Let $K/\mathbb Q$ be a finite extension such that $\mathrm{Aut}(K)$ acts transitively on the prime ideals that are not ramified above the same prime $p\in\mathbb N$, for every $p$. Is $K$ Galois?

Thanks in advance.


1 Answer 1


I think the answer has to be yes. In fact, say that $[K\colon \mathbb Q]=n$. If $K/\mathbb Q$ is not Galois, then $m=|\text{Aut}(K)|<n$. Now take $L$ as the normal closure of $K$, so that $L/\mathbb Q$ is Galois. Then there exists one prime (in fact there are infinitely many) $p$ such that $p$ does not ramify in $L$ (and so it doesn't ramify in $K$) and it splits completely in $L$. Since inertia and ramification degrees are multiplicative, this implies that $p$ also splits completely in $K$, so that there are $n$ distinct primes of $K$ which lie above $p$. But now by hypothesis $\text{Aut}(K)$ acts transitively on this set of primes, which is a contradition with the fact that $m<n$. In fact, if $G$ is a finite group which acts transitively on a finite set $S$, necessarily $|G|\geq |S|$.

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    $\begingroup$ Actually you don't need the assumption $m <n$. You have a direct proof of $m=n$. $\endgroup$ May 24, 2014 at 18:08

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