TITLE: Infinite direct product of semisimple rings is semisimple?
QUESTION [5 upvotes]: It's clear that the finite direct product of semisimple rings in again semisimple because in the finite case the direct producto looks isomorphic to the direct sum, but I want to know if there is a problem considering an infinite direct product.

REPLY [3 votes]: There are tons of ways to see this directly.
Given $R=\prod_{i\in \kappa} S_i$ where each $S_i$ is semisimple ($\kappa$ is an infinite well-ordered index set), then $I_i=\oplus \{S_j\mid j \leq i\}$ is obviously an infinite, strictly increasing chain of right ideals of $R$. So the ring is not right Noetherian.
You can also build a strictly decreasing chain of right ideals in the a similar way, showing it is not right Artinian.
Finally, you can show that $\bigoplus_{i\in \kappa}S_i$ is a proper essential right ideal of $R$. (In semisimple rings, the only essential right ideal is the trivial ideal consisting of the entire ring. Essential right ideals cannot split out of the ring.)

but I still don't see why a semisimple ring will be noetherian. 

If $R$ is a semisimple ring, then every right ideal splits out of $R$, and is equivalent to saying that it is generated by a single idempotent. Thus every right ideal is finitely generated (in fact, every right ideal is cyclic.)
Alternatively, you could cite the Hopkins-Levitzki theorem that right Artinian rings are right Noetherian.
Alternatively, you could note that after you have written $R$ as (necessarily) finite direct sum of simple right ideals, this makes a composition series for the ring as a module over itself, and we all know that modules with composition series are Artinian and Noetherian.