrak{p}K(R)$ sets up an order-preserving bijection $\text{Ass}(R) \rightiso \text{Ass}(K(R))$: indeed, if $\mathfrak{p} \ni t$ for some $t \in T$, then $\mathfrak{p}$ cannot belong to $\text{Ass}(R)$ because the union of $\text{Ass}(R)$ equals $R \smallsetminus T$. 89 | 90 | \begin{lemma}\label{prop:K-vs-localization} 91 | Let $R$ be reduced. Then $K(R) \rightiso \prod_{\mathfrak{p}} K(R/\mathfrak{p})$ as $R$-algebras, where $\mathfrak{p}$ ranges over the minimal prime ideals of $R$. For any multiplicative subset $S \subset R$ there is a canonical isomorphism of $R[S^{-1}]$-algebras 92 | \begin{align*} 93 | K(R[S^{-1}]) & \rightiso K(R)[S^{-1}]. 94 | \end{align*} 95 | \end{lemma} 96 | In other words, the formation of total fraction ring commutes with localizations. 97 | \begin{proof} 98 | Each element in $K(R) = R[T^{-1}]$ is either a zero-divisor or invertible. The set of zero-divisors of $K(R)$ is the union of minimal prime ideals $\mathfrak{p}_i K(R)$ of $K(R)$ (where $\text{Ass}(R) = \{ \mathfrak{p}_1, \ldots, \mathfrak{p}_m\}$ by an earlier discussion), therefore each prime ideal of $K(R)$ must equal some $\mathfrak{p}_i K(R)$, by prime avoidance (Proposition \ref{prop:prime-avoidance}). Hence $\mathfrak{p}_1 K(R), \ldots, \mathfrak{p}_m K(R)$ are also the maximal ideals in $K(R)$, with zero intersection. Chinese Remainder Theorem entails that $K(R) \simeq \prod_{i=1}^m K(R)/\mathfrak{p}_i K(R)$. To conclude the first part, notice that $K(R)/\mathfrak{p}_i K(R) = (R/\mathfrak{p}_i)[T^{-1}]$; this is a field in generated by an isomorphic copy of $R/\mathfrak{p}_i$ since $\mathfrak{p}_i \cap T = \emptyset$, hence equals $\text{Frac}(R/\mathfrak{p}_i)$. 99 | 100 | As for the second part, one decomposes $K(R[S^{-1}])$ and $K(R)[S^{-1}]$ by the previous step, noting that 101 | \begin{compactitem} 102 | \item $R[S^{-1}]$ is reduced; 103 | \item $\text{Ass}(R[S^{-1}]) = \left\{ \mathfrak{p}_i R[S^{-1}]: 1 \leq i \leq m, \; \mathfrak{p}_i \cap S = \emptyset \right\}$ consists of minimal primes; 104 | \item $K\left( R[S^{-1}] \big/ \mathfrak{p}_i R[S^{-1}]\right) \simeq K(R/\mathfrak{p}_i) = K(R/\mathfrak{p}_i)[S^{-1}]$ when $\mathfrak{p}_i \cap S = \emptyset$, by the arguments above; 105 | \item $K(R/\mathfrak{p}_i)[S^{-1}] = \{0\}$ when $\mathfrak{p}_i \cap S \neq \emptyset$. 106 | \end{compactitem} 107 | A term-by-term comparison finishes the proof. 108 | \end{proof} 109 | 110 | We use Lemma \ref{prop:K-vs-localization} to interchange $K(\cdot)$ and localizations in what follows. 111 | \begin{lemma}\label{prop:intersection-ass} 112 | Suppose $R$ is reduced. Then $x \in K(R)$ belongs to $R$ if and only if its image in $K(R)_{\mathfrak{p}} = K(R_{\mathfrak{p}})$ belongs to $R_{\mathfrak{p}}$ for every prime $\mathfrak{p}$ associated to a non zero-divisor. 113 | \end{lemma} 114 | \begin{proof} 115 | Only the ``if'' direction requires a proof. Write $x = a/t$ with $t$ not a zero-divisor. Suppose that $a \notin (t)$, i.e. $a$ does not map to zero in $R/(t)$. Lemma \ref{prop:zero-test-ass} asserts there exists $\mathfrak{p} \in \text{Ass}(R/(t))$ such that $a$ does not map to $0 \in (R/(t))_{\mathfrak{p}} = R_{\mathfrak{p}}/t R_{\mathfrak{p}}$. It follows that the image of $a/t$ in $K(R_{\mathfrak{p}})$ does not lie in $R_{\mathfrak{p}}$. 116 | \end{proof} 117 | 118 | \section{On normality} 119 | Fix a Noetherian ring $R$. 120 | 121 | \begin{exercise} 122 | Suppose $R$ is a domain, $K := \text{Frac}(R)$. If $R = \bigcap_{i \in I} R_i$ where $\{ R_i \subset K\}_{i \in I}$ are subrings such that $\text{Frac}(R_i) = K$ and $R_i$ is normal, for each $i$, then $R$ is normal. 123 | \end{exercise} 124 | 125 | \begin{proposition}\label{prop:normality-principal} 126 | Let $R$ be a Noetherian domain. Then $R$ is normal if and only if for every principal ideal $(t) \subset R$ and every $\mathfrak{p} \in \mathrm{Ass}(R/(t))$, the ideal $\mathfrak{p}R_{\mathfrak{p}}$ is principal. 127 | \end{proposition} 128 | \begin{proof} 129 | Assume the conditions above. To prove the normality of $R$, it suffices to use $R = \bigcap R_{\mathfrak{p}}$ where $\mathfrak{p}$ ranges over the primes associated to nonzero principal ideals (consequence of Lemma \ref{prop:intersection-ass}). Indeed, each $R_{\mathfrak{p}}$ is regular, hence normal by Proposition \ref{prop:reg-local-dvr}, therefore so is their intersection by the previous exercise. 130 | 131 | Conversely, assume $R$ is normal and let $\mathfrak{p} \in \text{Ass}(R/(t))$ with $t \neq 0$, we have to show $\mathfrak{p}R_{\mathfrak{p}}$ is principal. Upon replacing $R$ by $R_{\mathfrak{p}}$ and recalling how associated primes behave under localization, we may even assume $R$ is local with maximal ideal $\mathfrak{p}$. Express $\mathfrak{p}$ as the annihilator of some $\bar{x} \in R/(t)$ with $x \in R$. Define the fractional ideal 132 | \[ \mathfrak{p}^{-1} := \{ y \in \text{Frac}(R) : y\mathfrak{p} \subset R \}. \] 133 | It is an $R$-submodule of $\text{Frac}(R)$ containing $R$. Define the $R$-submodule $\mathfrak{p}^{-1}\mathfrak{p}$ of $\text{Frac}(R)$ in the obvious way. Clearly $\mathfrak{p} \subset \mathfrak{p}^{-1}\mathfrak{p} \subset R$. By maximality of $\mathfrak{p}$, exactly one of the $\subset$ is equality. If $\mathfrak{p}\mathfrak{p}^{-1}=\mathfrak{p}$, every element of $\mathfrak{p}^{-1}$ is integral over $R$, hence $\mathfrak{p}^{-1} \subset R$ by normality (integrality is ``witnessed'' by the module $\mathfrak{p}$). From $\mathfrak{p}x \subset (t)$ we see $x/t \in \mathfrak{p}^{-1} = R$; this would imply $\bar{x} = 0$ and $\mathfrak{p}=R$, which is absurd. 134 | 135 | Therefore we must have $\mathfrak{p}\mathfrak{p}^{-1} = R$. This implies that $\mathfrak{p}y \not\subset \mathfrak{p}$ for some $y \in \mathfrak{p}^{-1}$, hence $\mathfrak{p}y = R$ since $R$ is local. Hence $\mathfrak{p} = y^{-1}R \simeq R$ is principal. 136 | \end{proof} 137 | 138 | \begin{corollary}\label{prop:normality-dvr} 139 | The following are equivalent for a local domain $R$: 140 | \begin{enumerate}[(i)] 141 | \item $R$ is normal of dimension $1$; 142 | \item $R$ is a regular local ring of dimension $1$; 143 | \item $R$ is a discrete valuation ring. 144 | \end{enumerate} 145 | \end{corollary} 146 | \begin{proof} 147 | (iii) $\implies$ (i). We have seen that discrete valuation rings are principal ideal rings of dimension $1$, therefore also normal by unique factorization property. 148 | 149 | (i) $\implies$ (ii). Under the normality assumption, choose any $t \in R \smallsetminus \{0\}$. Since $\dim R = 1$ and $\{0\}$ is a prime ideal, the associated prime of $(t)$ can only be the maximal ideal $\mathfrak{m}$, which is principal by Proposition \ref{prop:normality-principal}. This shows that $R$ is regular local. 150 | 151 | (ii) $\implies$ (iii) is included in Proposition \ref{prop:reg-local-dvr}. 152 | \end{proof} 153 | 154 | \begin{corollary} 155 | Let $R$ be a Noetherian normal domain. Then 156 | \[ R = \bigcap_{\mathrm{ht}(\mathfrak{p}) = 1} R_{\mathfrak{p}} \] 157 | inside $\mathrm{Frac}(R)$. 158 | \end{corollary} 159 | \begin{proof} 160 | Evidently $\subset$ holds. By Lemma \ref{prop:intersection-ass} together with Proposition \ref{prop:normality-principal}, $R$ can be written as an intersection of $R_{\mathfrak{p}}$ where $\mathfrak{p}$ is associated to some non zero-divisor, such that $\mathfrak{p}R_{\mathfrak{p}}$ is principal; it suffices to show $\text{ht}(\mathfrak{p})=1$. From $\mathfrak{p} \neq \{0\}$ we see $\text{ht}(\mathfrak{p}) \geq 1$; on the other hand, by Hauptidealsatz or by the discussion on regular local rings, we see $\text{ht}(\mathfrak{p}) = \text{ht}(\mathfrak{p}R_{\mathfrak{p}}) \leq 1$. 161 | \end{proof} 162 | 163 | \section{Serre's criterion} 164 | 165 | \begin{lemma}\label{prop:reduced-test} 166 | Let $R$ be a Noetherian ring. Suppose that 167 | \begin{compactitem} 168 | \item the primes in $\mathrm{Ass}(R)$ are all minimal, and 169 | \item $R_{\mathfrak{p}}$ is a field for every minimal prime ideal $\mathfrak{p}$, 170 | \end{compactitem} 171 | then $R$ is reduced. 172 | \end{lemma} 173 | \begin{proof} 174 | Take a minimal primary decomposition $\{0\} = I_1 \cap \cdots \cap I_m$ with $\text{Ass}(R/I_j) = \{\mathfrak{p}_j = \sqrt{I_j} \}$ and $\text{Ass}(R) = \{\mathfrak{p}_1, \ldots, \mathfrak{p}_m \}$. By the properties of primary ideals, $\mathfrak{p}_j = \sqrt{I_j} \supset I_j$ for all $j$. By assumption each $\mathfrak{p}_j$ is minimal, and $R_{\mathfrak{p}_j}$ is a field. From the uniqueness of non-embedded components in primary decompositions, $I_j = \Ker\left[R \to R_{\mathfrak{p}_j}\right]$ is a prime contained in $\mathfrak{p}_j$, hence $I_j = \mathfrak{p}_j$. We deduce that $\{0\} = \bigcap_{j=1}^m \mathfrak{p}_j$, thereby showing $\sqrt{0_R} = \{0\}$. 175 | \end{proof} 176 | 177 | \begin{theorem}[J.-P. Serre]\label{prop:Serre-criterion}\index{Serre's criterion} 178 | A Noetherian ring $R$ is a finite direct product of normal domains if and only if the following two conditions hold. 179 | \begin{description} 180 | \item[R1] The localization of $R$ at every prime ideal of height $1$ (resp. $0$) is a discrete valuation ring (resp. a field). 181 | \item[S2] For every non zero-divisor $t$ of $R$, the primes in $\mathrm{Ass}(R/(t))$ are all of height $1$; the primes in $\mathrm{Ass}(R)$ are all of height $0$. 182 | \end{description} 183 | \end{theorem} 184 | The condition \textbf{R1} means regularity in codimension $\leq 1$. The condition \textbf{S2} is often rephrased in terms of \emph{depth}, which will be discussed in Proposition \ref{prop:S2}. 185 | 186 | \begin{proof} 187 | We begin with the $\implies$ direction. Suppose $R = R_1 \times \cdots \times R_n$ where each $R_i$ is a normal domain. As is well-known, $\Spec(R) = \bigsqcup_{i=1}^n \Spec(R_i)$ as topological spaces: to be precise, the elements of $\Spec(R)$ take the form $\mathfrak{p} = R_1 \times \cdots \times \mathfrak{p}_i \times \cdots \times R_n$, where $\mathfrak{p}_i \in \Spec(R_i)$. We have 188 | \[ \text{ht}(\mathfrak{p}) = \text{ht}(\mathfrak{p}_i), \quad R_{\mathfrak{p}} \simeq (R_i)_{\mathfrak{p}_i}. \] 189 | Furthermore, one easily checks that 190 | \[ \text{Ass}(R/(t)) = \bigsqcup_{i=1}^n \text{Ass}(R_i/(t_i)), \quad t = (t_1, \ldots, t_n) \in R \] 191 | compatibly with the description above. 192 | 193 | This reduces the verification of \textbf{S2} to the case of normal domains, which is addressed in Proposition \ref{prop:normality-principal}. The condition \textbf{R1} is implied by Corollary \ref{prop:normality-dvr} since normality is preserved under localizations. 194 | 195 | Assume conversely \textbf{R1} and \textbf{S2}. They imply the conditions of Lemma \ref{prop:reduced-test}, hence $R$ is reduced. Now for every prime $\mathfrak{p}$ associated to a non zero-divisor, we have $\text{ht}(\mathfrak{p})=1$ and $R_{\mathfrak{p}}$ is a normal domain by \textbf{R1} $\wedge$ \textbf{S2}. By Lemma \ref{prop:intersection-ass} (as $R$ is reduced), $R$ is integrally closed in $K(R)$: indeed, if $x \in K(R)$ is integral over $R$, so is its image in $K(R_{\mathfrak{p}}) = K(R)_{\mathfrak{p}}$ for every $\mathfrak{p}$ as above, therefore lies in $R_{\mathfrak{p}}$. Decompose $K(R) = \prod_{i=1}^m K(R/\mathfrak{p}_i)$ as in Lemma \ref{prop:K-vs-localization}. The idempotents $e_i \in K(R)$ associated to this decomposition are trivially integral over $R$: $e_i^2 - e_i = 0$, hence $e_i \in R$ for all $i$. It follows that $R = Re_1 + \cdots + Re_m \prod_{i=1}^m Re_i \subset K(R)$ and one easily checks that $Re_i = R/\mathfrak{p}_i$. 196 | 197 | Finally, since $R$ is integrally closed in $K(R)$, the decomposition above implies $R/\mathfrak{p}_i$ is integrally closed in $K(R/\mathfrak{p}_i)$. All in all, we have written $R$ as a direct product of normal domains. 198 | \end{proof} 199 | 200 | \begin{exercise} 201 | Recall that for an $R$-module $M$, a prime ideal $\mathfrak{p} \in \text{Ass}(M)$ is called \emph{embedded} if $\mathfrak{p}$ is not a minimal element in $\text{Ass}(M)$. Show that for $M=R$, embedded primes are primes in $\text{Ass}(R)$ with height $> 0$. For $M=R/(t)$ where $t$ is not a zero-divisor, embedded primes are primes in $\text{Ass}(R/(t))$ with height $> 1$. Use this to rephrase \textbf{S2} as follows: there are no embedded primes in $\text{Ass}(R/(t))$ ($t$ not a zero-divisor) or in $\text{Ass}(R)$. 202 | \end{exercise} 203 | 204 | \begin{exercise} 205 | Suppose a ring $R$ is isomorphic to a direct product $\prod_{i \in I} R_i$. Show that $R$ is a domain if and only if $|I|=1$ (say $I=\{i_0\}$), and $R_{i_0}$ is a domain. 206 | \end{exercise} 207 | 208 | \begin{corollary} 209 | A Noetherian domain $R$ is normal if and only if \textbf{R1} and \textbf{S2} hold for $R$. 210 | \end{corollary} 211 | \begin{proof} 212 | Immediate from the previous exercise and Theorem \ref{prop:Serre-criterion}. 213 | \end{proof} 214 | 215 | \section{Introduction to depth} 216 | Based on \cite{Bour98}, we give a brief account on the notion of depth. Let $R$ be a ring and $M$ be an $R$-module, $M \neq \{0\}$. Recall the $\Ext$-functors 217 | \[ \Ext^n_R(X,Y) := H^n(R\mathcal{H}\text{om}(X,Y)) = \text{Hom}_{D^+(R\dcate{Mod})}(X, Y[n]). \] 218 | 219 | \begin{definition}[Depth of a module]\index{depth} 220 | Let $I$ be a proper ideal of $R$. We define the \emph{depth} of $M$ relative to $I$ as 221 | \[ \text{depth}_I(M) := \inf \left\{n \geq 0: \Ext^n_R(R/I, M) \neq 0 \right\} \] 222 | with values in $\Z_{\geq 0} \sqcup \{+\infty\}$. 223 | \end{definition} 224 | 225 | \begin{proposition}\label{prop:depth-zero} 226 | For $I$, $M$ as above, the following are equivalent: 227 | \begin{enumerate}[(i)] 228 | \item $\mathrm{depth}_I(M)=0$; 229 | \item for all $x \in I$, the homomorphism $M \xrightarrow{x} M$ is not injective; 230 | \item $\mathrm{Ass}(M) \cap V(I) \neq \emptyset$. 231 | \end{enumerate} 232 | \end{proposition} 233 | \begin{proof} 234 | In each case we have $M \neq \{0\}$. If (i) holds, then $M \xrightarrow{x} M$ vanishes on the image of some nonzero $R/I \to M$, hence (ii). If (ii) holds, the union of $\text{Ass}(M)$ will cover $I$, and (iii) follows by prime avoidance. Finally, suppose $\mathfrak{p} \in \text{Ass}(M) \cap V(I)$, there is an embedding $R/\mathfrak{p} \hookrightarrow M$, which yields a non-zero $R/I \to M$. 235 | \end{proof} 236 | 237 | \begin{definition}\index{regular sequence} 238 | A sequence $x_1, \ldots, x_n \in R$ is called an $M$-\emph{regular sequence} of length $n$ if $(x_1, \ldots, x_n)M \subsetneq M$ and 239 | \[ 0 \to M/(x_1, \ldots, x_{k-1}) \xrightarrow{x_k} M/(x_1, \ldots, x_{k-1}) \] 240 | is exact for all $1 \leq k \leq n$. 241 | \end{definition} 242 | 243 | \begin{lemma} 244 | Let $M$ be an $R$-module, $x_1, \ldots, x_r$ be an $M$-regular sequence lying in an ideal $I \subsetneq R$. We have $\mathrm{depth}_I(M) = r + \mathrm{depth}_I(M/(x_1, \ldots, x_r)M)$. 245 | \end{lemma} 246 | \begin{proof} 247 | The case $r=1$ follows by staring at the long exact sequence attached to $0 \to M \xrightarrow{x_1} M \to M/x_1 M \to 0$. The general case follows by induction on $r$. 248 | \end{proof} 249 | 250 | \begin{theorem}\label{prop:regular-vs-depth} 251 | Assume $R$ Noetherian, $M$ finitely generated and $I \subsetneq R$. 252 | \begin{enumerate}[(i)] 253 | \item $\mathrm{depth}_I(M)$ is the supremum of the lengths of $M$-regular sequences with elements in $I$. 254 | \item Suppose $\mathrm{depth}_I(M) < +\infty$. Every $M$-regular sequence with elements in $I$ can be extended to one of length $\mathrm{depth}_I(M)$. 255 | \item The depth of $M$ relative to $I$ is finite if and only if $V(I) \cap \Supp(M) \neq \emptyset$, or equivalently $IM \neq M$. 256 | \end{enumerate} 257 | \end{theorem} 258 | \begin{proof} 259 | To prove (i) and (ii), by the previous Lemma we are reduced to show that $\text{depth}_I(M) > 0$ implies the existence of $x \in I$ which is not a zero-divisor of $M$; this follows from Proposition \ref{prop:depth-zero}. 260 | 261 | Now pass to the word ``equivalently'' in (iii). We have $M \neq IM$ if and only if $(M/IM)_{\mathfrak{p}} = M_{\mathfrak{p}}/ I_{\mathfrak{p}} M_{\mathfrak{p}} \neq 0$ for some prime ideal $\mathfrak{p}$. That quotient always vanishes when $\mathfrak{p} \not\supset I$, in which case $I_{\mathfrak{p}} = R_{\mathfrak{p}}$. On the other hand, when $\mathfrak{p} \in V(I)$ we have $I_{\mathfrak{p}} \subset \text{rad}(R_{\mathfrak{p}})$, thus the non-vanishing is equivalent to $M_{\mathfrak{p}} \neq \{0\}$ by Nakayama's Lemma \ref{prop:NAK}. 262 | 263 | The ``if'' direction of (iii) is based on the following fact 264 | \[ IM \neq M \implies \text{depth}_I(M) < +\infty \] 265 | which will be proved in the next lecture (Theorem \ref{prop:Koszul-depth}) using \emph{Koszul complexes}. As for the ``only if'' direction, $V(I) \cap \Supp(M) = \emptyset$ implies $I + \text{ann}(M) = R$, but the elements in $I + \text{ann}(M)$ annihilate each $\Ext^n(R/I, M)$, hence $\text{depth}_I(M)=+\infty$. 266 | \end{proof} 267 | 268 | \begin{figure}[h] 269 | \centering \vspace{1em} \includegraphics[height=220pt]{Jean-Louis_Koszul.jpg} \\ \vspace{1em} 270 | \begin{minipage}{0.7\textwidth} 271 | \small Jean-Louis Koszul (1921---2018) created the Koszul complexes in order to define a cohomology theory for Lie algebras; this device turns out to be a general, convenient construction in homological algebra, which will be discussed in the next lecture. The study of ``Koszulness'' in a broader (eg. operadic) context is now an active area of research. J.-L. Koszul is also a second-generation member of the Bourbaki group. Source: by Konrad Jacobs - \href{http://owpdb.mfo.de/detail?photo_id=2273}{Oberwolfach Photo Collection}, ID 2273. 272 | \end{minipage} 273 | \end{figure} 274 | 275 | \begin{corollary} 276 | With the same assumptions, let $(x_1, \ldots, x_r)$ be an $M$-regular sequence with $x_i \in I$. It is of length $\mathrm{depth}_I(M)$ if and only if $\mathrm{Ass}(M/(x_1, \ldots, x_r)M) \cap V(I) \neq \emptyset$. 277 | \end{corollary} 278 | \begin{proof} 279 | The sequence has length $\text{depth}_I(M)$ if and only if $R$-module $M/(x_1, \ldots, x_r)M$ has depth zero, so it remains to apply Proposition \ref{prop:depth-zero}. 280 | \end{proof} 281 | 282 | \begin{corollary} 283 | Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$, and $M \neq \{0\}$ a finitely generated $R$-module, then $\mathrm{depth}_{\mathfrak{m}}(M) \leq \dim M$. 284 | \end{corollary} 285 | \begin{proof} 286 | Consider the following situation: $x \in \mathfrak{m}$ is not a zero-divisor for $M \neq \{0\}$. In the discussion of dimensions, we have seen that $d(M) \geq d(M/xM) \geq d(M/xM)-1$, where $d(\cdot)$ is the degree of Hilbert--Samuel polynomial; on the other hand, since the alternating sum of Hilbert--Samuel polynomials in $0 \to M \xrightarrow{x} M \to M/xM \to 0$ has degree $< d(M)$, we infer that $d(M/xM) = d(M) - 1$. Hence $\dim(M/xM) = \dim(M) - 1$. By relating depth to $M$-regular sequences, we deduce $\text{depth}_{\mathfrak{m}}(M) \leq \dim M$. 287 | \end{proof} 288 | 289 | \begin{definition}[Cohen--Macaulay modules]\index{Cohen--Macaulay module} 290 | Let $R$ be a Noetherian ring. A finitely generated $R$-module $M$ is called \emph{Cohen--Macaulay} if 291 | \[ \mathrm{depth}_{\mathfrak{m}A_{\mathfrak{m}} }(M_{\mathfrak{m}}) = \dim M_{\mathfrak{m}} \] 292 | for every $\mathfrak{m} \in \MaxSpec(R) \cap \Supp(M)$. We say $R$ is a Cohen--Macaulay ring if it is Cohen--Macaulay as a module. 293 | \end{definition} 294 | 295 | \begin{example} 296 | Regular local rings are Cohen--Macaulay, although we do not prove this here. Another important class of Cohen--Macaulay rings is the algebra of invariants $A^G$ where $A$ is the algebra of regular functions on an affine $\Bbbk$-variety $\mathcal{X}$ with \emph{rational singularities} (eg. $A = \Bbbk[X_1, \ldots, X_n]$) with an action by a reductive $\Bbbk$-group $G$ (finite groups allowed), and we assume $\text{char}(\Bbbk)=0$. Here is the reason: Boutot \cite{Bou87} proved the GIT quotient $\mathcal{X}/\!/G$ has rational singularities as well, hence is Cohen--Macaulay; in characteristic zero this strengthens an earlier theorem of Hochster--Roberts. These algebras are interesting objects from the geometric, algebraic or even combinatorial perspectives. 297 | \end{example} 298 | 299 | \begin{exercise} 300 | Show by using Proposition \ref{prop:depth-zero} that $\text{depth}_{\mathfrak{m}}(R) = 0$ if $R$ is local with maximal ideal $\mathfrak{m}$ and has dimension zero. 301 | \end{exercise} 302 | 303 | \begin{proposition}\label{prop:S2} 304 | The condition \textbf{S2} in Theorem \ref{prop:Serre-criterion} is equivalent to 305 | \[ \mathrm{ht}(\mathfrak{p}) \geq i \implies \mathrm{depth}_{\mathfrak{p}R_{\mathfrak{p}}}(R_{\mathfrak{p}}) \geq i \] 306 | for all $\mathfrak{p} \in \Spec(R)$ and $i \in \{1,2\}$. 307 | \end{proposition} 308 | \begin{proof} 309 | Assume the displayed conditions. We first show that every $\mathfrak{p} \in \text{Ass}(R)$ has height zero. Upon localization we may assume $R$ local with maximal ideal $\mathfrak{p}$, thus $\mathfrak{p}$ has depth zero by Proposition \ref{prop:depth-zero}; our conditions force $\text{ht}(\mathfrak{p})=0$. Next, consider $\mathfrak{p} \in \text{Ass}(R/(t))$ with $t$ non zero-divisor. Note that $t$ remains a non zero-divisor in $R_{\mathfrak{p}}$, and $\mathfrak{p}R_{\mathfrak{p}} \in \text{Ass}(R_{\mathfrak{p}}/(t))$. Hence $\text{depth}(R_{\mathfrak{p}}) = \text{depth}(R_{\mathfrak{p}}/(t)) + 1 = 1$ since $\text{depth}_{\mathfrak{p}}(R_{\mathfrak{p}}/(t)) = 0$ by Proposition \ref{prop:depth-zero}. Our conditions force $\text{ht}(\mathfrak{p}) \leq 1$ and $t \in \mathfrak{p}$ implies $\text{ht}(\mathfrak{p}) > 0$. 310 | 311 | Conversely, assume \textbf{S2}. Suppose $\text{ht}(\mathfrak{p}) \geq 1$. If $R_{\mathfrak{p}}$ has depth zero then $\text{Ass}(R_{\mathfrak{p}}) \cap V(\mathfrak{p}R_{\mathfrak{p}}) \neq \emptyset$, which implies $\mathfrak{p}R_{\mathfrak{p}} \in \text{Ass}(R_{\mathfrak{p}})$ thus $\mathfrak{p} \in \text{Ass}(R)$, contradicting \textbf{S2}. Next, suppose $\text{ht}(\mathfrak{p}) \geq 2$. If $R_{\mathfrak{p}}$ has depth $\leq 1$, the standard property 312 | \[ \forall i \geq 0, \quad \Ext^i_{R_{\mathfrak{p}}}(X_{\mathfrak{p}}, Y_{\mathfrak{p}}) \simeq \Ext^i_R(X,Y)_{\mathfrak{p}} \] 313 | valid for Noetherian $R$ and finitely generated $R$-modules $X, Y$, yields 314 | \[ 0 \leq \text{depth}_{\mathfrak{p}}(R) \leq \text{depth}_{\mathfrak{p}R_{\mathfrak{p}}}(R_{\mathfrak{p}}) \leq 1. \] 315 | If $\text{depth}_{\mathfrak{p}}(R)=0$, there exists of $\mathfrak{p}' \supset \mathfrak{p}$ and $\mathfrak{p}' \in \text{Ass}(R)$ by Proposition \ref{prop:depth-zero}, but \textbf{S2} implies $\text{ht}(\mathfrak{p}) \leq \text{ht}(\mathfrak{p}') = 0$ which is absurd. If $\text{depth}_{\mathfrak{p}}(R)=1$, there exists a non zero-divisor $x$ in $R$ such that $\text{depth}_{\mathfrak{p}}(R/(x)) = 0$. The same argument furnishes $\mathfrak{p}' \supset \mathfrak{p}$ such that $\mathfrak{p}' \in \text{Ass}(R/(x))$. Now \textbf{S2} implies $\text{ht}(\mathfrak{p}) \leq \text{ht}(\mathfrak{p}')=1$, again a contradiction. 316 | \end{proof} 317 | 318 | Now the conditions \textbf{R1} and \textbf{S2} can be generalized to arbitrary $k \in \Z_{\geq 0}$: 319 | \begin{description} 320 | \item[Rk] $\text{ht}(\mathfrak{p}) \leq k$ implies $R_{\mathfrak{p}}$ is a regular local ring, for every $\mathfrak{p} \in \Spec(R)$; 321 | \item[Sk] $\text{depth}(R_{\mathfrak{p}}) \geq \min\{ k, \text{ht}(\mathfrak{p})\}$ for every $\mathfrak{p} \in \Spec(R)$. 322 | \end{description} 323 | One readily checks their compatibility with Proposition \ref{prop:S2}. Note that \textbf{S0} is trivial, whilst $R$ is Cohen--Macaulay if and only if it satisfies \textbf{Sk} for all $k$. 324 | 325 | \begin{example} 326 | From Theorem \ref{prop:Serre-criterion}, we see that any finite direct product of normal domains of dimension $\leq 2$ is Cohen--Macaulay. 327 | \end{example} 328 | 329 | \begin{exercise} 330 | Show that \textbf{S1} holds if and only if there are no embedded primes in $\text{Ass}(R)$. 331 | \end{exercise} -------------------------------------------------------------------------------- /YAlg3-8.tex: -------------------------------------------------------------------------------- 1 | % To be compiled by XeLaTeX, preferably under TeX Live. 2 | % LaTeX source for ``Yanqi Lake Lectures on Algebra'' Part III. 3 | % Copyright 2019 李文威 (Wen-Wei Li). 4 | % Permission is granted to copy, distribute and/or modify this 5 | % document under the terms of the Creative Commons 6 | % Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) 7 | % https://creativecommons.org/licenses/by-nc/4.0/ 8 | 9 | % To be included 10 | \chapter{Some aspects of Koszul complexes} 11 | 12 | This lecture is a faithful replay of the relevant sections of \cite{Bour80} and \cite{Bour98}; the main aim here is to complete the proof of Theorem \ref{prop:regular-vs-depth}. In what follows, we work with a chosen ring $R$. We impose no Noetherian or finiteness conditions here. 13 | 14 | \section{Preparations in homological algebra} 15 | For any $R$-module $L$, denote its \emph{exterior algebra}\index{exterios algebra} over $R$ by 16 | \[ \bigwedge L := \bigoplus_{n \geq 0} \bigwedge^n L. \] 17 | It is the quotient of the tensor algebra $T(L) = \bigoplus_{n \geq 0} (T^n(L) := L^{\otimes n})$ by the graded ideal generated by the pure tensors 18 | \[ \cdots \otimes x \otimes x \otimes \cdots, \quad x \in L. \] 19 | The multiplication operation in $\bigwedge L$ is written as $\wedge$. Note that $\bigwedge^0 L = T^0(L) = R$ by convention. The traditional notion of exterior algebras encountered in differential geometry is recovered when $\Q \subset R$. 20 | 21 | Given $u \in \Hom_R(L, R)$, one can define the corresponding \emph{contractions} $i_u: \bigwedge^{n+1} L \to \bigwedge^n L$, given concretely as 22 | \[ i_u (x_0 \wedge \cdots \wedge x_n) = \sum_{i=0}^n (-1)^i u(x_i) \cdot x_0 \wedge \cdots \widehat{x_i} \cdots \wedge x_n \] 23 | where $\widehat{x_i}$ means $x_i$ is omitted. It is routine to check that $i_u$ satisfies $i_u \circ i_u = 0$, thereby giving rise to a chain complex. 24 | 25 | \begin{definition}\label{def:Koszul-general}\index{Koszul complex} 26 | Let $L$ and $u$ be as above. Define the corresponding \emph{Koszul complex} as $K_\bullet(u) := (\bigwedge^\bullet L, i_u)$. For any $R$-module $M$, put 27 | \begin{align*} 28 | K_\bullet(u; M) & := M \dotimes{R} K_\bullet(u), \\ 29 | K^\bullet(u; M) & := \Hom_A(K_\bullet(u), M) 30 | \end{align*}\index{$K^\bullet(u; M)$} 31 | which is naturally a chain (resp. cochain) complex in positive degrees; here one regards $M$ as a complex in degree zero. These definitions generalize to the case of any complex $M$, and are functorial in $M$. 32 | \end{definition} 33 | 34 | The reader might have encountered the following result in differential geometry. 35 | \begin{proposition}[Homotopy formula] 36 | For any $x \in L$ and $\omega \in \bigwedge^n L$, we have 37 | \[ i_u (x \wedge \omega) + x \wedge (i_u(\omega)) = u(x) \omega. \] 38 | \end{proposition} 39 | \begin{proof} 40 | Consider $\omega = x_1 \wedge \cdots \wedge x_n$. Put $x_0 := x$. The left-hand side equals 41 | \[ \sum_{i=0}^n (-1)^i \cdot u(x_i) \cdots \wedge \widehat{x_i} \wedge \cdots \] 42 | whereas the right-hand side equals 43 | \[ \sum_{i=1}^n (-1)^{i+1} \cdot u(x_i) x_0 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots. \] 44 | The terms with a $\widehat{x_i}$ (non-existant --- hopefully this won't generate metaphysical issues) where $i > 0$ cancel out. We are left with $u(x_0) x_1 \wedge \cdots \wedge x_n = u(x) \omega$. 45 | \end{proof} 46 | Obviously, the same formula extends to all $\omega \in \bigwedge L$ by linearity. 47 | 48 | \begin{proposition}\label{prop:Koszul-u-vanishing} 49 | Set $\mathfrak{q} := u(L)$, which is an ideal of $R$. Then $\mathfrak{q}$ annihilates each homology (resp. cohomology) of $K_\bullet(u; M)$ (resp. $K^\bullet(u; M)$). Again, this generalizes to general complexes $M$. 50 | \end{proposition} 51 | \begin{proof} 52 | Given $t \in \mathfrak{q}$, the homotopy formula implies that the endomorphism $\omega \mapsto t\omega$ of $K_\bullet(u)$ is homotopic to zero, hence so are the induced endomorphisms of $K_\bullet(u; M)$ and $K^\bullet(u; M)$ by standard homological algebra. 53 | \end{proof} 54 | 55 | \begin{proposition}\label{prop:Koszul-long-exact-sequence} 56 | Suppose $L$ is projective over $R$ and $0 \to M' \to M \to M'' \to 0$ is exact. Then there is a natural short exact sequence of complexes 57 | \[ 0 \to K^\bullet(u; M') \to K^\bullet(u; M) \to K^\bullet(u; M'') \to 0 \] 58 | which gives rise to a long exact sequence of cohomologies of the Koszul complexes in question. 59 | \end{proposition} 60 | \begin{proof} 61 | Standard. It suffices to note that $L$ is projective implies each graded piece $\bigwedge^n L$ of $\bigwedge L$ is projective as well. 62 | \end{proof} 63 | 64 | \begin{exercise} 65 | Justify the assertion above concerning the projectivity of $\bigwedge^n L$. Hint: suppose $L$ is a direct summand of a free module $F$, show that $\bigwedge^n L$ is a direct summand of $\bigwedge^n F$ and $\bigwedge^n F$ is free. 66 | \end{exercise} 67 | 68 | Similar properties hold for the homological version when $L$ is flat over $R$. One needs the property that $\bigwedge^n L$ is flat if $L$ is. 69 | \begin{exercise} 70 | Prove the assertion above concerning flatness of $\bigwedge^n L$. Consult the proof in \cite[p.15]{Bour80} if necessary. 71 | \end{exercise} 72 | 73 | \section{Auxiliary results on depth} 74 | We fix an ideal $I \subset R$. 75 | \begin{proposition}\label{prop:depth-prod} 76 | For every family $\{M_\beta\}_{\beta \in \mathcal{B}}$ of $R$-modules, we have 77 | \[ \mathrm{depth}_I\left(\prod_\beta M_\beta \right) = \inf_\beta \mathrm{depth}_I(M_\beta). \] 78 | \end{proposition} 79 | \begin{proof} 80 | This follows from $\Ext^n_R(R/I, \prod_i M_i) = \prod_{i \in I} \Ext^n_R(R/I, M_i)$, as is easily seen by taking a projective resolution of $R/I$ and using the fact the $\Hom_R$ preserves direct products in the second variable. 81 | \end{proof} 82 | 83 | \begin{remark} 84 | By stipulation, the empty product is $0$, the zero object of the category $R\dcate{Mod}$. In parallel we define $\inf\emptyset := \infty$, so that Proposition \ref{prop:depth-prod} remains true in this case, since the zero module has infinite depth. 85 | \end{remark} 86 | 87 | \begin{proposition} 88 | Suppose $N$ is an $R$-module annihilated by some $I^m$, where $m \geq 1$. Then $\Ext^i_R(N,M)=0$ whenever $i < \mathrm{depth}_I(M)$. 89 | \end{proposition} 90 | \begin{proof} 91 | To show $\Ext^i_R(N,M)=0$ for $i < \text{depth}_I(M)$, we begin with the case $m=1$. This case follows by a dimension-shifting argument based on the short exact sequence 92 | \[ 0 \to K \to (R/I)^{\oplus I} \to M \to 0, \] 93 | together with induction on $i$ (note that $\Ext^{<0}_R(N,M)=0$ for trivial reasons). The case of general $m$ follows by a standard \emph{dévissage} using 94 | \[ 0 \to JN \to N \to N/JN \to 0 \] 95 | and the associated long exact sequence. 96 | \end{proof} 97 | 98 | \begin{lemma}\label{prop:depth-power} 99 | Suppose $J$ is an ideal satisfying $J \supset I^m$ for some $m \geq 1$. Then $\mathrm{depth}_I(M) \leq \mathrm{depth}_J(M)$. 100 | \end{lemma} 101 | \begin{proof} 102 | From the previous Proposition, we have $\Ext^i_R(R/J, M) = 0$ for $i < \text{depth}_I(M)$ since $I^m$ annihilates $R/J$. The assertion follows upon recalling the definition of depth. 103 | \end{proof} 104 | 105 | The following technical results will be invoked in the proof of Theorem \ref{prop:Koszul-depth}. 106 | \begin{proposition}\label{prop:depth-vanishing-aux} 107 | Let $C^\bullet$ be a cochain complex of $R$-modules such that $n \ll 0 \implies C^n = 0$. Let $h \in \Z$ such that for all integers $n \leq k \leq h$, the depth of $C^n$ with respect to $J_k := \mathrm{ann}(H^k(C^\bullet))$ is $> k-n$, then 108 | \[ H^{\leq h}(C^\bullet)=0. \] 109 | \end{proposition} 110 | In what follows, we write $H^n = H^n(C^\bullet) = Z^n/B^n$ in the usual notation for homological algebra. 111 | \begin{proof} 112 | Assume on the contrary that there exists $k \leq h$ with $H^{< k} = 0$ whereas $H^k \neq 0$. Write $J = J_k$. As $J$ annihilates the nonzero $R$-module $H^k$, the criterion of depth-zero modules (Proposition \ref{prop:depth-zero}) implies that $\text{depth}_J(H^k) = 0$. By assumption $\text{depth}_J(C_k) > k-k = 0$, it follows that $\text{depth}_J(Z_k) > 0$ since $Z_k \subset C_k$, by applying the aforementioned criterion of depth-zero. From the short exact sequence 113 | \[ 0 \to B^k \to Z^k \to H^k \to 0 \] 114 | we deduce distinguished triangles in $D(R\dcate{Mod})$ 115 | \begin{gather*} 116 | \mathcal{H}\text{om}(R/J, B^k) \to \mathcal{H}\text{om}(R/J, Z^k) \to \mathcal{H}\text{om}(R/J, H^k) \xrightarrow{+1}, \\ 117 | \mathcal{H}\text{om}(R/J, H^k)[-1] \to \mathcal{H}\text{om}(R/J, B^k) \to \mathcal{H}\text{om}(R/J, Z^k) \xrightarrow{+1}. 118 | \end{gather*} 119 | As the leftmost and rightmost terms of the last line are in $D^{\geq 1}(R\dcate{Mod})$, we see 120 | \[ \text{depth}_J(B^k) \geq 1; \] 121 | moreover, the piece $\Ext_R^0(R/J, Z^k) \to \Ext_R^0(R/J, H^k) \to \Ext_R^1(R/J, B^k)$ from the long exact sequence shows that $\text{depth}_J(B^k) = 1$. Now for $n < k$ we have short exact sequences 122 | \[ 0 \to \underbracket{B^n}_{= Z^n} \to C^n \to B^{n+1} \to 0. \] 123 | Again, one infers from the distinguished triangle 124 | \[ \mathcal{H}\text{om}(R/J, B^{n+1})[-1] \to \mathcal{H}\text{om}(R/J, B^n) \to \mathcal{H}\text{om}(R/J, C^n) \xrightarrow{+1} , \] 125 | the assumption $\text{depth}_J(C^n) > k-n$ and descending induction on $n$ that $n < k \implies \text{depth}_J(B_n) = k-n+1$. This is impossible since $B_{\ll 0} = 0$ has infinite depth. 126 | \end{proof} 127 | 128 | \begin{corollary}\label{prop:depth-vanishing} 129 | Let $I \subset R$ be an ideal, $C^\bullet$ be a cochain complex with $n \ll 0 \implies C_n=0$, and $h \in \Z$. Suppose that $n \leq h$ implies $I \cdot H^n(C^\bullet) = 0$ and $\mathrm{depth}_I(C^n) > h-n$. Then $H^{\leq h}(C^\bullet) = 0$. 130 | \end{corollary} 131 | \begin{proof} 132 | For $k \leq h$ we have $J_k := \text{ann}(H^k(C^\bullet)) \supset I$. Hence Lemma \ref{prop:depth-power} entails that 133 | \[ n \leq k \leq h \implies \text{depth}_{J_k}(C^n) \geq \text{depth}_I(C^n) > h-n \geq k-n. \] 134 | Now apply Proposition \ref{prop:depth-vanishing-aux}. 135 | \end{proof} 136 | 137 | \section{Koszul complexes and depth} 138 | The simplest Koszul complexes are defined as follows. Given $x \in R$, we form the cochain complex in degrees $\{0,1\}$ 139 | \[ K(x) := \left[ R \xrightarrow{x} R \right]. \] 140 | More generally, for any $R$-module $M$, viewed as a complex concentrated in degree zero, and a family $\mathbf{x} = (x_\alpha)_{\alpha \in \mathcal{A}}$ of element of $R$, we define the associated \emph{Koszul complex}\index{Koszul complex} as 141 | \[ K^\bullet(\mathbf{x}; M) := K^\bullet(u; M), \] 142 | where we take $u: R^{\oplus \mathcal{A}} \to R$ corresponding to $\mathbf{x}$ in Definition \ref{def:Koszul-general}. Unfolding definitions, we have \index{$K^\bullet(\mathbf{x};M)$} 143 | \[ K^h(\mathbf{x}; M) = \begin{cases} 144 | \Hom_R\left( \bigwedge^h (R^{\oplus \mathcal{A}}), M \right), & h \geq 0 \\ 145 | 0, & h < 0. 146 | \end{cases} \] 147 | Therefore $K^h(\mathbf{x}; M)$ consists of families $m(\alpha_1, \ldots, \alpha_h) \in M$ that are alternating in the variables $\alpha_1, \ldots, \alpha_h \in \mathcal{A}$. The differential is 148 | \begin{align*} 149 | \partial^h: K^h(\mathbf{x}; M) & \longrightarrow K^{h+1}(\mathbf{x}; M) \\ 150 | m & \longmapsto \left[ (\alpha_0, \ldots, \alpha_h) \mapsto \sum_{j=0}^h (-1)^j x_{\alpha_j} \cdot m(\ldots, \widehat{\alpha_j}, \ldots) \right]. 151 | \end{align*} 152 | We shall abbreviate the cohomologies of $K^\bullet(\mathbf{x}; M)$ as the \emph{Koszul cohomologies}. 153 | 154 | When $\mathcal{A} = \{1, \ldots, n\}$ we revert to 155 | \[ K^\bullet(x_1, \ldots, x_n; M) := M \otimes K(x_1) \otimes \cdots \otimes K(x_n) \] 156 | with the well-known sign convention. Also note that $R^{\oplus \mathcal{A}}$ is projective, hence the Proposition \ref{prop:Koszul-long-exact-sequence} can always be applied to Koszul cohomologies. 157 | 158 | Let $I$ denote the ideal generated by $\{x_\alpha: \alpha \in \mathcal{A} \}$. The reader is invited to verify that 159 | \begin{itemize} 160 | \item $K^\bullet\left( \mathbf{x}, \prod_{\beta \in \mathcal{B}} M_\beta \right) = \prod_{\beta \in \mathcal{B}} K^\bullet(\mathbf{x}; M_\beta)$ for any family $\{M_\beta\}_{\beta \in \mathcal{B}}$ of $R$-modules, and same for their cohomologies; 161 | \item the $0$-th cohomology of $K^\bullet(\mathbf{x}; M)$ is $\Hom_R(R/I, M) = \{x \in M: Ix=0 \}$, the $I$-torsion part of $M$; 162 | \item if $|A| = n$, the $n$-th cohomology of $K^\bullet(\mathbf{x}; M)$ is $M/IM$. 163 | \end{itemize} 164 | These facts will cast in our later arguments. 165 | 166 | \begin{lemma}\label{prop:Koszul-coho-ann} 167 | Each cohomology of $K^\bullet(\mathbf{x}; M)$ is annihilated by $I$. 168 | \end{lemma} 169 | \begin{proof} 170 | Apply Proposition \ref{prop:Koszul-u-vanishing} by observing that $\mathfrak{q} = I$ in our setting. 171 | \end{proof} 172 | 173 | \begin{theorem}\label{prop:Koszul-depth}\index{depth} 174 | Let $M$ be an $R$-module and $\mathbf{x} = \{x_\alpha \}_{\alpha \in \mathcal{A}}$ be a family of elements of $R$, which generate an ideal $I$ of $A$. Then $\mathrm{depth}_I(M)$ equals 175 | \[ \inf \left\{n \geq 0: H^n(K^\bullet(\mathbf{x}; M)) \neq 0 \right\} \in \Z_{\geq 0} \sqcup \{\infty \} . \] 176 | Consequently, if $I \subset R$ is an ideal generated by elements $x_1, \ldots, x_n$ and $IM \neq M$, then we have 177 | \[ \mathrm{depth}_I(M) \leq n. \] 178 | \end{theorem} 179 | \begin{proof} 180 | Let $d := \text{depth}_I(M)$. Since $K^h(\mathbf{x}; M)$ is a direct product of copies of $M$ (possibly the empty product $=0$), by Proposition \ref{prop:depth-prod} its depth equals either $d$ or $\infty$. Combining Lemma \ref{prop:Koszul-coho-ann} and Corollary \ref{prop:depth-vanishing} (with $h = d-1$), we see that $H^{< d}(K^\bullet(\mathbf{x}; M)) = 0$. It remains to show $H^d(K^\bullet(\mathbf{x}; M)) \neq 0$ provided that $d < \infty$, which we assume from now onwards. 181 | 182 | The case $d=0$ is clear since there exists $\mathfrak{p} \in \text{Ass}(M) \cap V(I)$, therefore $R/\mathfrak{p} \hookrightarrow M$ and $\exists x \in M$ with $\mathfrak{p} x \supset Ix = \{0\}$, whence $H^0(K^\bullet(\mathbf{x}; M)) \neq 0$. Now suppose $d \in \Z_{\geq 1}$ and assume $H^d(K^\bullet(\mathbf{x}; M)) = 0$. Take a free resolution $F_\bullet \to R/I \to 0$ and put $C^\bullet := \Hom_R(F_\bullet, M)$, so that 183 | \[ H^i(C^\bullet) \simeq \Ext^i_R(R/I, M), \quad i \geq 0. \] 184 | Hence for $i < d$ we have short exact sequences 185 | \[ 0 \to B^i \to C^i \to B^{i+1} \to 0 \] 186 | as usual; recall $B^i, Z^i \subset C^i$. Note that each $C^i$ is a direct product of copies of $M$, hence $H^{\leq d}(K^\bullet(\mathbf{x}; C^i)) = 0$. Indeed, $H^{< d}(K^\bullet(\mathbf{x}; M))=0$ has been settled in the first step, whilst $H^d(K^\bullet(\mathbf{x}; M))$ is the hypothesis to be refuted. It then follows from the long exact sequence for Koszul cohomologies (Proposition \ref{prop:Koszul-long-exact-sequence}) that 187 | \[ (s \leq d) \wedge (i < d) \implies H^s(K^\bullet(\mathbf{x}; B^{i+1})) \hookrightarrow H^{s+1}(K^\bullet(\mathbf{x}; B^i)). \] 188 | Suppose $i < d$. As $B^0 = 0$, an iteration yields 189 | \[ H^{d-i}(K^\bullet(\mathbf{x}; B^{i+1})) \hookrightarrow \cdots \hookrightarrow H^{d+1}(K^\bullet(\mathbf{x}; B^0)) = 0. \] 190 | In particular, $i=d-1$ gives rise to $H^1(K^\bullet(\mathbf{x}; B^d)) = 0$. Hence the short exact sequence $0 \to B^d \to Z^d \to H^d \to 0$ (for $C^\bullet$) together with Proposition \ref{prop:Koszul-long-exact-sequence} give rise to 191 | \[ H^0(K^\bullet(\mathbf{x}; Z^d)) \twoheadrightarrow H^0(K^\bullet(\mathbf{x}; H^d)). \] 192 | As $H^d := H^d(C^\bullet) \simeq \Ext^d_R(R/I, M)$ is nonzero and annihilated by $I$, the right-hand side is nonzero. Hence $H^0(K^\bullet(\mathbf{x}; Z^d)) \neq 0$ and then $H^0(K^\bullet(\mathbf{x}; C^d)) \neq 0$, as the zeroth Koszul cohomology is nothing but the $I$-torsion part. 193 | 194 | Finally, recall that $C^d \neq \{0\}$ is a direct product of copies of $M$, consequently 195 | \[ H^0(K^\bullet(\mathbf{x}; M)) \neq 0. \] 196 | However, this contradicts the earlier result that $H^{, baseline=-.75ex] 22 | \draw (0,0) -- node[below=-2pt] {\box\xratbelow} 23 | node[above] {\box\xratabove} 24 | (\xratlen,0) ;}} 25 | % 重定义 \xlefttarrow[below]{above} 26 | \renewcommand{\xleftarrow}[2][]{% 27 | \setbox\xratbelow=\hbox{\ensuremath{\scriptstyle #1}}% 28 | \setbox\xratabove=\hbox{\ensuremath{\scriptstyle #2}}% 29 | \pgfmathsetlengthmacro{\xratlen}{max(\wd\xratbelow, \wd\xratabove) + .6em}% 30 | \mathrel{\tikz [<-, baseline=-.75ex] 31 | \draw (0,0) -- node[below] {\box\xratbelow} 32 | node[above] {\box\xratabove} 33 | (\xratlen,0) ;}} 34 | % 重定义 \xleftrightarrow[below]{above} 35 | \renewcommand{\xleftrightarrow}[2][]{% 36 | \setbox\xratbelow=\hbox{\ensuremath{\scriptstyle #1}}% 37 | \setbox\xratabove=\hbox{\ensuremath{\scriptstyle #2}}% 38 | \pgfmathsetlengthmacro{\xratlen}{max(\wd\xratbelow, \wd\xratabove) + .6em}% 39 | \mathrel{\tikz [<->, baseline=-.75ex] 40 | \draw (0,0) -- node[below] {\box\xratbelow} 41 | node[above] {\box\xratabove} 42 | (\xratlen,0) ;}} 43 | % 重定义 \xhookrightarrow[below]{above}, 使用 tikz-cd 的 hookrightarrow 44 | \renewcommand{\xhookrightarrow}[2][]{% 45 | \setbox\xratbelow=\hbox{\ensuremath{\scriptstyle #1}}% 46 | \setbox\xratabove=\hbox{\ensuremath{\scriptstyle #2}}% 47 | \pgfmathsetlengthmacro{\xratlen}{max(\wd\xratbelow, \wd\xratabove) + .6em}% 48 | \mathrel{\tikz [baseline=-.75ex] 49 | \draw (0,0) edge[commutative diagrams/hookrightarrow] node[below] {\box\xratbelow} 50 | node[above] {\box\xratabove} 51 | (\xratlen,0) ;}} 52 | % 重定义 \xhooklefttarrow[below]{above}, 使用 tikz-cd 的 hookleftarrow 53 | \renewcommand{\xhookleftarrow}[2][]{% 54 | \setbox\xratbelow=\hbox{\ensuremath{\scriptstyle #1}}% 55 | \setbox\xratabove=\hbox{\ensuremath{\scriptstyle #2}}% 56 | \pgfmathsetlengthmacro{\xratlen}{max(\wd\xratbelow, \wd\xratabove) + .6em}% 57 | \mathrel{\tikz [baseline=-.75ex] 58 | \draw (0,0) edge[commutative diagrams/hookleftarrow] node[below] {\box\xratbelow} 59 | node[above] {\box\xratabove} 60 | (\xratlen,0) ;}} 61 | 62 | % 重定义 \xmapsto[below]{above}, 使用 tikz-cd 的 mapsto 63 | \renewcommand{\xmapsto}[2][]{% 64 | \setbox\xratbelow=\hbox{\ensuremath{\scriptstyle #1}}% 65 | \setbox\xratabove=\hbox{\ensuremath{\scriptstyle #2}}% 66 | \pgfmathsetlengthmacro{\xratlen}{max(\wd\xratbelow, \wd\xratabove) + .6em}% 67 | \mathrel{\tikz [baseline=-.75ex] 68 | \draw (0,0) edge[commutative diagrams/mapsto] node[below] {\box\xratbelow} 69 | node[above] {\box\xratabove} 70 | (\xratlen,0) ;}} 71 | 72 | % 定义 \xlongequal[below]{above}, 使用 tikz-cd 的等号 73 | \newcommand{\xlongequal}[2][]{% 74 | \setbox\xratbelow=\hbox{\ensuremath{\scriptstyle #1}}% 75 | \setbox\xratabove=\hbox{\ensuremath{\scriptstyle #2}}% 76 | \pgfmathsetlengthmacro{\xratlen}{max(\wd\xratbelow, \wd\xratabove) + .6em}% 77 | \mathrel{\tikz [baseline=-.75ex] 78 | \draw (0,0) edge[commutative diagrams/equal] node[below] {\box\xratbelow} 79 | node[above] {\box\xratabove} 80 | (\xratlen,0) ;}} 81 | \makeatother 82 | 83 | \endinput 84 | -------------------------------------------------------------------------------- /mycommand.sty: -------------------------------------------------------------------------------- 1 | % Copyright 2018 李文威 (Wen-Wei Li). 2 | % Permission is granted to copy, distribute and/or modify this 3 | % document under the terms of the Creative Commons 4 | % Attribution 4.0 International (CC BY 4.0) 5 | % http://creativecommons.org/licenses/by/4.0/ 6 | 7 | % Some custom commands that I prefer 8 | \NeedsTeXFormat{LaTeX2e} 9 | \ProvidesPackage{mycommand}[2018/02/20 Package for my own commands] 10 | 11 | \newcommand\hmmax{0} % default 3, Increase the capacity of fonts 12 | \newcommand\bmmax{0} % default 4, Increase the capacity of fonts... 13 | \RequirePackage{bm} % Bolface + other functionalities 14 | 15 | \renewcommand*\arraystretch{1.5} % Increase the space in arrays 16 | 17 | \RequirePackage{euscript} % "Euler script" fonts 18 | 19 | % Customize the list structures (using paralist) 20 | \setdefaultitem{$\diamond$}{}{}{} 21 | \renewcommand{\descriptionlabel}[1]{\hspace{\labelsep} $\vartriangleright$\enskip {\bfseries #1} \;} 22 | \renewcommand{\paradescriptionlabel}[1]{\normalfont \bfseries #1 \enskip} 23 | 24 | % Some commands that I am used to. 25 | % Well-known algebraic structures 26 | \newcommand{\N}{\ensuremath{\mathbb{N}}} 27 | \newcommand{\Z}{\ensuremath{\mathbb{Z}}} 28 | \newcommand{\Q}{\ensuremath{\mathbb{Q}}} 29 | \newcommand{\R}{\ensuremath{\mathbb{R}}} 30 | \newcommand{\CC}{\ensuremath{\mathbb{C}}} 31 | \newcommand{\F}{\ensuremath{\mathbb{F}}} 32 | \newcommand{\A}{\ensuremath{\mathbb{A}}} 33 | 34 | 35 | % Algebra 36 | 37 | \newcommand{\gr}{\operatorname{gr}} 38 | \newcommand{\Ass}{\operatorname{Ass}} 39 | \newcommand{\topwedge}{\ensuremath{\bigwedge^{\mathrm{max}}}} 40 | \newcommand{\rank}{\operatorname{rk}} 41 | \newcommand{\Aut}{\operatorname{Aut}} 42 | \newcommand{\Isom}{\operatorname{Isom}} 43 | \newcommand{\Hm}{\operatorname{H}} % Homology/cohomology 44 | \newcommand{\Tr}{\operatorname{Tr}} % trace 45 | \newcommand{\Nm}{\operatorname{N}} % norm 46 | \newcommand{\Ann}{\operatorname{Ann}} 47 | \newcommand{\Resprod}{\ensuremath{{\prod}'}} 48 | \newcommand{\Sym}{\operatorname{Sym}} 49 | \newcommand{\ord}{\operatorname*{ord}} 50 | \newcommand{\trdeg}{\operatorname{tr.deg}} 51 | \newcommand{\Gras}{\ensuremath{\mathbf{G}}} % Grassmannians 52 | \newcommand{\WittV}{\operatorname{W}} % Witt vectors 53 | 54 | % Analysis 55 | \newcommand{\dd}{\mathop{}\!\mathrm{d}} 56 | \newcommand{\champ}[1]{\ensuremath{\frac{\partial}{\partial #1}}} 57 | \newcommand{\norme}[1]{\ensuremath{\| #1 \|}} 58 | \newcommand{\normeL}[2]{\ensuremath{\| #2 \|_{L^{#1}}}} 59 | \newcommand{\normeLs}[3]{\ensuremath{\| #3 \|_{L^{#1}, #2}}} 60 | 61 | % General things... 62 | \newcommand{\ceil}[1]{\ensuremath{\lceil #1 \rceil}} 63 | \newcommand{\lrangle}[1]{\ensuremath{\left\langle #1 \right\rangle}} 64 | \newcommand{\mes}{\operatorname{vol}} 65 | \newcommand{\sgn}{\operatorname{sgn}} 66 | \newcommand{\Stab}{\operatorname{Stab}} 67 | \newcommand{\pr}{\ensuremath{\mathbf{pr}}} % projection morphism 68 | 69 | % Categorical Terms (in my view) 70 | \newcommand{\Obj}{\operatorname{Ob}} % Objects 71 | \newcommand{\Mor}{\operatorname{Mor}} % Morphisms 72 | \newcommand{\cate}[1]{\ensuremath{\mathsf{#1}}} % Font series for categories 73 | \newcommand{\dcate}[1]{\ensuremath{\text{-}\mathsf{#1}}} % Categories with a pre-dash 74 | \newcommand{\cated}[1]{\ensuremath{\mathsf{#1}\text{-}}} % Categories with a post-dash 75 | \newcommand{\identity}{\ensuremath{\mathrm{id}}} 76 | \newcommand{\prolim}{\ensuremath{\underleftarrow{\lim}}} 77 | \newcommand{\indlim}{\ensuremath{\underrightarrow{\lim}}} 78 | \newcommand{\Hom}{\operatorname{Hom}} 79 | \newcommand{\iHom}{\ensuremath{\EuScript{H}\mathrm{om}}} 80 | \newcommand{\End}{\operatorname{End}} 81 | \newcommand{\rightiso}{\ensuremath{\stackrel{\sim}{\rightarrow}}} 82 | \newcommand{\longrightiso}{\ensuremath{\stackrel{\sim}{\longrightarrow}}} 83 | \newcommand{\leftiso}{\ensuremath{\stackrel{\sim}{\leftarrow}}} 84 | \newcommand{\longleftiso}{\ensuremath{\stackrel{\sim}{\longleftarrow}}} 85 | \newcommand{\utimes}[1]{\ensuremath{\overset{#1}{\times}}} 86 | \newcommand{\dtimes}[1]{\ensuremath{\underset{#1}{\times}}} 87 | \newcommand{\dotimes}[1]{\ensuremath{\underset{#1}{\otimes}}} 88 | \newcommand{\dsqcup}[1]{\ensuremath{\underset{#1}{\sqcup}}} 89 | \newcommand{\munit}{\ensuremath{\mathbf{1}}} % unit in a monoidal category 90 | \newcommand{\Yinjlim}{\ensuremath{\text{\textquotedblleft}\varinjlim\text{\textquotedblright}}} % injective limit in the Yoneda category 91 | \newcommand{\Yprojlim}{\ensuremath{\text{\textquotedblleft}\varprojlim\text{\textquotedblright}}} % projective limit in the Yoneda category 92 | 93 | % Homological Algebra 94 | \newcommand{\Ker}{\operatorname{ker}} 95 | \newcommand{\Coker}{\operatorname{coker}} 96 | \newcommand{\Image}{\operatorname{im}} 97 | \newcommand{\Coim}{\operatorname{coim}} 98 | \newcommand{\Ext}{\operatorname{Ext}} 99 | \newcommand{\Tor}{\operatorname{Tor}} 100 | \newcommand{\otimesL}{\ensuremath{\overset{\mathrm{L}}{\otimes}}} 101 | 102 | % Geometry 103 | \newcommand{\Der}{\operatorname{Der}} 104 | \newcommand{\Lie}{\operatorname{Lie}} 105 | \newcommand{\Ad}{\operatorname{Ad}} 106 | \newcommand{\ad}{\operatorname{ad}} 107 | \newcommand{\Frob}{\operatorname{Fr}} 108 | \newcommand{\Spec}{\operatorname{Spec}} 109 | \newcommand{\MaxSpec}{\operatorname{MaxSpec}} 110 | \newcommand{\PP}{\ensuremath{\mathbb{P}}} 111 | \newcommand{\mult}{\operatorname{mult}} 112 | \newcommand{\divisor}{\operatorname{div}} 113 | \newcommand{\Gm}{\ensuremath{\mathbb{G}_\mathrm{m}}} 114 | \newcommand{\Ga}{\ensuremath{\mathbb{G}_\mathrm{a}}} 115 | \newcommand{\Pic}{\operatorname{Pic}} 116 | \newcommand{\Supp}{\operatorname{Supp}} 117 | \newcommand{\Res}{\operatorname{Res}} 118 | 119 | % Groups 120 | \newcommand{\Gal}{\operatorname{Gal}} 121 | \newcommand{\GL}{\operatorname{GL}} 122 | \newcommand{\SO}{\operatorname{SO}} 123 | \newcommand{\Or}{\operatorname{O}} 124 | \newcommand{\GSpin}{\operatorname{GSpin}} 125 | \newcommand{\Spin}{\operatorname{Spin}} 126 | \newcommand{\UU}{\operatorname{U}} 127 | \newcommand{\SU}{\operatorname{SU}} 128 | \newcommand{\PGL}{\operatorname{PGL}} 129 | \newcommand{\PSL}{\operatorname{PSL}} 130 | \newcommand{\SL}{\operatorname{SL}} 131 | \newcommand{\Sp}{\operatorname{Sp}} 132 | \newcommand{\GSp}{\operatorname{GSp}} 133 | \newcommand{\PSp}{\operatorname{PSp}} 134 | \newcommand{\gl}{\ensuremath{\mathfrak{gl}}} 135 | \newcommand{\sli}{\ensuremath{\mathfrak{sl}}} 136 | \newcommand{\so}{\ensuremath{\mathfrak{so}}} 137 | \newcommand{\spin}{\ensuremath{\mathfrak{spin}}} 138 | \newcommand{\syp}{\ensuremath{\mathfrak{sp}}} 139 | \newcommand{\Ind}{\operatorname{Ind}} 140 | --------------------------------------------------------------------------------