Generalized Matrix Artin Algebras

Question

Let $A$ and $B$ be Artin algebras, $M$ a finitely generated $(B$-$A)$-bimodule, and $N$ a finitely generated $(A$-$B)$-bimodule. Prove that \begin{bmatrix} A& N \\ M & B\end{bmatrix} is a generalized matrix Artin algebra.

Answer 1

A reasonable elementary strategy for showing it is right Artinian might go this way. Denote $R=\begin{bmatrix}A&N\\M& B\end{bmatrix}$ for brevity.

1. Here's an easy lemma we'll apply: if $A$ and $B$ are Artinian and $K$ is a finitely generated $(A,B)$ (or $(B,A)$) bimodule, then $K$ is an Artinian bimodule. (It is not a very sharp statement, but it's all we need.)
2. First, note $R=\begin{bmatrix}A&N\\0&0\end{bmatrix}\oplus \begin{bmatrix}0&0\\M& B\end{bmatrix}$ as right $R$ modules. Denote these two pieces as $T_1$ and $T_2$ respectively. If we can prove both of these are right Artinian $R$ modules, then we have shown $R$ is right Artinian.
3. Consider a submodule $T$ of the right $R$ module $T_1$. If $\begin{bmatrix}a&n\\0&0\end{bmatrix}\in T$, then so are $\begin{bmatrix}a&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&n\\0&0\end{bmatrix}$. (Can you see why?) From this, you can argue that $T=\begin{bmatrix}A'&N'\\0&0\end{bmatrix}$ where $A'$ is a right ideal of $A$ and $N'$ is a sub-bimodule of $N$.
4. Use the last point and the fact that $A$ and $N$ are Artinian to show that $T_1$ is Artinian.
5. Describe the analogous situation of $T_2$ using the same approach, and conclude it is Artinian as well. At this point $R$ has been shown to be right Artinian.
6. Say "The same argument mutatis mutandis shows that $R=\begin{bmatrix}A&0\\M&0\end{bmatrix}\oplus \begin{bmatrix}0&N\\0& B\end{bmatrix}$ is a decomposition of $R$ into two Artinian left $R$-modules, therefore $R$ is left Artinian."