Prime ideal in $\mathbb{Z}[X]$ with prime constant polynomials.

Question

Let $P ◅ \mathbb{Z}[X]$ be a prime ideal with $P \cap \mathbb{Z} = p\mathbb{Z}$, for a prime number $p$. I have proved by looking at the natural homomorphism $$\phi : \mathbb{Z}[X] \longrightarrow \mathbb{F}_p[X] $$ that takes all coefficients modulo $p$, that $\mathbb{Z}[X]/(p) \cong \mathbb{F}_p[X]$. As $\mathbb{F}_p[X]$ is an integral domain, $(p) ⊆ \mathbb{Z}[X]$ is a prime ideal. Suppose $P ≠ (p)$, and let $\bar{P} := \phi[P]$. Then, as $\mathbb{F}_p[X]$ is a PID, there exists an $\overline{f_p} ∈ \mathbb{F}_p[X]$ such that $\bar{P} = (\overline{f_p})$. I would like this $\overline{f_p}$ to be irreducible (or, equivalently, $f_p$ to be irreducible modulo $p$), and that $f_p$ together with $p$ generate $P$. What am I missing?

Answer 1

The map $\phi$ you have constructed is surjective. Thus, the image of prime ideals is prime (see A proof that shows surjective homomorphic image of prime ideal is prime). This means (as you say) that $\overline{P} = (\overline{f_p})$ is a prime ideal in $\mathbb{F}_p[x]$ and thus $f_p$ is irreducible over $\mathbb{F}_p$.

Now, let $\psi: \mathbb{F}_p[x] \to \mathbb{F}_p[x]/(f_p(x))$ denote the canonical projection. Consider the composition $\psi \circ \phi: \mathbb{Z}[x] \to \mathbb{F}_p[x]/(f_p(x))$. This map is surjective and has kernel $(p,f_p(x))$. Note that the image of $P$ under $\psi \circ \phi$ is trivial in the quotient. Thus, $P \subseteq (p,f_p(x))$. Moreover, since $p,f_p(x) \in P$ we have $(p,f_p(x)) = P$