diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index 499f599..3a760ed 100644 Binary files a/commutative-algebra-notes.pdf and b/commutative-algebra-notes.pdf differ diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex index 60eb50b..1b35444 100644 --- a/commutative-algebra-notes.tex +++ b/commutative-algebra-notes.tex @@ -117,8 +117,12 @@ $x \in A$ 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}.. + $x \in A$ 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. @@ -171,11 +175,16 @@ $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$ \end{defn} -\begin{defn}{}[local ring] - A \emph{local ring} has a unique maximal ideal. +\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} \subsection{Z and K[X], two Principal Ideal Domains} @@ -1388,7 +1397,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie Thus \begin{align*} - k^n &\longleftrightarrow m-Spec A\\ + k^n &\longleftrightarrow m-Spec~ A\\ (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n) \end{align*} \end{cor} @@ -1427,7 +1436,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie Thus there is a one-to-one correspondence: - points in $k^n ~~~ \longleftrightarrow~~~ m-Spec A$ (maximal ideals in $k[X_1, \ldots, X_n]$ + 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} @@ -1456,7 +1465,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie Therefore, $\exists$ a one-to-one correspondence \begin{align*} - V(X) &\longleftrightarrow m-Spec A\\ + 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} @@ -1490,11 +1499,11 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie \vspace{0.4cm} Thus,\\ every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie. - $$m-Spec A \longleftrightarrow k^n$$ + $$m-Spec~ A \longleftrightarrow k^n$$ 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)\\ + m-Spec~ A &\longleftrightarrow V(J)\\ \text{maximal spectrum} &\longleftrightarrow \text{variety} \end{align*} \end{proof} @@ -1580,7 +1589,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$ - $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)} g_i) + \text{(polynomial)} \cdot (1 - Yf)$$ + $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$ which, since $1 \in J'$, $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$ @@ -1633,7 +1642,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie \end{align*} Therefore, - $$Spec k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$ + $$Spec~ k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$ \end{corollary} \end{cor} @@ -1643,10 +1652,10 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie 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 - $$Spec A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$ + $$Spec~ A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$ \end{prop} \begin{proof} - By definition, $Spec A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$. + By definition, $Spec~ A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$. By Corollary \ref{cor.5.8}: $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$ @@ -1689,10 +1698,276 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$. Therefore, - $$\mathfrak{P} \in Spec A \longleftrightarrow X \subseteq V(J)$$ + $$\mathfrak{P} \in Spec~ A \longleftrightarrow X \subseteq V(J)$$ where $X = V(\mathfrak{P})$. \end{proof} + + +\section{Rings of fractions $S^{-1}A$ and localization} + +\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 $A \times S$: + $$(a, s) \sim (b, t) \Longleftrightarrow \exist y \in S ~\text{such that}~ u(at - bs) = 0$$ + + (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{eg}{ } + TODO +\end{eg} + +\begin{lemma}{6.2} \label{6.2} + For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong A[X]/(Xf-1)$. + + Then + $$A_f \cong \frac{A[X]}{(Xf-1)}$$ +\end{lemma} +\begin{proof} + 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 $h \in (Xf-1)~ \foracll~ h \in ker(\psi)$, 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), $a\f^n = 0$ iff $\exist~ k$ 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 $\exist~ \frac{b}{t} \in A_P$ such that + $\frac{a}{s} \cdot \frac{b}{t} = 1$. + + By definition of equality in localization, $\frac{ab}{st}=1$ means $\exist + u \in S$ such that + \begin{equation} + u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust + \tag{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 \eq{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{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} + + + \newpage \section{Exercises}