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.)


1 Answer 1


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.

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

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