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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 is very impolite to post in the imperative (giving orders), simply quoting your homework problem. You should give some context (what course is this for), and what you have tried or why you are stuck. "Help needed" is hardly sufficient.2011-05-02
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    @ Arturo, I'm sorry, this is my first time on here. I was directed here from someone else. I apologize for any trouble.2011-05-02
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    It's no trouble. But kindly edit the question to take out the unintended rudeness. Next time, you'll know to do it from the get go. (-:2011-05-02
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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