Let $K/\mathbb{Q}$ be a Galois extension, and $\bar{K}$ the algebraic closure of $K$.

If we consider the following grups:

  • $G \cong Gal(\bar{K}/\mathbb{Q})$
  • $H \cong Gal(\bar{K}/K)$

Then $H$ is a normal subgroup of $G$, and applying the main theorem of Galois theory,

$$Gal(K/\mathbb{Q}) \cong G/H$$

Is this correct?

Thank you in advance.

  • $\begingroup$ Warning: you need to be careful when using infinite extensions and Galois theory $\endgroup$ Apr 3, 2016 at 15:24

1 Answer 1



Let $\psi:Gal(\bar K/\mathbb Q)\longrightarrow Gal(K/\mathbb Q)$ be the restriction, i.e. $\psi(\sigma )=\sigma |_K$. It's easy (more or less... it's in fact important to prove this result) to see that $\psi$ is surjective and that $\ker \psi=Gal(\bar K/K)$, what gives you the isomorphism.


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