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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?

Loye94
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  • If you explain your comment about what you tried with the W-A theorem, your question might not get close votes for lack of context. What did you see? – rschwieb May 07 '19 at 20:35
  • I don't know hoy to use it. I only get that $R$ is isomorphic to $\mathbb{M_m(D)}$ with $D$ a division algebra and the same for $\mathbb{M_n}(R)$ but I don't know how to use this onfomation and if this is the correct way to solve the prblem. – Loye94 May 07 '19 at 21:15

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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 $f_{xy}=f_y\circ f_x$. $\Box$

freakish
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  • +1 This approach is sound but there are a few errata. The first sentence makes it sound like any simple algebra has only one isotype of simple left module, but it's worth knowing that this is not true. Finite dimensionality (of course the context here) is critical for that to hold. Then "isomorphic to a minimal ideal" should read "isomorphic to a minimal left ideal." (That of course also required finite dimensionality.) Finally, rather than "$f(v)=\lambda v$, it should read $f(v)=v\lambda$, according to the matrix equation referenced. – rschwieb May 08 '19 at 15:45
  • @rschwieb I've fixed "simple algebra" and "minimal ideal" issues. But I don't agree with $f(v)=v\lambda$ notation: everything here is acting from left, including scalar multiplication of $D$ on $D^n$. I'm not sure why would you multiply by scalar from right? – freakish May 08 '19 at 15:52
  • Remember, $\lambda$ is from a division ring, so the order matters. According to your equation, $f\left(\begin{bmatrix}x_1\x_2\\vdots\x_n\end{bmatrix}\right)=\begin{bmatrix}x_1&0&\ldots\ x_2&0&\ldots\ \vdots \ x_n&0&\ldots\end{bmatrix}\begin{bmatrix}\lambda \ 0 \ 0 \ \vdots\end{bmatrix}=\begin{bmatrix}x_1\lambda\x_2\lambda\\vdots \x_n\lambda\end{bmatrix}$. – rschwieb May 08 '19 at 17:35
  • @rschwieb ah, yes, you are right. – freakish May 08 '19 at 17:41
  • @rschwieb hmm, actually now $\tau$ is an antiisomorphism or am I missing something again? $f_{xy}=f_yf_x$. Maybe I should write $D^{op}$ instead. Are they isomorphic? – freakish May 08 '19 at 17:50
  • Hmm... yeah, that could be. It's easy to mix these two cases up. It is definitely not always the case that a division ring is isomorphic to its opposite. But either way you're fine, because both algebras would be isomorphic to the same ring. – rschwieb May 08 '19 at 18:15