Galois group - extend homomorphism to automorphism Let $K \subset L$ be a finite Galois extension, $M$ a field with $K \subset M \subset L$ and $G := \text{Aut}(L/K)$.
I want to show that if 
$\sigma \, \colon M \longrightarrow L$ is a $K$-homomorphism there exists a $\tau \in G$ with $\tau\vert_M=\sigma$.
$K$-homomorphism means that $\sigma\vert_K=\text{id}$.
$\\ \\ $
$\textbf{My ideas:}$
I thought to use the Isomorphism extension theorem, which I know in the following form:
Let $\varphi \, \colon K \longrightarrow K'$ be a field isomorphism, $f \in K[X]$ a polynomial with a root $x$ in a field extension of $K$ and $x'$ a root of $\varphi(f)$ in a field extension of $K'$.
Then there exists exactly one field isomorphism $\phi \, \colon K[x] \longrightarrow K'[x']$ with $\phi(x)=x'$ and $\phi\vert_K=\varphi$. 
But I think this doesn't help in this situation.
 A: You can use the key theorem you state, if $L/M$ is finite then $L = M(x_1,\dots,x_n)$ for some $x_i$ in $L$. Let $M_0 = M$ and $M_{i+1} = M_i(x_{i+1})$. 
The crucial part of this question is that $L/K$ is Galois, which for finite extensions is equivalent to being normal and separable. This means that the minimal polynomial of each of the $x_i/K$ splits completely into linear factors in $L$. Thus the minimal polynomial of each of the $x_i$ over $M_{i-1}$ also splits completely in $L$.
So, if we let $f_i(X)$ be the minimal polynomial for $x_i$/$M_{i-1}$, then for example $f_1(X)$ $\sigma(f_1)$ splits completely in $L$*. Thus we can extend $\sigma$ to some $\sigma_1:M_1\rightarrow L$ by the theorem, because we have shown that we have a root of $\sigma(f_1)$ in $L$. We can repeat this process iteratively until we have extended $\sigma$ up through each extension to $L$.
As pointed out in the comments, my original argument for this was invalid. We can prove this by taking $g_i(X)\in K[X]$ to be the minimal polynomial of $x_i$ over $K$, $f_i|g_i$, $g_i$ splits completely in $L$ and is fixed by $\sigma$ since it lies in $K[X]$. Thus $\sigma(f_i)|g_i$ also, so it too splits completely in $L$.
A: Suppose all degrees are finite.
We first prove that there are at most $\lvert M : K \rvert$ $K$-homomorphisms $M \to L$. Proceed by induction on $\lvert M : K \rvert$; suppose we know that there are $\le \lvert N : K \rvert$ $K$-homomorphisms $N \to L$. Let $\alpha \in L \setminus N$, with minimal polynomial $g \in N[X]$ of degree $k > 1$ over $N$. Then $\lvert N(\alpha) : N \rvert = k$, and any $\sigma : N \to L$ can be extended to at most $k$ $K$-homomorphisms $N(\alpha) \to L$, as $\alpha$ can only be mapped in one of the $k$ roots of $g$. Hence there are at most $\lvert N(\alpha) : N \rvert \cdot \lvert N : K \rvert = \lvert N(\alpha) : K \rvert$ $K$-homomorphisms $N(\alpha) \to L$.
Now consider the restriction of the elements $\tau \in Aut(L/K)$ to $M$. (Of course these are not necessarily automorphisms of $M$, but rather $K$-homomorphisms $M \to L$.) The restrictions of $\tau, \tau'$ coincide on $M$ iff $\tau Aut(L/M) = \tau' Aut(L/M)$. So by Lagrange there are
$$
\frac{\lvert Aut(L/K) \rvert}{\lvert Aut(L/M) \rvert}
=
\frac{\lvert L:K \rvert}{\lvert L : M  \rvert}
=
\lvert M : K  \rvert
$$
such restrictions. 
So there are exactly $\lvert M : K  \rvert$ $K$-homomorphisms $M \to L$, and each of these is the restriction to $M$ of an element of $Aut(L/K)$.
