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"Let $(K,v)$ be a valued field, and $(L,w)$ be a finite extension of degree $n$ of $(K,v)$. Let $e=e(w\mid v)$ and $f=f(w\mid v)$. Then $ef\le n$, and if $v$ is discrete and $L/K$ is separable we have the fundamental identity $ef=n$".

This is proved by Neukirch's Algebraic Number Theory assuming that $(K,v)$ is henselian. However, the inequality $ef\le n$ apparently does not need this hypothesis in the proof. I wonder if the second part is also true, assuming that $w$ is discrete.

The only passage in the solution that uses $K$ henselian is to conclude that the valuation ring $B$ of $L$ is a finitely generated module over the valuation ring $A$ of $K$. When $K$ is henselian, this holds because in this case $B$ is the integral closure of $A$ in $L$. I wonder if we can conclude that $B$ is a finitely generated $A$-module without the hypothesis that $(K,v)$ is henselian (of course the extension $w$ may not be unique, but assume $w$ is fixed).

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    $\begingroup$ In fact $B$ is integral (and hence finite if $L/K$ is separable) over $A$ iff $v$ extends uniquely to $L$ (at least when $v$ is discrete, not sure if it also works in general) which is e.g. the case if the field is henselian. $\endgroup$
    – leoli1
    Commented Sep 18, 2021 at 20:18

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No, there is no chance of it being true in general, since if $w_1,\dots,w_r$ are non-equivalent extensions ov $v$ to $L$ then $\sum_i e(w_i|v)f(w_i|v)\leqslant n$, so as long as there are at least two different extensions the equality cannot hold.

The fact that $v$ is henselian guarantees that this cannot happen.

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