Milne's notes on class field theory has the following corollary to the Chebotarev density theorem:

If a polynomial $f(X)\in K[X]$ splits into linear factors modulo $\mathfrak{p}$ for all but finitely many prime ideals $\mathfrak{p}$, then it splits in $K[X]$.

Here $K$ is a number field. Milne's proof is the following:

Apply the Chebotarev density theorem to the splitting field.

I've managed to prove this by first showing that the polynomial must be separable modulo all but finitely many primes $\mathfrak{p}$. If $\mathfrak{P}\mid \mathfrak{p}$, we may assume $f(X)$ splits and is separable modulo the prime. It follows that $G(\mathfrak{P})=1$ for all but finitely many primes, since $f$ must split in the completion of $K$ at $\mathfrak{p}$. Hence, almost all primes split, so the splitting field equals $K$.

I'm wondering if there's a shorter or even trivial proof of this since, Milne didn't bother writing down anything at all? The problem as I see it is that one can't use Chebotarev directly since the factorization of $f(X)$ does not necessarily give the factorization of a prime $\mathfrak{p}$. This would be because we don't know what the generators of the ring of integers look like.

  • 1
    $\begingroup$ Sorry, I'm new to the site (which is great!). I'll go check my old questions. $\endgroup$
    – pki
    Nov 6, 2011 at 0:19
  • $\begingroup$ Thanks! I don't think such a comment has ever had such an effect in such a short time :-) $\endgroup$
    – joriki
    Nov 6, 2011 at 7:20
  • $\begingroup$ Related: math.stackexchange.com/questions/608919 $\endgroup$
    – Watson
    Dec 2, 2018 at 9:39

3 Answers 3


As you note, the assumption that $f(x)$ splits into linear factors modulo almost all $\mathfrak p$ implies that the decomposition group at $\mathfrak P$ is trivial for almost all $\mathfrak P$. (This is standard algebraic number theory, and presumably Milne expects the reader to realize this without any explanation being necessary.)

With this fact in hand, Milne's claim follows directly from Cebotarev:

Let $G$ be the splitting field of $L$ over $K$. Cebotarev says that any element of $G$ can be realized as a Frobenius element (in particular as an element of the decomposition group $G({\mathfrak P})$) for infintely many $\mathfrak P$. But since $G(\mathfrak P)$ is trivial for all but finitely many $\mathfrak P$, we see that every element of $G$ must be trivial, hence that $G$ is trivial, i.e. that $L = K$, i.e. that $f(x)$ splits over $K$.

(If this is the argument you had in mind, then I'm not sure what you're asking.)

A side remark:

I'm not sure what you mean when you write that the factorization of $f(X)$ does not necessarily give the factorization of a prime $\mathfrak p$. As I mentioned in my answer to this question of yours, if $f(x)$ is any element of $K[x]$, then we may write $f(x) \in \mathcal O_K[1/a][x]$, for some $a \in \mathcal O_K\setminus \{0\}.$ Choosing $a$ appropriately, we may furthermore assume that the discrminant of $f$ is invertible in $\mathcal O_K[1/a]$, so that for every prime $\mathfrak p$ of $\mathcal O_K[1/a]$, the reduction of $f(x)$ modulo $\mathfrak p$ is separable.

The factorization of any prime $\mathfrak p$ of $\mathcal O_K[1/a]$ in the splitting field $L$ of $f(x)$ then corresponds precisely to the factorization of $f(x)$ modulo $\mathfrak p$.

Of course, the primes of $\mathcal O_K[1/a]$ are precisely the primes of $\mathcal O_K$ which do not divide $a$, and so consist of all but finitely many prime of $\mathcal O_K$. Thus if one is willing to throw away finitely many primes (which is obviously harmless in the case of e.g. Milne's claim), then one can just work with the ring $\mathcal O_K[1/a]$ and apply Kummer's lemma in this ring. Everything proceeds just as if $f(x)$ has integral coefficients (and there is no need whatsoever to worry about generators of the ring of integers in either $K$ or $L$; e.g., since the discriminant of $f(x)$ is invertible in $\mathcal O_K[1/a]$, the ring $\mathcal O_L[1/a]$ is generated by the roots of $f(X)$ over $\mathcal O_K$).

  • $\begingroup$ Alright, so there seems to be a few things I don't follow. First you say that since the ring $\mathcal{O}_L[1/a]$ is generated by roots of $f$, then we can use Kummer's lemma. However, the only form of it that I'm aware of is the one where the ring of integers is defined by precisely one element. Am I missing something here? $\endgroup$
    – pki
    Nov 6, 2011 at 2:39
  • $\begingroup$ I might add that the Chebotarev argument after noticing that almost every prime splits is the one I had in mind. I guess to me it seemed like a pretty big step to realize that the decomposition groups would be trivial, so I thought there would be something easier... I guess I'm wrong. However, your side note seems to spawn some interesting questions that I had not thought about. Thanks. $\endgroup$
    – pki
    Nov 6, 2011 at 2:44

There is a "trivial" proof based on the fact that the Dedekind zeta function of a number field has a simple pole at $s=1$.

Let $K_1$ be the splitting field of $f$. Then $f$ splitting completely at almost all primes is the same as almost all primes in $K$ splitting completely in $K_1$. The proof is completed by showing that the Dedekind zeta function of $K_1$ will have a pole of order equal to the degree of the extension $K_1/K$.

Say the degree of the extension is $n$. Thus there are $n$ primes in $K_1$ lying over almost all primes of $K$. Moreover, if $P$ lies over $p$, then $N_{K_1/\mathbb Q}(P)=N_{K/\mathbb Q}(p)$. The Dedekind zeta function of $K_1$ is $$\zeta_{K_1}(s)=\prod_{P}(1-N_{K_1/\mathbb Q}(P)^{-s})^{-1}$$ Let $S$ be the set of primes of $K$ that do not split completely. Then $$\zeta_{K_1}(s)=\prod_{P|p\atop p\in S}(1-N_{K_1/\mathbb Q}(P)^{-s})^{-1}\prod_{P|p\atop p\notin S}(1-N_{K_1/\mathbb Q}(P)^{-s})^{-1}$$ $$=\prod_{P|p\atop p\in S}(1-N_{K_1/\mathbb Q}(P)^{-s})^{-1}\prod_{p\notin S}(1-N_{K/\mathbb Q}(p)^{-s})^{-n}$$ $$=\prod_{P|p\atop p\in S}(1-N_{K_1/\mathbb Q}(P)^{-s})^{-1}\prod_{p\in S}(1-N_{K/\mathbb Q}(p)^{-s})^{n}\zeta_K(s)^n$$ Since $\zeta_K$ has a simple pole at $s=1$, this will have a pole of order $n$ at $s=1$. Since it is equal to $\zeta_{K_1}$, which has a simple pole, $n$ must be $1$.

This is trivial in the sense that the proof is contained in one sentence: if all but finitely many primes split completely in an extension of degree $n$, then the Dedekind zeta function will have a pole of order $n$.


Refer to milne's ANT(p146,version3.03).
$\mathfrak{P}\in Spl(L/K) => \mathfrak{P}\in Spl(K/K)$ for almost many primes.
So, $Spl(L/K) = Spl(K/K)$.


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