Is this matrix ring semisimple?

Question

I'm trying to solve the following problem.

Let $\mathbb{C}$ be the complex numbers and $\mathbb{C}[x]$ the polynomial ring over $\mathbb{C}$. Let $\alpha\in\mathbb{C}$. For which $n\geq 1$ is the ring $R_n:=\begin{bmatrix} \frac{\mathbb{C}[x]}{((x-\alpha)^n)} && \frac{\mathbb{C}[x]}{((x-\alpha)^n)} \\ \frac{\mathbb{C}[x]}{((x-\alpha)^n)} && \frac{\mathbb{C}[x]}{((x-\alpha)^n)} \end{bmatrix}$ semisimple?

Here the definition of a semisimple ring:

A ring $R$ is called semisimple if $R$ as a module over itself is a sum of simple $R$-submodules of $R$. A module $M$ is called simple if the only submodules of $M$ are $(0)$ and $M$.

So far I have that it is semisimple for $n=1$. The submodules of $R_1=\begin{bmatrix} \frac{\mathbb{C}[x]}{(x-\alpha)} && \frac{\mathbb{C}[x]}{(x-\alpha)} \\ \frac{\mathbb{C}[x]}{(x-\alpha)} && \frac{\mathbb{C}[x]}{(x-\alpha)} \end{bmatrix}$ are of the form $\begin{bmatrix} I && 0 \\ I && 0 \end{bmatrix}$ and $\begin{bmatrix} 0 && I \\ 0 && I \end{bmatrix}$ where $I$ is an ideal of $\frac{\mathbb{C}[x]}{(x-\alpha)}$. Since $\frac{\mathbb{C}[x]}{(x-\alpha)}\simeq \mathbb{C}$ and $\mathbb{C}$ is a field, we know that $I$ is either $(0)$ or $\frac{\mathbb{C}[x]}{(x-\alpha)} $. Thus, $I$ is simple and

$\begin{bmatrix} \frac{\mathbb{C}[x]}{(x-\alpha)} && \frac{\mathbb{C}[x]}{(x-\alpha)} \\ \frac{\mathbb{C}[x]}{(x-\alpha)} && \frac{\mathbb{C}[x]}{(x-\alpha)} \end{bmatrix}\simeq \begin{bmatrix} \frac{\mathbb{C}[x]}{(x-\alpha)} && 0 \\ \frac{\mathbb{C}[x]}{(x-\alpha)} && 0 \end{bmatrix} + \begin{bmatrix} 0 && \frac{\mathbb{C}[x]}{(x-\alpha)} \\ 0 && \frac{\mathbb{C}[x]}{(x-\alpha)} \end{bmatrix}$. Hence, the ring $R_1$ is semi simple.

Is this correct so far? And how can I proceed for $n>1$? I know that $\frac{\mathbb{C}}{((x-\alpha)^n)}$ is not a field since the ideal is not maximal but I don't know what I should do next.

Thanks.

Answer 1

By Wederburn's theorem, if your ring is semisimple, it is isomrphic to $\prod M_{n_i}(D_i)$ where the $D_i$'s are division algebras over $\Bbb{C}$. But the only division algebra over the complex numbers is $\Bbb{C}$.

So your ring is semisimple iff it is isomrphic to $\prod M_{n_i}(\Bbb{C})$. But for $n>1$, these rings have non-isomorphic center.

So your ring is semisimple iff $n=1$.

Answer 2

Set $A=\mathbb{C}[x]/((x-\alpha)^n)$. Then $A$ is a commutative artinian local ring, with maximal ideal $\mathfrak{m}=(x-\alpha)/((x-\alpha)^n)$. So your question can be generalized to whether the matrix ring $$ M_2(A)=\begin{bmatrix} A & A \\ A & A \end{bmatrix} $$ is semisimple, where $A$ is a commutative artinian local ring, with maximal ideal $M$. Let's show that the ideal $$ M_2(\mathfrak{m})=\begin{bmatrix} \mathfrak{m} & \mathfrak{m} \\ \mathfrak{m} & \mathfrak{m} \end{bmatrix} $$ consists of left quasi-regular elements; an element $r$ in a ring $R$ is left quasi-regular if $1-r$ is left-invertible. So, suppose $u,v,x,y\in \mathfrak{m}$ and let's look for an inverse of $$ \begin{bmatrix} 1-u & -v \\ -x & 1-y \end{bmatrix} $$ Since $1-u-y+uy-vx$ is invertible in $A$, because $u+y-uy+vx\in \mathfrak{m}$, we're done because the matrix has invertible determinant.

Thus the given ideal is contained in the Jacobson radical of the ring; a semisimple ring has zero Jacobson radical, so a necessary condition for the ring to be semisimple is that $\mathfrak{m}=0$. In your case, it boils down to $n=1$.

---

Perhaps simpler: the ideal $M_2(\mathfrak{m})$ is not a direct summand. If it were, it would be generated (as a left or right module) by an idempotent $e$; but, as we saw, $1-e$ would be an idempotent invertible, so $1-e=1$ and $e=0$.