\documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsfonts} % \usepackage{yfonts} % WIP \usepackage{amsthm} \usepackage{amsmath} \usepackage{enumerate} \usepackage{hyperref} \usepackage{amssymb} \usepackage{tikz} % diagram \begin{filecontents}[overwrite]{commutative-algebra-notes.bib} @misc{am, author = {M. F. Atiyah and I. G. MacDonald}, title = {{Introduction to Commutative Algebra}}, year = {1969} } @misc{reid, author = {Miles Reid}, title = {{Undergraduate Commutative Algebra}}, year = {1995} } @misc{mit-course, author = {Steven Kleiman}, title = {{Commutative Algebra - MIT OpenCourseWare}}, year = {2008}, note = {\url{https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}}, url = {https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/} } \end{filecontents} \nocite{*} \theoremstyle{definition} \newtheorem{innerdefn}{Definition} \newenvironment{defn}[1] {\renewcommand\theinnerdefn{#1}\innerdefn} {\endinnerdefn} \newtheorem{innerthm}{Theorem} \newenvironment{thm}[1] {\renewcommand\theinnerthm{#1}\innerthm} {\endinnerthm} \newtheorem{innerlemma}{Lemma} \newenvironment{lemma}[1] {\renewcommand\theinnerlemma{#1}\innerlemma} {\endinnerlemma} \newtheorem{innerprop}{Proposition} \newenvironment{prop}[1] {\renewcommand\theinnerprop{#1}\innerprop} {\endinnerprop} \newtheorem{innercor}{Corollary} \newenvironment{cor}[1] {\renewcommand\theinnercor{#1}\innercor} {\endinnercor} \newtheorem{innereg}{Example} \newenvironment{eg}[1] {\renewcommand\theinnereg{#1}\innereg} {\endinnereg} \newtheorem{innerex}{Exercise} \newenvironment{ex}[1] {\renewcommand\theinnerex{#1}\innerex} {\endinnerex} \newcommand{\aA}{\mathfrak{a}} % TODO: use goth font \newcommand{\mM}{\mathfrak{m}} \title{Commutative Algebra notes} \author{arnaucube} \date{} \begin{document} \maketitle \begin{abstract} Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$. The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding. \end{abstract} \tableofcontents \section{Ideals} \subsection{Definitions} % WIP \subsection{Lemmas, propositions and corollaries} \begin{thm}{1.X}{Zorn's lemma} \label{zorn} TODO \end{thm} \begin{thm}{1.3} \label{1.3} Every ring $A \neq 0$ has at lleast one maximal ideal. \end{thm} \begin{proof} By Zorn's lemma \ref{zorn}. \end{proof} \begin{cor}{1.4} \label{1.4} if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$. \end{cor} \begin{cor}{1.5} \label{1.5} Every non-unit of $A$ is contained in a maximal ideal. \end{cor} \begin{defn}{Jacobson radical} The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$. Denoted $Jac(A)$. $Jac(A)$ is an ideal of $A$. \end{defn} \begin{prop}{1.9} \label{1.9} $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$. \end{prop} \begin{proof} Suppose $1-xy$ not a unit. By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal. But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$. Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit. Conversely:\\ Suppose $x \not\in \mM$ for some maximal ideal $\mM$. Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$. Hence $1 -xy \in \mM$, and is therefore not a unit. \end{proof} \section{Modules} \subsection{Modules} \subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries} \begin{prop}{2.4}(Cayley-Hamilton Theorem) \label{2.4} Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$. Then $\psi$ satisfies $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$ with $a_i \in \aA$. \end{prop} \begin{proof} Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\ By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$. Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$ with $a_{ij} \in \aA$. Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations: $$ \left. \begin{aligned} \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\ \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\ \ldots\\ \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n \end{aligned} \right\} \begin{aligned} &\text{n elements $\psi(x_i) \in \aA M$ which}\\ &\text{are linear combinations of the}\\ &\text{$n$ generators of $M$} \end{aligned} $$ Next step: rearrange in order to use matrix algebra. Observe that each row equals $0$, and rearranging the elements at each row we get \begin{align*} &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\ &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\ &\ldots\\ &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0 \end{align*} Then, group the $x_i$ terms together; as example, take the row $i=1$: $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$ for $i=2$: $$-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0$$ So, $\forall i \in [n]$, as a matrix: $$ \begin{pmatrix} \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\ -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\ \vdots\\ -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ \vdots\\ 0 \end{pmatrix} $$ Kronecker delta: $\delta_{ij} = \begin{cases} 1 & \text{if } i = j,\\ 0 & \text{otherwise} \end{cases}$ With the Kronecker delta, $\psi(x_i)$ can be expressed as $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$ so the previous matrix ccan be characterized as $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$ The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$) \begin{itemize} \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$ \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$ \end{itemize} Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$, the ring of operators $A[\psi]$ is a commutative ring. $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant. We're interested in the determinant because it is the only way to turn a system of multiple equations in a single scalar-like equation that describes the endomorphism $\psi$.\\ $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\ $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$ $$det(M) \cdot x_i = 0~~ \forall i$$ where $x_i$ are the generators of $M$. Since $det(M)$ kills every generator, it must kill every element in $M$\\ $\Longrightarrow~~ det(M)$ is the zero map. Leibniz formula of the determinant of an $n \times n$ matrix: $$ det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)} $$ so, $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$ expanding it, \begin{itemize} \item highest power is $1 \cdot \psi^n$ \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\ where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$ \item the rest of coefficients of $\psi^k$ are also elements in $\aA$ \end{itemize} So we have $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$ with $a_i \in \aA$. Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so: $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$ with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem. \end{proof} \begin{cor}{2.5} \label{2.5} Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$. Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$. \end{cor} \begin{proof} take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}): \begin{align*} &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\ \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\ \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0 \end{align*} apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$ with $a_i \in \aA$. \begin{enumerate}[\text{part} i.] \item $xM=0$:\\ Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$. \item $x \equiv 1 \pmod \aA$:\\ $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\ then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\ so that $x=(1+b) \in \aA$.\\ Then $x-1 = (1+b)-1 = b \in \aA$\\ so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated. \end{enumerate} \end{proof} \begin{prop}{2.6}{Nakayama's lemma} \label{2.6} Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$. Then $\aA M = M$ implies $M=0$. \end{prop} \begin{proof} By \ref{2.5}: since $\aA M = M$, we have $x M =0$ for some $x \equiv 1 \pmod {Jac(A)}$. (notice that at \ref{2.5} is $\pmod \aA$ but here we use $\pmod {Jac(A)}$, since we have $\aA \subseteq Jac(A)$). By \ref{1.9}, $x$ is a unit in $A$. Hence $M = x^{-1} \cdot \underbrace{x \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$. Thus, if $\aA M = M$ then $M=0$. \end{proof} \begin{cor}{2.7} \label{2.7} Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal. Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$. \end{cor} \begin{proof} The idea is to apply Nakayama (\ref{2.6}) to $M/N$. Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module. Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too. By definition, $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$ where $m_i$ are the generators of $M$. Then, for $M/N$, $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$ observe that $a_i(m_i+N)= a_i m_i +N$, thus $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$ Hence, \begin{equation} \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N \label{eq:2.7.1} \end{equation} By definition, if we take $\frac{\aA M + N}{N}$, then $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$ thus every $y \in \aA M +N$ can be written as $$y=x+n,~~ \text{with} x \in \aA M,~ n\in M$$ which comes from \eqref{eq:2.7.1}. Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient. Hence, every element of $\frac{\aA M +N}{N}$ has the form $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$ as in \eqref{eq:2.7.1}. Thus \begin{equation} \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N} \label{eq:2.7.2} \end{equation} By the Collorary assumption, $M = \aA M + N$; quotient it by $N$: \begin{equation} \frac{M}{N} = \frac{\aA M +N}{N} \label{eq:2.7.3} \end{equation} So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}: $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$ thus, $\aA (\frac{M}{N}) = \frac{M}{N}$. By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$ Note that $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$ (the zero element in $\frac{M}{N}$ is the coset $N=0+N$) Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$. Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$. Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$. So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$. Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$. \end{proof} \begin{prop}{2.8} \label{2.8} Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vecctor space. Then the $x_i$ generate $M$. \end{prop} \begin{proof} Let $N$ submodule $M$, generated by the $x_i$. Then the composite map $N \longrightarrow M \longrightarrow \frac{M}{m M}$ maps $N$ onto $\frac{M}{m M}$, hence $N + \aA M = M$, which by \ref{2.7} implies $N = M$. \end{proof} \subsection{Noetherean rings} \section{Exercises} For the exercises, I follow the assignements listed at \cite{mit-course}. The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}. \subsection{Exercises Chapter 1} \begin{ex}{R.1.1} Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$? \end{ex} \begin{proof} Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$. $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\ Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$. Thus $I \cup J$ is not closed under addition; thus is not an ideal. Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum: $$I+J = \{ a+b | a \in I, b \in J \}$$ $gcd(2,3)=1$, so $I+J = \mathbb{Z}$. Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$. \end{proof} \begin{ex}{R.1.5} let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\ In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$. \end{ex} \begin{proof} (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$). Let $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$ The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$. \begin{enumerate}[i.] \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\ $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$). If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so $$\psi(a-b)= \psi(a) - \psi(b) \in P$$ hence $a-b \in \psi^{-1}(P)$. If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\ Thus $ra \in \psi^{-1}(P)$. $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$. \item show that $\psi^{-1}(P)$ is prime:\\ $\psi^{-1}(P) \neq A$, since if $\psi^{-1}(P)=A$, then $1_A \in \psi^{-1}(P)$, so $\psi(1_A)=1_B \in P$, which would mean that $P=B$, a contradiction since $P$ is prime ideal of $B$. Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$. Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\ Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$. Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$. \end{enumerate} \end{proof} \begin{ex}{R.1.6} prove or give a counter example: \begin{enumerate}[a.] \item the intersection of two prime ideals is prime \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$ \end{enumerate} \end{ex} \begin{proof} \begin{enumerate}[a.] \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime. Then $I \cap J = (2) \cap (3) = (6)$. The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$. Thus the intersection of two primes can not be prime. \item $P_1=(2),~ P_2=(3)$, both prime. Then, $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$ $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$. (recall: principal= generated by a single element) So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$. The whole ring is not a prime ideal (by the definition of the prime ideal), so $P_1+P_2$ is not a prime ideal. Henceforth, the sum of two prime ideals is not necessarily prime. \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$. Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$. Then \begin{align*} \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\ \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A \end{align*} But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field. Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal. \end{enumerate} \end{proof} \subsection{Exercises Chapter 2} \bibliographystyle{unsrt} \bibliography{commutative-algebra-notes.bib} \end{document}