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5 fixed a typo

Let $X$ be a smooth proper toric variety, $Z\subseteq X$ a smooth subvariety.

Is the fundamental class $[Z] \in H_\ast(X) = A_\ast(X)$ nonzero?

Background

If $X$ is a Kaehler variety, this is of course true, since the intersection of $Z$ with some power of the Kaehler class is positive. But our $X$ is not Kaehler if it is not projective (the Hodge structure is trivial + Kodaira embedding theorem).

I don't think that the statement is true for general complex algebraic varieties, for example the "Hironaka twist" (an example of a $3$-dimensional non-projective smooth proper algebraic variety) has two disjoint smooth curves $M_1$, $M_2$ with the property that $M_1 + M_2$ is numerically zero. So the union of these curves should give a zero class in homology. EDIT. This wouldn't be possible if $M_1$ and $M_2$ intersected - see Dustin Cartwright's and ulrich's comments below.

Chow rings (or homology) of smooth proper toric varieties is very well understood: it is generated by the boundary divisors, who which are smooth toric varieties themselves, intersecting transversely. Relations are obvious from the fan description on $X$. In particular, we need to prove that $Z$ has non-zero intersection with some intersection of the boundary divisors. This could allow for an induction-on-dimension argument, if only $Z$ intersected the boundary nicely. We cannot hope for that, but maybe again there is a "perturbation" argument that may ensure this.

EDIT (continued). Dustin Cartwright's comment below shows that homology classes of irreducible curves in smooth proper varieties are nonzero. Therefore I am tempted to ask more than in the original question:

Let $X$ be a smooth proper variety, $Z\subseteq X$ a smooth subvariety. Is the fundamental class $[Z] \in H_*(X, \mathbb{Q})$ nonzero?

4 added 14 characters in body

Let $X$ be a smooth proper toric variety, $Z\subseteq X$ a smooth subvariety.

Is the fundamental class $[Z] \in H_\ast(X) = A_\ast(X)$ nonzero?

Background

If $X$ is a Kaehler variety, this is of course true, since the intersection of $Z$ with some power of the Kaehler class is positive. But our $X$ is not Kaehler if it is not projective (the Hodge structure is trivial + Kodaira embedding theorem).

I don't think that the statement is true for general complex algebraic varieties, for example the "Hironaka twist" (an example of a $3$-dimensional non-projective smooth proper algebraic variety) has two disjoint smooth curves $M_1$, $M_2$ with the property that $M_1 + M_2$ is numerically zero. So the union of these curves should give a zero class in homology. EDIT. This wouldn't be possible if $M_1$ and $M_2$ intersected - see Dustin Cartwright's comment and ulrich's comments below.

Chow rings (or homology) of smooth proper toric varieties is very well understood: it is generated by the boundary divisors, who are smooth toric varieties themselves, intersecting transversely. Relations are obvious from the fan description on $X$. In particular, we need to prove that $Z$ has non-zero intersection with some intersection of the boundary divisors. This could allow for an induction-on-dimension argument, if only $Z$ intersected the boundary nicely. We cannot hope for that, but maybe again there is a "perturbation" argument that may ensure this.

EDIT (continued). Dustin Cartwright's comment below shows that homology classes of irreducible curves in smooth proper varieties are nonzero. Therefore I am tempted to ask more than in the original question:

Let $X$ be a smooth proper variety, $Z\subseteq X$ a smooth subvariety. Is the fundamental class $[Z] \in H_*(X, \mathbb{Q})$ nonzero?

3 deleted 71 characters in body

Let $X$ be a smooth proper toric variety, $Z\subseteq X$ a smooth subvariety.

Is the fundamental class $[Z] \in H_\ast(X) = A_\ast(X)$ nonzero?

Background

If $X$ is a Kaehler variety, this is of course true, since the intersection of $Z$ with some power of the Kaehler class is positive. But our $X$ is not Kaehler if it is not projective (the Hodge structure is trivial + Kodaira embedding theorem).

I don't think that the statement is true for general complex algebraic varieties, for example the "Hironaka twist" (an example of a $3$-dimensional non-projective smooth proper algebraic variety) has two disjoint smooth curves $M_1$, $M_2$ intersecting transversely and with the property that $M_1 + M_2$ is numerically zero. So the union of these curves should give a zero class in homology. This union is not smooth, but maybe can be made smooth by some perturbation. EDIT. This is actually not wouldn't be possible if $M_1$ and $M_2$ intersected - see Dustin Cartwright's comment below.

Chow rings (or homology) of smooth proper toric varieties is very well understood: it is generated by the boundary divisors, who are smooth toric varieties themselves, intersecting transversely. Relations are obvious from the fan description on $X$. In particular, we need to prove that $Z$ has non-zero intersection with some intersection of the boundary divisors. This could allow for an induction-on-dimension argument, if only $Z$ intersected the boundary nicely. We cannot hope for that, but maybe again there is a "perturbation" argument that may ensure this.

EDIT (continued). Dustin Cartwright's comment below shows that homology classes of irreducible curves in smooth proper varieties are nonzero. Therefore I am tempted to ask more than in the original question:

Let $X$ be a smooth proper variety, $Z\subseteq X$ a smooth subvariety. Is the fundamental class $[Z] \in H_*(X, \mathbb{Q})$ nonzero?

2 edit due to Dustin Cartwright's comment
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