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?
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.