$[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $

Does the following product rule holds true:

If $F \subset L \subset K$ are fields, then

a) $[K : F]_s = [K : L]_s [L : F]_s $

b) $[K : F]_i = [K : L]_i [L : F]_i $

where $[:]_s$ means separable extension and $[:]_i$ means inseparable extension .

Solution 1:

Perhaps it depends on your definition of $[K:k]_s$ and $[K:k]_i$. Lang shows that for finite extensions, $[K:k]_s$ divides $[K:k]$, and then defines the inseparable degree as the quotient. If that’s your definition too, then you need only show that the separable degree is multiplicative.

For multiplicativity of the separable degree, if you define this to be the number of distinct $k$-morphisms of $K$ into an algebraic closure $\Omega$ of $k$, the multiplicativity is easy enough, at least in concept: each of the $[F:L]_s$ morphisms of $L$ into $\Omega$ can be extended in $[L:K]_s$ ways. The total number is certainly the product of the two separable degrees, since all the morphisms so found are distinct. As I recall, there are irritatingly many technical details that need to be worked out; that’s one reason I don’t want to go into the matter more deeply here.

My recollection of the last time I taught Galois Theory is that I used a different definition of the inseparable degree, however. If that’s so, I certainly don’t recall what that definition may have been.