# Generators of prime ideals and factorization of polynomials

Let $$\alpha \in \mathbb{C}$$ be a zero of a monic irreducible polynomial $$f \in \mathbb{Z}[X]$$. Define $$K = \mathbb{Q}[\alpha]$$ and $$A := \mathbb{Z}[\alpha]$$, then $$A \subseteq O_K$$, where the latter denotes the ring of integers of the number field $$K$$.

Let $$p$$ be a prime number, and assume that $$f$$ factors modulo $$p$$ as $$f \equiv g_1^{\alpha_1} \cdots g_k^{\alpha_k},$$ where the $$g_i$$ are monic and irreducible modulo $$p$$ (i.e., irreducible as elements in $$(\mathbb{Z}/p\mathbb{Z})[X]$$). Suppose further that there exist prime ideals $$\mathfrak{q}_1,\ldots, \mathfrak{q}_k \subseteq O_K$$ such that $$p O_K = \mathfrak{q}_1^{\alpha_1} \cdots \mathfrak{q}_k^{\alpha_k}$$ and such that $$\text{each } \mathfrak{q}_i \text{ has degree } \deg(g_i), i.e., [O_K/\mathfrak{q}_i:\mathbb{Z}/p\mathbb{Z}] = \deg(g_i).$$ (EDIT: If necessary, one may assume here that the above holds for every prime number $$p$$.)

Is it then true$$^\star$$ that
$$\mathfrak{q}_i = \langle p, g_i(\alpha) \rangle_{O_K} := p \cdot O_K + g_i(\alpha) \cdot O_K \text{ ?}$$

$$\big($$ $$^\star$$ This is meant up to ''reasonable'' permutation. E.g., if $$g_i$$ and $$g_j$$ have the same degree and $$\alpha_i = \alpha_j$$, then it could of course happen that $$\mathfrak{q}_j = \langle p, g_i(\alpha) \rangle_{O_K}$$ and $$\mathfrak{q}_i = \langle p, g_j(\alpha) \rangle_{O_K}$$.$$\big)$$

Of course, my claim holds for all but finitely many prime numbers $$p$$, which is precisely the Kummer-Dedekind-Theorem. To be concrete, it holds for all $$p$$ not dividing the index $$[O_K : A]$$. Hence I am asking here whether the generators of the prime ideals $$\mathfrak{q}_i$$ are nevertheless as predicted by Kummer-Dedekind, if we assume that $$f$$ and $$pO_K$$ have the same factorization pattern.

A few observations:

1. The ideals $$\mathfrak{p_i} := \langle p, g_i(\alpha) \rangle_A = p \cdot A + g_i(\alpha) \cdot A$$ are precisely the prime ideals in $$A$$ containing $$p$$, which is due to the isomorphisms $$A/pA \cong \mathbb{Z}[X]/(f,p) \cong (\mathbb{Z}/p \mathbb{Z})[X]/(f).$$
2. The ideals $$\mathfrak{q}_i$$ are precisely the prime ideals in $$O_K$$ containing $$p$$.
3. Since $$O_K \mid A$$ is an integral ring extension, for every $$\mathfrak{p}_i$$ there exists a prime ideal in $$O_K$$ lying over it which (necessarily) contains $$p$$. In particular, by 1. and 2., there is a permutation $$\sigma \in S_k$$ such that $$\mathfrak{p}_i = \mathfrak{q}_{\sigma(i)} \cap A$$. However, (to my knowledge) this is not strong enough to conclude that $$\mathfrak{q}_{\sigma(i)} = \mathfrak{p}_i \cdot O_K = \langle p, g_i(\alpha) \rangle_{O_K}$$. At least, we may conclude that $$\mathfrak{p}_i \cdot O_K$$ is a power of $$\mathfrak{q}_{\sigma(i)}$$.
4. By the isomorphisms from 1., we know that $$|A/\mathfrak{p}_i| = |(\mathbb{Z}/p\mathbb{Z})[X]/(g_i)| = p^{\deg(g_i)}$$. Further, our assumptions imply that $$|O_K/\mathfrak{q}_i| = p^{\deg(g_i)}$$. In the notation of 3., we have a canonical injection $$A/\mathfrak{p}_i \to O_K/\mathfrak{q}_{\sigma(i)}$$, hence $$\deg(g_i) \leq \deg(g_{\sigma(i)})$$ for every $$1 \leq i \leq k$$, hence these must be equalities, thus even $$A/\mathfrak{p}_i \cong O_K/\mathfrak{q}_{\sigma(i)}$$ for every $$1 \leq i \leq k$$.
5. Let's now involve $$p$$-adic numbers. By Hensel's Lemma, the factorization $$f \equiv g_1^{\alpha_1} \cdots g_k^{\alpha_k}$$ modulo $$p$$ can be lifted to a decomposition of $$f$$ as a product $$f = G_1 \cdots G_k \text{ in } \mathbb{Z}_p[X],$$ where each $$G_i$$ is monic and reduces to $$g_i^{\alpha_i}$$ modulo $$p$$, in particular $$\deg(G_i) = \alpha_i \cdot \deg(g_i)$$.
6. Using the isomorphism $$K \otimes_\mathbb{Q} \mathbb{Q}_p \cong \prod_{\substack{\mathfrak{p} \in \mathrm{Spec}(O_K) \\ \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}}} K_\mathfrak{p},$$ where $$K_\mathfrak{p}$$ denotes the completion of $$K$$ with respect to the $$\mathfrak{p}$$-adic topology, the factorization $$p O_K = \mathfrak{q}_1^{\alpha_1} \cdots \mathfrak{q}_k^{\alpha_k}$$ (togehter with the nice properties of étale algebras) implies that the factorization of $$f$$ over $$\mathbb{Z}_p$$ consists of precisely $$k$$ (monic) irreducible factors, whose degrees are given by $$\deg(g_i) \cdot \alpha_i$$, $$1 \leq i \leq k$$. This proves that $$G_1, \ldots, G_k$$ are irreducible.

Note that this is not a homework problem, hence my claim could possibly be wrong. Any help is appreciated!

• The example that I describe in an answer at mathoverflow.net/questions/21247/… does not directly address your question but it does show that the way a ramified prime number factors in $\mathcal O_K$ need not correspond to the way some element of the Galois group of the Galois closure of $K$ over $\mathbf Q$ permutes a primitive root of $K/\mathbf Q$ and its $\mathbf Q$-conjugates. Because of that phenomenon, I would not be surprised if your question has a negative answer. Oct 3 '19 at 4:41

$$\newcommand{\Z}{\mathbb{Z}}$$No.

It is easier to construct a counter-example by starting from the local case.

Let $$f\in\Z_p[X]$$ be monic irreducible of degree $$e>1$$ such that $$F = \mathbb{Q}_p[X]/f$$ is totally ramified, let $$\alpha = X \bmod f$$, and let $$v$$ be the normalised valuation on $$F$$.

Assume that $$f = X^e \bmod p$$, i.e. $$\alpha$$ is nilpotent mod $$p$$, i.e. $$v(\alpha)\ge 1$$. The local question corresponding to yours is: is $$(p,\alpha)$$ the maximal ideal of $$\Z_F$$? Equivalently, is $$\alpha$$ a uniformiser, i.e. $$v(\alpha)=1$$?

From this local formulation it is easy to find a counter-example, for instance $$f(X) = X^3-p^2$$, which then also works globally.

• Thank you for providing intuition and example. Of particular interest is the case p=3, because then both $\mathrm{disc}(K)$ (for $K = \mathbb{Q}[X]/(f)$) and $\mathrm{disc}(f)$ are (distinct) powers of $3$, thus also the index $[O_K : \mathbb{Z}[X]/(f)]$ is a (non-trivial) power of $3$, as indicated by @KConrad here: mathoverflow.net/questions/21247/…. (In particular, $O_K$ is not generated by $f$ over $\mathbb{Z}$.) Thus 3 is the only prime not covered by the Kummer-Dedekind-Theorem, and as you indicate, ... Oct 3 '19 at 11:07
• ... the factorization patterns of $f$ (mod $3$) and $3O_K$ agree, but the unique prime lying over $3$ in $O_K$ (namely, the principal ideal ($3^{1/3}$) is not of the desired shape. Oct 3 '19 at 11:09
• Yes, and if you want examples with the property you mention for all $p$, you can take $X^4-8$ for $p=2$ and $X^p-p^2$ for $p$ odd. Oct 3 '19 at 13:07

As pointed out by @KConrad and utilized by @Aurel, my claim can be disproved by considering primes $$p$$ ramified in $$K$$ and dividing $$[O_K : \mathbb{Z}[\alpha]]$$. Finally, let us use this observation (and SageMath) to construct further counterexamples to my claim, with the additional assumption that $$K = \mathbb{Q}[X]/(f)$$ is a Galois extension of $$\mathbb{Q}$$.

To be more concrete, let $$f = X^3 + b \cdot X^2 + c \cdot X + d \in \mathbb{Z}[X]$$, $$K := \mathbb{Q}[X]/(f)$$ and $$p$$ be a prime number such that

1. $$f \in \mathbb{Q}[X]$$ is irreducible of degree $$3$$,
2. $$\mathrm{disc}(f)$$ is a power of $$p$$ with even exponent,
3. $$\Delta_K < \mathrm{disc}(f)$$,
4. $$f \equiv g^3$$ mod $$p$$ for some polynomial $$g \in \mathbb{Z}[X]$$ of degree $$1$$.

Then 1. and 2. imply that $$K$$ is a Galois extension of $$\mathbb{Q}$$ of degree $$3$$ with Galois group $$A_3$$. Further, 3. implies that $$[O_K : \mathbb{Z}[\alpha]]$$ is a non-trivial (i.e., $$>1$$) power of $$p$$, in particular $$O_K \supsetneq \mathbb{Z}[\alpha]$$. Since $$\Delta_K > 1$$ and $$\Delta_K \mid \mathrm{disc}(f)$$, this implies that there is precisely one ramified prime dividing $$[O_K : \mathbb{Z}[\alpha]]$$, namely $$p$$. Since $$[K:\mathbb{Q}] = 3$$ is prime and $$K$$ is Galois, there is only one possible splitting behaviour for $$p$$ in $$O_K$$, namely, $$pO_K = \mathfrak{p}^3$$ for some $$\mathfrak{p} \in \mathrm{Spec}(O_K)$$ of degree $$1$$. By 4., we see that $$f$$ mod $$p$$ and $$p$$ in $$O_K$$ hence have the same splitting behaviour. By a similar argument as in @Aurel's answer (and having in mind that $$g = X+n$$ for some $$0 \leq n < p$$), we conclude that $$f$$ provides a counterexample to my claim.

For instance, let $$p=3$$, and assume that $$0 \leq a,b,c \leq 100$$. Using SageMath, we obtain the following $$10$$ possible choices: $$f = X^3 + 9X^2 + 18X + 9$$ $$f = X^3 + 12X^2 + 39X + 19$$ $$f = X^3 + 12X^2 + 39X + 37$$ $$f = X^3 + 15X^2 + 48X + 17$$ $$f = X^3 + 15X^2 + 66X + 71$$ $$f = X^3 + 15X^2 + 66X + 89$$ $$f = X^3 + 18X^2 + 81X + 27$$ $$f = X^3 + 18X^2 + 81X + 81$$ $$f = X^3 + 21X^2 + 66X + 19$$ $$f = X^3 + 27X^2 + 54X + 27$$

Additional remark: Let $$f = X^3 + 18X^2 + 81X + 27$$, then (since $$f$$ appears in the above list) we have $$O_K \supsetneq \mathbb{Z}[\alpha]$$, where $$\alpha$$ is (as usually) a zero of $$f$$. However, according to SageMath, it turns out that $$O_K = \mathbb{Z}[\alpha/3]$$.