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If $L/K$ is a Galois extension, then any prime $\mathfrak{p}$ of $K$ splits into a product ${\mathfrak P}_1^e\cdots {\mathfrak P}_g^e$ of primes in $L$, and the exponents on the primes are equal since the Galois group acts transitively on the primes dividing $\mathfrak{p}$.

Question: Is the converse true? Namely, if $L/K$ is an extension such that every prime of $K$ has a single associated exponent in this manner, then is $L/K$ necessarily Galois?

I believe the answer is yes but I was wondering in particular if there was an easy way to see this.

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  • $\begingroup$ You might be interested in my question here: mathoverflow.net/questions/34180, though I was requiring that the residue degrees also agreed for all primes above $\mathfrak{p}$. $\endgroup$
    – Zev Chonoles
    Commented Apr 30, 2012 at 23:08
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    $\begingroup$ Just demanding that all the exponents are equal isn't enough. Just take any non-Galois extension unramified everywhere -- e.g. take a number field $A$ whose Hilbert class field has a non-trivial Hilbert class field $B$ (these exist), and then take a non-normal subgroup of $Gal(B/A)$ giving an extension $C/A$, non-Galois but with every prime unramified. $\endgroup$
    – Kevin Buzzard
    Commented Apr 30, 2012 at 23:33
  • $\begingroup$ Zev's question is more reasonable since all but finitely many primes in $K$ are unramified in $L$, hence most $e$'s are 1 anyway. A Galois extension would have all residue field degrees over each $\mathfrak p$ equal, not just all ramification indices, so the converse is natural to ask with the residue field degrees taken into account also. $\endgroup$
    – KConrad
    Commented May 1, 2012 at 2:05
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    $\begingroup$ Jon: I changed the primes over $\mathfrak p$ in your question to ${\mathfrak P}_i$ instead of $\beta_i$: $\mathfrak P$ is a capital $P$, not anything like $B$. Don't $\mathfrak P$ and $\mathfrak B$ look completely different? :) $\endgroup$
    – KConrad
    Commented May 1, 2012 at 2:07

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If you throw in the residue degrees as well, then you get Zev's question. Otherwise, the converse is not true. As an example, consider a quadratic field that has an unramified everywhere $A_5$ Galois extension (see e.g. this MO thread). Since $A_5$ is simple, any proper intermediate extension will not be Galois over the quadratic. But the ramification indices will all be 1, since they are all 1 in the big extension.

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