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Let $K$ be a number field and $f(X)\in K[X]$. Let $E$ be the splitting field of $K$, so that we know that the set of primes splitting in $E$ has density $1/[E:K]$.

Milne uses this as the argument that if $f(X)$ splits modulo almost every prime $\mathfrak{p}$ in $K$, then $f(X)$ splits in $K$ i.e. $E=K$.

What he leaves out is apparently the claim that if $f(X)$ splits modulo $\mathfrak{p}$, then $\mathfrak{p}$ splits in $E$. I don't see how this works if we don't even assume $f(X)$ to have coefficients in $\mathcal{O}_K$. Anyone care to elaborate?

EDIT: I guess the trick is not to use Kummer's theorem, but looking at completions, we have instead $E_\mathfrak{p}=K_\mathfrak{p}$, so that $G(\mathfrak{p})\simeq Gal(E_\mathfrak{p}/K_\mathfrak{p})$ which is trivial, so $\mathfrak{p}$ splits.

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up vote 2 down vote accepted

$f(X)$ has only finitely many denominators, so lies in $\mathcal O_K[1/a]$ for some $a \in \mathcal O_K\setminus \{0\}.$ Now apply whatever argument you had in mind with $\mathcal O_K$ replaced by $\mathcal O_K[1/a]$.

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