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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?

Jos van Nieuwman
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  • Please write the full question in the text, not only in the title (at least for parts of it). Also, please state explicitly what we have and what we want to show, then the own thoughts how to get the result. In other words, please isolate the result that gives rise to the question, than the question in a compact manner, than the own thoughts. – dan_fulea Nov 14 '19 at 17:39
  • Have you looked at this previous answer? Does it answer your question? – Arturo Magidin Nov 15 '19 at 05:45

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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 $(p,f_p(x)) = P$

Mike
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  • @Arturo Magidin It seems like prime ideals that satisfy this property are necessarily maximal? – Mike Nov 15 '19 at 05:50
  • No: $(p)$ has the property, but is not maximal. But other prime ideals, yes, because they will contain $(p)$, and prime ideals in $\mathbb{Z}[x]$ have height at most 2. – Arturo Magidin Nov 15 '19 at 05:51
  • @Arturo Magidin True, so the polynomial you pick to generate $\overline{P}$ could be $0$ – Mike Nov 15 '19 at 05:53