Prime ideals in $\mathbb{Z}[X], \mathbb{Q}[X], \mathbb{F}_{19}[X]$.

Question

Old exam question

Consider the following ideals :

$I = (X^{2018}+3X+15)$;

$J = (X^{2018}+3X+15, X-1)$;

$K = (X^{2018}+3X+15, 19)$.

Determine whether they are prime ideals in $\mathbb{Z}[X], \mathbb{Q}[X], \mathbb{F}_{19}[X]$, respectively.

As $\mathbb{Z}[X]$ is a UFD, $X^{2018}+3X+15$ satisfies Eisenstein's criterion at $p=3$, so is it irreducible in $\mathbb{Z}[X]$. Now for PIDs, we know that all irreducibles are prime, but as $\mathbb{Z}[X]$ is not a PID, we cannot invoke the equivalence $$(p) \text{ is a prime ideal} \iff p \text{ is a prime element} \iff p \text{ is irreducible}.$$ Is there another way to determine that $I$ is prime? If so, would that imply it is irreducible in $\mathbb{Q}[X]$? For $\mathbb{F}_{19}[X]$ I'm not sure whether it's different then for $\mathbb{Z}[X]$, as "reduction modulo 19" does nothing in this case.

I do see that for $K$, the case $\mathbb{F}_{19}[X]$ reduces to the same case for $I$, as $\bar{19} = \bar{0}$, so this does not add anything to the ideal.

For $J$, I see no feasible strategy at all.

Answer 1

In fact, the implication irreducible $\implies$ prime is true for UFDs, as discussed in this question. (Moreover, assuming every element factors into irreducibles in a domain $R$, then $R$ is a UFD iff every irreducible is prime. See this page of Stacks Project for a proof.)

Hint for $J$: Divide $X^{2018} + 3X + 15$ by $X-1$ or, better yet, realize that you can find its remainder by simply plugging in $X=1$. This shows that $X^{2018} + 3X + 15 = q(X) (X-1) + 19$ for some $q(X) \in \mathbb{Z}[X]$, so $J = (X^{2018}+3X+15, X-1) = (X-1, 19)$.

Answer 2

A characterisation of U.D.s is this

A ring is a U.F.D. if and only it satisfies the following two conditions:

1. Every ascending sequence of principal ideals satisfies the ascending chain condition (i.e. every weakly ascending sequence of principal ideals is ultimately constant).
2. Irreducible elements are prime.

Also, it is known that a polynomial in $\mathbf Z[X]$ is irreducible in that ring if and only if ithe g.c.d. of its coefficients is $1$ and it is irreducible in $\mathbf Q[X]$ (Gauß' lemma).