Is an automorphism of a normal extension determined by its image of the maximal separable sub extension? Let $L / K$ be a normal, algebraic field extension. Suppose that the maximal separable sub- extension $M/K$ is finite, $K \subseteq M \subseteq L$. By the primitive element theorem, $M=K(x)$ for some $x \in L$. Is it true that any automorphism of $L$ over $K$ is determined by its image on $x$? If yes, why?
 A: Let $L$ and $F$ be algebraic extensions of a field $K$ with $L_s$ and $F_s$ the $K$-separable parts. Then the restriction map
$\mathrm{Hom}_K(L,F)\rightarrow\mathrm{Hom}_K(L_s,F_s)$ 
is injective (these are $K$-algebra homomorphisms). Note that if $\sigma:L\rightarrow F$ is an element of the source, then for any $\alpha\in L_s$, $\sigma(\alpha)$ has the same minimal polynomial, and therefore is also separable over $K$, i.e., $\sigma(\alpha)\in F_s$.
The assertion is tautological if $K$ has characteristic zero, so I'll assume the characteristic is $p>0$. Suppose $\sigma\vert_{L_s}=\tau\vert_{F_s}$. Given $\alpha\in L$, there exists $n\geq 0$ such that $\alpha^{p^n}\in L_s$. Then 
$\sigma(\alpha)^{p^n}=\sigma(\alpha^{p^n})=\tau(\alpha^{p^n})=\tau(\alpha)^{p^n}$,
so $\sigma(\alpha)=\tau(\alpha)$ because the map $\beta\mapsto\beta^{p^n}$ is injective on any field of characteristic $p$ (being a ring map).
In your setup, $L=F$, and $L_s=M=F_s$. Since a $K$-algebra endomorphism of $M$ is determined entirely by where $x$ goes (it can go to any other root of its minimal polynomial in $M$), the above shows that a $K$-algebra endomorphism of $L$ is determined by the image of $x$.
