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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?

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  • $\begingroup$ 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. $\endgroup$ Sep 25, 2023 at 11:44
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    $\begingroup$ @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. $\endgroup$
    – Margaret
    Sep 25, 2023 at 11:53
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    $\begingroup$ 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? $\endgroup$ Sep 25, 2023 at 13:57
  • $\begingroup$ Just double-checking. And maybe there is a better argument… $\endgroup$
    – Margaret
    Sep 25, 2023 at 16:47
  • $\begingroup$ @Margaret, there are books which do not assume commutativity. They usually have the words "non commutative" in their title! $\endgroup$ Sep 25, 2023 at 21:05

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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.

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