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

$R$ 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!!

• I think it would help if you can show that $R[x]/xR[x] \cong R$. – matt stokes Sep 13 '18 at 15:16
• Related – rschwieb Sep 14 '18 at 14:01

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

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.