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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!

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Can you exhibit any nontrivial elements of ker(e_r)? – Qiaochu Yuan Dec 7 '10 at 5:08
For a in R, the polynomial f(x) = ax-ar is in the kernel of e_r – Bey Dec 7 '10 at 5:17
No! For $f(x)= ax-ar$ to be in $(x)$, there has to be a polynomial $g(x)$ such that $f(x)=xg(x)$. And there may not be one (depending on what $a$ is and on $R$). Why do you say it is "certainly in $(x)$"? – Arturo Magidin Dec 7 '10 at 5:21
Haha, I feel horrible, but I edited my comment just before I saw this (you posted 56 seconds ago). I realized my mistake and was hoping to take it out before anyone saw. Whoops! – Bey Dec 7 '10 at 5:23
up vote 6 down vote accepted

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.

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I started thinking in this direction while considering Qiaochu's comment above. I didn't feel quite right just picking and choosing where to evaluate, but I'm not sure why. Thank you for your clarification. =) – Bey Dec 7 '10 at 5:25
@Bey: Oh, you can pick and choose to your heart's content. To show that $R[x]/(x)\cong R$, you just need to find a homomorphism $f\colon R[x]\to R$ that is onto and has kernel $(x)$; it can be as simple or as complicated as you want, but you can certainly choose whichever one you prefer. It was a very good idea to look at the evaluation maps, since $R$ is commutative. It was just a matter of looking at the right evaluation map. – Arturo Magidin Dec 7 '10 at 5:28
The factor theorem is true in every ring. Indeed $\rm x-r\ |\ f(x)-f(r)\ $ via $\rm\ x-r\ |\ x^n-r^n\ $ and linearity. In particular, in your example $\rm\ 2\ (x-3) = 2\ x \ \in \ \mathbb Z/6\ $. – Bill Dubuque Dec 7 '10 at 5:39
@Bill: Of course. If the ring doesn't have a 1, then you're in trouble, but if the leading coefficient is a unit, then I should have known better. Thanks for the smack in the back of the head. – Arturo Magidin Dec 7 '10 at 5:42
@Bey: it seems to me when some students encounter these types of arguments, they think it is too easy, so they don't believe it can work. People are somehow convinced that math ought to be hard. Not true at all! Math is very easy compared to some of the other things people try to do, like macroeconomics. – Qiaochu Yuan Dec 7 '10 at 5:51

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.

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