Denote $C=\bar{K}$ algebraic closure of field $K$ and $E/K$ some finite field extension. Since the separable degree of field extension $E/K$ is defined as $|Hom(E/K, C/K)|$ instead of $|End(E/K)|$ where both $Hom,End$ means the set $K-$algebra homomorphism and endomorphisms, I would like to compute the simple extension's endormorphism and check whether it is trivial.
Suppose $E=K(a)$ where $a$ algebraic over $K$. So $E\cong K^n$ for some $n\in N$. $End(E/K)\cong K^{n^2}$ as $K$ vector space.
- However, not all of them are $K-$algebra homomorphism.
Clearly $Id_{K(a)}\in End(K(a)/K$. Any endomorphism of field extension is injecitve and thus isomorphism between $K(a)/K$ and $K(a)/K$. So some element $b\in K(a)$ is being sent to $a\in K(a)/K$. So $K(a)=K(b)$. $a$ and $b$ must differ by a non-zero element of $K$.(i.e $f(a)=k'a,k'\in K^{\star}$). So $|End(E/K)|=|K^{\star}|$ where $K^{\star}$ are non-zero elements.
How do I identify the morphism between $End(E/K)$ and $K^{\star}$? Identify them as group? Does this happen to be in Abelian category?
I should expect $|Hom(E/K,C/K)|\geq |End(E/K)|$. For simple finite extension I do not see this unless I have made error above.
The right way to look at your situation is to consider the minimal polynomial for $a$, say $g(X)\in K[X]$, and to ask how many other roots of $g$ lie in $K(a)$. Depending on $a$ (equivalently, on $g$) there may be no other roots lying in $K(a)$, or all roots of $g$ may be there. This is the core of the concept of normality of an extension.