# If $K$ is Galois over $k$, so is $\mathfrak o_K/\mathfrak P$ over $\mathfrak o_k/\mathfrak p$

Let $k\subset K$ be number fields and $\mathfrak o_k\subset \mathfrak o_K$ their respective algebraic integers. Also let $\mathfrak P\neq 0$ be a prime (and hence maximal ideal) of $\mathfrak o_K$ lying above $\mathfrak p$. Can we reason that $\mathfrak o_K/\mathfrak P$ is Galois over $\mathfrak o_k/\mathfrak p$, if $K$ is Galois over $k$? [$\mathfrak o_k/\mathfrak p$ is associated with $\{\alpha + \mathfrak P\mid \alpha\in \mathfrak o_k\}\subset \mathfrak o_K/\mathfrak P$.]

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## 1 Answer

Sure, because every finite degree extension of finite fields is Galois.

There is a more general setup where the conclusion is not so trivial. Namely, suppose $R$ is an integrally closed domain with fraction field $K$, $L$ is a finite Galois extension of $K$ and $S$ is the integral closure of $R$ in $L$. Then for $\mathfrak{p}$ a nonzero prime ideal of $R$ and a prime ideal $\mathcal{P}$ of $S$ lying over $\mathfrak{p}$, one can ask whether the finite degree field extension $(S/\mathcal{P}) / (R/\mathfrak{p})$ is Galois. It turns out that it is always normal, so if $R/\mathfrak{p}$ is perfect then it is necessarily also separable and thus Galois.

See for instance Theorem 6 of these notes.

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okay, thanks. I feel awkward for even asking. Both results you mentioned are well known to me. –  Peter Patzt Jul 5 '11 at 18:43
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