Product of semisimple rings is semisimple.

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A semisimple ring $R$ with $1$ (but not necessarily a commutative one) considered as left-$R$ module is a direct sum

$$R\cong L_1\oplus L_2 \oplus \cdots\oplus L_n$$ such that for some $e_i$ are in $R$

  • $L_i=Re_i$ simple left ideal of $R$
  • $e_ie_j=0$ for different $i,j$
  • $e^2_j=e_j$
  • $\sum e_i=1$ the unit of the ring

Now I want to show $R_1,R_2$ are semisimple rings with $1$ if and only if $R_1\times R_2$ is semisimple.

My problem is how to see their structure, if $R_1,R_2$ semisimple how to construct and see the decomposition of $R_1\times R_2\cong L_1\oplus L_2 \oplus \cdots\oplus L_n\times K_1\oplus K_2 \oplus \cdots\oplus K_m$

$L$s for $R_1$, $K$s for $R_2$ but these summands have $R_1$ and $R_2$ structre.

By the way how to solve this problem in more efficient and clever way, can we use homological algebra, another decompositions?

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$L$s for $R_1$, $K$s for $R_2$ but these summands have $R_1$ and $R_2$ structure.

That's just fine, because you can define an $R$ structure on $L_i$ as $(r_1,r_2)x=r_1x$ and an $R$ structure on $K_i$ as $(r_1,r_2)x=r_2x$ and it turns out that with this action they are all simple as $R$ modules too.

By the way how to solve this problem in more efficient and clever way

I think this is about as efficient as it gets. If by clever you mean "fancy" or "sophisticated" then I think that is almost always possible.