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Let $V$ an affine variety. A principal open set is a set of the form $V(f) = V \setminus\{f=0\} $. A well known theorem states that all such sets are affine varieties, and moreover (Shafarevich, p.50) have coordinate ring $k[V(f)]=k[V][f^{-1}]$.

Now, I am interested in a more general situation - consider again an affine variety $V$, but now look at $$V_{f_1,\dots,f_t}=V\setminus \{f_1 = \dots = f_t =0\}$$ These are quasiprojective varieties, since $V_{f_1,\dots,f_t}=\bigcup_{i=1}^{t}V_{f_i}$.

  1. How can it be shown that such a set is not an affine variety (if it is indeed the case)? In general, what tools are used to show that a given quasiprojective variety is not affine?
  2. Are such sets projective varieties? If so, they are automatically not affine, since the only variety which is affine and projective is one point.
  3. What about the ring of regular functions of $V_{f_1,\dots,f_t}$? Does it equal to $K[V][f_1^{-1}]\dots[f_t^{-1}]$? If not, what can be said about it?
  4. After finding it, can this ring be used to show that the variety is not affine (the equations may be very unpleasant)?
Mike
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2 Answers2

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1) Your varieties $V_{f_1,\dots,f_t}$ are almost never affine, except in very degenerate cases (like for instance $f_1=\dots=f_t$ !).
The basic tool is the following very general and powerful theorem:

Given an arbitrary subvariety $Y\subset X$ of the arbitrary variety $X$, if the complement $X\setminus Y$ is affine, then $\operatorname {codim} _X(Y)=1$

Notice that we don't assume $X$ affine here.
This is a scandalously underappreciated result due to Goodman (Ann. of Math, vol.89, page 162), which doesn't seem to be mentioned in the standard books .
The codimension $1$ condition is of course not sufficient as witnessed by the example $Y=\{*\} \times \mathbb P^1\subset X=\mathbb P^1 \times \mathbb P^1$

2) An open subset $U\subset X$ of an affine variety $X$ of positive dimension is never projective because the regular functions on $U$ obtained by restriction from the regular functions on $X$ separate the points of $U$ while the regular functions on a projective variety are constant.

3) As for the ring of regular functions on $U=X\setminus Y$ in the case $\operatorname {codim} _X(Y)\geq 2$, it is easy to determine under a mild hypothesis (cf. Matsumura's Commutative Algebra, theorem 38, page 124):

If $X$ is normal, then the restriction map $\mathcal O(X)\to \mathcal O(U)$ is bijective

This is the algebraic version of an extraordinary result discovered in 1906 by Hartogs in the analytic setting.

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Try thinking about a simple case. Let $V=\mathbb{C}^2$, an affine variety with co-ordinates $x,y$. Then $\mathbb{C}^2-\{0\}=V_{x,y}$ in your notation. Show that $\mathbb{C}[V_{x,y}]=\mathbb{C}[x,y]$, thus showing that $V_{x,y}$ is not affine and answering your 3 negatively. For 2, no quasi affine varieties are projective except in the trivial case of finitely many points. Similarly, 4 can occur too, in the sense the coordinate ring of suitable quasi affine open sets can be say, non-Noetherian.

Mohan
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  • To complete the example, how can it be shown that $\mathbb{C}^2 \not\cong \mathbb{C}^2 \setminus {(0,0)}$? Also, I do not understand your answer to 2 - every projective variety is in particular quasi-projective. – Mike Jul 25 '15 at 17:52
  • If $\mathbb{C}^2-{0}$ is affine, the argument will say that the natural inclusion of this into $\mathbb{C}^2$ is an isomorphism, which is absurd, since it is not surjective. I said that no quasi AFFINE variety is projective except when it is a finite set of points. – Mohan Jul 25 '15 at 19:03
  • Another way to think about it is that if the punctured plane were affine then the maximal ideal $(x, y)$ would have to cut out a point, but instead it cuts out nothing. – Hoot Jul 25 '15 at 23:42
  • @Hoot, I do not understand your argument. Why should two regular functions cut out a non-empty subset in general? – Mohan Jul 26 '15 at 00:26
  • There's a correspondence between points and maximal ideals in the affine case that's explicit, right? To go from the ideal to the point you looking at where it vanishes. I don't think I'm saying anything deep; my sense is that you're much more expert than I so I'm a little nervous now. – Hoot Jul 26 '15 at 00:36
  • Mike, here are three proofs that the punctured plane is not affine. – Georges Elencwajg Jul 26 '15 at 05:54