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I will say zero is effective on a complete, smooth variety $X$ if some positive linear combination of irreducible varieties is rationally equivalent to zero. In other words, zero is effective if there exist (not necessarily distinct) irreducible closed subvarieties $V_1, \dotsc, V_k$ of $X$, with $k \geq 1$, such that $$[V_1] + \dotsb + [V_k] = 0$$ in the Chow ring. (If you know how to answer this question with rational equivalence replaced by numerical or algebraic equivalence, please do so.)

If $X$ is projective, then zero is not effective on $X$ since any subvariety $V$ has positive degree after $X$ is embedded in $\mathbb P^n$.

[More intrinsically, no positive linear combination of points can be rationally equivalent to zero; and if $D$ is an ample divisor on $X$ and $V$ is a dimension-$k$ variety in $X$, then $[D]^k . [V]$ is a positive linear combination of points, hence nonzero in the Chow ring; whereas $[D]^k . [0] = 0$ in the Chow ring.]

At one point, I assumed naively that zero would not be effective on any complete smooth variety; the requirement that $X$ be projective was surely an artifact of the proof. However, there is the example of a non-projective complete smooth threefold (due to Hironaka) that is described in Hartshorne Appendix B Example 3.4.1 is shown to be non-projective precisely by producing sum of two curves on the variety rationally (or at least algebraically) equivalent to zero--showing that on this variety, zero is (algebraically) effective. [Note that Hironaka's construction can easily be performed so as to ensure rational effectiveness.]

The next most naive conjecture would be that the effectiveness of zero characterizes non-projective smooth complete varieties among all smooth complete varieties. If true, this would give what I consider a nicer characterization for projectivity than simply the existence of an ample divisor. Philosophically, it seems that zero should not be effective on any "nice" variety, so a characterization like this would make it somehow less annoying when proofs use a hypothesis of projectivity and not only completeness.

Chances are the statement "next most naive" conjecture in the preceeding paragraph is false. However, even if I could somehow invent a possible counterexample, I would have no idea how to prove it was not projective, since the only way I know to give such a proof is to show that zero is effective.

Hence, the question:

Does there exist a smooth, complete, non-projective variety on which zero is not effective (in the sense of being numerically equivalent to an effective algebraic cycle)?

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In Hironaka's example zero is not effective since the curve there is not irreducible. In fact, it is elementary that zero is not effective on any smooth proper threefold since given any irreducible subvariety of such a variety there exists a subvariety of complementary dimension which intersects it in a finite set of points. So in your question it is probably better to remove the condition that V be irreducible. –  ulrich Jul 5 '13 at 2:59
    
@ulrich: I think I was mislead by a typo in Harshorne's writeup. (There's an $l_0$ that should probably be an $l_0'$. I'm interested in your comment about threefolds--you are quoting as "elementary" a fact that I have never heard of before. Could you elaborate a bit and/or provide a reference? –  Charles Staats Jul 5 '13 at 3:55
    
@ulrich: Also, I'm taking your suggestion about modifying the definition of "zero is effective." –  Charles Staats Jul 5 '13 at 3:56
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What I said works for curves and divisors in arbitrary dimension and was discussed in an earlier question on MO. Let $C$ be an irreducible curve and $U$ an affine open set intersecting $C$. There exists an irreducible divisor $D'$ in $U$ such that $D' \cap C \cap U$ is a non-empty finite set. Let $D$ be the Zariski closure of $D'$ in $X$. $D \cap C$ is finite and non-empty since it is a non-empty proper subset of $C$. –  ulrich Jul 5 '13 at 4:13
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