Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

This is an exercise in Hartshorne's book.

For a quasi projective variety $Y$ with dimension $\geq 2$ and $p \in Y$ a normal point, if $f$ is regular on $Y-\{p\}$ then $f$ can be extended to a regular function on $Y$.

I want to get some hints to prove this problem....

share|cite|improve this question

Since the problem is local around $p$, you can assume that $Y=Spec(A)$ where $A$ is a noetherian domain (quasi-projectiveness is irrelevant).

Clearly, every point $\mathfrak q \in Spec(A)$ of height one is distinct from $p$ (since $\mathfrak q$ it corresponds to a subvariety of codimension $1$). So every function $f$ defined on $Spec(A)\setminus \lbrace p\rbrace $ is defined at $\mathfrak q $.
You can then conclude that $f\in A$, that is $f$ extends regularly through $p$, thanks to the formula valid for a noetherian normal domain (Matsumura, Commutative ring theory, Theorem 11.5, page 81)

$$ A=\bigcap_{ht(\mathfrak q)=1} A_ \mathfrak q $$

A general result in this vein is that if $X$ is a locally noetherian normal integral scheme and $Y\subset X$ a closed subset of codimension $\geq 2$, the restriction morphism $\mathcal O_X(X)\to O_X(X\setminus F)$ is bijective.

share|cite|improve this answer
    
Thank Georges, But, This problem is one of exercise of Hartshorne's book "Algebraic geometry" chapterI.1.3.... Maybe there is s proof with out scheme theory...?? – Sang Cheol Lee Mar 8 '12 at 15:36
    
@Georges Elencwajg : Could you please show me where to find the proof of the above general result ? – Arsenaler Jan 27 '13 at 18:46

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.