Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am a bit stuck with the following assertion:

Let $X$ be a separated integral scheme. Then to every (schematic) point $x \in X$ we can correspond its local ring, and look at it as a subring of the field of rational functions on $X$. How do we show that different points give rise to different such subrings?

If $X$ is affine, it is clear, since we can find a function which is zero on one point and non-zero on another. If $X$ is not affine, I guess that for two points which do not lie in same affine open subscheme we should choose affine neighbourhoods and consider their intersection, which is affine by separatedness. But what to do then?

Thanks, Sasha

share|improve this question

1 Answer 1

This follows from the valuative criterion for separatedness: for every valuation ring $O$ of $K(X)$ there exists at most one point $x\in X$ such that $O$ is a local ring extension of the local ring $O_{X,x}$.

Assume $O_{X,x}=O_{X,y}=:O^\prime$. By the so-called $x-\frac{1}{x}$-Lemma there exists a valuation ring $O$ of $K(X)$ such that the extension $O^\prime\subseteq O$ is local. Hence $x=y$.

share|improve this answer
    
The proof for the uniqueness in case of valuation rings of $K(X)$ works mutatis mutandis for any integral domain with field of fractions $K(X)$. This would avoid finding valuation rings $O$ dominating the local rings. But the existence of $O$ is sometimes helpful in more advanced problems. –  user18119 Nov 25 '11 at 9:02

Your Answer

 
discard

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.