Let $A$ be a semisimple $k$ algebra and $k$ is of characteristic zero. Let $G$ be a linearly reductive group over $k$ acting on $A$. Is $A^G$ semisimple ?
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1$\begingroup$ this seems like homework on semi-simple algebras $\endgroup$– VenkataramanaCommented Jan 24, 2014 at 15:04
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$\begingroup$ The question isn't research-level, and lacks context and motivation. Presumably it's assumed that $A$ is finite dimensional over $k$? Then, as Jeremy points out, the most basic theory applies and there is no need to restrict the characteristic. $\endgroup$– Jim HumphreysCommented Jan 24, 2014 at 15:58
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$\begingroup$ @JimHumphreys: The OP has posted some related questions recently that, read together, provide some context. Though it would have been better to reference the previous questions and be more explicit about the motivation (something to do with equivariant strongly exceptional collections for varieties, I'm guessing?). $\endgroup$– Jeremy RickardCommented Jan 24, 2014 at 16:59
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Yes (and characteristic zero isn't necessary).
Since $G$ is linearly reductive, $A$ is a direct sum $A^G\oplus B$ as $G$-modules, where $B$ is the sum of all non-trivial irreducible $G$-submodules of $A$, and it's easy to see that $A^G$ and $B$ are both $A^G$-bimodules.
So if $M$ is any $A^G$-module, then $M=M\otimes_{A^G}A^G$ is a direct summand of $M\otimes_{A^G}A$ as an $A^G$-module, but $M\otimes_{A^G}A$ is projective as an $A$-module, since every $A$-module is, and is therefore projective on restriction to $A^G$. So every $A^G$-module is projective, and so $A^G$ is semisimple.
answered Jan 24, 2014 at 15:36