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In this post a semi-simple ring means a ring $R$ with $1$ s.t. the module $_RR$ is semisimple (each submodule is a direct summand).

Summary: If $R$ is semisimple, $I$ is a left ideal with an identity. (That is, exists $e\in I$, $ex=x=xe$ for all $x\in I$.), then is $I$, viewed as a ring, semi-simple?


Motivation: It is easy to give an instance that a semi-simple ring can have a subring not semi-simple. For instance let $k$ be a field. $k(x)$ is a field, and a fortiori semi-simple. But its subring $k[x]$ is not semi-simple, because $k[x]$ is not artinian (consider $\{\langle x^n\rangle\}$), and semi-simple rings are artinian. So, what if we strengthen "subring" to "a left ideal with an identity"?

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It is a well-known theorem that, if $R$ is a semisimple ring then every $R$ module is semisimple. In particular, $I$ is a semisimple $R$ module. So given an ideal $J$ of $I,$ we note that $J$ is $R$ submodule of $I,$ hence a direct summand of $I$ say $I = J \oplus J'.$ Then $J'$ is an ideal of $I,$ thus $I$ is a semisimple ring.

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  • $\begingroup$ Could you please explain why $J$ is a submodule (i.e. left ideal) of $R$? I think we can only obtain $IJ\subseteq J$, not $RJ\subseteq J$. $\endgroup$
    – Asigan
    Dec 17, 2023 at 2:37

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