\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{2026} \begin{document} \maketitle \begin{abstract} Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. For the exercises, I follow the assignments listed at \cite{mit-course}. 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} \begin{defn}{}{ideal} $I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\ \hspace*{2em} ie. $I$ absorbs products in $R$. \end{defn} \begin{defn}{}{prime ideal} if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$. \end{defn} \begin{defn}{}{principal ideal} generated by a single element, $(a)$. $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$. \end{defn} \begin{defn}{}{maximal ideal} $\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$. \end{defn} \begin{defn}{}{unit} $x \in A$ such that $xy=1$ for some $y \in A$. ie. element \emph{which divides 1}. \end{defn} \begin{defn}{}{zerodivisor} $x \in A$ such that $\exists 0 \neq y \in A$ such that $xy=0 \in A$. ie. $x$ \emph{divides 0}.. If a ring does not have zerodivisors is an integral domain. \end{defn} \begin{defn}{}{prime spectrum - $Spec(A)$} set of prime ideals of $A$. ie. $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$ \end{defn} \begin{defn}{}{integral domain} Ring in which the product of any two nonzero elements is nonzero. ie. no zerodivisors. ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$. Every field is an integral domain, not the converse. \end{defn} \begin{defn}{}{principal ideal domain - PID} integral domain in which every ideal is principal. ie. ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$. \end{defn} \begin{defn}{}{nilpotent} $a \in A$ such that $a^n=0$ for some $n>0$. \end{defn} \begin{defn}{}{nilrad A} set of all nilpotent elements of $A$; is an ideal of $A$. if $nilrad A = 0 ~\Longrightarrow$ $A$ has no nonzero nilpotents. $$nilrad A = \bigcap_{P \in Spec(A)} P$$ \end{defn} \begin{defn}{}{idempotent} $e \in A$ such that $e^2=e$. \end{defn} \begin{defn}{}{radical of an ideal} $$rad I = \{ f \in A | f^n \in I~ \text{for some} n \}$$ $rad I$ is an ideal. $nilrad A = rad 0$ $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$ \end{defn} \begin{defn}{}{local ring} A \emph{local ring} has a unique maximal ideal. Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$: $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$ \end{defn} \subsection{Z and K[X], two Principal Ideal Domains} \begin{lemma}{} $\mathbb{Z}$ is a PID. \end{lemma} \begin{proof} Let $I$ a nonzero ideal of $\mathbb{Z}$. Since $I \neq \{0\}$, there is at least one nonzero integer in $I$. Choose the smallest element of $I$, namely $d$. Observe that $(d) \subseteq I$, since $d \in I$. Then, every multiple $nd \in I$, since $I$ is an ideal. Take $a \in I$. By the Euclidean division algorithm in $\mathbb{Z}$, $a=qd+r$, with $q,r \in \mathbb{Z}$ and $0 \leq r \leq d$. Then $r = a - qd \in I$, but $d$ was chosen to be the smallest positive element of $I$, so the only possibility is $r=0$. Hence, $a=qd$, so $a \in (d)$, giving $I \subseteq (d)$. Since we had $(d) \subseteq I$ and now we got $I \subseteq (d)$, we have $I = (d)$, so every ideal of $\mathbb{Z}$ is principal. Thus $\mathbb{Z}$ is a Principal Ideal Domain(PID). \end{proof} \begin{lemma}{} $K[X]$ is a PID. \end{lemma} \begin{proof} This proof follows very similarly to the previous proof.\\ Let $K$ be a field, $K[X]$ a polynomial ring. Take $\{0\} \neq I \subseteq K[X]$. Since $I \neq \{0\}$, there is at least one non-zero polynomial in $I$. Let $p(X) \in I$ be of minimal degree among nonzero elements of $I$. Observe that $(p(X)) \subseteq I$, because $p(X) \in I$ and $I$ is an ideal. Let $f(X) \in I$. By Euclidean division algorithm in $K[X]$, $\exists q, r \in K[X]$ such that $f(X) = q(X) \cdot p(X) + r(X)$ with eithr $r(X)=0$ or $deg(r) < deg(p)$. Since $f,p \in I$, then $r(X) = f(X) - q(X)\cdot p(X) \in I$ If $r(X) \neq 0$, then $deg(r) < deg(p)$, which contradicts the minimality of $deg(p)$ in $I$. Therefore, $r(X)=0$, thus $f(X)=q(X)\cdot p(X)$, hence $f(X) \in (p(X))$. Henceforth, $I \subseteq (p(X))$. Then, since $(p(X)) \subseteq I$ and $I \subseteq (p(X))$, we have that $I = (p(X))$. So every ideal of $K[X]$ is principal; thus $K[X]$ is a PID. \end{proof} \subsection{Zorn's lemma and Jacobson radicals} Let $\Sigma$ be a partially orddered set. Given subset $S \subset \Sigma$, an \emph{upper bound} of $S$ is an element $u \in \Sigma$ such that $s2$ follows by induction. \item if $M$ finite, then $\exists~$ surjective homomorphism $$A^r \longrightarrow M \longrightarrow 0$$ for some $r$, so that $M$ is a quotient $$M \cong A^r / N$$ for some submodule $N \subset A^r$. $A^r$ is a Noetherian module by i., so $M$ is Noetherian due Proposition \ref{R.3.4.P}. Conversely, $M$ Noetherian implies $M$ finite. item as in previous implications:\\ $M$ finite and $A$ Noetherian $\Longrightarrow$ $M$ is Noetherian,\\ $\Longrightarrow$ since $N \subseteq M$, then $N$ is Noetherian too\\ $\Longrightarrow$ which implies that $N$ is a finite $A$-module. \item $B$ is Noetherian as an $A$-module; but ideals of $B$ are submodules of $B$ as an $A$-submodule, so that $B$ is a Noetherian ring. \end{enumerate} \end{proof} \vspace{0.5cm} \begin{thm}{R.3.6}{Hilbert basis theorem} \label{hilbert-basis} if $A$ a Noetherian ring, then so is the polynomial ring $A[x]$. \end{thm} \begin{proof} Prove that any ideal $I \subset A[x]$ is fingen. Define auxiliary sets $J_n \subset A$ by $$J_n = \{ a \in A ~|~ \exists f \in I ~\text{s.th.}~ f = a x^n + b_{n-1}x^{n-1} + \ldots b_0 \}$$ ie. $J_n$ is the set of leading coefficients of $I$ of degree $n$. $J_n$ is an ideal, since $I$ is an ideal. $J_n \subset J_{n+1}$, since for $f \in I$ also $x f \in I$. Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.\\ Using the assumption that $A$ is Noetherian, deduce that $J_n = J_{n+1}$ for some $n$. For each $m \leq n, ~~ J_m \subset A$ is fingen, ie. $$J_m = (a_{m,1}, \ldots a_{m, r_m})$$ By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\ $\exists$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$. $$\Longrightarrow~~ \{ f_{m,j} \}_{m0$ we have two cases: \begin{itemize} \item $y_1, \ldots, y_n$ are algebraically independent over $K$, then $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself. \item $y_1, \ldots, y_n$ are algebraically dependent over $K$, $$\exists 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$ \end{itemize} Goal: is to change variables so that $f$ becomes monic in one of the variables; this allows to express one generator as an integral element over the others. Following from Lemma \ref{R.4.6.L}, define new variables $y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~\text{and}~ A=A^*[y_n]$$ By inductive hypothesis on $A^*,~~ \exists~ z_1, \ldots, z_m \in A^*$ algebraically independent over $K$ and with $A^*$ finite over $B=K[z_1, \ldots, z_m]$. Since $y_n$ integral over $A^* ~~\Longrightarrow~ A^*[y_n]$ is finite over $A^*$.\\ Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required. \end{proof} \begin{eg}{ } $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$. This corresponds to the fact that the hyperbola $XY=1$ has the line $X=0$ as an asymptotic line (so that its projection to the $X$-axis misses a root over $X=0$). Take $X' = X- \epsilon Y$ as the element of $A$ instead of $X$; then the relation becomes $(X' + \epsilon Y) Y=1$, monic in $Y$ if $\epsilon \neq 0$. This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymtotic line. \end{eg} \newpage \section{Exercises} For the exercises, I follow the assignments listed at \cite{mit-course}. The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}. \subsection{Exercises Chapter 1} \begin{ex}{R.1.1} Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$? \end{ex} \begin{proof} Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$. $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\ Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$. Thus $I \cup J$ is not closed under addition; thus is not an ideal. Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum: $$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$$ The claim is that $\psi^{-1}(P)$ is prime ideal of $A$. \begin{enumerate}[i.] \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\ $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$). If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so $$\psi(a-b)= \psi(a) - \psi(b) \in P$$ hence $a-b \in \psi^{-1}(P)$. If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\ Thus $ra \in \psi^{-1}(P)$. $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$. \item show that $\psi^{-1}(P)$ is prime:\\ $\psi^{-1}(P) \neq A$, since if $\psi^{-1}(P)=A$, then $1_A \in \psi^{-1}(P)$, so $\psi(1_A)=1_B \in P$, which would mean that $P=B$, a contradiction since $P$ is prime ideal of $B$. Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$. Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\ Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$. Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$. \end{enumerate} \end{proof} \begin{ex}{R.1.6} prove or give a counter example: \begin{enumerate}[a.] \item the intersection of two prime ideals is prime \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$ \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$ to maximal ideals of $A$ \end{enumerate} \end{ex} \begin{proof} \begin{enumerate}[a.] \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime. Then $I \cap J = (2) \cap (3) = (6)$. The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$. Thus the intersection of two primes can not be prime. \item $P_1=(2),~ P_2=(3)$, both prime. Then, $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$ $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$. (recall: principal= generated by a single element) So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$. The whole ring is not a prime ideal (by the definition of the prime ideal), so $P_1+P_2$ is not a prime ideal. Henceforth, the sum of two prime ideals is not necessarily prime. \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$. Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$. Then \begin{align*} \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\ \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A \end{align*} But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field. Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal. \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal. Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3). By the isomorphism theorems, $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$ Since $\frac{(A/I)}{M}$ is a field, the quotient $\frac{A}{\psi^{-1}(M)}$ is a field, so $\psi^{-1}(M)$ is a maximal ideal of $A$. $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal. \end{enumerate} \end{proof} \begin{ex}{R.1.12.a} if $I,J$ ideals and $P$ prime ideal, prove that $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$ \end{ex} \begin{proof} assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$, then $$x = \sum_{k=1}^n a_k b_k$$ with $a_k \in I,~ b_k \in J$. Each $a_k \in I \subseteq P$. Since $P$ an ideal, $$\sum_{k=1}^n a_k b_k \in P$$ thus $x \in P$, hence $IJ \subseteq P$. So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$. \vspace{0.5cm} Conversely,\\ assume $P$ prime and $IJ \subseteq P$. Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$. \begin{itemize} \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$ \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$ \end{itemize} Since $a \in I,~ b \in J,~~ ab \in IJ \subseteq P$, but $P$ is prime, so $ab \in P$ implies that $a \in P$ or $b \in P$. This contradicts $a,b$ being taken outside of $P$. Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false. \vspace{0.3cm} So both directions are proven, hence $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$ \end{proof} \begin{ex}{R.1.18} Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal. \end{ex} \begin{proof} Let $P$ prime ideal of $R$. $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$ Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$. $P \subset S$, so $S$ is nonempty. Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion. Define $$Q_C = \bigcap_{Q \in C} Q$$ Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$. Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals. Intersection of a decreasing chain of prime ideals is again a prime ideal: \begin{itemize} \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$ \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$ \end{itemize} If there were some $Q_1,~ Q_2 \in C$ with $a \in Q_1$ and $b \not\in Q_2$, then by total ordering, either $Q_1 \subseteq Q_2$ or $Q_2 \subseteq Q_1$. In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded. Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$. Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$. Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\ Therefore, $S$ has a minimal element $P_{min}$. By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$. So $P_{min}$ is a minimal prime ideal, contained in $P$. \end{proof} \begin{ex}{R.1.10} \end{ex} \begin{proof} \end{proof} \begin{ex}{R.1.11} \end{ex} \begin{proof} \end{proof} \begin{ex}{R.1.4} \end{ex} \begin{proof} \end{proof} \subsection{Exercises Chapter 2} \begin{ex}{R.2.9} $0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$. \end{ex} \begin{proof} Denote the generators of $L$ and $N$ respectively as \begin{align*} \{l_1, \ldots, l_k \} &\subseteq L\\ \{n_1, \ldots, n_p \} &\subseteq N \end{align*} By s.e.s. definition, \begin{itemize} \item[-] $\alpha$ is injective (one-to-one), so $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$ \item[-] $\beta$ is surjective (onto), so $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$ \end{itemize} We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite: Let $m \in M$, then $\beta(m) \in N$, and $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$ Take $m' \in M$, with $m' = \sum a_j y_j$, then $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$ Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus $$(m-m') \in ker(\beta)$$ By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$. Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$. Since $\{l_i\}_k$ generate $L$, $$l = \sum^k b_i l_i$$ thus $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$ Rearrange, $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$ So, $L$ provides $k$ generators for the kernel part of $M$, $N$ provides $p$ "lifts" for the quotient part of $M$; thus $M$ is generated by $k+p$ elements.\\ Thus $M$ is finitely generated over $A$. \end{proof} \subsection{Exercises Chapter 3} \begin{ex}{R.3.2} $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space. Prove that $A$ is Noetherian and Artinian. \end{ex} \begin{proof} $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$. \begin{enumerate} \item Noetherian:\\ let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$. Since each $I_i$ is a subspace, we have $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$ where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Noetherian. \item Artinian:\\ Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$. then $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$ where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Artinian. \end{enumerate} \end{proof} \begin{ex}{R.3.5} Let $0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ an exact sequence. Let $M_1, M_2 \subseteq M$ be submodules of $M$. Prove if the following holds or not: $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$ \end{ex} \begin{proof} Counterexample showing that it does not hold: Let $K$ a field, $M = K \oplus K~, L=K,~N=K$. Set, for $l \in L,~ (m_1, m_2) \in M$, \begin{align*} \alpha:~ &l \longmapsto (l, 0)\\ \beta:~ &(m_1, m_2) \longmapsto m_2 \end{align*} So we have $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$ Then, \begin{align*} M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\ M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)} \end{align*} (Geometric interpretation: $M_1,~ M_2$ are the \emph{diagonal line} and \emph{y-axis} respectively; and $\alpha,~\beta$ capture information about the \emph{vertical} components (x-axis, y-axis respectively), but not about the \emph{diagonal} way a submodule is embedded in $M$). Then, \begin{align*} \beta(M_1) &= \{ x ~|~ x \in K\} = K \\ \beta(M_2) &= \{ x ~|~ x \in K\} = K \end{align*} thus, $\beta(M_1) = \beta(M_2)$. For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\ for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\ thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$. So we've seen that \begin{align*} \beta(M_1) = \beta(M_2)\\ \alpha^{-1}(M_1)=\alpha^{-1}(M_2) \end{align*} while having $M_1 \neq M_2$. \end{proof} \begin{ex}{R.3.3} Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring. Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian. \end{ex} \begin{proof} \begin{enumerate}[i.] \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian. $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian. \item Take the canonical homomorphism $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$ by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$. $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$. Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective. Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$. We know that any submodule of a Noetherian module is Noetherian, thus, since \begin{itemize} \item $A/I_i$ is Noetherian by hypothesis of the exercise \item $A \cong im(\phi)$ \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$ \end{itemize} then, $A$ is Noetherian. \end{enumerate} \end{proof} \begin{ex}{R.3.4} Prove that if A is a Noetherian ring and M a finite A-module, then there exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$. That is, M has a presentation as an A-module in terms of finitely many generators and relations. \end{ex} \begin{proof} since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$. Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$: \begin{align*} \beta: A^p &\longrightarrow M\\ (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i \end{align*} Let $K=ker(\beta)$. By the 1st Isomorphism Theorem, $$M \cong A^p / K$$ Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module. Every submodule of a Noetherian module is fingen. $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen. Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$. Define $\psi: A^q \longrightarrow K$. Compose it with the inclusion map $i: K \longrightarrow A^p$, $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$ So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where \begin{itemize} \item $\beta$ is surjective \item $im(\alpha)=ker(\beta)$ \end{itemize} so that it is a exact sequence, thus, $M$ has a finite presentation. \end{proof} \bibliographystyle{unsrt} \bibliography{commutative-algebra-notes.bib} \end{document}