Why is an absolute value generated by a simple subvariety of a variety V well-behaved? 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!
 A: 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
A: Another approach might be the following:
Let $k$ be a field and $V/k$ a variety, i.e. a seperated $k$-scheme
of finite type which is geometrically integral. Let $F$ be its function
field. Let $x\in V$ be a point whose local ring ${\cal O}_x$ is a d.v.r.
Denote by $v$ the discrete valuation of $F$ corresponding to this d.v.r.
Let $F_v$ be the completion of $F$ at $v$.
For your question the following fact seems to be important. $V$ is
an excellent scheme, because it is of finite type over a field. (c.f. EGA IV.2.7.8).
This implies that $F_v/F$ is a separable field extension.
Let $E/F$ be a finite (not necessarily separable) field extension. Then the canonical map
$E\otimes_F F_v\to \prod_{w/v} E_w$ is bijective. (Note that $E\otimes_F F_v$ is reduced, because $F_v/F$ is separable.)
This shows that $v$ is well-behaved (if I understood correctly what well-behaved means).
