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Let $K$ be a number field such that $K/\mathbb{Q}$ is Galois. I am asked to show there are infinitely many primes that split completely. I've already showed that for a monic polynomial $f(x) \in \mathbb{Z}[x]$ there are infinitely primes p such that $f(x)$ has a root mod p.

Because of the primitive element theorem I can say $K=\mathbb{Q}(\alpha)$, I would like to say that $\alpha \in O_K$ but I am not sure if I can say this.

Assuming that $\alpha \in O_K$ then I would use that for a monic polynomial $f(x) \in \mathbb{Z}[x]$ there are infinitely primes p such that $f(x)$ has a root mod p and because there are only finitely many primes that ramify there would have to be infinitely many primes that split completely but then I wouldn't be using that the extension is Galois.

NuKexZ
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    Only in Galois extensions unramified and one root $\bmod p$ implies splits completely. Usually to know the density of each factorization type we look at the Dedekind zeta function $\zeta_K(s) = \sum_{I \subset \mathcal{O}K} N(I)^{-s} = \prod{p} \prod_{\mathfrak{p} | p} \frac{1}{1-N(\mathfrak{p})^{-s}}=\prod_{p} \frac{1}{(1-p^{-sf_p})^{g_p}}$ where $g_p = n/f_p$ for all but finitely many (ramified) $p$. There are infinitely many $p$ with $f_p = 1$ because otherwise $\zeta_K$ wouldn't have a pole at $s=1$. – reuns Oct 15 '17 at 23:50
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    For any $\alpha \in K$ there is a $b \in \mathbb{Z}$ (the leading coefficients of its minimal polynomial) such that $b \alpha \in \mathcal{O}_K$. Thus $\mathbb{Z}[b \alpha] \cong \mathbb{Z}[x]/(f(x))$ has finite index in $\mathcal{O}_K$ and for $p \nmid \Delta(b \alpha) N(f'(\alpha))$ you can look at the roots of $f \bmod p$ – reuns Oct 16 '17 at 01:35
  • Duplicate of https://math.stackexchange.com/questions/2071675 – Watson Nov 23 '18 at 12:37

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