\classheader{2012-11-21}

\begin{theorem} Let $A$ be a finite-dimensional algebra over $k$. 
Then
\begin{enumerate}
\item $J(A)$ is the maximal nilpotent ideal of $A$, i.e., $J(A)$ is a nilpotent
ideal and it contains any other nilpotent ideal.
\item $A/J(A)$ is a semisimple algebra.
\item $A$ has only finitely many maximal two-sided ideals.
\item $J(A) = \bigcap$ maximal two-sided ideals. 
\end{enumerate}
\end{theorem}
\begin{proof}[Proof of 1] Let $J:=J(A)$. We have a decreasing chain of two-sided ideals
$A \supset J \supset J^2 \supset J^3 \supset \dotsb$, which must stabilize since
$A$ is finite-dimensional (hence Artinian). Therefore there exists $N \gg 0$
such that $J^{N+1} = J^N$. Since $J^N$ is a finitely generated $A$-module, 
Nakayama's lemma implies that $J^N=0$, so $J$ is nilpotent.
\end{proof}
\begin{proof}[Proof of 2]
$J(A/J(A))=0$ so by Problem 6 of Homework 8, we know $A/J(A)$ is semisimple.
\end{proof}
\begin{proof}[Proof of 3]
Wedderburn's theorem implies that $A/J(A) = A_1 \oplus \dotsb \oplus A_n$
where $A_i = \M_{r_i}(D_i)$ is a simple algebra. Any two-sided ideal 
$I \subset \bigoplus A_i$ has the form $I = I_1 \oplus \dotsb \oplus I_n$
where $I_n$ is a two-sided ideal in $A_i$. The maximal
ideals in $A/J(A)$ are $A_1 \oplus \dotsb \oplus A_{i-1} \oplus 0 \oplus
A_{i+1} \oplus \dotsb \oplus A_n$. 

We claim that if $\mathfrak a$ is a maximal two-sided ideal in $A$, then
$J(A) \subset \mathfrak a$. If this were not the case, then $J(A) + \mathfrak a = A$. Hence $1 = j+a$ for some $j\in J(A), a \in 
\mathfrak a$. But then $a= 1-j$ is invertible, a contradiction.

In general, we have a bijection 
\[\{\text{maximal two-sided ideals of }A\text{ containing }J(A)\}\longleftrightarrow \{\text{maximal two-sided ideals
of }A/J(A)\}.\] Thus, the number of maximal two-sided ideals is finite.
\end{proof}
\begin{proof}[Proof of 4]
This is clear from our description of maximal ideals in $A/J(A)$ from part 3.
\end{proof}

\begin{remark} Let $A$ be a ring, $J \subset A$ a two-sided ideal, 
and $M$ a left (resp.~right) $A$-module. Then $M/JM = (A/J) \otimes_A M$ (resp.
$M/MJ=M \otimes_A (A/J)$) is a left (resp.~right) $(A/J)$-module.
\end{remark}

\begin{theorem} Let $A$ be a finite-dimensional algebra over an algberaically closed field $k$. Let $J=J(A)$, and define $\bar{A} := A/J$. Then 
$A \cong (T_{\bar{A}} (J/J^2))/I$ where $I$ is a two-sided ideal in the tensor
algebra satisfying \[ T^{\ge 2}(J/J^2)\supseteq I \supseteq
T^{\ge N}(J/J^2) \] for sufficiently large $N$.
\end{theorem}

\begin{corollary} Let $A$ be finite-dimensional over an algebraically closed field $k$ such that
\[\bar A = A/J(A) = \underbrace{k \oplus \dotsb \oplus k}_{n\text{ times}}.\] Then 
there exists a finite quiver $Q$ with vertex set $\{1,\dotsc,n\}$
such that $A = kQ/I$ with $(kQ)_{\ge 2} \supseteq I \supseteq (kQ)_{\ge N}$
for some sufficiently large $N$.
\end{corollary}
\begin{proof}
By the remark, $E:=J/J^2$ is an $A/J$-bimodule. Since $A/J$ is a direct
sum of fields, we have a decomposition $E = \bigoplus E_{ij}$. Define
the quiver $Q$ so that $E$ corresponds to the paths. 
\end{proof}

\begin{proof}[Proof of Theorem 2]
The proof proceeds in several steps.

{\it Step 1:} We claim that there exists a section $\varepsilon : \bar A \to A$, i.e. that
there exists a subalgebra $A' \subset A$ such that $A' \hookrightarrow A
\to \bar A$ is an isomorphism. \vspace{-0.1in}
\begin{quotation}
\begin{lemma}[Lifting of idempotents] If $J \subset A$ is a nilpotent
two-sided ideal, let $\bar e_1,\dotsc,\bar e_n$ be
a collection of orthogonal idempotents, i.e., $\bar e_i \cdot\bar e_j = 
\delta_{ij} \bar e_i$. Then there exist orthogonal idempotents $e_i \in A$
such that $e_i \bmod J = \bar e_i$ for all $i=1,\dotsc,n$. 
\end{lemma}
\end{quotation}
The above is Corollary 7.5 of Etingof's \emph{Introduction to representation theory}
and we omit the proof. 

Since $\bar A$ is semisimple over an algebraically closed field, it is a direct sum of
matrix algebras $\bigoplus_\ell \M_{n_\ell}(k)$. Let $\bar e_{\ell,ij}$ denote
the matrix with a $1$ in the $(i,j)$ position as an element of $\M_{n_\ell}(k)$.
Then $\{\bar e_{\ell,ii}\}$ forms a collection of orthogonal idempotents 
in $\bar A$, so by the lemma we can lift them to orthogonal idempotents in
$e_{\ell,ii} \in A$. Since $\sum_{\ell,i} e_{\ell,ii} \in 1+J$ is idempotent
and invertible, it is $1$. Let $e_\ell = \sum_i e_{\ell,ii}$. 
Then $e_\ell A e_\ell$ are orthogonal subalgebras of $A$ for distinct $\ell$. 
Hence it suffices to lift each $\M_{n_\ell}(k)$ to $e_\ell A e_\ell$. 
We therefore drop the $\ell$ subscript and assume $\sum e_{ii} = 1$. 

Suppose that for $I$ a two-sided square-zero ideal of $A$, we can lift
a matrix subalgebra of $A/I$ to $A$. Then applying this to the
sequence $A = A/J^N \to A/J^{N-1} \to \dotsb \to A/J = \bar A$ with
$I = J^i/J^{i+1} \subset A/J^{i+1}$, we are done. To prove the square-zero case, 
i.e., $\bar A = A/I$ and $I^2=0$: first take an arbitrary lift $e_{i,i+1} \in A$ of
$\bar e_{i,i+1} \in \bar A$. By multiplying on the left by $1-e_{jj}$ for 
$j \ne i$ and on the right by $1-e_{jj}$ for $j \ne i+1$, we get
$e_{jj} e_{i,i+1} = \delta_{ij}$ and $e_{i,i+1} e_{jj} = \delta_{i+1,j}$.
Do the same for $e_{i+1,i}$. Then $e_{i,i+1} e_{i+1,i} - e_{ii} \in I$
implies 
\[ (e_{i,i+1} e_{i+1,i} - e_{ii})^2 = e_{i,i+1} (r-e_{i+1,i}) + e_{ii} = 0 \]
where $r = e_{i+1,i} e_{i,i+1} e_{i+1,i} - e_{i+1,i} \in I$. 
Thus $e_{i,i+1}(e_{i+1,i} - r) = e_{ii}$. The analogous computation
shows that $(e_{i+1,i}-r)e_{i,i+1} = e_{i+1,i+1}$. 
Replacing $e_{i+1,i}$ with $e_{i+1,i}-r$, we can assume that 
\[ e_{i+1,i} e_{i,i+1} = e_{i+1,i+1} , \quad e_{i,i+1} e_{i+1,i} = e_{ii}. \]
Now set $e_{ij} = e_{i,i+1} e_{i+1,i+2} \dotsb e_{j-1,j}$ 
and $e_{ji} = e_{j,j-1} \dotsb e_{i+1,i}$ for $i<j$. One sees that
the span of $\{e_{ij}\}$ is isomorphic to $\M_n(k)$ as algebras.

\medskip

{\it Step 2:} The projection $J \to J/J^2$ is an $\bar A$-bimodule map,
where $\bar A$ acts on $J$ via $\varepsilon$. We want to show that there exists
an $\bar A$-bimodule section $J/J^2 \to J$. 

Let $M = J/J^2$.
Keeping the notation from Step 1, it suffices to lift $e_\ell M e_m$
to $e_\ell J e_m$. Note that an $(\M_{n_\ell}(k), \M_{n_m}(k))$-bimodule is the
same as an $\M_{n_\ell}(k) \otimes \M_{n_m}(k)^\op \cong \M_{n_\ell}(k) \otimes
\M_{n_m}(k) \cong \M_{n_\ell n_m}(k)$-module. Any such module is a direct sum
of the simple module $k^{n_\ell n_m}$. Decompose $M$ into a direct sum of 
simple modules and lift one vector from each summand. The 
$\bar A \otimes \bar A^\op$-span of these lifts gives the desired section.

\medskip
{\it Step 3:} Let $f : \widetilde A \to A$ be an algebra homomorphism. 
Let $\widetilde J \subset \widetilde A$ and $J \subset A$ be two-sided nilpotent
ideals such that (i) $f(\widetilde J) \subset J$, 
	   (ii) $\widetilde A / \widetilde J \to A/J$
is surjective, and (iii) $\widetilde J/\widetilde J^2 \to J/J^2$ is surjective.
Then $f$ is surjective.

This claim just follows by induction (think $J$-adic completion). 

\medskip

{\it Finishing the proof:} Steps 1 and 2 give maps $\varepsilon : \bar A \to A$
and $J/J^2 \hookrightarrow J \hookrightarrow A$. The universal property
of tensor algebras\footnote{The universal property says that if $\bar A \to A$ 
is an algebra homomorphism, $M$ an $\bar A$-bimodule, and $f : M \to A$ an
$\bar A$-bimodule map, then there exists a unique algebra map $T_{\bar A} M \to A$
extending $f$.} gives an algebra map $f : T:= T_{\bar A}(J/J^2) \to A$. 
Then $f(T^{\ge 1}) \subset J$ implies $f(T^{\ge i}) \subset J^i$ for all $i\ge 1$.
In particular, since $J^N=0$ for some $N$, we get $f(T^{\ge N}) \subset J^N=0$. 
Hence $T^{\ge N} \subset \ker(f) =: I$. On the other hand, our
construction of $f$ gives a commutative diagram

\begin{center}
\begin{tikzcd}
T/T^{\geq 2} \ar{r}{f} \ar{d}[swap]{\cong} & A/J^2 \arrow[equals]{r} & (\bar{A}\oplus J)/J^2 \ar{d}{\cong} \\
\bar{A}\oplus(J/J^2) \ar{rr}[swap]{\id} & &\bar{A}\oplus (J/J^2)
\end{tikzcd}
\end{center}

%\[ \xymatrix{ T/T^{\ge 2} \ar[r]^f \ar[d]^\cong & A/J^2 \ar@{=}[r] & (\bar A \oplus J)/J^2 
%	\ar[d]^\cong \\ \bar A \oplus J/J^2 \ar[rr]^{\mathrm{id}} && \bar A \oplus J/J^2 }\]

which shows that if $a \in I$, then $f(a\bmod T^{\ge 2}) =0$ implies
$a \bmod{T^{\ge 2}} = 0$, i.e. that $a \in T^{\ge 2}$. Hence $I \subset T^{\ge 2}$.

Lastly, apply Step 3 to $\widetilde A = T$ and $\widetilde J = T^{\ge 1}$ 
to conclude the theorem.
\end{proof}