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!

  • 2
    $\begingroup$ 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. $\endgroup$
    – KConrad
    Oct 3, 2019 at 4:41

2 Answers 2



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.

  • $\begingroup$ 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, ... $\endgroup$
    – Algebrus
    Oct 3, 2019 at 11:07
  • $\begingroup$ ... 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. $\endgroup$
    – Algebrus
    Oct 3, 2019 at 11:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – Aurel
    Oct 3, 2019 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 $O_K = \mathbb{Z}[\alpha/3]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.