5
$\begingroup$

While reading algebraic number theory, I came across the following statement: Let $K$ be a galois extension over $\mathbb{Q}$ and $H$ be the Hilbert class field (maximal unramified abelian extension) of $K$. As $H$ is maximal, $H$ is also galois over $\mathbb{Q}$. Can anyone explain why maximality implies $H$ is a galois extension over $\mathbb{Q}$?

$\endgroup$
  • 2
    $\begingroup$ Suppose it weren't Galois, and consider one of its conjugates. Composit that guy with $K$ and you have a larger abelian unramified extension, contradicting maximality. $\endgroup$ – Cam McLeman Feb 6 '14 at 6:00
  • $\begingroup$ More details in the comments to my answer at math.stackexchange.com/questions/142236/… $\endgroup$ – Cam McLeman Feb 6 '14 at 6:00
  • $\begingroup$ I did not get your comment. Let $\sigma$ be an embedding of $H$ that fixes $\mathbb{Q}$ pointwise. Why would the compositum of $K$ and $\sigma(H)$ be unramified and abelian? Where are we using the fact that $K$ is galois over $\mathbb{Q}$? $\endgroup$ – user92156 Feb 6 '14 at 7:17
4
$\begingroup$

Sorry, my comment had a typo which made it unhelpful (the $K$ should be an $H$).

Here is the structure of the argument (with some details left to be filled in). Suppose $H/\mathbb{Q}$ were not Galois. Then there exists a conjugate $H^\sigma$ of $H$ distinct from $H$, and $H^\sigma$ is an unramified abelian extension of $K^\sigma$. Since $K/\mathbb{Q}$ is Galois, $K^\sigma=K$, and so both $H$ and $H^\sigma$ are both unramified abelian extensions of $K$. Thus their compositum $HH^\sigma$ is as well, and by assumed distinctness, properly contains $H$. This contradicts that $H$ was the maximal unramified abelian extension of $K$.

$\endgroup$
  • $\begingroup$ Sure thing -- sorry for the earlier confusion. $\endgroup$ – Cam McLeman Feb 7 '14 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.