I'm confused about the conclusion of Wedderburn's structure theorem for semisimple rings.  

Let's consider the special case where $R=M^n$ as modules for some simple module $M$.
Wedderburn's theorem says that $R=M\_n(D)$ for $D=\text{End}\_R(M)^{o}$.
Thus, $R=(D^n)^n$ as modules. I see that $D^n$ is a simple $R$-module.

So, since all simple $R$-modules appear in the decomposition of the $R$-module $R$ as a direct sum of simple $R$-modules, I conclude that $M\cong D^n$. In particular, $M$ can be seen as a $D$-module and $n=\dim\_D(M)$ (we can talk about dimension since $D$ is a division ring by Schur's lemma).

My question:

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> Remembering that $D=\text{End}\_R(M)^{o}$, is there a natural way to view $M$ as a $D$-module? What is the ring homomorphism $\text{End}\_R(M)^o\rightarrow\text{End}\_{\mathbb{Z}}(M)$ that describes the action of $D$ on $M$?
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There is an obvious way to view $M$ as an $\text{End}\_R(M)$-module since $\text{End}\_R(M)$ is a subring of $\text{End}\_{\mathbb{Z}}(M)$, but since $D$ is the opposite ring relative to $\text{End}\_R(M)$ and not $\text{End}\_R(M)$ itself, I am confused.

**Motivation**: I'd like to be able to look at a semisimple ring and ask how many times a simple module $M$ appears in its decomposition by looking at $M$ as an $\text{End}\_R(M)^o$-module and computing the dimension.