port rest of ch 6 notes (localization, etc)

commutative-algebra-notes.tex

@@ -175,7 +175,7 @@
   $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$
 \end{defn}
 
-\begin{defn}{}[local ring]
+\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 $K=A/\mM$:
@@ -1397,7 +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}
@@ -1436,7 +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}
@@ -1465,7 +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}
@@ -1499,11 +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}
@@ -1589,7 +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 $1 \in J'$,
   $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
@@ -1642,7 +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}
 
@@ -1652,10 +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}:
     $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
@@ -1698,14 +1698,17 @@ 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}
 
-\begin{defn}{6.1}[ring of fractions]
+\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 $st \in S
   ~\forall s, t \in S$).
 
@@ -1717,10 +1720,23 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   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 usua arithmetic op'ns on fractions:
+  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}
@@ -1808,6 +1824,149 @@ By the 1st isomorphism theorem:
 \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}
+
+
 
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