If $K$ is an imaginary quadratic field and $M$ is an unramified Abelian extension of $K$, the prove that $M$ is Galois over $\mathbb{Q}$

Let see... If $L$ is the Hilbert class field of $K$, then $L$ is the maximal extension unramified of $K$, then $\mathbb{Q} \subset K \subset M \subset L$ and $L$ is Galois over $K$...


  • $\begingroup$ See the answer by Hurkyl, where he stated a general theorem that the Hilbert class field of $\mathbb Q(\sqrt m)$ is formed by adjoining square roots of factors of $m$, hence galois over $\mathbb Q$. Then, $M$ being a sub-field of $L$, we might be able to prove some assertions about the forms of $M$, hence obtaining the galoisity of $M/\mathbb Q$. Hope this helps. $\endgroup$ – awllower Aug 8 '13 at 9:50
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    $\begingroup$ @awllower: Dear awllower, This description of the HCF is not correct. (See the comment I added to Hurkyl's answer.) Regards, $\endgroup$ – Matt E Aug 26 '13 at 2:46
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    $\begingroup$ Dear P.M.O., As I commented on another question of yours related to class field theory, it would help to have a little more context and background on where you are coming from, so as to know what would constitute a satisfactory answer. Regards, $\endgroup$ – Matt E Aug 26 '13 at 2:48
  • $\begingroup$ @MattE I have partially answered the question..I hope to go for right direction. $\endgroup$ – P. M. O. Aug 30 '13 at 3:45
  • $\begingroup$ Dear P.M.O., As I said before, it would help to know a bit more about your background and where this question is coming from. (Are you reading a book on class field theory that this problem came from. Which one? What do you know about class field theory so far?) Regards, $\endgroup$ – Matt E Aug 30 '13 at 11:07

Partial answer:

We have $K \subset L$, $M \subset L$ and $\mathbb{Q} \subset K$, Galois extensions. Also, $K \subset M$ Galois, because $Gal(L/M)$ is normal subgroup of $Gal(L/K)$.

What can we say about $Aut(M/\mathbb{Q})$?

$|Aut(M/\mathbb{Q})|= 2|Gal(M/K)|$?


The following simple argument seems to work for any $K$ that is Galois over $\mathbb{Q}$ (could it be that $K$ is specified to be imaginary quadratic in order to ensure one understands that $M/K$ is unramified not only at the finite but also at the infinite primes?): since $M$ is unramified, abelian, it must be contained in $CF(K)$. As $CF(K)/K$ is Galois and abelian, all the intermediate fields must be Galois and abelian. So $M/K$ is Galois. Since $K/\mathbb{Q}$ is also Galois, $M/\mathbb{Q}$ must be Galois as well.

  • $\begingroup$ Actually, almost everything is said in P.M.O.'s answer. The only thing I should have added is that a Galois extension of a Galois extension is again Galois. $\endgroup$ – streetcar277 Feb 7 '14 at 15:27
  • $\begingroup$ Galois extensions of Galois extensions need not be Galois: consider ${\mathbb Q}(\sqrt[4]{2})$ $\endgroup$ – franz lemmermeyer Mar 15 '16 at 17:44

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