# Multiplicative nature of the separability degree

In what follows, let $E / F$ be an algebraic extension, $h(x),f(x)\in F[x]$ polynomials, $h(x)$ irreducible.

Definitions.

1. We say $h(x)$ is separable if it has not repeated factors. We say $f(x)$ is separable if all its irreducible factors are separable.

2. We say $\alpha\in E$ is separable over $F$, if its minimal polynomial over $F$ is.

3. The extension $E/F$ is separable if all the elements of $E$ are separable over $F$.

4. We define the field $$\newcommand\sclo{\operatorname{Scl}}\sclo_F E= \{\alpha\in E : \text{ \alpha is separable over F }\}$$ as the separable closure of $F$ in $E$.

5. The separability degree of the extension $E/F$ is defined as $$[E:F]_s = [\sclo_F E:F].$$

Now, suppose the extension $E/F$ is finite.

My question is: how can one proof that with these definitions, if $B$ is an intermediate field of the extension $E / F$, then $$[E:F]_s = [E:B]_s[B:F]_s?$$

I know that the purpose of the separability degree is to count in how many ways we can extend certain injective homomorphism to an automorphism. If one starts defining separability degree this way, as done in the book by Lang, the result follows without much difficulty.

I wonder how to achieve it in this way.

Note that your definition, you don't even know whether separable closure is a field, i.e., if $\alpha,\beta$ are separable over $F$, then so is $\alpha+\beta$ and $\alpha\beta$, this is non-trivial to prove using your definition.
Nonetheless, regarding your question, multiplicativity of separable degree can indeed be proven using your definition. See Fields and Galois Theory by Patrick Morandi, Section 4 proves separable closure is indeed a field (he shows this by proving $E/F$ is separable if and only if some extension of it is Galois), section 8 shows the multiplicativity (by some cumbersome fiddling with multiple fields and their separable closures). Both proofs are very long and ad hoc.