# Is the left ideal of a semi-simple ring with identity a semi-simple ring?

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"?

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.
• 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 $RJ\subseteq J$. Dec 17, 2023 at 2:37