Abstract Algebra – Semisimple Commutative Ring and 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

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.