# A PID is a semisimple ring iff it is a field

I am trying to prove that a PID $$R$$ is a semisimple ring iff it is a field. Clearly any field is semisimple. I am not sure about the converse. By Artin-Wedderburn, $$R$$ is a product of matrix rings over divison rings. But there can only appear one factor (since $$R$$ is a domain), and the matrix ring must be a ring of $$1$$ times $$1$$ matrices (again since $$R$$ is a domain). In other words, $$R$$ is isomorphic to a division ring. Thus $$R$$ is a field. Is my argument correct?

• If you are not assuming that $R$ is commutative, you should specify exactly what you mean by PID. If not, then all but the final step of your argument works - but why does R being a division ring imply that it is a field? In fact, the quaternions should be a counterexample: all its ideals are principal, and it is simple, thus semisimple, but not a field. Sep 25, 2023 at 11:44
• @SomeCallMeTim I am assuming that a PID is commutative. Is there any textbook/article that doesn’t? A PID to me is an integral domain in which every ideal is principal. An integral domain is a non-zero commutative ring without zero divisors. Sep 25, 2023 at 11:53
• You are correct, those are the most common definitions, but sometimes you want to extend such things a little. But with this assumption, your proof seems very obviously true, so what are your doubts? Sep 25, 2023 at 13:57
• Just double-checking. And maybe there is a better argument… Sep 25, 2023 at 16:47
• @Margaret, there are books which do not assume commutativity. They usually have the words "non commutative" in their title! Sep 25, 2023 at 21:05

This argument is fine but it's possible to avoid Artin-Wedderburn, as follows. Let $$r \in R$$ and consider the ideal $$(r)$$. This is an $$R$$-submodule of $$R$$ so by semisimplicity it has a complement, which is another $$R$$-submodule and hence another ideal, which is principal; call it $$(s)$$. Then $$R$$ is the direct sum of $$(r)$$ and $$(s)$$, meaning that every element has a unique representation as a linear combination $$rx + sy$$.
But $$rs = r(s) = s(r)$$, which violates uniqueness unless $$rs = 0$$. Since $$R$$ is a domain $$r = 0$$ or $$s = 0$$. The conclusion is that $$(r)$$ is either zero or the unit ideal and hence that every nonzero element of $$R$$ is invertible, so $$R$$ is a field.