# Show that $[K:F]_{s} = [K:L]_{s}[L:F]_{s}$ and $[K:F]_{i} = [K:L]_{i}[L:F]_{i}$.

Problem. If $F \subset L \subset K$ are fields, then show that $$[K:F]_{s} = [K:L]_{s}[L:F]_{s}$$ and $$[K:F]_{i} = [K:L]_{i}[L:F]_{i}.$$

I've found a resolution here and here, but use different definitions of my professor. Since this question is part of the list of preparatory exercises for my exam, I want to solve using only what was defined in class. My professor defines it like this:

Definition. Let $K$ be a finite extension of $F$. If $S$ and $I$ are the separable and purely inseparable closures of $F$ in $K$, respectively, we define the $$separable degree $[K:F]_{s}$ of $K/F$ to be $[S:F]"$ and the $$inseparable degree $[K:F]_{i}$ to be $[K:S]"$.

We can see that

$$[K:F] = [K:S][S:F] = [K:F]_{s}[K:F]_{i}.$$

It has already been proven that if $K/F$ is normal, $S/F$ is Galois, $K/I$ is Galois and $K=SI$. Hence

$$[K:I] = [S:F]$$

and so

$$[K:S] = [I:F].$$

These are all the information I can conclude from the definition, but I couldn't see how I can use this to solve the problem. I appreciate any hint!

• The definition doesn't use $I$. – Arnaud Mortier May 19 '18 at 0:24
• @ArnaudMortier, yes, because using $[K:S]$ rather than $[I:F]$, we have a better measure of how far $K/F$ if from begin separable. – Corrêa May 19 '18 at 0:28
• I am doubtful that this can be done with just these definitions and results. It seems to me that we crucially use the fact that if $K/F$ is a finite extension, then the separable degree $[K:F]_s$ equals the number of $F$-embeddings of $K$ into $\bar{K}$. – user279515 May 31 '19 at 13:41
• Also, I guess you are referring to Patrick Morandi's textbook, in which case you could take a look at my answer here: math.stackexchange.com/q/1480158/279515. Basically, he sometimes poses problems earlier than we are expected to solve them. – user279515 May 31 '19 at 13:44
• Also related: math.stackexchange.com/q/591077/279515 – user279515 May 31 '19 at 14:24