Particularity of Galois-Extensions Let $L/K$ be a Galois-Extension and $G=Aut_K(L)$ its Galois-Group. My book ("Algebra" by Siegfried Bosch) states that, "An important property of Galois-Extensions $L/K$ is manifested in the fact that $K$ is the fixed field of the Galois-Group $G$. This means, that $K$ consists of all elements of $L$ which are fixed under all elements of $G$." Now, I don't understand what is so special about this. A $K$-Automorphism is defined to be an extension of the identity map of $K$. So I of course expect $K$ to be the fixed field. I can only imagine that there are no other elements fixed under $G$, because otherwise it would not make sense.

> I can only imagine that there are no other elements fixed under G, because otherwise it would not make sense.

Yes, exactly: the field fixed by $G$ will be bigger if the extension is not Galois. For instance, let $L = \mathbb{Q}(\sqrt[3]{2})$ and consider the non-Galois extension $L/\mathbb{Q}$. Then $G = \operatorname{Aut}_\mathbb{Q}(L) = \\{\operatorname{id}\\}$, so the fixed field of $G$ is all of $L$.

The problem is that the other roots of $x^3-2$ do not lie in $L$, so there is no place to send $\sqrt[3]{2}$ (except to itself). This hints at the equivalence of two definitions of _Galois_ for finite extensions:

* the fixed field of $\operatorname{Aut}_K(L)$ is exactly $K$; and
* $L$ is the splitting field of a separable polynomial.