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 $\mathbb{Q}$?

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    $\begingroup$ 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

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 $K$.

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

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