Number of F-homomorphism from K to E

Question

Let $ E/F $ be a finite normal field extension and let $ F \subset K \subset E $ be an intermediate field.
Show that the number of $F$-homomorphisms from $K$ to $E$ is $[K:F]$ iff $K/F$ is separable.
What i got that the number of such homomorphisms is at most $[K:F]$, and that $E/K$ is also normal, but i am not sure how to advance from here.

Answer 1

Use the fact that degree and degree of separability (which is exactly the number of homomorphisms you need to find) are both multiplicative. Then choose a chain of simple intermediate extensions

$$F\subseteq F(a_1)\subseteq\dots\subseteq F(a_1,\dots,a_k)=E.$$

Now show that the degree of separability of a simple extensions is equal to the degree iff the extension is separable using the extension lemma: Let $F$ be a field, $a$ algebraic over $F$ with minimal polynomial $f=\sum_{i=0}^n b_ i X^i$ and let $\bar F$ be an algebraic closure. Then for every homomorphism $\sigma: F\to \bar F$ the extensions of $\sigma$ to $F(a)$ are uniquely determined by the image $\sigma(a)$ of $a$, and the allowed images of $a$ are exactly the roots of $f_\sigma$, where $f_\sigma=\sum_{i=0}^n\sigma(b_i)X^i$.

Keep in mind that if $\sigma$ is an $F$-homomorphism, $f_\sigma=f$, so the last bit becomes easier to handle: $a$ can be sent to any of the roots of its minimal polynomial, which is where separability will play a role.

Answer 2

Hint: The primitive element theorem implies that $K=F(a)$. Let $f$ be the minimal polynomial of $a$, $f(x)=(X-a_1)...(X-a_n)$ and $g$ an automorphism of $K$, $g(a)$ is a root of $f$ implies that there are at most $n$ extensions. There are $n$ extensions since the roots of $f$ are distinct.