\documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsfonts} % \usepackage{yfonts} % WIP \usepackage{amsthm} \usepackage{amsmath} \usepackage{enumerate} \usepackage{hyperref} \usepackage{amssymb} \usepackage{tikz} % diagram \usepackage{changepage} % for adjustwidth \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{innerremark}{Remark} \newenvironment{remark}[1] {\renewcommand\theinnerremark{#1}\innerremark} {\endinnerremark} \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{cor}{1.8} \label{1.8} $A = A^{\times} \sqcup \bigcup m$ (where $\sqcup$ denotes ``disjoint union"), ie. $f \in A$ is either a unit or it is contained in a maximal ideal, not both. \end{cor} \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}{1.13}[local ring] \label{1.13} A ring is \emph{local} if it has a unique maximal ideal. Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$: $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$ \vspace{0.3cm} By Corollary \ref{1.8}, $A$ is local\\ \hspace*{4em}$\Longleftrightarrow$ $A$ has only one maximal ideal.\\ \hspace*{4em}$\Longleftrightarrow$ all the nonunits of $A$ form an ideal. \end{defn} \begin{prop}{(R/m=Field)}\label{maximal-quotient-field} Let $R$ be a commutative ring with identity, and let $m \subsetneq R$ be a maximal ideal. Then the quotient ring $R/m$ is a field. \end{prop} \begin{proof} Let $x + m$ be a nonzero element of $R/m$. Since $$x + m \neq m$$ we have $x \not\in m$. Consider the ideal $m + (x)$, where $(x)$ is the principal ideal generated by $x$. Since $x \not\in m$, this ideal strictly contains $m$: $$m \subsetneq m + (x)$$ Because $m$ is maximal, there are no proper ideals strictly between $m$ and $R$. Therefore, $m + (x) = R$ In particular, $1 \in m + (x)$. Hence, there exist $a \in m$ and $y \in R$ such that $a + yx = 1$. Passing to the quotient modulo $m$, we obtain $$ (y+m)(x+m) = yx+m = 1+m, $$ because $a \in m$ implies $a+m=m$. Thus, $y+m$ is the multiplicative inverse of $x+m$. Therefore, every nonzero element of $R/m$ is invertible, and hence $R/m$ is a field. \end{proof} \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 either $r(X)=0$ 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} \subsection{Hilbert basis} \begin{thm}{R.3.6}[Hilbert basis theorem] \label{hilbert-basis} If $A$ a Noetherian ring, then so is the polynomial ring $A[x]$. ie. Every ideal in a polynomial ring over a field has a finite set of generators (is fingen). \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 $a_{m, j}$. $$\Longrightarrow~~ \{ f_{m,j} \}_{m0$ we have two cases: \begin{itemize} \item[-] $y_1, \ldots, y_n$ are algebraically independent over $K$, then by definition \ref{R.4.6.D} $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself, with $m=n$. \item[-] $y_1, \ldots, y_n$ are algebraically dependent over $K$, $$\exists~ 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$ \end{itemize} Want $f$ to be \emph{monic}, so that $y_n$ is integral over new defined variables $y_1^*, \ldots, y_{n-1}^*$. In other words, want some polynomial like $$y_n^d+ a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0~~~~~~~a_i \in K[y_1^*, \ldots, y_{n-1}^*]$$ \hspace*{2em} ie. monic, so that by definition (\ref{R.4.1}), $y_n$ is integral over $K[y_1^*, \ldots, y_{n-1}^*]$. $~\longrightarrow~$ Change variables so that $f$ becomes monic in one of the variables ($y_n$); this allows to express one generator ($y_n$) as an integral element over the others. \vspace{0.3cm} Following from Lemma \ref{R.4.6.L}, define the new variables $y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over $$A^* = K[y^*_1, \ldots, y^*_{n-1}], ~~\text{and}~ A=A^*[y_n]$$ Setting $y_i^* = y_i - y_n^{r_i}$, so that $y_i = y_i^* + y_n^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$. Use those new variables at $f(y_1, \ldots, y_n)=0$: $$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_{n-1}^* + y_n^{r_{n-1}}, y_n) = 0$$ Then the highest power of $y_n$ in each term of $f$ will look like $y_n^{(\sum a_i r_i)}$, and with $r_i$ growing fast enough we ensure that each monomial in $f$ produces a unique power of $y_n$. Then we have $c \cdot y_n^D + \text{(terms with lower powers of $y_n$)} = 0$ with $c \in K \setminus \{0\}$. So that dividing by $c$ we get the shape $y_n^D + \ldots =0$, thus $y_n$ is integral over $A^* = K[y_1^*, \ldots, y_{n-1}^*]$. \vspace{0.3cm} Induction:\\ Since $y_n$ integral over $A^* ~~\Longrightarrow~ A=A^*[y_n]$ is finite over $A^*=K[y_1^*, \ldots, y_{n-1}^*]$ (by \ref{integral-implies-finite}). 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]$. Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required. \end{proof} \vspace{0.5cm} \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 asymptotic line. \end{eg} \vspace{1cm} \subsection{Weak Nullstellensatz} \begin{prop}{R.4.9}\label{R.4.9} let $A \subset B$ be an integral extension of integral domain,\\ then $A$ is a field $\Longleftrightarrow~ B$ is a field. \end{prop} \begin{proof} $\Longrightarrow$:\\ let $0 \neq x \in B$, then $\exists~~ x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0 ~~~~ a_i \in A$, monic. Since $A$ is a field, $\exists$ inverse, observe that: \begin{align*} x^n &+ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\\ x(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = - a_0\\ -a_0^{-1}(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = x^{-1} \in B \end{align*} thus there exists inverse in $B$, so $B$ is a field too. $\Longleftarrow$:\\ if $B$ is a field and $0 \neq x \in A$, then $x^{-1} \in B$, so $x^{-1}$ is integral over $A$. So there is a relation of the form $$(x^{-1})^n + a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 =0$$ Therefore \begin{align*} (x^{-1})^n &+ a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 = 0\\ (x^{-1})^n &= -a_{n-1} (x^{-1})^{n-1} - \ldots - a_0\\ x^{-n} &= -a_{n-1} x^{-n+1} - \ldots - a_0 ~~\text{(mult by $x^{n-1}$)}\\ x^{-n+(n-1)} &= -a_{n-1} x^{-n+1+(n-1)} - \ldots - a_0 x^{n-1}\\ x^{-1} &= -a_{n-1} - \ldots - a_0 x^{n-1} \in A \end{align*} thus there exists inverse in $A$, so $A$ is a field too. \end{proof} \begin{thm}{4.10.prev}[Zariski's lemma] \label{zariski} $k$ a field, $L$ fingen $k$-algebra and a field. Then $L$ is a finite algebraic extension of $k$. If $k$ is algebraically closed, then $L=k$. \end{thm} \vspace{0.4cm} From Zariski's lemma we can see that:\\ let maximal ideal $m \subset k[X_1, \ldots, X_n]$, then $k[X_1, \ldots, X_n]/m$ is a (quotient) field. It's a fingen $k$-algebra, so by Zariski's lemma, $k[X_1, \ldots, X_n]/m$ is algebraic over $k$. Since $k[X_1, \ldots, X_n]/m$ is algebraically closed, it must equal $k$; ie. $$\frac{k[X_1, \ldots, X_n]}{m}=k$$ Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, we have the natural map $$\psi: k \longrightarrow \frac{k[X_1, \ldots, X_n]}{m}$$ so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $k[X_1, \ldots, X_n]/m$ must be an element of $k$).\\ Then, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$. Thus, $$a_i = X_i +m$$ $$\Longrightarrow X_i - a_i \in m$$ % So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$. % $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$ % $$\Longrightarrow x_i - a_i \in m$$ The ideal generated by these terms is a subset of $m$: $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$ Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown also at \ref{5.2}). \vspace{0.4cm} Then: For $k$ algebraically closed, every maximal ideal of $k[X_1, \ldots, X_n]$ has the form $(X_1 - a_1, \ldots, X_n - a_n)$ for some $a_i \in k$. Equivalently, $V(I) = \emptyset ~~\Longleftrightarrow 1 \in I$. \vspace{0.4cm} \begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski2} let $k$ a field, $K$ a $k$-algebra which \begin{enumerate} \item is finitely generated as a $k$-algebra \item is a field \end{enumerate} Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field extension. That is, $[K:k] < \infty$. \end{thm} \begin{proof} $K=k[z_1, \ldots, z_m]$ a field; want to show that $K$ is algebraic over $k$. \vspace{0.3cm} Since $K$ is a fingen $k$-algebra, by Noether normalization lemma (\ref{noether-normalization}),\\ $\exists~ z_1, \ldots, z_m \in K$ such that \begin{itemize} \item are algebraically independent \item $K$ is integral over the polynomial ring $A=k[z_1, \ldots, z_m]$ (which by \ref{integral-implies-finite} is finite) \end{itemize} \\ Now we're at the situation of \ref{R.4.9}: $A \subset K$ is integral, $K$ is a field $~~\Longrightarrow~$ therefore $A$ is a field. \vspace{0.2cm} Since $z_1, \ldots, z_m \in K$ are algebraically independent,\\ \hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$ (since in $k[z_1]$ the element $z_1$ is not invertible, since $1/z_1$ is a rational function, not a polynomial). So $A=k$; which by Noether normalization we saw that $K$ is integral over $A=k$, and by \ref{integral-implies-finite} that it is finite, thus $K$ is finite over $k$, ie. $[K:k],\infty$, and $K$ is algebraic over $k$. \end{proof} \vspace{1cm} \section{Nullstellensatz} \subsection{Corollary from Weak Nullstellensatz} Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue field $K=k[X_1, \ldots, X_n]/m$ satisfies the Zariski's lemma (\ref{zariski}), thus $K$ is a finite algebraic extension of $k$. \vspace{0.3cm} \begin{cor}{5.2} \label{5.2} $k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots, X_n]$ is of the form $$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$ The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$. Thus \begin{align*} k^n &\longleftrightarrow m-Spec~ A\\ (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n) \end{align*} \end{cor} \begin{proof} let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal. By fundamental property of maximal ideals, $K=A/m$ is a field. Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $\psi(x_i) = x_i' = x_i +m$ (where $\psi: k \longrightarrow A/m$). By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$. Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\ \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$. \vspace{0.3cm} Recall that since $A$ is a $k$-algebra, we have the natural map $\psi: k \longrightarrow A/m$, so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $A/m$ must be an element of $k$).\\ Also, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$. Thus, $$a_i = X_i +m$$ $$\Longrightarrow X_i - a_i \in m$$ The ideal generated by these terms is a subset of $m$: $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$ Since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$ \vspace{0.3cm} Let \begin{align*} \phi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\ \phi: x_i &\longmapsto a_i \end{align*} Which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$. \vspace{0.2cm} Since $\phi$ is a $k$-algebra homomorphism, then $\forall f \in A$: $$\phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n))= f(a_1, \ldots, a_n)$$ Thus there is a one-to-one correspondence: points in $k^n ~~~ \longleftrightarrow~~~ m-Spec~ A$ (maximal ideals in $k[X_1, \ldots, X_n]$ $(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$ \end{proof} \vspace{0.5cm} \subsection{Variety} \begin{defn}{5.3}[Variety] A \emph{variety} $V \subset k^n$: $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$ \begin{itemize} \item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$ \item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations. \end{itemize} \end{defn} \begin{prop}{5.3} \label{5.3} $k$ an algebraically closed field, and $A=k[X_1, \ldots, X_n]$ a fingen $k$-algebra of the form $A=k[X_1, \ldots, X_n]/J$, where $J$ is an ideal of $k[X_1, \ldots, X_n]$. (notation: $x_i = X_i \pmod J$) Then every maximal ideal of $A$ is of the form $$(x_1 -a_1, \ldots, x_n - a_n)$$ for some point $(a_1, \ldots, a_n) \in V(J)$. Therefore, $\exists$ a one-to-one correspondence \begin{align*} V(X) &\longleftrightarrow m-Spec~ A\\ \text{given by}~~~(a_1, \ldots, a_n) &\longleftrightarrow (x_1 -a_1, \ldots, x_n - a_n) \end{align*} \end{prop} \begin{proof} the ideals of $A$ are given by ideals of $k[X_1, \ldots, X_n]$ containing $J$, since for $Q=R/I$, $\exists$ one-to-one correspondence between ideals of $Q$ and ideals of $R$ that contain $I$, ie. \begin{align*} m \subset A \longleftrightarrow &m' \subseteq k[X_1, \ldots, X_n]\\ &\text{s.th.}~ J \subseteq m' \end{align*} Thus, every maximal ideal of $A$ is of the form $$\underbrace{(x_1 -a_1, \ldots, x_n - a_n)}_{i.} ~\text{such that}~ \underbrace{J \subset (X_1 - a_1, \ldots, X_n - a_n)}_{ii.}$$ \begin{enumerate}[i.] \item Since $k$ is algebraically closed, the maximal ideals of $k[X_1, \ldots, X_n]$ look like $m=(X_1 - a_1, \ldots, X_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$. Which when projected to the quotient ring $A$, $X_i \longmapsto x_i$ (residue class), giving $(x_1 - a_1, \ldots, x_n - a_n)$. \item for $m$ to exist in $A$, the corresponding $m'$ must contain $J$; since if it didn't contain $J$ it wouldn't "survive" the quotient process. \end{enumerate} \vspace{0.3cm} However, since $(X_1 - a_1, \ldots, X_n - a_n)$ is the kernel of the evaluation map $f \longmapsto f(a_1, \ldots, a_n)$ $\Longrightarrow~$ means that $m'=(X_1 - a_1, \ldots, X_n - a_n)$ consists of all polynomials that vanish at point $P=(a_1, \ldots, a_n)$. If $J \subseteq m'$, then $\forall~ f \in J$ must vanish at $P$. By definition, the set of points where all polynomials in $J$ vanish is the \emph{variety}, $V(J)$. \vspace{0.4cm} Thus,\\ every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie. $$m-Spec~ A \longleftrightarrow k^n$$ The condition that the ideal belongs to the quotient ring $A=k[X_1, \ldots, X_n]/J$ forces that point to lie in $V(J)$, so \begin{align*} m-Spec~ A &\longleftrightarrow V(J)\\ \text{maximal spectrum} &\longleftrightarrow \text{variety} \end{align*} \end{proof} \begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5} A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$). So $V$ gives a map:\\ $$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$ correspondence going the other way: $$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$ defined by taking a subset $X \subset k^n$ into the ideal $$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$ \vspace{0.4cm} $V,~I$ satisfy reverse inclusions: $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$ \end{prop} \vspace{0.4cm} \subsection{Nullstellensatz} \vspace{0.5cm} \begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz} Let $k$ algebraically closed field. \begin{enumerate}[a.] \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$ \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$, $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$ \end{enumerate} \end{thm} \begin{proof} \begin{enumerate}[a.] \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\ Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\ Now, \begin{itemize} \item[-] since $m$ maximal, $L= k[X_1, \ldots, X_n]/m$ is a field. \item[-] since $k[X_1, \ldots, X_n]$ is a fingen $k$-algebra, $L$ is a fingen $k$-algebra. \end{itemize} $\Longrightarrow~$ thus $L$ is a finite field extension of $k$. (Recall: if $k$ algebraically closed and $L$ a fingen field extension of $k$, then $L \cong k$.) Therefore, $$L=\frac{k[X_1, \ldots, X_n]}{m} \cong k$$ Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, $\exists$ a surjective homomorphism $$\psi: k[X_1, \ldots, X_n] \longrightarrow k$$ Let $a_i = \psi(x_i)$. Then $x_i - a_i \in ker(\psi) = m ~\forall~ i \in [n]$. Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$. (as in \ref{5.2}) Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$, $$f \in m ~\Longleftrightarrow~ f(P)=0$$ Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$, ie. every element of $J$ vanishes at $P$.\\ $\Longrightarrow~$ therefore $P \in V(J)$, and thus $V(J) \neq \emptyset$. \item $I(V(J)) = rad J$:\\ \begin{align*} I(V(J)) &= rad J\\ \text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety} \end{align*} where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$. Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$. Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$ Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie. $$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$ Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$ a contradiction. Therefore, $V(J') = \emptyset$. \vspace{0.4cm} Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\ \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$. Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$ $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$ which, since $1 \in J'$, $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$ substitute $Y=1/f$, $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$ thus $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$ multiply by $f^n$, $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$ thus $f^n$ is a linear combination of $g_i$.\\ Thus $f^n \in J$, so $f \in rad~J$. \end{enumerate} \end{proof} \vspace{0.4cm} \subsection{Irreducible varieties} \begin{defn}{5.7}[Irreducible variety] a variety $X \subset k^n$ is \emph{irreducible} if it is nonempty and not the union of two proper subvarieties; that is, if $$X = X_1 \cup X_2 ~~\text{for varieties}~ X_1, X_2 ~\Longrightarrow~ X= X_1 ~\text{or}~ X_2$$ \end{defn} \begin{prop}{5.7} \label{5.7} a variety $X$ is irreducible iff $I(X)$ is prime. \end{prop} \begin{proof} set $I=I(X)$. if $I$ not prime, then $f,g \in A \setminus I$ be such that $fg \in I$. Define new ideals $$J_1=(I, f) ~~\text{and}~~ J_2=(I, g)$$ Then, since $f \not\in I(X)$, it follows that $V(J_1) \subsetneq X$ \hspace*{2em}$\Longrightarrow$ so $X=V(J_1) \cup V(J_2)$ is reducible. The converse is similar. \end{proof} \begin{cor}{5.8} \label{cor.5.8} let $k$ algebraically closed field. Then $V$ and $I$ induce one-to-one correspondences \begin{align*} \{ \text{radical ideals}~J~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{varieties}~ X \subset k^n \}\\ \{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \} \end{align*} Therefore, $$Spec~ k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$ \end{cor} \begin{prop}{5.8} \label{5.8} let $A= k[x_1, \ldots, x_n]$ a fingen $k$-algebra ($k$ an algebraically closed field). Write $J$ for the ideal of relations holding between $x_1, \ldots, x_n$, so that $A=k[X_1, \ldots, X_n]/J$. Then there is a one-to-one correspondence $$Spec~ A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$ \end{prop} \begin{proof} By definition, $Spec~ A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$. By Corollary \ref{cor.5.8}: $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} \longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$ About varieties: \begin{itemize} \item $I(X)$ in $R=k[X_1, \ldots, X_n]$ is the ideal of the variety $X$\\ \hspace*{2em}ie. the set of all polynomials that vanish on every point of $X$. \item $I(X)$ in $A=k[X_1, \ldots, X_n]/J$, we're not looking at all possible polynomials but at the residue classes. \end{itemize} If $P$ is prime in $A~~\Longrightarrow~$ it must correspond to some prime ideal $\mathfrak{P}$ in $R$ (also $J \subset R$). Then from Nullstellensatz (\ref{nullstellensatz}), every prime ideal $\mathfrak{P}$ in $R$ is the ideal of some irreducible variety $X$.\\ \hspace*{2em}$\Longrightarrow~~ \mathfrak{P}=I(X)$. Since we're restricted to the ring $A=k[X_1, \ldots, X_n]/J$, the ideal $P$ are the elements of $I(X)$ viewed through the lens of the quotient\\ $$\Longrightarrow~~ P = \{ f+J ~|~ f \in I(X) \}~~ = I(X) ~\text{mod}~J$$ Now, $J$ is the set of equations defining our "universe" $V(J)$. Since $A=R/J ~~ \Longrightarrow~$ we thus have $J \subseteq R$. Also we have a correspondence between $A=R/J$ and $R$. Let $\mathfrak{P}$ be the preimage of $P$ in $R=k[X_1, \ldots, X_n]$, ie. \begin{align*} R &\longrightarrow A=R/J\\ \mathfrak{P} &\longmapsto P\\ I(X) &\longmapsto I(X) ~\text{mod}~J \end{align*} $\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$. Now, $J \subset \mathfrak{P} = I(X)$;\\ by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so $$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$ $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$. Therefore, $$\mathfrak{P} \in Spec~ A \longleftrightarrow X \subseteq V(J)$$ where $X = V(\mathfrak{P})$. \end{proof} \subsection{Again: Zariski, Weak Nullstellensatz, Nullstllensatz} This subsection proves again Zariski's lemma, Weak Nullstellensatz and Nullstellensatz, in a more interconnected way than the proofs described previously. \begin{lemma}{}[Zariski's lemma] $k$ a field, $K$ a fingen $k$-algebra and a field. Then $K$ is a finite algebraic extension of $k$. If $k$ algebraically closed, then $K=k$. \end{lemma} \begin{proof} $K$ a fingen $k$-algebra $\Longleftrightarrow$ $\exists~ x_1,\ldots,x_n\in K$ such that $$K=k[x_1,\ldots,x_n].$$ Want to show that $K$ is algebraic over $k$. Since $K$ is a fingen $k$-algebra, by Noether normalization \ref{noether-normalization},\\ $\exists~ \{y_1, \ldots, y_r\} \subset K$ such that \begin{itemize} \item the set $\{y_1, \ldots, y_r\}$ is algebraically independent over $k$ \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \subset K$. \end{itemize} By \ref{R.4.9}:\\ if $A \subset B$ integral extension of integral domains,\\ then $A$ is a field $\Longleftrightarrow~ B$ is a field. which applied to our case: $$B:=K ~\text{(a field)},~~ A:= R=k[y_1, \ldots, y_r]$$ since $K$ is integral over $R$ and $K$ a field $\Longrightarrow$ by \ref{R.4.9} $R$ must be a field. \vspace{0.3cm} Since $y_i \in K$ are algebraically independent\\ $\Longrightarrow~ R=k[y_1, \ldots, y_r]$ is a polynomial ring in $r$ indeterminates. Then, if $r \geq 1$, we have at least one variable $y_1$ (with $k[y_1]$);\\ but the only invertible elements in a polynomial ring over a field are the non-zero constants of the field $k$, so $y_1$ is not invertible in $R$. (alternatively: the element $y_1$ is not invertible in the polynomial ring, since $1/y_1$ is a rational function, not a polynomial). Therefore, $R=k[y_1, \ldots, y_r]$ can be a field iff $r=0$, and we know that it is a field, therefore $r=0$ and $R=k$. By Noether normalization, at the previous step we knew that $K$ is integral over $R$ ($=k$). \vspace{0.3cm} By \ref{integral-implies-finite}, a fingen algebra that is integral over its base field is a finite vector space over that field. Thus $[K:k]<\infty$, and $K$ is algebraic over $k$. If $k$ is algebraically closed, every element $\alpha\in K$ is algebraic over $k$, and its minimal polynomial over $k$ must have degree one. Therefore $\alpha\in k$, and thus $K=k$. \end{proof} \vspace{0.5cm} This brings us to the \emph{Weak Nullstellensatz}: \begin{thm}{4.10}[Weak Nullstellensatz]\label{weak-nullstellensatz-again} $k$ a field, $K$ a $k$-algebra with \begin{itemize} \item $K$ fingen as a $k$-algebra \item $K$ a field \end{itemize} Then $K$ is finite-dimensional over $k$, and hence algebraic over $k$; that is $[K:k] < \infty$. If $k$ is algebraically closed, then $K=k$. \end{thm} \begin{proof} proven at previous proof. \end{proof} \vspace{0.8cm} Now, assume that $k$ is algebraically closed, and let $m\subset k[X_1, \ldots, X_n]$ be a maximal ideal. Let $A=k[X_1, \ldots, X_n]$. Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{A}{m}$ is a field (by basic ideal quotient theory \ref{maximal-quotient-field}). Moreover, $\frac{A}{m}$ is a fingen $k$-algebra, generated by the classes $$X_1+m,\ldots,X_n+m.$$ Therefore, by Zariski's lemma, $\frac{A}{m}$ is algebraic over $k$. And since $k$ is algebraically closed $\Longrightarrow~ \frac{A}{m}\cong k$. Let $$\pi:A\longrightarrow \frac{A}{m}$$ be the quotient map. By definition, $\pi$ is the surjective homomorphism that sends each polynomial to its coset modulo $m$, and the kernel of this map is exactly $ker(\pi)=m$. By Zariski's lemma, $\frac{A}{m}$ is a finite field extension of $k$, and since $k$ is algebraically closed, by Zariski's lemma, this extension must be $k$ itself; denote this relation by $$\theta:\frac{A}{m}\stackrel{iso}{\longrightarrow} k$$ the $k$-algebra isomorphism. Define the composite map $$\phi=\theta\circ\pi:A\longrightarrow k.$$ which is a $k$-algebra homomorphism. Thus, $\forall~ i\in[n]$, define $$a_i=\phi(X_i)\in k.$$ Since $\phi$ is a $k$-algebra homomorphism, it must map any scalar $c \in k$ to itself, ie. $\phi(c) = c$. Therefore $$\phi(X_i - a_i) = \phi(X_i) - \phi(a_i) = a_i - a_i = 0$$ Since $\phi = \theta \circ \pi$, and $\theta$ is an isomorphism (by definition injective), the only way for $\phi(f)$ to be $0$ is if $\pi(f)=0$: \begin{align*} &\phi(X_i-a_i)=\theta(\pi(X_i - a_i)) = 0\\ &\Longrightarrow \pi(X_i-a_i)=0\\ \end{align*} Since $ker(\pi)=m$, any element that maps to $0$ under $\pi$ must belong to $m$, thus $$\pi(X_i - a_i) =0 ~~\Longrightarrow~ X_i-a_i\in m$$ Alternative way: by definition of a quotient ring, and elements of $\frac{A}{m}$ are cosets of the form $f+m$. $\pi(x_i)$ sends $X_i$ to the coset $X_i+m$, ie. $\pi(x_i)=X_i+m$. Thus, if $a_i$ is the image of $X_i$, then $$X_i \equiv a_i \pmod m,$$ which by the definition of congruence modulo ideal, this means $X_i -a_i \in m$. \vspace{0.5cm} Since $X_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$, since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal. \begin{adjustwidth}{4em}{0pt} (Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$. \vspace{0.4cm} We had $k \stackrel{iso}{\cong} \frac{A}{m} = \frac{k[X_1, \ldots, X_n]}{m}$. We have the map \begin{align*} \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\ X_i &\mapsto a_i \in k \end{align*} which is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$. Since $\phi$ is a $k$-algebra homomorphism, then $$\forall~ f \in A,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(X_1), \ldots, \phi(X_n)) = f(a_1, \ldots, a_n).$$ Thus, $\exists$ a one-to-one correspondence: \begin{align*} \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\ (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n) \end{align*} \end{adjustwidth} \vspace{0.5cm} Now let $J\subsetneq A$ be an arbitrary proper ideal. Since every proper ideal is contained in a maximal ideal, $\exists$ a maximal ideal $m$ such that $$J\subseteq m$$ We have seen that $$m=(X_1-a_1, \ldots, X_n-a_n)$$ for some $P=(a_1,\ldots,a_n)\in k^n$. So $P$ is a zero for every polynomial in $m$, ie. $$\forall~ f \in m ~\Longrightarrow~ f(P)=0.$$ Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$. $\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states: \begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz-again} Let $k$ be an algebraically closed field and let $J\subseteq k[X_1,\ldots,X_n]$ be an ideal. \begin{enumerate}[a.] \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$ \item $I(V(J)) = rad(J)~~(=\sqrt{J})$, in other words, for $f \in k[X_1, \ldots, X_n]$, $$f(P)=0 ~\forall~ P \in V(J) ~~\Longleftrightarrow~~ f^N \in J ~\text{for some $N\geq 1$.}$$ \end{enumerate} \end{thm} \begin{proof} \begin{enumerate}[a.] \item already proven in the previous steps. \item $I(V(J)) = \sqrt{J}$\\ ie. vanishing ideal of a variety = radical of the ideal defining the variety where $$\sqrt{J} = \{ f \in A ~|~ f^N \in J ~\text{for some}~ N>0 \}.$$ First, want to show that $$\sqrt{J}\subseteq I(V(J)).$$ Let $f\in\sqrt{J}$. Then $f^N\in J$ for some $N\geq 1$. Thus, for every $P\in V(J)$, $$f(P)^N=f^N(P)=0,$$ and therefore $f(P)=0$. Hence $f\in I(V(J))$, so $$\sqrt{J}\subseteq I(V(J)).$$ Now, want to show that if a polynomial vanishes at all points where $J$ vanishes, then $f \in \sqrt{J}$. Since $k[X_1,\ldots,X_n]$ is Noetherian, write $$J=(g_1,\ldots,g_m).$$ Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$. Suppose there is a point $(a_1, \ldots, a_n,b)$ that is a zero of $J'$. ie. $$\exists~ (a_1, \ldots, a_n,b) \in V(J').$$ Since $$g_i(a_1,\ldots,a_n)=0,$$ our hypothesis says $$f(a_1,\ldots,a_n)=0.$$ However, the last generator $(1-Yf)$ requires $$1 - b f(a_1,\ldots,a_n) = 0 ~~\Longrightarrow~ 1 - b \cdot 0 = 0 ~\Longrightarrow~ 1=0,$$ a contradiction. Therefore, $V(J') = \emptyset$. \vspace{0.4cm} By part a., applied to $k[X_1,\ldots,X_n,Y]$,\\ \hspace*{2em} if $V(J')=\emptyset$ then $J'=k[X_1,\ldots,X_n,Y]$, so $1 \in J'$. Every element in an ideal is a linear combination of its generators:\\ $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$ $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf).$$ which, since $1 \in J'$, $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X)).$$ Now work in the localization $$A_f=A[f^{-1}],$$ where $f$ is invertible, and apply the homomorphism $$A[Y]\longrightarrow A_f,\qquad Y\longmapsto f^{-1}.$$ Then $$1= \left( \sum_{i=1}^m p_i(X,f^{-1}) g_i(X) \right) +q(X,f^{-1})\cdot \underbrace{(1-f^{-1}f(X))}_{0},$$ thus $$1= \sum_{i=1}^m p_i(X,f^{-1}) g_i(X).$$ Since there are finitely many polynomials $p_i$, $\exists~ N\geq 0$ such that $$f^N p_i(X,f^{-1})\in A ~~\forall~ i.$$ Multiplying by $f^N$, $$f^N= \sum_{i=1}^m A_i(X) g_i(X),$$ where $$A_i(X)=f^N p_i(X,f^{-1})\in A.$$ Thus $f^N$ is a linear combination of $g_i$.\\ Therefore $f^N \in J$, so $f \in \sqrt{J}$. Hence $$I(V(J))\subseteq\sqrt{J}.$$ Combining both inclusions, $$I(V(J))=\sqrt{J}.$$ \end{enumerate} \end{proof} \section{Rings of fractions $S^{-1}A$ and localization} \subsection{Small recap} \paragraph{principal ideal:} generated by a single element, ie. for $a \in A$, $(a)=aA=\{ra ~|~ r\in A\}$. \paragraph{unit:} $u \in A$ is a unit if $\exists v \in A$ with $uv=1$. \paragraph{proper ideal:} $I \subsetneq A$, equivalently $1 \not\in I$. Therefore, a proper ideal does not have units. \paragraph{maximal ideal:} $\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$. \begin{itemize} \item is proper \item for $m$ maximal ideal, $\frac{A}{m}$ is a field. \item every maximal ideal is prime \item elements of a maximal ideal are always non-units\\ \hspace*{2em} since if $x \in m$ were a unit, then $1=x^{-1}x \in m$, a contradiction since if $1 \in m$ then $m=A$, which would mean that it is not proper; and it is maximal, hence proper. \end{itemize} \paragraph{prime ideal:} proper ideal $P$ such that $$ab \in P ~~\Longrightarrow~ a \in P ~\text{or}~ b \in P$$ for $P$ prime, $\frac{A}{P}$ is an integral domain. \paragraph{Spec(A):} prime spectrum, $Spec(A)=\{ P \subset A ~|~ P ~\text{is prime}\}$ \vspace{0.6cm} \subsection{Rings of fractions $S^{-1}A$} \begin{defn}{6.1}[ring of fractions] \label{def.6.1} let $A$ a ring, $S \subset A$ a multiplicative set ($1 \in S$, and $st \in S ~\forall s, t \in S$). Introduce th following relation $\sim$ on $A \times S$: $$(a, s) \sim (b, t) \Longleftrightarrow \exists y \in S ~\text{such that}~ u(at - bs) = 0$$ (write $a/s$ for the equivalence class of $(a, s)$. Then, the \emph{ring of fractions of $A$ with respect to $S$} is $$S^{-1}A = (A \times S)/\sim$$ with ring op'ns defined by the usual arithmetic op'ns on fractions: $$\frac{a}{s} \pm \frac{b}{t} = \frac{(at \pm bs)}{st} ~~\text{and}~~ \frac{a}{s} \cdot \frac{b}{t} = \frac{ab}{st}$$ \end{defn} \begin{prop}{6.1} \label{prop.6.1} \begin{enumerate}[i.] \item $\sim$ is an equivalence relation \item the ring op'ns are well defined, and $S^{-1}A$ is a ring \item \begin{align*} \psi: A &\longrightarrow S^{-1}A\\ a &\longmapsto a/1 \end{align*} is a ring homomorphism. \end{enumerate} \end{prop} \begin{lemma}{6.2} \label{6.2} For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$. Then $$A_f \cong \frac{A[X]}{(Xf-1)}$$ \end{lemma} \begin{proof} (adjoining a variable $X$ and imposing $Xf=1$ is the algebraic way of saying "add $f^{-1}$".)\\ Define the homomorphism \begin{align*} \psi: A[X] &\longrightarrow A_f\\ a &\longmapsto a/1\\ X &\longmapsto 1/f \end{align*} By the 1st isomorphism theorem: \begin{tikzpicture}[node distance=1.5cm, auto] \node (G) {$A[X]$}; \node (H) [right of=G] {$A_f$}; \node (GmodK) [below of=G, xshift=0.75cm] {$A[X]/\ker(\psi)$}; \draw[->] (G) to node {$\psi$} (H); \draw[->] (G) to node [swap] {$\phi$} (GmodK); \draw[->] (GmodK) to node [swap] {$\eta$} (H); \end{tikzpicture} Thus we want to prove that $ker(\psi) = (Xf-1)$, so that $\frac{A[X]}{ker(\psi)} = \frac{A[X]}{(Xf-1)}$, and the lemma is proven. \vspace{0.3cm} First, observe that $\psi(Xf-1) = \psi(X)\psi(f)-\psi(1) = \frac{1}{f} \frac{f}{1} - 1 = 1-1=0$, so $Xf-1 \in ker(\psi)$, ie. $(Xf-1) \subseteq ker(\psi)$. Now, we want to prove that $ker(\psi) \subseteq (Xf-1)$. Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \forall~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$. \vspace{0.3cm} Want to prove that $h(X)$ is a multiple of $(Xf-1)$. Let $$h(X)=a_n X^n + a_{n-1} X^{n-1} + \ldots + a_1 X + a_0$$ % Since $h \in ker(\psi) ~~\Longrightarrow~~ \psi(h)=0$, so % $$\psi(h) = \frac{a_n}{f^n} + \frac{a_{n-1}}{f^{n-1}} + \ldots \frac{a_1}{f} + \frac{a_0}{1} = 0 \in A_f$$ multiply $h(X)$ by $f^n$: $$f^n \cdot h(X)=a_n (f^n X^n) + a_{n-1} f (f^{n-1} X^{n-1}) + a_{n-2} f^2 (f^{n-2} X^{n-2}) \ldots$$ Note that since $\forall~ i \geq 1,~~ f^i X^i = (Xf -1)\cdot (f^{i-1} X^{i-1} + f^{i-2} X^{i-2} + \ldots + 1)$, then $f^i X^i \equiv 1 \pmod{Xf-1}$. So, $$f^n \cdot h(X)=\underbrace{a_n (1) + a_{n-1} f (1) + a_{n-2} f^2 (1) + \ldots + a_0 f^n}_{C~~\text{(constant)}} \pmod{Xf-1}$$ $$\Longrightarrow~~ f^n \cdot h(X)=C \pmod{Xf-1}$$ $$\Longleftrightarrow~~ f^n \cdot h(X)=Q(X) \cdot (Xf-1) + C$$ Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of fractions), $\frac{a}{f^n} = 0$ iff $\exists~ k$ such that $f^k \cdot a = 0$ in $A$. So, multiply both sides by $f^k$: $$f^k \cdot f^n \cdot h(X)=f^k \cdot (Q(X) \cdot (Xf-1) + C)$$ $$\underbrace{f^k f^n}_{f^{nk}} \cdot h(X)=\underbrace{f^k Q(X)}_{Q'(X)} \cdot (Xf-1) + \underbrace{f^k C}_{0}$$ $$\Longrightarrow~~ f^{n+k} \cdot h(X)=Q'(X) \cdot (Xf-1)$$ $$\Longleftrightarrow~~ f^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$ multiply it by $X^{n+k}$: $$X^{n+k} \cdot f^{n+k} \cdot h(X) \equiv X^{n+k} \cdot 0 \pmod{Xf-1}$$ $$(Xf)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$ Now, since we had $Xf \equiv 1$ in $\frac{A[X]}{(Xf-1)}$, $$(1)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$ $$\Longrightarrow~~ h(X) \equiv 0 \pmod{Xf-1}$$ By definition this is saying $h(X) \in (Xf-1) ~\forall~ k \in ker(\psi)$. Thus $ker(\psi) \subseteq (Xf-1)$. Initially we saw that $(Xf-1) \subseteq ker(\psi)$. Therefore $ker(\psi)=(Xf-1)$. Hence, $$A_f \cong \frac{A[X]}{(Xf-1)}$$ \end{proof} \vspace{0.4cm} Given a ring homomorphism $\psi: A \longrightarrow B$, there is a correspondence $$e: \{ \text{ideals of } A \} \longrightarrow \{ \text{ideals of } B \}$$ given by $e(I) = \psi(I)B = IB$ (called \emph{extension}), and $$r: \{ \text{ideals of } B \} \longrightarrow \{ \text{ideals of } A \}$$ given by $r(J) = \psi^{-1} J$ (called \emph{restriction}, written $A \cap J$). Set $B= S{-1}A$. Then $S^{-1}I = e(I) = \psi(I)B$. \begin{prop}{6.3} \label{prop.6.3} \begin{enumerate}[a.] \item $\forall$ ideal $J$ of $S^{-1}A$, $e(r(J))=J$ \item $\forall$ ideal $I$ of $A$, $$r(e(I)) = \{ a \in A ~|~ as \in I ~\text{for some}~ s \in S \}$$ \item if $P$ prime and $P \cap S = \emptyset$,\\ then $e(P)=S^{-1}P$ is a prime ideal of $S^{-1}A$. \end{enumerate} \end{prop} \begin{proof} \end{proof} \begin{cor}{6.3} \label{cor.6.3} for an ideal $I$ of $A$, the necessary and sufficient condition for $r(e(I))=I$ is $$as \in I \Longrightarrow a \in I~~ \forall s \in S$$ \end{cor} \subsection{Localization} If $P$ prime ideal, then $S=A \setminus P$ is a multiplicative set. Define the set $A_P=S^{-1}A$. \begin{prop}{6.4} \label{6.4} $a/s \in A_P$ is a unit of $A_P ~\Longleftrightarrow~ a \not\in P$. Therefore $A_P$ is a \emph{local ring}, with maximal ideal $e(P)=P A_P$. The local ring $(A_P, PA_P)$ is called the \emph{localization} of $A$ at $P$. \end{prop} \begin{proof} \begin{enumerate} \item[$\Longleftarrow$] (if $a \not\in P ~\Longrightarrow~ a/s \in A_P$ is a unit)\\ if $a \not\in P$, then by definition of $S,~~ a \in S$. In the localization $A_P$, every element of $S$ is invertible. The inverse of $\frac{a}{s}$ is $\frac{s}{a}$\\ \hspace*{2em}$\longrightarrow~~ \frac{a}{s} \cdot \frac{s}{a} = 1$, thus $a/s \in A_P$ is a unit. \item[$\Longrightarrow$] (if $a/s$ unit $~\Longrightarrow~ a \not\in P$)\\ Suppose $a/s$ a unit; then $\exists~ \frac{b}{t} \in A_P$ such that $\frac{a}{s} \cdot \frac{b}{t} = 1$. By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that \begin{equation} u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust \label{eq:6.4} \end{equation} with $u,s,t \in S$. Since $S=A \setminus P$ and $P$ is a prime ideal,\\ the products of elements outside $P$ must also be outside of $P$. $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$ At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$. If $uab \not\in P$, then $a \not\in P$. Next we will prove that $A_P$ is a local ring.\\ A ring is local if it has exactly one maimal ideal (\ref{1.13}); so we show that the set of non-units forms an ideal. $a/s$ is not a unit iff $a \in P$. Let $m= \{ a/s | a \in P,~ s \not\in P \} = PA_P$. Want to show that $PA_P$ is the unique maximal ideal: \begin{itemize} \item it's an ideal: let $\frac{a}{s}, \frac{b}{t} \in PA_P$ with $a,b \in P$. Then $\frac{a}{b} + \frac{b}{t} \in PA_P$ with numerator in $P$. If multiply a fraction in $PA_P$ by any fraction in $A_P$, the numerator stays in $P$, thus $PA_P$ is an ideal. \item every element outside of $PA_P$ is a unit (proven at the beginning of this proposition's proof ($\Longleftarrow$)). \item maximality: every ideal strictly larger than $PA_P$ must contain a unit.\\ If an ideal contains a unit, it must be the entire ring $A_P$.\\ \hspace*{2em}$\Longrightarrow$ therefore, $PA_P$ is the unique maximal ideal. \end{itemize} \end{enumerate} \end{proof} In other words: \emph{Localization}: formal way to 'force' certain elements to have inverses. It's the smallest ring that contains $A$ and makes all the elements of $S$ invertible. There is a natural map $\psi: A \longrightarrow S^{-1}A$, by $f(a)=\frac{a}{1},~ f(s)=\frac{1}{s}~~ \text{for}~ s \in S$. \subsection{Localization commutes with taking quotients} Let $A$ ring, $S$ multiplicative set, $I$ ideal. Write $T$ for the image of $S$ in $A/I$. Then $S^{-1}I=I \cdot S^{-1}A$ is an ideal of $S^{-1}A$, can take the quotient ring $S^{-1}A / S^{-1} I$. $\longleftrightarrow$ can take the quotient $A/I$ and then localize to get $T^{-1}(A/I)$. \begin{cor}{6.7} \label{6.7} $$T^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$ In particular, for $P$ prime ideal, $$\frac{A_P}{PA_P} = k(P) = Frac(\underbrace{A/P}_{\substack{\text{integral}\\\text{domain}}})$$ From \ref{6.4}, $A_P$: local ring, $PA_P$: unique maximal ideal. $k(P)$ and $Frac(A/P)$ are field of fractions. \end{cor} \begin{proof} the quotient ring $A/I$ can be viewed as an $A$-module and $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$ The \ref{cor.6.6}.i, gives an isomorphism of modules $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$ it's easy to see that this is a ring homomorphism. \end{proof} % TODO 6.6 (proposition and corollary) \vspace{1.0cm} \subsection{Key ideas} Main idea: make some choosen elements invertible. For localization, it is important to not get confused between "localization at an element" and "localization at a prime ideal". \paragraph{Localization at an element $f$.} Let $S=\{1, f, f^2, f^3, \ldots \}$ a multiplicative set. Then $S^{-1}A$ is the ring of fractions, and $$A_f= S^{-1}A = A \big[ \frac{1}{f} \big] = \{\frac{a}{f^n} ~|~ a \in A, n \geq 0\}$$ ie. "make $f$ invertible", then $\frac{1}{f} \in A_f$.\\ $\Longrightarrow~ A_f$ is the ring obtained from $A$ by making every $f \in S$ invertible. Lemma \ref{6.2} will show that $A_f \cong \frac{A[X]}{(Xf-1)}$. \paragraph{Localization at a prime ideal P.} For $P \in Spec(A)$ (ie. $P$ is a prime ideal), $$A_P = ( \underbrace{A \setminus P}_{S} )^{-1} A = \{ \frac{a}{s} ~|~ a \in A, s \not\in P \}$$ ie. "make every element of $A$ not in $P$ invertible". Since $p \in P$, then $\frac{1}{p} \not\in A_P$.\\ Given the ideal $I \subseteq A$, its extension to the localized ring is $$I A_P = \{ \frac{a}{s} ~|~ a \in I, s \not\in P \}$$ \vspace{0.6cm} The only nonunits are fractions whose numerator lies in $P$: $\{ \frac{a}{s} ~|~ a \in P, s \not\in P \} = P A_P$ Hence, $A_P$ has one maximal ideal: $P A_P$. $(A_P, P A_P)$ denotes the local ring $A_P$ together with $P A_P$, often called the localization of $A$ at $P$ as a local ring. \vspace{1.0cm} \section{Tensor Product of Modules} \subsection{Constructing the Tensor Product} We use vector spaces but the same applies for modules. Let $Z$ be the vector space with basis $\{ e_{v,w} \} ~\forall~ (v,w) \in V \times W$. Want to introduce the relation $a = b \in Z$. Consider the element $a-b \in Z$; take the span $Span\{a-b\}$, that is, all of multiples of $a-b$. Consider the quotient space $\frac{Z}{Span\{a-b\}}$. Anything in the span of $a-b$ is equal to zero. $\Longrightarrow$ This is how we use relations to get vector spaces out of $Z$. \paragraph{Direct Sum, $V \oplus W$}. Want \begin{enumerate}[i.] \item $(v_1, w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2)$ \item $\lambda (v, w) = (\lambda v, \lambda w)$ \end{enumerate} Take the quotient $$Z / Span \left\{ \substack{ (v_1,w_1)+(v_2,w_2)-(v_1+v_2,w_1+w_2)\\ \lambda(v,w)-(\lambda v,\lambda w) } \right\} $$ \paragraph{Tensor Product, $V \otimes W$} relations: \begin{enumerate}[i.] \item distribuitivity law: $$a (b+c) = ab + ac$$ $$(a+b)c = ac + bc$$ \item scalar multiplication: $$\lambda(a \cdot b) = (\lambda a) b = a (\lambda b)$$ \end{enumerate} $\Longrightarrow$ \begin{align*} &a (b+c) = ab + ac\\ &(a+b)c = ac + bc\\ &\lambda (a \cdot b) = (\lambda a) b\\ &\lambda (a \cdot b)= a (\lambda b) \end{align*} then we can define the tensor product as $$V \otimes W = Z / Span \left\{ \substack{ a \otimes (b+c) - a \otimes b - a \otimes c\\ (a+b) \otimes c - a \otimes c - b \otimes c\\ \lambda (a \otimes b) - (\lambda a) \otimes b\\ \lambda (a \otimes b) - a \otimes (\lambda b) } \right\} $$ \subsection{Universal Property of Tensor Products} \begin{prop}{}[Universal Property of Tensor Products] For every bilinear map $g: V \times W \longrightarrow U$, $\exists !$ linear map $f: T \longrightarrow U$ such that $f \circ \tau = g$. \begin{tikzpicture}[node distance=2.5cm, auto] \node (VW) {$V \times W$}; \node (T) [right of=VW] {$T = V \otimes W$}; \node (U) [below of=VW] {$U$}; \draw[->] (VW) to node {$\tau$} (T); \draw[->] (VW) to node [swap] {$g$} (U); \draw[->, dashed] (T) to node [swap] {$f$} (U); \end{tikzpicture} \end{prop} \begin{proof} \begin{enumerate}[i.] \item uniqueness:\\ suppose we have two different T.P., $(T, \tau)$ and $(T', \tau')$. Want to prove $T \stackrel{iso}{\cong} T'$. On the right is the same relation as on the left, but switching $T$ by $T'$: \begin{minipage}[t]{0.48\textwidth} \centering \begin{tikzpicture}[node distance=2.0cm, auto] \node (VW) {$V \times W$}; \node (T) [right of=VW] {$T$}; \node (T2) [below of=VW] {$T'$}; \draw[->] (VW) to node {$\tau$} (T); \draw[->] (VW) to node [swap] {$\tau'$} (T2); \draw[->, dashed] (T) to node [swap] {$f$} (T2); \end{tikzpicture} \vspace{0.5em} by the U.P., $f \circ \tau = \tau'$. \end{minipage} \hfill \begin{minipage}[t]{0.48\textwidth} \centering \begin{tikzpicture}[node distance=2.0cm, auto] \node (VW) {$V \times W$}; \node (T) [right of=VW] {$T$}; \node (T2) [below of=VW] {$T'$}; \draw[->] (VW) to node {$\tau$} (T); \draw[->] (VW) to node [swap] {$\tau'$} (T2); \draw[->, dashed] (T) to node [swap] {$f$} (T2); \end{tikzpicture} \vspace{0.5em} by the U.P., $f \circ \tau = \tau'$. \end{minipage} \vspace{0.3cm} $$\text{thus,}~~ f \circ f' \circ \tau' = \tau'$$ Want to show that $f$ has an inverse, so that it is an isomorphism $T \stackrel{iso}{\cong} T'$. we just saw that $$f \circ f' \circ \tau' = \tau'$$ also we can get it the other way $$f' \circ f \circ \tau = \tau$$ we can see that $f \circ f'$ seems to be the identity, need more info to prove it. What if we apply the U.P. to the same module on both sides? \begin{tikzpicture}[node distance=2.0cm, auto] \node (VW) {$V \times W$}; \node (T) [right of=VW] {$T$}; \node (T2) [below of=VW] {$T$}; \draw[->] (VW) to node {$\tau$} (T); \draw[->] (VW) to node [swap] {$\tau$} (T2); \draw[->, dashed] (T) to node {$? = id_T$} (T2); \end{tikzpicture} thus, $id_T \circ \tau = \tau$, hence \begin{tikzpicture}[node distance=2.0cm, auto] \node (VW) {$V \times W$}; \node (T) [right of=VW] {$T$}; \node (T2) [below of=VW] {$T$}; \draw[->] (VW) to node {$\tau$} (T); \draw[->] (VW) to node [swap] {$\tau$} (T2); \draw[->, dashed] (T) to node {$id_T,~ f' \circ f$} (T2); \end{tikzpicture} and by uniqueness of the U.P., we can not have two different functions ($id_T$ and $f' \circ f$) and must be that $f' \circ f = id_T$. Similarly for $f \circ f' = id_{T'}$. Therefore, $f'$ is the inverse of $f$, and $T \stackrel{iso}{\cong} T'$, which means that the tensor prodduct is unique. % TODO (ii.) existence \end{enumerate} \vspace{1.0cm} This is stated in the Atiyah and MacDonald book (\cite{am}) for multiple nodes and multilinear maps: \begin{prop}{2.12} Let $M_1, \ldots, M_r$ be $A$-modules. Then $\exists ~ (T, g)$ with $T$ an $A$-module, and $g$ a multilinear map $g: M_1 \times \ldots \times M_r \longrightarrow T$, with the following property: Given any $A$-module $P$ and any $A$-multilinear map $f: M_1 \times \ldots \times M_r \longrightarrow T$, $\exists !$ an $A$-homomorphism $f': T \longrightarrow P$ such that $f' \circ g = f$. Moreover, if $(T,g)$ and $(T',g')$ are two pairs with this property, then $\exists !~ j: T \stackrel{iso}{\longrightarrow} T'$ such that $j \circ g = g'$. \end{prop} \subsection{Usefulness of Tensor Product's Universal Property} Anytime we define a function on the T.P. we need to know that it is well defined and linear. This is hard in general because of all the relations of the elements of the T.P.. But using the U.P. of T.P. we can remove all of the extra work and just look at a bilinear map on the cartesian product, which is a much easier object to work with. \paragraph{Using the U.P. to show that a function on the T.P. is well defined and linear:} \begin{enumerate}[i.] \item define a map $\phi$ on the Cartesian product $$(x,y) \in M \times N$$ \item prove bilinearity of $\phi$:\\ show that $\phi$ is $A$-linear in each variable separately. ie. check \begin{align*} \phi(x + x', y) &= \phi(x, y) + \phi(x', y)\\ \phi(x, y + y') &= \phi(x, y) + \phi(x, y')\\ \phi(ax, y) &= a \phi(x, y) = \phi(x, ay) \end{align*} \item invoke the Universal Property.\\\ Once bilinearity is proven on $\phi$, the U.P. guarantees the existence of a unique map $f$ on the tensor product $M \otimes N$ \item automatic well-definedness and linearity: $f$ is linear by definition. And since it is a single map originating from the bilinear map $\phi$ on the product, it is automatically well-defined on the tensor product (ie. automatically "respects" the relations that define tensors). \end{enumerate} \end{proof} %%%%%%%%%%%%%%%%%%%%%%%% %%%%%% %%%%%% end of notes, %%%%%% beginning of exercises %%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \newpage \section{Exercises} For the exercises, I follow the assignments listed at \cite{mit-course}. The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}. \subsection{Exercises Chapter 1} \begin{ex}{R.1.1} Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$? \end{ex} \begin{proof} Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$. $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\ Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$. Thus $I \cup J$ is not closed under addition; thus is not an ideal. Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum: $$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$$ So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where \begin{itemize} \item $\beta$ is surjective \item $im(\alpha)=ker(\beta)$ \end{itemize} so that it is a exact sequence, thus, $M$ has a finite presentation. \end{proof} \subsection{Exercises Chapter 4} \begin{ex}{R.4.1.a} $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$. \end{ex} \begin{proof} $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts: $$f(X) = p(X^2) + X \cdot q(X^2)$$ with $p(X^2),~ q(X^2) \in k[X^2]$, and\\ $p(X^2)$: sum of all terms with even exponents\\ $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$. \vspace{0.3cm} (Observation: this is used in FRI cryptographic protocol\\ \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf}) Rearrange it \begin{align*} f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\ (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\ f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\ f(X)^2 \underbrace{-[2 p(X^2)]}_{a_1} f(X) &+ \underbrace{[p(X^2)^2 - X^2 \cdot q(X^2)^2]}_{a_0} = 0 \end{align*} Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$. \vspace{0.2cm} The integral dependence relation for any $f \in k[X]$ is given by the monic polynomial from \ref{R.4.1}.ii, in this case $T^2 + a_1 T + a_0 = 0$ with $a_i \in k[X^2]$. We have that \begin{align*} a_1 &= -2 p(X^2)\\ a_0 &= p(X^2)^2 - X^2 q(X^2)^2 \end{align*} So for example, for $f(X)= X^3 + X^2 + X + 1$: \begin{align*} f(X) = (X^2 + 1) &+ X (X^2 + 1)\\ (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\ (f(X) - p(X))^2 &= X^2 (q(X))^2 \end{align*} \end{proof} \begin{ex}{R.4.5} Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$. \end{ex} \begin{proof} $A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$: Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then \begin{align*} (tX)^2 &= X^2 + X^3\\ t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\ t^2 &= 1+X, ~\text{thus}\\ X&=t^2-1 \in k[X] \end{align*} Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$. Hence $X, Y \in k[X]$. Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$. \vspace{0.4cm} By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$. From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$. $(1+X) \in A$, so $t$ satisfies the monic polynomial $$P(T) = T^2 - (1+X) \in A[T]$$ Thus $t$ is integral over $A$. Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$. Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed). $Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$ and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$. Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$. \end{proof} \begin{ex}{R.4.9} $k$ a field, $A= k[X,Y,Z]/(X^2- Y^3 -1, XZ - 1)$, find $\alpha, \beta \in k$ such that $A$ is integral over $B = k[X + \alpha Y + \beta z]$, and write a set of generators of $A$ as a $B$-module. \end{ex} \begin{proof} (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$) The relations defining $A$ are \begin{align*} X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\ XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1} \end{align*} Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$. Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$. $Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$. Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\ set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$. Let $w=X+Z$; since $XZ=1$, we have $$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$ which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$. Since $Z=w-X$, $Z$ is also integral. \vspace{0.4cm} Generators of $A$ as a $B$-module: we hadd $B=k[w]$ with $w=X+Z$. From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$. Thus any polynomial in $X$ can be reduced to a linear form $b_1 X + b_0$ with $b_i \in k[w]$. Hence it's partial basis is $\{1, X\}$. Fitting $X^2$ into $(*)$, \begin{align*} X^2 &- Y^3 -1 =0\\ Y^3 &= X^2-1\\ Y^3 &= wX -2 \end{align*} thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$). So its partial basis is $\{1, Y, Y^2\}$. For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$. \vspace{0.2cm} Combining the previous partial basis, the generators are $$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$ \end{proof} \subsection{Exercises Chapter 6} \begin{ex}{R.6.3.a} Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$. \end{ex} \begin{proof} ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$. Localization map \begin{align*} \psi: A &\longrightarrow A'\\ (a', a'') &\longmapsto a' \end{align*} note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$. Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$. Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$. In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$. Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$, with $s \in S$, so $s=(1,0)$, hence $$(1,0)\cdot(a',a'')=(0,0)$$ $$\Longrightarrow~ (a', 0) = (0, 0)$$ \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$. $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$ $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$ By the 1st isomorphism theorem, \begin{tikzpicture}[node distance=1.5cm, auto] \node (G) {$A$}; \node (H) [right of=G] {$A'$}; \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$}; \draw[->] (G) to node {$\psi$} (H); \draw[->] (G) to node [swap] {$\phi$} (GmodK); \draw[<->] (GmodK) to node [swap] {$\eta$} (H); \end{tikzpicture} $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$ \vspace{0.5cm} Take the localization map \begin{align*} \phi: A &\longrightarrow S^{-1}A\\ a &\longmapsto a/1 \end{align*} $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$ and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$ $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$ which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$. \vspace{0.2cm} $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel, thus $A' \cong S^{-1}A$. \end{proof} \begin{ex}{R.6.4} \begin{enumerate}[a.] \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$. \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$ \end{enumerate} \end{ex} \begin{proof} \begin{enumerate}[a.] \item (\emph{saturated sets})\\ General rule of saturation:\\ two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation. Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$: $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$ \vspace{0.4cm} Example 1: $A=\mathbb{Z}$,\\ $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$ localization: $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$ $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$ Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\ since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$. \vspace{0.4cm} Example 2: $A= k[x]$, $$S = \{ x^n ~|~ n \geq 0 \}$$ $$T = \{ (2x)^n ~|~ n \geq 0 \}$$ $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$ \item Show that $T$ is a multiplicative set: (must contain $1$ and be closed under multiplication)\\ Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$. Let $t_1,~ t_2 \in T$; by definition of $T$, $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$ Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\ rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\ Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$. Thus $T$ is a multiplicative set. Next, we show that $T$ is maximal: Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$. Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit. Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\ \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\ \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\ $\Longrightarrow~ s'ua=s's \in S$ since $s,s' \in S$, their product is also in $S$. Let $b=s'a$, then $ub \in S$. By definition of $T$, $u \in T$. Therefore, $U \subseteq T$; thus $T$ is maximal. \end{enumerate} \end{proof} \bibliographystyle{unsrt} \bibliography{commutative-algebra-notes.bib} \end{document}