Characterization of normal and separable extensions in terms of embeddings Let $E/F$ be a finite field extension.  Let $\text{Emb}(E/F)$ denote the set of field homomorphisms $E \to \overline{F}$ that fix $F$.  Here, $\overline{F}$ is the algebraic closure of $F$.
My understanding is that the following things are true, but I haven't seen any explicit statements in the literature:


*

*$E/F$ is normal $\iff$ $\text{Aut}(E/F) = \text{Emb}(E/F)$

*$E/F$ is separable $\iff$ $|\text{Emb}(E/F)| = [E:F]$

*In general: $|\text{Emb}(E/F)| \leq [E:F]$.


Is all of the above correct?

Second Question: Is there any relation of the above facts (if they're true) to the following one?  (Dummit+Foote: Sec 14.1)


*

*Fact: Let $\varphi \colon F \to F'$ be an isomorphism.  Let $E$ be the splitting field of $f \in F[x]$, and $E'$ the splitting field of $\varphi(f) \in F'[x]$.  Then there are $\leq [E:F]$ extensions of $\varphi$ to an isomorphism $E \to E'$.  If $f$ is separable, then equality holds.

 A: All of the cited statements are true. You can find them (with different notation) in chapter $\mathsf{V}$ of Lang's Algebra, revised 3rd edition, which is on algebraic extensions.
From p.237-238:


(proof of Theorem 3.3)


From p.239-240:


(argument that this doesn't depend on $\sigma$)

(other stuff)


Now for your second question, let $L$ be an algebraic closure of $F'$ containing $E'$. Note that $F$ and $E$ naturally have $F'$-algebra structures, via $\varphi^{-1}$. Let $\Phi:F\to L$ denote the map $\varphi:F\to F'$ considered as a map into $L$.
Any $F'$-algebra homomorphism $f:E\to L$ extending $\Phi$ produces a field $F'\subseteq f(E)\subseteq L$ in which $\Phi(f)=\varphi(f)$ splits. Because $E'$ is the splitting field of $\varphi(f)$ in $L$, we must have $f(E)\supseteq E'$; but $$[E:F]=[f(E):f(F)]=[f(E):F']$$ is equal to $[E':F']$ so that we would have to have $f(E)=E'$. Thus, any $F'$-algebra homomorphism $f:E\to L$ extending $\Phi$ will have image $E'$.
Therefore, an $F'$-algebra homomorphism $f:E\to L$ is equivalent to an $F'$-algebra homomorphism $E\to E'$ (by restricting or extending the codomain as necessary). Note that any $F'$-algebra homomorphism $E\to E'$ is necessarily an isomorphism, by a degree argument again.
Thus, the set of isomorphisms $E\to E'$ extending $\varphi$ has the same cardinality as the set of $F'$-algebra homomorphisms $f:E\to L$ extending $\Phi$, which is (almost, but not quite, by definition) equal to the separable degree $[E:F]_s$, which is less than or equal to $[E:F]$, and we have equality (by definition) when $E/F$ is separable, which is the case precisely when $f$ is separable.
