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I am reading Iwasawa's article On $\Gamma$-extensions of algebraic number fields. In paragraph 7.3 : " We now take as M the maximal unramified abelian $p$-extension of $L$ in $\Omega$ , i.e. the $p$-part of the Hilbert's class field over $L$; $M/K$ is then obviously a Galois extension. " with $\Omega$ an algebraic closure of $K$ and $L/K$ a $\Gamma$-extension (i.e. $Gal(L/K)=\Gamma \simeq \mathbb{Z}_p$ )

My question : Why is $M/K$ a Galois extension?

Thanks.

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  • $\begingroup$ I misread your question, as you pointed out. Since I also don't see why it should be obvious that $M/K$ is Galois, I deleted my answer. $\endgroup$ Feb 14, 2014 at 3:37
  • $\begingroup$ That's my fault, I am sorry, I made a mistake when copying the article. I edited my post. $\endgroup$
    – gpst
    Feb 14, 2014 at 12:52

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This is because of the maximal property of M : if f is any element of Gal(L/K) and if we extend it to a K-automorphism g of Omega, then g(M) is unramified over L since M is so, and g(M) will be contained in M because of the maximality of M; this means that M is stable by all the g's, and this is one of the many equivalent conditions for M to be normal over K. You will notice that we did not use the particular form of Gal(L/K).

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