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Factorization in the ring $\mathbb{Z}[x]/(x^2+1)\mathbb{Z}[x]\cong \mathbb{Z}[i]$ is well known. For instance, $5$ and $13$ (and any prime $\equiv 1\pmod{4}$) are no longer prime.

The factorization of $5$ lifts to $\mathbb{Z}[x]/((x^2+1)^2)\mathbb{Z}[x]$, but it isn't a simple consequence of Hensel's lemma. The trouble with using Hensel's lemma is that it requires a form of Bezout's identity, which does not generally hold over $\mathbb{Z}[x]$. Instead, for any two coprime polynomials $a,b\in \mathbb{Z}[x]$, all we know in $\mathbb{Z}[x]$ is that there exists some polynomials $c,d\in \mathbb{Z}[x]$ such that $ac+bd\in \mathbb{Z}-\{0\}$.

For example, $1+2x$ and $1-2x$ are relatively prime, but the "smallest" positive integer that is a linear combination is $$ (1+2x)+(1-2x) = 2. $$ Note that $5=(1+2x)(1-2x) + 4(1+x^2)$, and the coefficient on $(1+x^2)$ is divisible by $2$. This lucky coincidence is what allows us to lift the factorization of $5$.

The factorization of $13$ as $(2+3x)(2-3x)\pmod{(1+x^2)}$ also lifts (as pointed out in the comments below by Johan). Here, the smallest positive integer $\mathbb{Z}[x]$-linear combination of $2+3x$ and $2-3x$ is $4$, and that's not enough. But the smallest $\mathbb{Z}[x]$-linear combination of $2+3x$, $2-3x$, and $1+x^2$ is $1$, which is enough.

Question: Given a monic irreducible polynomial $q(x)\in \mathbb{Z}[x]$, of degree at least $2$, and an integer $n\geq 2$, is there some integer prime $p$ that factors in $\mathbb{Z}[x]/q(x)^n\mathbb{Z}[x]$?

This question arose thinking about this problem.

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  • $\begingroup$ What about $\mathbb{Z}[x]/(x^2)$? $\endgroup$
    – Luc Guyot
    Commented Oct 21, 2021 at 19:27
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    $\begingroup$ @LucGuyot Sorry, I thought it was obvious that I intended $\deg(q)\geq 2$. I've now made that clear. The point of my question is whether at least one prime factorization lifts. There are no (nontrivial) factorizations when $q$ is linear. So of course there can't be any such factorizations if we go from working mod $q$ to mod $q^2$. $\endgroup$
    – Pace Nielsen
    Commented Oct 21, 2021 at 21:03
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    $\begingroup$ @GerryMyerson The difference is that the smallest positive integer that is a $\mathbb{Z}[x]$-linear combination of $3+2x$ and $3-2x$ is $6$, which doesn't divide $4$. $\endgroup$
    – Pace Nielsen
    Commented Oct 21, 2021 at 22:12
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    $\begingroup$ $13 = (2 + 3*x + (1 + 3*x)*(1 + x^2)))*(2 - 3*x - (1 + x^2)) \bmod (1 + x^2)^2$ according to my pari/gp $\endgroup$
    – Johan
    Commented Oct 21, 2021 at 23:18
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    $\begingroup$ You are missing a quotient symbol / in the first sentence of the second paragraph. $\endgroup$
    – KConrad
    Commented Oct 21, 2021 at 23:34

1 Answer 1

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Assume that $\mathcal{O} = \mathbb{Z}[x]/(q(x))$ is the ring of integers of a number field $K$. One can search for a prime $p$ which is not inert in $\mathcal{O}$ and such that there is a principal prime ideal above $p$. I think such primes exist because the density of principal prime ideals is $1/h$, where $h$ is the class number of $\mathcal{O}$ (see this question), and because the prime ideals of $\mathcal{O}$ lying above an inert prime have density 0 (because their norm is $p^{[K:\mathbb{Q}]}$).

Write $p \mathcal{O} = \mathcal{P} \cdot I$ where $\mathcal{P} = (a)$ is a prime ideal, $I = (b) \neq \mathcal{O}$ is coprime to $\mathcal{P}$ and $p=ab$ in $\mathcal{O}$. Then you will have $p=a(x)b(x)+q(x)r(x)$ for some $a(x), b(x), r(x) \in \mathbb{Z}[x]$, and Hensel's lemma will work because $\langle a(x), b(x), q(x) \rangle = \mathbb{Z}[x]$. It will give a factorisation of $p$ in $\mathbb{Z}[x]/(q(x))^n$ for every $n \geq 2$.

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  • $\begingroup$ In general, $\mathbb{Z}[x]/(q)$ won't be the full ring of integers, but that can probably be overcome. However, I still have a question. After Johan's comment, I also was thinking about showing $\langle a,b,q\rangle = \mathbb{Z}[x]$. How do you show that equality in general? $\endgroup$
    – Pace Nielsen
    Commented Oct 22, 2021 at 20:15
  • $\begingroup$ @PaceNielsen Here, this is because $\mathcal{P}+I=\mathcal{O}$. More generally, in a Dedekind domain $R$, you just need two coprime ideals $I,J$, this implies $I+J=R$. $\endgroup$
    – François Brunault
    Commented Oct 22, 2021 at 20:17
  • $\begingroup$ Unfortunately, generally we don't have a Dedekind domain. (For instance, try $q(x)=x^2-8$.) $\endgroup$
    – Pace Nielsen
    Commented Oct 22, 2021 at 20:32
  • $\begingroup$ Let $K=\mathbb{Q}[x]/(q)$. Let $\mathcal{O}_{K}$ be the full ring of integers, and let $\mathcal{O}=\mathbb{Z}[x]/(q)$, which is an order in the ring of integers. Let $m=[\mathcal{O}_K:\mathcal{O}]$. Let $p\in \mathbb{Z}$ be prime, and relatively prime to $m$, and not ramified in $\mathcal{O}_K$. These conditions only disqualify finitely many primes. Assume that $p=ab$ for some (nonunits) $a,b\in \mathcal{O}$. (They remain nonunits in $\mathcal{O}_K$, since any algebraic integer unit $u$ has norm $\pm 1$, and hence has an inverse that is a $\mathbb{Z}$-linear combination of powers of $u$.) $\endgroup$
    – Pace Nielsen
    Commented Oct 23, 2021 at 0:42
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    $\begingroup$ John Voight says: Yes, this is a consequence of the Chebotarev density theorem, which implies that the trivial class in (the finite abelian group) Pic(R) is represented by infinitely many primes of degree 1: i.e., there are infinitely many principal primes pp = pi.R in Pic(R). Taking norms, we get Nm(pp) = p = +/- Nm(pi). We have Nm(pi) = pi.mu with mu in R from Galois theory... Adjusting signs we get p = pi.mu with pi, mu in R. We can't have pi a unit (its norm is p), and taking norms, Nm(p) = p^d = p.Nm(mu) so as long as d = deg q(x) > 1, we have that mu is also not a unit. $\endgroup$
    – Pace Nielsen
    Commented Oct 23, 2021 at 2:54

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