About the number of $K$-homomorphisms 
Let $K\subseteq L\subseteq M$ be fields with $[L:K]=n$. I should prove that the number of $K$-homomorphisms from $L$ to $M$ (so field homomorphisms that fix pointwise $K$) is less or equal than $n$?

The case $L=K(u)$ is easy, because any $K$-homomorphism exchanges the roots of the minimal polynomial of $u$ over $K$. But what about the general case? I think that it is necessary an induction argument on $n$.
 A: We wish to prove for a given homomorphism $K \to M$ that the number of ways to extend it to a homomorphism of $L \to M$ is at most $[L:K]$.
This proof is a standard argument using induction. We induct on $n$. If $n = 1$ we have $L = K$ and there is nothing to prove, for the extension of any given homomorphism $K \to M$ to one of $L \to M$ is just the original homomorphism itself!  Now suppose in the general case that $[L:K] > 1$. Choose $\alpha \in L \notin K$ and consider the tower
$$\begin{array}{c} L \\ | \\ K(\alpha) \\ | \\ K. \end{array}$$
If $K(\alpha) = L$ then the number of homomorphisms of $L$ into $M$ that extend a given homomorphism $K \to M$ is just going to be the number of roots of the minimal polynomial of $\alpha$ over $K$ which is $\leq [L:K]$. If $K(\alpha) \neq L$, then by induction the number of homomorphisms $L \to M$ that extend a given homomorphism $K(\alpha) \to M$is $\leq [L : K(\alpha)]$, and the number of homomorphisms from $K(\alpha) \to M$ that extend a given $K \to M$ is $\leq [K(\alpha) : K]$. Thus the total number of homomorphisms $L \to M$ that extend  a given homomorphism $K \to M$  is $\leq [L:K(\alpha)][K(\alpha) : K] = [L:K]$.
A: I assume that you are familiar with the notion of sparability, in particular, separable extensions and separable closures and let us begin by an observation: a field homomorphism is either $0$, or an isomorphism into a subfield of the target. Hence, a $K$-field homomorphism is an isomorphism, if $K\not=0$.  

If $L/K$ is separable, then by the primitive element theorem, we know that $L=K(u)$ for some $u\in K$, which case you have solved.
  Now if $L/K$ is inseparable, $K$ must be an (infinite) field of characteristic $p\not= 0$. (And $K$ cannot be algebraically closed, otherwise $L=K$.)
  By taking separable closures, and by reason of the multiplicativity of the degrees of extensions, we can assume that $L/K$ is purely inseparable. Thus every $\alpha\in L$ satisfies $\beta=\alpha^{p^t}\in K$ for some $t\in\mathbb N$. Now $\alpha$ satisfies $x^{p^t}-\beta=(x-\alpha)^{p^t}=0$ so that every $K$-homomorphism of $L$ must leave $\alpha$ invariant. Since $\alpha\in L$ was arbitrary, it follows that the only such homomorphism is the identity.  

The proof is complete now.
If some inappropriate points occur, tell me, thanks.
