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I'm trying to teach myself Galois theory, and running into a seeming contradiction that I can't figure out. In the wikipedia page for Galois extensions, it's claimed that for any field of characteristic $0$, its algebraic closure is a Galois extension. This would seem to imply that $\mathbb{C}$ is a Galois extension of $\mathbb{R}$, and indeed, lots of other sources seem to back that up.

But then I saw this question, which seems to imply the existence of automorphisms of $\mathbb{C}$ that don't fix $\mathbb{R}$ - in particular, they (at minimum) only fix the rationals. So how can $\mathbb{C}$ be a Galois extension of $\mathbb{R}$, if it doesn't fix the base field? Is the theorem above wrong? Or am I confusing my definitions?

Nico A
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1 Answers1

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The theorem is right and $\mathbb{C}$ is a Galois extension of $\mathbb{R}$. There are two standard definitions of Galois extension and the answer will be different for each. The one I prefer is that an algebraic extension $K \to L$ is Galois if the group $G = \text{Aut}(L/K)$ of automorphisms of $L$ which fix $K$ has the property that $K$ is precisely its fixed subfield, or in symbols $K = L^G$. This definition makes no reference to any other automorphisms of $L$ that don't fix $K$, which can do whatever they want.

Qiaochu Yuan
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  • Thanks, that makes a lot of sense! Just to clarify - is the statement "$L/K$ is a galois extension iff the field fixed by every element of $\operatorname{Aut}(L/K)$ is precisely $K$" a wrong definition I have somehow picked up, or an alternative definition actually used by some people? – Nico A Sep 04 '20 at 20:20
  • @Nico: that's the same as the definition I just gave. Look more carefully at the definition of $\text{Aut}(L/K)$ (I've decided to switch notation from Gal to Aut so this is less confusing). – Qiaochu Yuan Sep 04 '20 at 20:38
  • Oh of course! I was thinking of $\operatorname{Aut}$ only in terms of the extension field. Thank you, that clears things up nicely. – Nico A Sep 04 '20 at 20:40