If $f(x)$ is irreducible over $\mathbb{Z}$ then prove that $\langle f(x) \rangle$ is not maximal ideal

Question

I know that if $f(x)$ is irreducible then $\langle f(x) \rangle$ is a prime ideal. Then I thought: is it maximal? Then I search about it, I find that it is not maximal ideal but cannot find any proof.

Any help is appreciated. Thanks.

Answer 1

If $f$ has degree 0, i.e. $f$ is a constant polynomial with a prime value $p$, then $\langle p \rangle \subsetneqq \langle p, x \rangle \subsetneqq \langle 1 \rangle$.

If $f$ has positive degree, let $p$ be a prime number not dividing the leading coefficient of $f$. Then $\langle f \rangle \subsetneqq \langle f, p \rangle \subsetneqq \langle 1 \rangle$.

Answer 2

An ideal $I \subset \mathbb Z [X]$ is maximal if and only if the quotient $\mathbb Z [X] / I$ is a field.

Suppose $f \in \mathbb Z [X]$ is irreducible and let $I = (f)$ the ideal generated by $f$.

If $f$ is degree zero (constant) then $\mathbb Z[X] / (f) \simeq \mathbb Z/(f)[X]$. Do you see why this is not a field? (hint: find an inverse of $X$)

If $f$ is of nonzero degree then $\mathbb Z[X] / (f) \simeq \mathbb Z[\alpha]$ where $\alpha$ is a root of $f$. Do you see why this is not a field? (hint: find an inverse of 2 or 3 or any $n \in \mathbb Z$).