\classheader{2012-11-28}

Recall that $A$ is a nice commutative algebra. 
Let $\widehat{A}$ be the collection of algebra homomorphisms from $A$ into $\C$. 
We already showed
that $\widehat{A} \cong S_A$ given by $\chi \mapsto \C_\chi$. Let $\Nil(A)$ be the collection of
nilpotent elements of $A$. Let $\ev: A \to \C \{ \widehat{A} \}$ be the evaluation algebra
homomorphism $\ev(a)(\chi) = \chi(a)$. We will also denote $\widehat{a} = \ev(a)$.

\begin{theorem}
Let $A$ be a nice commutative algebra. Then 
\begin{enumerate}
\item $\ker(\ev) = \Nil(A)$.
\item $\im(\ev)$ is pointwise dense.
\item For $a \in A$, $\spec(a)$ is the set of values of $\widehat{a}$.
\end{enumerate}
\end{theorem}
\begin{proof}
We know parts 1 and 2, but not 3 yet. This holds if and only if $\spec(a) = \{ \chi(a) \mid \chi \in \widehat{A} \}$. For any $\lambda\in\C$,
$\lambda \in \spec(a)$ if and only if $a - \lambda$ is not invertible, which is the case if and only
if $A(a - \lambda)$ is contained in a maximal ideal (by Zorn's lemma), which we know is the kernel
of some $\chi$. Thus, $a - \lambda \in A(a - \lambda) \subseteq \ker(\chi)$ if and only if $0 = \chi(a - \lambda)$, if and only if $\lambda = \chi(a)$.
\end{proof}

\begin{theorem}[Hilbert's Nullstellensatz]
Let $A$ be a finitely generated commutative $\C$-algebra. Let $I \subseteq A$ be an ideal; then $a \in \bigcap \{ \text{maximal ideals of $A$ containing $I$} \}$ if and only if $a^n \in I$ for some sufficiently large $n$.
\end{theorem}
\begin{proof}
$a \bmod I$ is contained in the intersection of maximal ideals in $A/I$ if and only if it lies in $\Nil(A/I)$, i.e. $a^n$ is eventually in $I$.
\end{proof}

\begin{corollary}
Let $A = \C[x_1, ..., x_n]$. Let $V(I) = \{ c \in \C^n \mid p(c) = 0 \text{  for all } p \in I\}$. Then $f|_{V(I)} = 0$ if and only if $f^n \in I$ for some $n$.
\end{corollary}
\begin{proof}
$f|_{V(I)} = 0$ if and only if $\ev_c(f) = 0$ for all $c \in V(I)$. 
%, i.e. $\ev_c(I) = 0$. Thus, $\ev_c(f)=0$ 
The latter condition is equivalent to saying that $f \in \bigcap_{\{c\mid \ker(\ev_c) \supset I \}} \ker(\ev_c)= \bigcap \{\text{maximal ideals in $A$ containing $I$}\}$, which the theorem shows is the case if and only if $f^n\in I$ for some $n$.
\end{proof}

\subsection*{Topological Versions of These Results}

\begin{definition}
Let $A$ be a $\C$-algebra. A norm on $A$ is a function $|\cdot | : A \to \R_{\geq 0}$ such that
\begin{enumerate}
\item $|a| = 0$ if and only if $a = 0$.
\item $|\lambda a | = |\lambda| \cdot |a|$ for all $\lambda \in \C$.
\item $|a + b| \leq |a| + |b|$.
\item $|ab| \leq |a||b|$.
\end{enumerate}
\end{definition}

\begin{examples}
$\text{}$

\begin{enumerate}
\item $C(X)$ where $X$ is a compact topological space and $\abs{f}= \lVert f \rVert_\infty
= \max_{x\in X} \abs{f(x)}$.
\item $A = \End_\C(V)$ where $|a| = \sup_{|v| = 1} |av|$.
\end{enumerate}
\end{examples}

A \emph{Banach algebra} is an algebra $A$ with norm $| \cdot |$ which is complete as a metric space.

\begin{lemma}
Let $A$ be a Banach algebra. Fix an element $a \in A$. Then 
\begin{enumerate}
\item $\spec(a)$ is a closed subset of the disk of radius $|a|$.
\item The function $f: \C \setminus \spec(a) \to A$ given by $z \mapsto (a - z)^{-1}$ is a holomorphic function.
\item $|(a - z)^{-1}| \to 0$ as $|z| \to \infty$.
\end{enumerate}
\end{lemma}
\begin{proof}
The idea is just to factor $a$ out of the resolvent to get $a^{-1}(1 - a^{-1}z)^{-1} = \frac{1}{a}
\sum_k (a^{-1}z)^k$, which is absolutely continuous, hence convergent in the Banach algebra.
Therefore, it is continuous and one can differentiate termwise. This lemma is Homework 9, Problem 7.
\end{proof}

\begin{theorem}[Gelfand]
For any Banach algebra $A$ and $a \in A$, $\spec(a) \neq \varnothing$.
\end{theorem}
\begin{proof}
Suppose that $\spec(a) = \varnothing$; then $z \mapsto (a - z)^{-1}$ is a holomorphic function $\C \to A$. It is also bounded by (3). By Liouville, this must be constant, which is a contradiction.
\end{proof}

\begin{remark} All of our results on nice algebras in Lecture 24 only relied on
the fact that $A$ nice implies $\spec(a) \ne \varnothing$ 
(in particular, so we get Schur's Lemma). 
Thus all of those results also hold for Banach algebras by Gelfand's theorem.
\end{remark}

Now equip $\widehat{A}$ with the topology of pointwise convergence. This is called the \emph{Gelfand
representation}.

\begin{theorem}
Let $A$ be a commutative Banach algebra.
\begin{enumerate}
\item For all $a \in A$, $\widehat{a} = \ev(a)$ is a continuous function on $\widehat{A}$ and
$\sup_{\chi \in \widehat{A}} |\widehat{a}(\chi)| \leq |a|$.
\item $\widehat{A}$ is compact and $\ev: A \to C(\widehat{A})$ is a weak contraction.
\item $\ker(\ev)$ is the set of elements that are topologically nilpotent, i.e. $\lim \sup |a^n|^{1/n} \to 0$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1) The topology of pointwise convergence is (by definition) the weakest
topology such that $\widehat a$ is continuous for all $a \in A$. By Gelfand's theorem, $\spec(a)$ is a nonempty compact set contained in the disk of radius $|a|$
and is equal to $\ev_a (\widehat{A})$ by the remark.

(3) The kernel of the evaluation map is exactly $\{ a \in A | \chi(a) = 0 \text{ for all $\chi \in \widehat{A}$} \}$, which happens if and only if $\spec(a) = \{ 0 \}$, which is equivalent to $\lim \sup |a^n|^{1/n} = \max\{ |z|, z \in \spec(a) \} = 0$
by Homework 9, Problem 7.

(2) Let $D$ be the unit disk in $\C$ and let $D_A$ be the unit disk in $A$. For every $\chi \in
\widehat{A}$, $\chi$ maps $D_A$ to $D$ by part (1). Therefore $\chi$ is a weak contraction, hence
continuous; and $\widehat{A}$ is a pointwise closed subset in $\mathrm{Maps}(D_A, D) = D^{D_A}$,
which is a compact set by Tychonoff (pointwise convergence topology is exactly the product
topology). Therefore, $\widehat{A}$ is compact and the map is a weak contraction.
%jw -- he claimed this map is an isometry, but that is not the case if it is not injective
\end{proof}