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Rings of fractions & localization (#7)

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* add extended proof of lemma 6.2

* port rest of ch 6 notes (localization, etc)
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@@ -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}