I am going over a proof in the Fundamental Theorem of Galois Theory and I need a little clarification. I hope you can help.
Let $L/K$ be a finite extension with Galois group $G$ and let $M$ be an intermediate field. Denote by $M^*$ the group of all $M$-automorphisms of $L$. Then a part of the Fundamental Theorem of Galois Theory states:
If an intermediate field $M$ is a normal extension of $K$ then the Galois group of $M/K$ is isomorphic to the quotient group $G/M^*$.
Now the proof is rather simple just defining the map $\phi: G \to G'$ by $\phi(\tau)=\tau \mid_M$ where $\tau \in G$ and $G'$ is the Galois group of $M/K$. Then $\phi$ is a group homomorphism and surjective. The deciding point in the proof is the claim that the kernel of $\phi$ is $M^*$, i.e., the kernel of $\phi$ is the group of all $M$-automorphisms of $L$. Why is this so? It seems to me that if $\tau \in M^*$, $\phi(\tau)=\tau \mid_M$ and $\tau \mid_M$ is just the identity on $M$ (since it is an $M$-automorphism)?
If I can get the kernel to be $M^*$ then $G' \cong G/\text{ker}(\phi) = G/M^*$ and that would be the end of it.
If $\varphi$ is an automorphism of $L$ that fixes $M$, then it will restrict to the identity automorphism on the extension $M \subset K$ (since it fixes every element of M). Likewise, any element of the Galois group of $L$ over $K$ that restricts to the identity on $M$ must fix $M$ pointwise (mostly by definition here), so it follows that $\ker \phi = $ Gal$(L/M)$ (where $\phi$ is the restriction map).
As you said in your post, this implies that Gal$(M/K) \cong G/M^{\ast}$, where $G = $Gal($L/K$) and $M^{\ast}$ = Gal($L/M$).