MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
It would help if you provided Lang's definition of "well-behaved". – S. Carnahan Oct 4 '10 at 6:49
Also I do not know what a simple subvariety is. Furthermore I do not think there is a discrete valuation around, unless $W$ is a prime divisor, i.e. of codimension $1$. – Sebastian Petersen Oct 4 '10 at 10:10
yes, i forgot saying that W is of codimension 1. – vu viet Oct 4 '10 at 12:46
up vote 2 down vote accepted

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 $[E:K]$.


share|cite|improve this answer
For a more concrete approach, if you're variety is over an algebraically closed field, i believe you can prove this using Bezout's theorem, see exercise II.6.2 in Hartshorne. – Max Flander Oct 4 '10 at 11:38
Could you type your answer clearer, please? It's hard to understand. – vu viet Oct 4 '10 at 13:32

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).

share|cite|improve this answer
Your argument seems correct to me. However you are hiding the work to be done under the notion of excellence, which in particular includes the property of having finite normalisations in finite extensions. And this latter property is the only thing one really needs. Hagen – Hagen Oct 5 '10 at 13:13
I absolutely agree, Hagen. However, I consider the notation of excellent scheme and the basic theorems about it to be quite useful, when dealing with such matters. But this is probably a matter of taste. – Sebastian Petersen Oct 5 '10 at 14:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.