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!
Best Answer
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