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I know that the maximal ideals in $\mathbb{Z}[X]$ are given by the ideals $(p,f)$ such that $f$ is a monic integral polynomial which is irreducible in $\mathbb{F}_{p}[X]$ and $p$ any prime number. Let $g\in\mathbb{Z}[X]$ be primitive and irreducible in $\mathbb{Q}[X]$. I want to find all maximal ideals in $\mathbb{Z}[X]$ containing this $g$.

Suppose that we have a maximal ideal containing $g$, then $g = p\alpha + f\beta$ for some $\alpha,\beta\in\mathbb{Z}[X]$. Since $g$ is primitive and irreducible we see that $\beta\neq 0$. Then $g = f\beta \mod p$.

JanBakfiets1
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    Yes, $(p,f)\supseteq (g),$ in $\Bbb Z[x] \iff (f)\supseteq (g),$ in $,\Bbb F_p[x]$ $\iff f\mid g,$ in $,\Bbb F_p[x],,$ so they correspond to maximal (= irred) factors of $g$ in $\Bbb F_p[x].\ $ Generally see the Ideal Correspondence Theorem. – Bill Dubuque Mar 04 '20 at 14:59
  • @Bill Dubuque Does the fact that $f$ is monic also imply some extra restrictions? Since for instance if we look at $p=3$ and the irreducible primitive polynomial $5x+1$, then its irreducible factor is $2x+1$ in $\mathbb{F}_{3}[X]$. And consequently $(3,2x+1)$ is not a maximal ideal of $\mathbb{Z}[X]$ since $2x+1$ is not monic. – JanBakfiets1 Mar 04 '20 at 15:12
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    Over a field we can always normalize ideal generators to be monic by scaling by them by the inverse of the lead coef - see unit normalization, e.g. $,(2x+1) = (x+2),$ over $\Bbb F_3\ \ \ $ – Bill Dubuque Mar 04 '20 at 15:24

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