# Artin-wedderburn and simple modules

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.

• You're right, will amend. – Ted Jh May 5 at 13:34

## 1 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.

• Thanks for the explanation, that helps a lot. Don't suppose you also know of any references that discuss this too? – Ted Jh May 5 at 11:52
• References that discuss which aspect? – Max May 5 at 11:54
• References that discuss how the Artin-wedderburn theorem can be used to determine an algebras/rings simple modules, in the way you have given in your answer? As I have seen many texts on Artin-wedderburn, but none said what you just have.. – Ted Jh May 5 at 12:04
• Unfortunately no, I don't know the references; I got that from a course I followed. However, you could try to see what I said in a more general setting : try to look for instance at a general finite product of rings $\prod_i A_i$ and try to characterize in a similar way what its simple modules are. The catch is that it depends on the simple modules of each $A_i$. – Max May 5 at 13:02
• Ok thanks! I have actually just noticed the proof of the Artin-wedderburn theorem in some university course notes actually contains a fair amount of useful material on this point - so thought I'd point that out. – Ted Jh May 5 at 14:00