Let $R$ be a commutative ring with $1$ such that $R[x]$ is a PID. Show that $R$ is a field.

Question

Let $R$ be a commutative ring with $1$ such that $R[x]$ is a principal ideal domain. Prove $R$ is a field.

So I thought to take $a\in R$ and look at $aR[x]$ which is a principal ideal and obviously $a\in R[x]$ is a polynomial of degree $0$ so $\langle a\rangle=aR[x]$.

I don't have a clue how to continue. I wanted to somehow show that $a^{-1}\in R$ and so every element is invertible (because we took a random one) so $R$ is a field. But I can't seem to find a way to prove that $a^{-1}\in R$. Any ideas?

Answer 1

Your approach so far doesn't use that $R[x]$ is supposed to be a PID. You've just taken an ideal that is principle no matter what $R[x]$ is, and tried to do something with that. What you need to do is take an ideal with several generators which does not automatically simplify to a principle ideal, and then explore what happens if you assume that it does simplify.

Alternatively, you could do the contrapositive. Consider what happens to $(a,x)\subseteq R[x]$ if $a\neq0$ and $aR\neq R$. Is it principal?