A PID is a semisimple ring iff it is a field

Question

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?

Answer 1

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.