$f(x)\bmod p$ and decomposition of prime ideals

Question

While reading Serre's beautiful book Lectures on $N_X(p)$, I thought of a related question.

Let $f(x)\in \mathbb{Z}[x]$ be a monic irreducible polynomial with integer coefficients. Let $K$ be the splitting field of $f$ over $\mathbb{Q}$. Then we can consider the decomposition of $f(x)\bmod p$ in $\mathbb{F}_p[x]$, and on the other hand we have the decomposition of the ideal $p\mathcal{O}_K$ in $\mathcal{O}_K$. How are these two viewpoints related to each other?

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The point here is that I know there is such a theorem/fact:

• Let $K/F$ be any finite extension of number fields, with the rings of integers $\mathcal{O}_K\supset \mathcal{O}_F$. Then we can pick a primitive element $\alpha\in \mathcal{O}_K$ such that $K=F(\alpha)$. Let $h(x)\in \mathcal{O}_F[x]$ be the minimal polynomial of $\alpha$. For each prime ideal $\mathfrak{p}\subset \mathcal{O}_F$ with decomposition $$\mathfrak{p}\mathcal{O}_K=\mathfrak{P}_1^{e_1}\cdots \mathfrak{P}_g^{e_g},$$ we write $f_i:=[\mathcal{O}_K/\mathfrak{P}_i:\mathcal{O}_F/\mathfrak{p}]$, the degree of the extension of the residue fields.

• Meanwhile, we consider $(h(x)\bmod \mathfrak{p})\in (\mathcal{O}_F/\mathfrak{p})[x]$, and suppose it decomposes as $$(h(x)\bmod \mathfrak{p})=\overline{h}_1^{m_1}\cdots \overline{h}_k^{m_k},$$ where $\overline{h}_i\in (\mathcal{O}_F/\mathfrak{p})[x]$ are monic irreducible. Then, we have $g=k$, and up to a suitable permutation, $$m_i=e_i,\quad \deg(\overline{h}_i)=f_i.$$ (See, for example, Marcus, Number Fields, Theorem 27.)

This is an effective way to compute the ramification invariants, but one has to use the minimal polynomial of some primitive element, in particular it is of degree $[K:F]$. But for the above question, $f(x)$ may not be of degree $[K:F]$.

The following example seems to indicate that even if $f(x)$ is not the minimal polynomial of some primitive element, it is still enough to carry out the above effective algorithm:

• Example: $f(x)=x^3-x-1$, its discriminant is $\Delta=(-23)$, so that only $p=23$ is ramified in the splitting field $K$. The Galois group $\mathrm{Gal}(K/\mathbb{Q})\simeq S_3$ has three distinct conjugacy classes: $(1)$, $(12)$, $(123)$. For $p\neq 23$ unramified, the decomposition of $$(f(x)\bmod p)\in \mathbb{F}_p[x]$$ has three possibilities: (1) three linear factors; (2) one linear and one irreducible quadratic factor; (3) it is irreducible. It turns out that these three possibilities correspond precisely to the situation when the Frobenius conjugacy class of $p$ is $(1)$, $(12)$, or $(123)$. Thus for this example, although the splitting field is of degree $6$, it suffices to carry out the modulo $p$ computation of $f(x)$, rather than the minimal polynomial of some primitive element (of degree $6$).

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Thus to be more concrete what I want to ask is: can we use the decomposition for $f(x)\bmod p$ to determine the $e,f,g$ invariants of the decomposition of $p\mathcal{O}_K$ in general?

Thanks a lot in advance for any advice or discussion.

Answer 1

The theorem/fact you mention need not hold for all $\mathfrak p$, e.g., take $f(x) = x^2 + 9$. Then $f(x) \equiv x^2 \bmod 3$ but $(3) \not= \mathfrak P^2$ in $K = \mathbf Q(\sqrt{-9}) = \mathbf Q(i)$. A comment to your question points out that such a result can be made to work as long as you avoid primes dividing a certain index.

Back to your question over $\mathbf Q$, let $r$ be a root of $f(x)$, so $K \supset \mathbf Q(r) \supset \mathbf Q$. Since $K$ is a compositum of fields isomorphic to $\mathbf Q(r)$, $p$ is unramified in $K$ if and only if it is unramified in $\mathbf Q(r)$. The primes over $p$ in $K$ have a common residue field degree $f_p(K/\mathbf Q)$ over $\mathbf Z/(p)$ since $K/\mathbf Q$ is Galois. When $p$ is unramified in $\mathbf Q(r)$, which applies to all but finitely many primes, there is a formula for $f_p(K/\mathbf Q)$ in terms of the residue field degrees over $\mathbf Z/(p)$ at the primes over $p$ in $\mathbf Q(r)$: calling those residue field degrees $f_1, \ldots, f_g$, we have $f_p(K/\mathbf Q) = {\rm lcm}(f_1,\ldots,f_g)$. So as long as you focus on the prime numbers $p$ that are unramified in $K$, you can work out $f_p(K/\mathbf Q)$ by working out the residue field degrees over $p$ in $\mathbf Q(r)$.

When $p$ is an arbitrary prime number, with $g$ prime ideals over it in $\mathbf Q(r)$, the products $e_1f_1, \ldots, e_gf_g$ can not necessarily always be read off from how $f(x) \bmod p$ factors, but they can always be read off from how $f(x)$ factors in $\mathbf Q_p[x]$: the $e_if_i$'s are the degrees of the irreducible $p$-adic factors of $f(x)$. When $p$ is unramified in $\mathbf Q(r)$, all the $e_i$'s are $1$, so the $f_i$'s are the degrees of the irreducible factors of $f(x)$ in $\mathbf Q_p[x]$.