Multiplicity of the simple $R$-module $M$ in the semisimple ring $R$

Question

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:

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 $M$?

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.

Answer 1

An abelian group $M$ has a right $D$ module structure iff it has a left $D^{op}$ module structure.

If $M$ is a left $R$ module, then there is the natural operation of $D$ on the right of $M$. Moving over to the opposite ring, this becomes a left $D^{op}$ structure on $M$.

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Here's the way to see where the dimension $n$ comes from for a simple Artinian ring. It's easy to prove that all simple left $R$ modules are isomorphic to each other, so you can start by decomposing $R$ into minimal left ideals. Suppose $R=\oplus_{i=1}^n S$ is a direct sum of simple left $R$ modules. Then $D=End(_RS)$ is a division ring, and since $R^{op}\cong End(_RR)=End(\oplus_{i=1}^n S)\cong M_n(End(_RS))=M_n(D)\cong M_n(D^{op})^{op}$. Then we have that $M_n(D^{op})\cong R$ as rings. But clearly ($D^{op})^n$ is a simple left $M_n(D^{op})$ $(=R!)$ module, therefore $S\cong (D^{op})^n$ as left $R$ modules.

Notice that $D^{op}$ is embedded as the constant diagonal matrices in $R$, and multiplication by these matrices is the same as looking at $(D^{op})^n$ as a left $D^{op}$ vector space. Switching to the right, it is a right $D$ vector space.

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Now for general semisimple rings, you first just split the ring into simple rings. Then it's easy to show that the simple left modules are just exactly the simple modules for each simple ring in that factorization. So, you can reduce the general case to finitely many cases like the one above.

Answer 2

Here is my attempt to describe the left $\text{End}(_RM)^{op}$-module structure of $M$.

Wedderburn's theorem gives us and isomorphism: $R\cong\text{End}(_RR)^{op}\cong\text{End}(_RM^n)^{op}\cong M_n(\text{End}(_RM))^{op}\cong M_n(\text{End}(_RM)^{op})$ (the first isomorphism takes $r\in R$ to the endomorphism of right multiplication by $r$ and the last one is transposition).

The diagonal injection $\text{End}(_RM)^{op}\hookrightarrow M_n(\text{End}(_RM)^{op})$ gives any $M_n(\text{End}(_RM)^{op})$-module a structure of an $\text{End}(_RM)^{op}$-module.

Wedderburn's isomorphism gives a way to look at any $R$-module as an $\text{End}(_RM)^{op}$-module by the composite map $\text{End}(_RM)^{op}\hookrightarrow M_n(\text{End}(_RM)^{op})\cong R$. To describe it more concretely, we need to follow Wedderburn's isomorphism backwards. Since we're only looking at the image of the diagonal injection $\text{End}(_RM)^{op}\hookrightarrow M_n(\text{End}(_RM)^{op})$, the transposition does nothing. We will allow ourselves to represent elements of $R$ using coordinates corresponding to the module decomposition $R=M^n$. So, $d\in\text{End}(_RM)^{op}$ maps to the element $r\in R$ such that for any $(m_1,\dotsc,m_n)\in M^n$ it holds that $(m_1,\dotsc,m_n)r=(d(m_1),\dotsc,d(m_n))$. Choose $(e_1,\dotsc,e_n)\in M^n$ representing $1\in R$ and get $r=(d(e_1),\dotsc,d(e_n))$.

As I described it, looking at the $\text{End}(_RM)^{op}$-structure of $M$ is not a very useful way to determine $n$ because to understand it, one already need to know the module decomposition $R=M^n$, which includes knowing $n$.

Maybe adding additional structure to $R$ (like making it an algebra over a field) would make it possible to understand the dimension of $M$ over $\text{End}(_RM)^{op}$ without actually understanding the action itself. To do so, we'd probably have to check some of the homomorphisms above are algebra homomorphisms.

The idea behind the previous paragraph is that if $A$ is a division algebra over a field $k$ and $M$ is a module over $A$ then we get an induced $k$-module structure of $M$. It is then enough to understand the $k$-module structure of $A$ and the induced $k$-module structure of $M$ and use the equtation $[M:k]=[M:A]\cdot [A:k]$ to extract $n=[M:A]$.

To justify looking at $k \subset A \subset M$ as a tower of subspaces, we need to note that the algebra action homomorphism $k \hookrightarrow A$ coincides with $k \hookrightarrow \text{End}(_AM)^{op} \rightarrow A$.