I am reading "Fundamentals of Diophantine Geometry" by Serge Lang. 


Let V be a (absolute) variety, W be a simple subvariety of V. Then we know that the local ring of W is a discrete valuation ring, hence induces a discrete valuation v. But why is v well-behaved (in the sense of Lang's book)?


Can anyone help me find the answer or recommend me a book proving this assertion? Thanks in advance!



Why is an absolute value generated by a simple subvariety of a variety V well-behaved?
According to Lang the valuation $v$ of the field $K$ is well-behaved, if for every finite extension $E/K$ the equation


$
[E:K] =\sum\limits\_{w|v} [E\_w:K\_v]
$


holds, where the summation runs over all extension $w$ of $v$ to $E$, and $K\_v$, $E\_w$ are the completions of the fields $K$, $E$ with respect to the valuations $v$, $w$ respectively.


For a discrete valuation $v$ the completion $K\_v$ is equal to the field of fractions of the $M\_v$-adical completion $\widehat{O\_v}$ of the local ring $O\_v$, where $M\_v$ is the maximal ideal of $O\_v$.


The discrete valuation $v$ we are discussing here by assumption is a localization of an integral, finitely generated $k$-algebra($k$ the field over which the variety $V$ lives). Such an algebra has a finite normalisation in every finite extension of their field of fractions. This property is inherited by localisations of the algebra, thus $O\_v$ has this property too: the normalization $O\_v(E)$ of $O\_v$ in a finite extension $E$ of the field of fractions $K$ is a finitely generated, torsion-free $O\_v$-module. It is well-known that such modules over a factorial ring are free - and $O\_v$ is factorial. The rank of $O\_v(E)$ msut be $[E:K]$ - just localise at $0$.


There is a bijection between the valuation rings $O\_w$ of the extensions $w$ of $v$ to $E$ and the localisations of $O\_v(E)$ at maximal ideals $M$.


The product of all these maximal ideals is some ideal $I$. The $I$-adical completion $\widehat{O\_v(E)}$ of $O(E)$ satisfies:


$
\widehat{O\_v(E)} = \prod\limits\_{w|v}\widehat{O\_w} 
$


(Matsumura, Thm. 8.15).


Since a power of $I$ lies in the ideal $M\_vO\_v(E)$ and a power of $M\_vO\_v(E)$ lies in $I$, the completions of $O\_v(E)$ with respect to these two ideals coincide.


The completion of $O\_v(E)$ with respect to $M\_vO\_v(E)$ on the other hand equals the tensor product $
O(E)\otimes\_{O\_v}\widehat{O\_v} $. Since the extension $\widehat{O\_v}/O\_v$ is faithfully flat this tensor product is a free $\widehat{O\_v}$-module of rank $[E:K]$.


Altogether we see now:


$
\prod\limits\_{w|v}\widehat{O\_w}
$


is a free $\widehat{O\_v}$-module of rank $[E:K]$, from which we get


$
\prod\limits\_{w|v}E\_w
$


is a free $K\_v$-module of rank $[E:K]$.


Hagen


