Let $R$ be a non-zero commutative ring. Prove that the ideal $(x)$ of $R[x]$ is prime if and only if $R$ is an integral domain.
I'm working on the left-to-right direction right now. I know that $R[x]/(x)$ is an integral domain since $(x)$ is prime. So I want to fix $r\in R$ and use the evaluation map $e_r:R[x]\to R$ given by $f(x)\mapsto f(r)$, a surjective ring homomorphism, and apply the first isomorphism theorem. But I'm having trouble with the kernel of $e_r$.
First of all, is it even true that $\ker{(e_r)}=(x)$? If so, any hints you could drop would be great. Thanks!
Best Answer
No, the kernel of $e_r$ consists of all polynomials that are zero at $r$: remember, $\mathrm{ker}(e_r) = \{ f(x)\in R[x]\mid e_r(f(x)) = f(r) = 0\}$. It will contain $(x-r)$ (assuming your ring has an identity), but in general rings it may be all sorts of things (edit: as Bill Dubuque points out, the Factor Theorem will be on hand in any ring with $1$, but in rings without $1$ weird things may occur). But that does not really matter here, because you don't want to look at an arbitrary $r$ in $R$. You want to pick a particular $r$ that gives you the kernel you want.
And for you to get $(x)$ as the kernel, you should look at $e_0$, evaluation at $0$. Trivially, $(x)$ is contained in the kernel; showing that every element of the kernel is in $(x)$ should be straightforward.