# Non-projective smooth complete threefolds with a pair of points intersecting every surface

I've been learning about non-projective complete varieties and I am trying to get a handle on how crazy they can get.

$\textbf{Question:}$ Let $V$ be a complete threefold over $\mathbb{C}$. Given distinct points $P,Q$ on $V$, does there exist a complete surface $S\subset V$ which contains neither $P$ nor $Q$?

I would prefer it if $V$ were smooth, though I'd be surprised if that were relevant. I would also be interested in examples/proofs over fields of finite characteristic.

Thanks!

• Have you looked at Hironaka's example? It contains some pairs of points which are not contained in any affine open subset, they would be good candidates for your $P,Q$. – abx Nov 12 '13 at 7:09
• Hironaka's example satisfies this condition, because it maps onto a projective variety of positive dimension. (See my answer below). – Sándor Kovács Nov 12 '13 at 8:00
• Blow down some lines on a quintic threefold. This gives singular, proper, three-dimensional algebraic spaces that have the pathological property you describe. – Jason Starr Nov 12 '13 at 12:40

I do not know the answer to the question. However, let me point out that the answer is negative if you look for surfaces avoiding more points. More precisely, there exists a smooth complete threefold that does not satisfy Sándor's property $NC_{10}$: it contains 10 points that cannot be simultaneously avoided by a divisor.

In [Fujino - Payne, Smooth complete toric threefolds with no nontrivial nef line bundles], the authors construct varieties as indicated in the title. Their best example $X$ has Picard rank $5$, and they show that it is optimal. Now, the property that $X$ has no nontrivial nef line bundle is equivalent to the fact that the cone of effective curves is the whole vector space $N_1(X)$ of $1$-cycles up to numerical equivalence. Since this vector space is $5$-dimensional, it will be possible to find $10$ integral curves $C_1,\dots,C_{10}$ on $X$ spanning this vector space as a cone (see [Davis, Theory of positive linear dependance, Theorem 6.7]). Choose a point $P_i$ in $C_i$. I claim that no surface in $X$ avoids all the $P_i$. Indeed, such a surface $S$ would intersect non-negatively the $C_i$, hence any curve. This is obviously impossible: working in an affine neighbourhood of a point of $S$, one constructs a curve intersecting $S$ positively.

Note that in Fujino and Payne's precise example, it should be possible to find less than $10$ integral curves spanning $N_1(X)$ as a cone, but certainly not less than $6$ curves.

Finally, toric varieties are never going to provide counter-examples to your original question, as any two points in a toric variety have a common affine neighbourhood.

• very nice example! – Sándor Kovács Nov 14 '13 at 19:47
• Fantastic! This basically answers my question, I was never so focused on the number 2. Thanks a lot! – jacob Nov 15 '13 at 22:40

This is just a partial answer to restrict the search.

Condition $NC_q$ Let us say that $Z$ satisfies condition $NC_q$ if for any $z_1,\dots,z_q\in Z$ there exists $D=D_{z_1,\dots,z_q}\subset Z$, a codimension one subscheme of $Z$ such that $z_i\not\in D$, for any $i=1,\dots,q$.

Example If $Z$ is projective, then it saisfies $NC_q$ for any $q\in\mathbb N$.

Claim Suppose that $Z$ is a scheme that satisfies $NC_q$ and let $f:V\to Z$ be a dominant morphism. Then $V$ satisfies $NC_q$.

Proof Apply the condition for $f(v_1),\dots,f(v_q)$ and take $f^{-1}(D)\subset V$.

Corollary If $V$ is an arbitrary threefold that admits a dominant morphism to a projective variety of positive dimension, then it satisfies $NC_2$. In particular, Hironaka's example of a non-singular non-projective complete threefold does not provide a counter-example.

There are non-projective schemes with trivial Picard group and hence without non-trivial line bundles. Obviously these cannot admit a non-trivial morphism to a projective space (variety).

On the other hand, such a scheme has to be singular: Take an arbitrary dense affine subset and take an effective divisor there. The closure of this in the original scheme will give a non-trivial divisor which would then have to be a non-Cartier divisor, as otherwise the Picard group could not be trivial. But then the ambient scheme has to be singular.

Alternatively one could argue, using Chow's lemma that there are non-trivial divisors on any complete variety and hence the Picard group of a complete non-singular variety cannot be trivial.

So, an example of a complete non-singular threefold failing $NC_q$ would have the property that every linear system has base points. I don't know if that exists.

For a singular example you could do the following (this is motivated by Jason's comment): Take a projective threefold whose Picard group is $\mathbb Z$ and hence every effective divisor is ample on it. Contract some curves to get something non-projective. Now take any surface on the target. The pre-image of that is an effective divisor on the original threefold which then has to be ample, and hence it has to intersect the contracted curves. Therefore the chosen surface has to contain the image(s) of these curves, so if you choose your $P,Q$ among these image points, then you get a counterexample.

• Thaanks for your answer! I'm a little confused though: Can you just blow down curves and still get a scheme? your example can't work too well, because say X is a projective threefold with picard group Z and I blow down some curve C to get a scheme Y, where the image of C is some point P. Now take any surface on Y avoiding P (you can certainly do this, just take an affine open around P, take a surface there and then close up). Pulling this back to X gives you a contradiction. I realize of course I'm just doing your argument backwards! But doesn't this show that you can't contract such curves? – jacob Nov 14 '13 at 8:58
• @jacob : Jason's and Sándor's examples are indeed not schemes, but algebraic spaces (equivalently, if working over $\mathbb{C}$, compact Moishezon spaces). As for your question in the "smooth proper scheme" setting, toric varieties are not going to provide counterexamples either, as every pair of points have a common affine neighbourhood. I have not been able to make up my mind what the answer to your question should be: do you have a guess ? – Olivier Benoist Nov 14 '13 at 9:30
• Ah, Ok. I was just confused because I thought they meant singular schemes, rather than algebraic spaces. No, I don't really know what to expect either. I'm intrigued by the strategy proposed, though. Is it known which curves can be contracted on a threefold? If one can find a projective X with a bunch of curves $C_1,...,C_n$ all of which can be contracted, and such that their sum intersects each surface positively, then one will have made a threefold violating $NC_{n}$. Is such a thing feasible to carry out? – jacob Nov 14 '13 at 10:43
• @jacob: although, as Olivier pointed out, indeed I wasn't claiming that the result is a scheme, I am not sure that your argument in your first comment works. If the surface you pick through the image point $P$ is not $\mathbb Q$-Cartier, then what is its pull-back? In other words it appears that what you are proving is that the target cannot be $\mathbb Q$-factorial. – Sándor Kovács Nov 14 '13 at 19:44
• @SándorKovács: Since I'm assuming the surface I pick does not go through P, so pulling it back is easy as X\C is isomorphic to Y\P. Am i missing something? – jacob Nov 15 '13 at 22:43