$R$ is a commutative ring, show $xR[x]$ is prime iff $R[x]$ is an integral domain

Question

$R$ is a commutative ring, show $xR[x]$ is prime iff $R[x]$ is an integral domain and show $xR[x]$ is maximal iff $R[x]$ is a field.

Now, I'm trying to use the theorem that a quotient ring is an integral domain iff the ideal is prime, but this situation is a little different. Once I get one of the arguments down then i'm sure the only will follow by adopting a common strategy of how to turn the proof of why the quotient ring that is formed by using a maximal ideal is a field and adapting it to this situation. Any help is appreciated!!

Answer 1

Consider $f:R[X]\rightarrow R$ defined by $f(P)=P(0)$, it is a morphism and its kernel is $XR[X]$.

Answer 2

I'd encourage proving these lemmas, which are all worth knowing:

$R[x]/(x)\cong R$

and

$R/I$ is a domain iff $I$ is a prime ideal.

and

$R[x]$ is a domain iff $R$ is a domain

If you chain these together, you will have your answer to the first question.

The second question about $R[x]$ being a field is bogus. It is never a field since it always has the nonzero proper ideal $(x)$.

---

There is another plausible interpretation of your problem, if you perhaps made a mistake when writing it:

Prove the following

1. $R$ is a domain iff $(x)$ is prime in $R[x]$.

2. $R$ is a field iff $(x)$ is maximal in $R[x]$.

Those both also follow from the above lemmas.