# 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.

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 $$J = (X^{2018}+3X+15, X-1) = (X-1, 19)$$.
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).