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.

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

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.

  • $\begingroup$ Thanks for the explanation, that helps a lot. Don't suppose you also know of any references that discuss this too? $\endgroup$ – Ted Jh May 5 at 11:52
  • $\begingroup$ References that discuss which aspect? $\endgroup$ – Max May 5 at 11:54
  • $\begingroup$ 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.. $\endgroup$ – Ted Jh May 5 at 12:04
  • $\begingroup$ 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 $A_i$. $\endgroup$ – Max May 5 at 13:02
  • $\begingroup$ 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. $\endgroup$ – Ted Jh May 5 at 14:00

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