I'm reading the book Fesenko & Vostokov "Local fields and their extensions" and I can't prove a simple statement.
The setting is the following: we have a complete discrete valuation field $(F,v)$ (actually, in the book they assume $F$ only to be Henselian but I don't need this level of generality). Let $L$ be an algebraic extension of $k$ such that $L$ with the valuation $w$ that extends $v$ is a complete discrete valuation field. For my purposes, we can suppose $L \supset k$ to be a finite extension. Also, in the text it's supposed that $L^{alg}=F^{alg}.$ We denote the residue fields by $\bar{k},\bar{L}.$ In general, the extension $\bar{L} \supset \bar{k}$ is not separable. If it's separable I have no problems and everything works.
The proposition states that $L^{un} = L\cdot k^{un}$ where $L^{un},k^{un}$ stands for the maximal unramified extension of $L,k$ in an algebraic closure. In the proof, I can't verify the following sentence:
The compositum of the fields $\bar{k}^{sep},\bar{L}$ coincides with $\bar{L}^{sep}$ since $\bar{L}\supset\bar{k}$ is an algebraic extension.
Abstractly, consider and algebraic extension $K \supset k$ and their separables closures $K^{sep},k^{sep}.$
Why is $K^{sep}=K\cdot k^{sep}$?
The inclusion $K\cdot k^{sep} \supset K^{sep}$ is trivial and the problem is the other inclusion. My attempt was to consider the expression $k^{sep} = k(X)$ with $X$ the set of all roots of separable polynomials over $k.$ This way, we have that $K\cdot k^{sep} = K(X)$ but I can't prove that this field coincide with $K^{sep} = K(Y),$ with $Y$ the set of all roots of separable polynomials over $K.$
I've given the context since it could be helpful although I think my problem is only about fields.
Thanks in advance!
For a finite extension $K/k$ (you can extend to any algebraic extension),
$K^{sep}=K(k^{sep})$ follows from that when $a$ is separable over $k$ then $k(a^{p^n})=k(a)$.
(proof: $k(a)/k(a^{p^n})$ is purely inseparable, if it wasn't trivial then $k(a)/k(a^{p^n})/k$ wouldn't be separable)
Thus, $$K^{sep}=K((K^{sep})^{p^n})$$
With $p^n$ the largest power of the characteristic dividing $[K:k]$ then every element of $K^{p^n}$ is separable over $k$ so that $$(K^{sep})^{p^n}= (K^{p^n})^{sep}\subset k^{sep}$$ and hence $$K^{sep}=K((K^{sep})^{p^n})\subset K(k^{sep})$$