$I$ is ideal in $\mathbb Z[x]$. Does $\mathbb Z[x] / I$ being integral domain imply $\mathbb Z[x] / I$ is a field?

Question

$I$ is ideal in $\mathbb Z[x]$. Does $\mathbb Z[x] / I$ being integral domain imply $\mathbb Z[x] / I$ is a field? The other way around is correct.

But I'm not certain whether in this case, integral domain $\rightarrow$ field.

And when $\mathbb F[x]/ I$ being integral domain implies it's also a field?

Answer 1

Given a commutative ring $R$ and an ideal $I$ of $R$, $R/I$ is an integral domain iff $I$ is a prime ideal.

$R/I$ is a field iff $I$ is a maximal ideal.

Thus, $R/I$ being an integral domain implies that $R/I$ is a field when $I$ being prime implies that $I$ is maximal. In particular, this occurs in PIDs (Provided that $I$ is non-zero).

Note that $\mathbb Z[x]/(x) \cong \mathbb Z$, which is not a field, so this does not hold in this particular case (Because $(x)$ is prime, but not maximal.)

Also, for your second question, if $F$ is a field, this is true (Again, assuming that $I$ is non-zero), as the ring of polynomials over a field is a PID.

Answer 2

For your example, what about $I=(x)$?

$R/I$ is a field iff $I$ is maximal ideal, and $R/I$ is a domain iff $I$ is a prime ideal. Can you see the relationship between the two things now?

Answer 3

More generally, let $R$ be any PID. Then the quotient $R/I$ is an integral domain iff $R/I$ is a field.

One way to see it is this answer.