Given a finite-dimensional semisimple algebra $A$ over algebraically-closed field $k$, Artin-Wedderburn says it is isomorphic to the direct sum of matrix rings $M_{n_i}(k)$. If $\dim(A)=n$ then $\sum_i n_i^2=n$. Let $n'=\sum_i n_i$, then I can see why the column space $k^{n'}$ is a left $A$-module and why the subspaces of $k^{n'}$ with zero's everywhere except in the entries where the $M_{n_i}(k)$ block acts, would give simple $A$-submodules.
However I'd like to see an explanation (or reference with one) for why this determines all the simple $A$-modules. So that in particular the $n_i$ determines the dimensions of every simple $A$-module. It makes sort of sense that the simple pieces of $A$ would determine the simple $A$-modules, but I'm looking for a more solid explanation.
Best Answer
For this solution you have to know what simple $M_n(k)$-modules look like : there is only one up to isomorphism, and it's $k^n$ with the obvious action.
Let $M$ be a simple module. To simplify the notations, I'll write $A= \displaystyle\prod_i M_{n_i}(k)$ (a finite product - so $=$ instead of "isomorphic to") and I'll write $e_i$ the identity of $M_{n_i}(k)$ sitting in $A$, so that $\sum_i e_i = 1$ and $e_ie_j=\delta_{ij}e_i$; and also $xe_i = e_ix_i$ for $x=(x_i)_i\in A$
Then look at $e_iM$ for each $i$. Clearly, because of $xe_i = e_ix_i$ this is an $A$-submodule of $M$.
Moreover, if $i\neq j$, by $e_ie_j = 0$ and $e_i^2=e_i$, we get $e_iM\cap e_jM = 0$ so that (by simplicity) at most one $e_i M$ is nonzero; and since $M=\sum_i e_iM$, exactly one $e_iM$ is nonzero and $e_iM = M$.
Now by $s\cdot m := e_is m$ we get that $M$ is a $M_{n_i}(k)$-module, and it's easy to note (by $e_iM=M$) that it is a simple $M_{n_i}(k)$-module. It follows that $M\simeq k^{n_i}$ with the obvious action of $M_{n_i}(k)$; and so by tracing back we see that $A$ acts on $M$ by projecting onto $M_{n_i}(k)$ and then acting in the obvious way.