\classheader{2012-11-26}

Today, we'll discuss a trick that will let us extend some of our results about finite-dimensional algebras to some infinite-dimensional algebras.

\begin{definition}
An algebra is said to be ``nice'' (this isn't standard terminology) if it is a $\C$-algebra of at most countable dimension over $\C$.
\end{definition}

Clearly, a subalgebra or a quotient of a nice algebra is nice, and any finitely generated $\C$-algebra is nice because it is a quotient of some $\C\langle x_1,\ldots,x_n\rangle$, which has a countable basis.

Recall that for an algebra $A$ and $a\in A$, we defined $\spec(a)=\{\lambda\in\C\mid a-\lambda\text{ is not invertible}\}$.

\begin{theorem}[Spectral theorem]
Let $A$ be a nice algebra, and let $a\in A$. Then
\begin{enumerate}
\item $a$ is nilpotent $\iff$ $\spec(a)=\{0\}$.
\item $a$ is algebraic $\iff$ $\spec(a)$ is a non-empty finite set.
\item $a$ is non-alegbraic $\iff$ $\C\setminus\spec(a)$ is at most countable.
\end{enumerate}
\end{theorem}

Let $\C(t)$ be the field of rational functions in $t$.

\begin{lemma}
Let $\lambda_1,\ldots,\lambda_n\in\C$ be distinct. If $\sum \frac{c_i}{t-\lambda_i}=0$ in $\C(t)$, then all of the $c_i$ are 0.
\end{lemma}
\begin{proof}[Proof of lemma]
Clearing denominators, we get that $\frac{f(t)}{\prod (t-\lambda_i)}=0$ for some $f(t)\in\C[t]$, hence $f(t)=0$. But for any $i$, the fact that $f(\lambda_i)=0$ implies that $c_i=0$.
\end{proof}

\begin{proof}[Proof of theorem]
Given an $a\in A$, define
\[R_a=\{\tfrac{f}{g}\in\C(t)\mid f,g\text{ are coprime, and } g\text{ is non-zero on }\spec(a)\}.\]
Thus, for any $g_1,\ldots,g_n\in\C\setminus\spec(a)$, we have $\frac 1{g} \in R_a$ where 
\[g(t)=(t-g_1)\cdots(t-g_n)\in R_a\] %jw-- don't see the point of g_0 
and any denominator has this form.
Note that $R_a$ is a subring of $\C(t)$. Now, we can define an evaluation map $j_a:R_a\to A$ by
\[\frac{f}{g}\mapsto f(a)\cdot g(a)^{-1}=f(a)(a-g_1)^{-1}\cdots(a-g_n)^{-1}.\] %%g_0^{-1}(a-g_1)^{-1}\cdots(a-g_n)^{-1}.\]
Observe that if $a$ is not algebraic, then for any $\frac{f}{g}\in R_a$, if $j_a(\frac{f}{g})=0$, we have $f(a)g(a)^{-1}=0$, hence $f(a)=0$, hence $f=0$ because $a$ is not algebraic. Thus, $j_a$ is injective when $a$ is not algebraic.


Our lemma demonstrates that $\{\frac{1}{t-\lambda}\mid \lambda\in\C\setminus\spec(a)\}$ is a
$\C$-linearly independent subset of $R_a$, and our observation above implies that
$\{(a-\lambda)^{-1}\mid \lambda\in\C\setminus\spec(a)\}$ is linearly independent (when $a$ is not
algebraic). Because $\dim_\C(A)\leq\text{countable}$, we must have that $|\C\setminus\spec(a)|\leq\text{countable}$.

We also know that if $a$ is algebraic, then $\spec(a)=\{\text{roots of the minimal polynomial of }a\}$, which must be a finite set, and which is non-empty because $\C$ is algebraically closed.

It is impossible to have $\spec(a)$ be finite and $\C\setminus\spec(a)$ to be at most countable at the same time, so parts 2 and 3 follow.

Now we will prove part 1. We clearly know that $a$ is nilpotent $\iff$ $a$ is algebraic and $\spec(a)=\{0\}$, but we want to drop the assumption that $a$ is algebraic. We can do this because $\spec(a)=\{0\}$ implies that $\spec(a)$ is finite, hence $a$ is algebraic by part 2.
\end{proof}

\begin{corollary}
If $A$ is nice and $a\in A$, then $\spec(a)$ is non-empty.
\end{corollary}
\begin{proof}
Obviously, $a$ is either algebraic or not algebraic; apply the theorem.
\end{proof}
\begin{corollary}
If $A$ is a nice division algebra, then $A=\C$.
\end{corollary}
\begin{proof}
Let $a\in A\setminus\C$, so that $a-\lambda\neq 0$ for all $\lambda\in\C$. Because $A$ is a division algebra, this implies that $a-\lambda$ is invertible for all $\lambda\in\C$, hence $\spec(a)=\varnothing$, which is a contradiction.
\end{proof}
\begin{theorem}[Schur lemma for nice algberas]
If $A$ is a nice algebra and $M$ is a simple $A$-module, then $\End_A(M)=\C$.
\end{theorem}
\begin{proof}
Pick a non-zero $m\in M$. We have a diagram
\begin{center}
\begin{tikzcd}[column sep=0.8in]
A \ar{r}{a\mapsto am}[swap]{\ell} & M & \End_A(M)\ar{l}{i}[swap]{f(m) \mapsfrom f}
\end{tikzcd}
\end{center}
but because $M$ is simple, $\ell$ is surjective. Again because $M$ is simple, if $f(m)=0$ then $f(A\cdot m)=0$, hence $f=0$, so that $i$ is injective.

The fact that $A$ is nice then implies that $\End_A(M)$ is nice, hence $\End_A(M)$ is a nice division algebra, hence $\End_A(M)=\C$.
\end{proof}

\begin{proposition}
Let $A$ be a nice algebra. Then $J(A)$ is the unique maximal nil-ideal.
\end{proposition}
\begin{remark}
We know that if $A$ is finite-dimensional, then $J(A)$ is nilpotent.
\end{remark}
\begin{proof}
If $a\in J(A)$, then $1-xa$ is invertible for all $x\in A$, so  $1-\lambda a$ is invertible for all $\lambda\in\C$. Therefore, $a-\lambda$ is invertible for all $\lambda\neq 0$, so we must have $\spec(a)\subseteq\{0\}$. By the spectral theorem, $a$ is nilpotent. Thus, $J(A)$ is a nil-ideal. But we have proved in a previous class that any nil-ideal is contained in $J(A)$; this implies the claim.
\end{proof}


\begin{theorem}
Let $A$ be a nice algebra with trivial Jacobson radical, i.e. $J(A)=0$. The action map $A\to \prod_{M\in S_A}\End_\C(M)$ is an embedding of $A$ as a pointwise dense subalgebra, i.e. for any finite subset $S\subset S_A$, any $m_1,\ldots,m_n\in\bigoplus_{M\in S}M$, and any $f\in\bigoplus_{M\in S}\End_\C(M)$, there is some $a\in A$ such that $f(m_i)=am_i$ for all $i=1,\ldots,n$.
\end{theorem}
\begin{proof}
Because
\[J(A)=\bigcap_{M\in S_A}\Ann(M)=0,\]
we have that the action map $j$ is injective. Because $M$ is simple, $A_M^!=\End_A(M)=\C$, and thus $A_M^{!!}=\End_\C(M)$. Now the result follows from the Jacobson density theorem.
\end{proof}

%Now let's look at the case when the algebra $A$ is commutative.
\subsection*{Commutative case}

\begin{theorem}
Let $A$ be a nice commutative algebra. Then $J(A)=\{\text{nilpotent }a\in A\}$. Given an algebra homomorphism $\chi:A\to\C$, let $\C_\chi$ be $\C$ considered as an $A$-module via $\chi$. Then the natural maps
\begin{center}
\begin{tikzcd}[column sep=0.99in]
\{\text{\emph{algebra homomorphisms }}A\to\C\}\ar{r}{\chi\mapsto \C_\chi} & S_A \ar{r}{\;\;M\mapsto \Ann(M)\;\;} & \{\text{\emph{maximal ideals in }}A\}
\end{tikzcd}
\end{center}
are bijections.
\end{theorem}
\begin{proof}
If $a\in A$ is nilpotent, then $Aa$ is a nil ideal, so that $Aa\subseteq J(A)$. Conversely, any $a\in J(A)$ is nilpotent by our proposition.

Let $M\in S_A$ be a simple $A$-module. Since $A$ is commutative, $A_M\subset A_M^!$ ($A$ obviously commutes with itself). Because $M$ is simple, we have $A_M^!=\C$ by Schur's lemma, so that
$A_M=\C$, hence $\dim_\C(M)=1$, hence there is some $\chi$ such that $am=\chi(a)m$. 
%We'll do the other map next time.

Next we need to show that any maximal ideal $I$ in a commutative algebra $A$ is
the annihilator of a simple module. Take $M=A/I$ so that $\Ann(M)=I$. 
Then $I$ is a maximal left ideal iff $M$ is simple. 
\end{proof}
