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Let R and S be rings. Show that R x S is semisimple if and only if both R and S are semisimple.

I have the converse direction ($\Leftarrow$). It is the other direction ($\Rightarrow$), that I know I have use the Wedderburn-Artin Theorem. I just am stuck in that direction.

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    it was also impolite there too :)2011-05-02

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Show that $R\times S$ is a sum of simple $R\times S$-submodules, by noticing that each simple $R$-submodule of $R$ can be viewed as an $R\times S$-submodule of $R\times S$, and similarly for $S$-submodules. This will prove that $R\times S$ is semisimple when $R$ and $S$ are.

To prove the converse, show first that $R$ is an $R\times S$-module in a natural way, in such a way that $R$-submodules of $R$ and $R\times S$-submodules are exactly the same thing. Then, if we suppose that $R\times S$ is semisimple, then the $R\times S$-module $R$ is sum of simple $R\times S$-submodules, which are $R$-submodules: it follows that $R$ is semisimple; same thing with $S$, of course.

N.B.: in particular, there is absolutely no need to use Artin-Wedderburn for this!

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    @ Mariano, thank you.2011-05-02