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By an affine variety I mean a variety that is isomorphic to some irreducible algebraic set in $\mathbb A^n$ and by a quasi-projective variety I mean a locally closed subset of $\mathbb P^n$, with the usual Zariski topology and structure sheaf. (I am not quite familiar with the language of schemes.)

Assume that the underlying field is algebraically closed.

Let $X$ be a quasi-projective variety. Are there any effective ways to tell whether $X$ is affine?

Let me be more specific. If $X$ is affine, then $\mathcal O_X(X)$ is large enough so that it completely determines the variety. In particular, the Nullstellensatz holds, i.e. the map $$X\to \mathrm {spm}(\mathcal O_X(X))$$ $$x\mapsto \{f\in\mathcal O_X(X)|f(x)=0\}$$ is a 1-1 correspondence, where $\mathrm{spm}(\mathcal O_X(X))$ denotes the maximal ideals in $\mathcal O_X(X)$.

Now I'd like to pose the question: If $X$ is quasi-projective and the above map is a bijection, is $X$ necessarily affine?

Thanks!

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  • $\begingroup$ If it is an isomorphism of varieties, then yes – of course. The map you construct is automatically a morphism of varieties, so the question is whether it is enough that it be a bijection for it to be an isomorphism. Unfortunately, there are examples of morphisms that are bijective but nonetheless not invertible... $\endgroup$
    – Zhen Lin
    Jul 29, 2012 at 1:37
  • $\begingroup$ According to a theorem of Serre (Hartshorne Theorem III.3.7), it is necessary that $H^1(X,\mathcal I)=0$ for every coherent sheaf of ideals $\mathcal I$ on $X$. I don't know whether checking all such conditions is really feasible though... $\endgroup$
    – Andrew
    Jul 29, 2012 at 1:45
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    $\begingroup$ @ZhenLin: Yeah, I know that map. But that does not provide an immediate counterexample to my question, does it? $\endgroup$
    – YZhou
    Jul 29, 2012 at 1:58
  • $\begingroup$ Dear Andrew, Note that $\mathcal O(X)$ is not always f.g. over the ground ring $k$, so that maxspec $\mathcal O(X)$ is not always an affine variety. (See my answer for a little more on this point, and a link to an example.) Regards, $\endgroup$
    – Matt E
    Jul 30, 2012 at 2:57

1 Answer 1

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If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine.

Here is an argument:

By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite, and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$. Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward.

(I'm not sure that the full strength of ZMT is needed, but it is a natural tool to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.)


In fact, the argument shows something stronger: suppose that we just assume that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres, i.e. is quasi-finite and surjective. Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine, and since by assumption it is surjecive, it is an isomorphism.

Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis.

E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine, but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective.

E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) = \mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre over $0$.


Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective.

(Of course, one could work more generally with arbitrary schemes and Spec rather than maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.)

Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's

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  • $\begingroup$ Wooonderful arguement! Thanks so much! $\endgroup$
    – YZhou
    Aug 1, 2012 at 8:43
  • $\begingroup$ For the fact the that a quasi-affine variety satisfying my hypothesis must be affine, I only have a argument minicing the proof for that a distinguish open subset of a affine variety is affine. I didn't quite understand your "But it is standard that the map $X\to$ maxspec $\mathcal O(X)$ is an open immersion". Could you explain that further? Thanks! $\endgroup$
    – YZhou
    Aug 1, 2012 at 8:46
  • $\begingroup$ @Andrew: Dear Andrew, Here is a lemma (which I am calling standard, although I don't recall it being in Hartshorne --- so maybe it's not completely standard!): a scheme $X$ is quasi-affine if and only if the morphism $X \to $ Spec $\mathcal O(X)$ is an open immersion. The if direction is clear, since we then get an open immersion of $X$ into an affine scheme. The converse is a good exercise. If you have trouble with it, you could ask it as another question. Cheers, $\endgroup$
    – Matt E
    Aug 1, 2012 at 12:42
  • $\begingroup$ @MattE: Ok, let me have a try! Thank you, Matt! $\endgroup$
    – YZhou
    Aug 2, 2012 at 15:58

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