Isomorphic algebras of endormporphisms

Question

Let $R$ be a simple algebra, $M$ a simple $R$-module and $N$ a simple $\mathcal{M_n}(R)$-module (always considering finite dimension). Prove that $End_R(M)$ and $End_{\mathcal{M_n}(R)}(N)$ are isomorphic algebras.

I have tried to prove it as a consequence of Wedderburn-Artin theorem, but I have not got it. Could someone help me?

Answer 1

So the main observation is that there is only one simple module (up to isomorphism) over a simple (finite-dimensional) algebra.

If $R=\mathcal{M}_k(D)$ is simple then all simple modules are isomorphic to a minimal left ideal of $R$, which in turn is isomorphic (as an $R$-module) to $D^k$, see here. Note that $D^k$ is a left $R$-module via standard matrix multiplication.

Now since $\mathcal{M}_n(R)=\mathcal{M}_{kn}(D)$ is simple as well then the same holds for it. So all that you have to prove is that $End_R(D^k)$ is isomorphic to $End_{\mathcal{M}_n(R)}(D^{kn})$.

This can be simplified by proving the following:

Lemma. For any associative unital algebra $D$ and any $n$ we have an isomorphism of algebras $End_{\mathcal{M}_{n}(D)}(D^n)\simeq D^{op}$ where $D^{op}$ is the algebra created from $D$ by reversing multiplication.

Proof. If $f:D^n\to D^n$ is a $\mathcal{M}_n(D)$ endomorphism then

$$f\left(\begin{matrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{matrix}\right)= \left(\begin{matrix} x_1 & 0 & \cdots & 0 \\ x_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ x_n & 0 & \cdots & 0 \\ \end{matrix}\right) f\left(\begin{matrix} 1 \\ 0 \\ \vdots \\ 0 \end{matrix}\right) $$

meaning that $f$ is fully determined by value on $(1,0,\ldots,0)$. So let $f(1,0,\ldots,0)=(y_1,\ldots, y_n)$. It follows that

$$\left(\begin{matrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{matrix}\right)=f\left(\begin{matrix} 1 \\ 0 \\ \vdots \\ 0 \end{matrix}\right)= \left(\begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ \end{matrix}\right) f\left(\begin{matrix} 1 \\ 0 \\ \vdots \\ 0 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ \end{matrix}\right) \left(\begin{matrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{matrix}\right)= \left(\begin{matrix} y_1 \\ 0 \\ \vdots \\ 0 \end{matrix}\right) $$

and so for $\lambda=y_1$ we have that $f(v)=v \lambda$ for any $v\in D^n$. So it is a scalar multiplication (note that from the right, the order matters), denote it by $f_\lambda$. This shows that

$$\tau:D^{op}\to End_{\mathcal{M}_{n}(D)}(D^n)$$ $$\tau(\lambda)=f_\lambda$$

is a surjective function. I leave as an exercise that it is an injective algebra homomorphism. Note that we have to reverse multiplication in $D$ due to $f_{xy}=f_y\circ f_x$. $\Box$