Understanding the kernel of the evaluation map 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! 
 A: 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.
A: You are on the right track by applying: $\rm\ A/I\ $ is a domain iff $\rm\ I\ $ is prime. For $\rm\ A = R[x],\ \ I = (x)\ $ the problem reduces to showing that $\rm R[x]/(x)\cong R\:.\:$ As you surmise, this is easily achieved by exhibiting an evaluation map with kernel $\rm\ (x)\:,\:$ i.e. an evaluation which maps $\rm\ x\to \ \ldots$?   
Presumably you already know the existence (and uniqueness) of such evaluation maps for polynomial rings. In fact this universal mapping property characterizes polynomial rings. It can be employed to give slick conceptual proofs of results about polynomial rings. For a simple example see this proof that $\rm\:x\:$ is not invertible in $\rm\: R[x]\:.$ The importance of such universal characterizations will become clearer when you study universal algebra and category theory.
