Quotient of semi-simple rings I can't seem to find a reference for the following statement:

A quotient ring of a (left) semi-simple ring (left artinian with no nonzero nilpotent ideals) is a semi-simple ring. 

Does this follow from quotients of semi-simple modules being semi-simple?
A quotient ring of a left artinian ring is left artinian. So we only have to show that the quotient contains no nonzero nilpotent ideals. I'm looking for something more low-level than $\text{rad}(R/I) = \text{rad}(R) + I$, which I seem to remember to be true. Can I somehow use the characteristic that every left ideal is generated by an idempotent element?


Can I somehow use the characteristic that every left ideal is generated by an idempotent element?

Yes, that is a good approach. Let $T$ be a right ideal of $R$ containing $I$ properly. Then $T\setminus I$ contains a nonzero idempotent, and $e+I\in (T+I)^n$ shows that no right ideal is nilpotent.
Alternatively, you could use the Artin-Wedderburn theorem to note that an ideal is a product of simple components, and the quotient is the complementary set of simple components.
Alternatively you could note that $R$ and all of its quotients are Von Neumann regular, and an Artian VNR ring is semisimple.
