Commutative ring is semisimple iff it's isomorphic to a finite direct product of fields.

Question

I am trying to prove the following:

Let $R$ be a commutative ring. Prove that $R$ is semisimple if and only if it is isomorphic to a direct product of a finite number of fields.

Suppose $R$ is a semisimple commutative ring. By Wedderburn, $R \cong \bigoplus_{i=1}^n M_{n_i}(D_i)$, where $D_i$ is a division ring for each $i$. Since $R$ is commutative, each $M_i:=M_{n_i}(D_i)$ is a simple commutative ring. I'll show that each element has an inverse: take $a \in M_i$ and consider $aM_i$, then $aM_i=M_i$ so there is $b \in M_i : ab=1$. With the same argument, we get that there is $c \in M_i$ with $bc=1$, but then $a=a.1=a(bc)=(ab)c=1.c=c$. It follows each $M_i$ is a field.

I don't know how to show the other implication, any help would be appreciated.

Answer 1

One way is to show that every ideal of a finite direct product of fields is a direct summand. Can you see why?

Answer 2

A finite product of fields is clearly Artinian.

It's also then clearly Jacobson semisimple (since it has no nonzero nilpotent elements, and the radical is nilpotent.)

Since Jacobson radical zero and Artinian combine to make "semisimple," we're done.