4
$\begingroup$

$X$ is a variety and there are $m$ points $x_1,x_2,\cdots,x_m$ on $X$. Can we find an open affine set which contains all $x_i$s?

  • 1
    I think a more accurate title would be "Open affine neighborhood of *finite subsets*".2012-12-19

1 Answers 1

9

A such variety is sometimes called FA-scheme (finite-affine). Quasi-projective schemes over affine schemes (e.g. quasi-projective varieties over a field) are FA.

On the other hand, there are varieties which are not FA. Kleiman proved that a propre smooth variety over an algebraically closed field is FA if and only if it is projective.

Some more details can be found in § 2.2 in this paper.

There is an easy proof for projective varieties $X$ over a field. Just take a homogeneous polynomial $F$ which doesn't vanish at any of $x_1,\dots, x_m$. Then the principal open subset $D_+(F)$ is an affine open subset containing the $x_i$'s. The existence of $F$ is given by the graded version of the classical prime avoidance lemma:

Edit

Let $R$ be a graded ring, let $I$ be a homogeneous ideal generated by elements of positive degrees. Suppose that any homogenous element of $I$ belongs to the union of finitely many prime homogeneous ideals $\mathfrak p_1, \dots, \mathfrak p_m$. Then $I$ is contained in one of the $\mathfrak p_i$'s.

A (sketch of) proof can be found in Eisenbud, § 3.2. For the above application, take $\mathfrak p_i$ be the prime homogeneous ideal corresponding to $x_i$ and $J=R_+$ be the (irrelevant) ideal generated by the homogeneous elements of positive degrees. As $R_+$ is not contained in any $\mathfrak p_i$, the avoidance lemma says there exists a homogeneous $F\in R_+$ not in any of the $\mathfrak p_i$'s.

This method can be used to prove that any quasi-projective variety $X$ is FA (embed $X$ into a projective variety $\overline{X}$ and take $J$ be a homogeneous ideal defining the closed subset $\overline{X}\setminus X$. let $F\in J$ be homogeneous and not in any $\mathfrak p_i$, then $D_+(F)$ is affine, contains the $x_i$'s and is contained in $X$).

  • 0
    Dear QiL, could you please state the graded version you mention of the prime-avoiding lemma ?2012-12-13
  • 0
    @GeorgesElencwajg: yes sure.2012-12-13
  • 0
    Dear QiL, thanks a lot for your answer, but it seems to me that what you need is the second part of Eisenbud's Lemma 3.3 and not the result overlined in grey of your edit, which is the usual prime-avoidance lemma. And actually you use that second part of Lemma 3.3 in the application to Hezudao's question , and not your overlined statement2012-12-13
  • 0
    @GeorgesElencwajg: you are absolutely right. Thanks !2012-12-13
  • 0
    By the way I didn't know this beautiful application of the (graded) prime-avoidance lemma: as usual I'm learning a lot by reading you!2012-12-14
  • 0
    @GeorgesElencwajg, merci c'est trop gentil !2012-12-14