Prime ideals in $\mathbb{Z}[X], \mathbb{Q}[X], \mathbb{F}_{19}[X]$. 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. 
 A: 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 $J = (X^{2018}+3X+15, X-1) = (X-1, 19)$.
A: A characterisation of U.D.s is this

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

*

*Every ascending sequence of principal ideals satisfies the ascending chain condition (i.e. every weakly ascending sequence of principal ideals is ultimately constant).

*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).
