# Fundamental Identity - Extension of Valuations

"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).

• 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. Commented Sep 18, 2021 at 20:18

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.