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.

Given an integral scheme $X$, let $K(X)=\mathrm{Frac}(R)$ be its function field, where $\mathrm{Spec}(R)$ is some non-empty open affine subscheme of $X$. Take the maximal ideal $P$ of some DVR of $K(X)$ (i.e. a place $P$). How can one associate a closed point of $X$? Does it exist a bijection between closed points of $X$ and places of $K(X)$?

share|improve this question
3  
Why should there be? The DVR corresponding to a place would be a $1$-dimensional local ring of $X$, which means it corresponds to a codimension $1$ subvariety of $X$. Unless $X$ is itself $1$-dimensional, this won't be a point. –  Zhen Lin Dec 14 '12 at 12:32
    
In fact, any valuation determines uniquely a point of the scheme (not a closed point, though) as its center if and only if the scheme is separated. –  Mauro Porta Dec 14 '12 at 22:20
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.