My definition of an extension $M/K$ to be Galois is that $Gal(M/K)$ only fixes things in K. I'm trying to prove that this is equivalent to $M/K$ being normal and separable.
I know that fact that if $M/K$ is normal and separable then there are exactly $d = [M:K]$ homomorphisms $j_i: M \to M$ that extend inclusion $K \to M$.
Now the proof claims that this then shows $|Gal(M/K)| = d$. However I only know how to show that the $j_i$ are homomorphisms whereas surely here we need them to be isomorphisms? I know that a field homomorphism is always injective but is there any other reason that allows us to ensure the $j_i$ are also surjective?
Thanks for any help here!
An injective linear map from a finite dimensional vector space to itself (such as M over K) is necessarily surjective. These homomorphisms are K linear since they extend the inclusion map, and so are the identity on K.
To check this, if $a,b\in K$ and $u,v\in M$, and $\sigma$ is our map, then $\sigma(au+bv)=\sigma(a)\sigma(u)+\sigma(b)\sigma(v)=a\sigma(u)+b\sigma(v)$, since $a,b\in K$ so $\sigma(a)=a$ and $\sigma(b)=b$.