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 $\sigma\vert_K=\text{id}$.
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$\textbf{My ideas:}$
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.
Best Answer
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 $L$.