# $L|F$ and $F|K$ unramifed imply $L|K$ unramified

Let $$K$$ be a Henselian field with respect to a (non-Arch) valuation $$v$$ (not necessarily discrete). Let $$L|K$$ be an infinite algebraic extension and let $$F|K$$ be any subextension. Then $$v$$ extends uniquely to $$F$$ and $$L$$. Let $$v_L$$ be its extension to $$L$$.

Suppose that $$F|K$$ and $$L|F$$ are unramified, then can we show that $$L|K$$ is unramified?

(We say that a finite extension $$L|K$$ is unramified if $$[L:K]=[l:k]$$ and $$l|k$$ is separable, where $$l$$ (resp. $$k$$) denotes the residue field of $$v_L$$ and $$v$$. An algebraic extension $$L|K$$ is unramified, if all its finite subextensions are unramified.

If $$L|K$$ is finite, then we can indeed show this by computing all involved degrees. So far it is not clear that why we can pass from an algebraic extension to a finite one.)

If $$L/F$$ and $$F/K$$ are infinite and unramified then take an arbitrary finite subextension $$A/K$$, let $$B$$ be the normal closure of $$A$$ in $$\overline{L}$$, let $$C=B\cap L$$ and $$D=C\cap F$$.

For simplicity assume that $$CF = F(c)$$ has a primitive element with $$c\in C$$ (the idea stays the same if there is no primitive element). The $$F$$-minimal polynomial of $$c$$ is in $$D[x]$$.

So $$[CF:F]=[C:D]$$.

$$CF/F$$ is unramified so $$[CF:F]=[\kappa(CF):\kappa(F)]$$.

We know that $$[\kappa(C):\kappa(D)] \ge [\kappa(CF):\kappa(F)]$$.

Whence $$[\kappa(C):\kappa(D)] = [C:D]$$ and $$C/D$$ is unramified.

Since $$D/K$$ is unramified as well we get that $$[C:K]=[C:D][D:K]=[\kappa(C):\kappa(D)][\kappa(D):\kappa(K)]=[\kappa(C):\kappa(K)]$$ ie. $$C/K$$ is unramified which proves that $$L/K$$ is unramified.

• @Tian Ok, I edited my answer Commented Sep 12, 2022 at 19:14
• Thanks a lot for your wonderful argument! (Actually this question has confused me for quite a long time...)
– Tian
Commented Sep 13, 2022 at 0:58