# A question on effective divisors

Let $$X$$ be a projective variety with two morphisms $$f:X\rightarrow Y$$ and $$g:X\rightarrow Z$$ with irreducible fibers of positive dimension. Assume that $$Pic(X) = f^{*}Pic(Y)\oplus g^{*}Pic(Z)$$. Then if $$D$$ is a divisor on $$X$$ we can write $$D = f^{*}D_Y + g^{*}D_Z$$, where $$D_Y,D_Z$$ are divisors on $$Y$$ and $$Z$$ respectively.

If $$D$$ is effective are then $$D_Y$$ and $$D_Z$$ effective as well?

This holds for instance when $$X = \mathbb{P}^n \times \mathbb{P}^m$$ is a product and $$f,g$$ are the projections onto the factors.

• Just a comment that if $X = Y \times Z$, then $Pic(X)$ is not in general isomorphic to $Pic(Y) \oplus Pic(Z)$. Indeed, if $Y$ and $Z$ are smooth projective curves of positive genus, then the extra divisors on $Y \times Z$ come from correspondences between $Y$ and $Z$. Commented Jul 16, 2021 at 8:43
• You are right. I corrected the last sentence. In your example is it true that the Neron-Severi space of $X$ is the direct sum of the Neron-Severi spaces of $Y$ and $Z$? Commented Jul 16, 2021 at 8:58
• If $X = Y\times Z$, by Kunneth formula for sheaf cohomology, $D$ is effective if and only if both $D_Y$ and $D_Z$ are effective. Commented Jul 16, 2021 at 9:59
• @Friedrich: I don't think so. If E is an elliptic curve, then $NS(E \times E)$ will have rank $3$ or $4$ depending on whether $E$ has complex multiplication or not, as the graph of the complex multiplication provides an extra divisor. Commented Jul 16, 2021 at 10:44

This is false. The original post, included below, had some mistakes.

The example of @Pop is better, and in fact that example is where I started. It is straightforward to modify that example into an example satisfying the constraints. If @Pop wants to add an answer, then I am happy to delete this answer.

Let $$Y$$ be the projective plane. Fix a point $$p$$ in $$Y$$, and denote by $$Z'$$ the $$1$$-dimensional projective space parameterizing lines $$L$$ in $$Y$$ that contain $$p$$. Let $$Z$$ be the product of $$Z'$$ with a projective space $$W$$ of dimension $$\geq 1$$. Let $$X'$$ be the parameter of ordered pairs $$(q,L)$$ of a points of $$Y$$ and a line $$L$$ in $$Y$$ that contains both $$p$$ and $$q$$. Let $$X$$ be the product of $$X'$$ with $$W$$. Then, as a subvariety of the product $$Y\times Z$$ considered as a projective space bundle (of relative dimension $$2$$) over $$Z$$, the variety $$X$$ is a projective space subbundle over $$Z$$ (of relative dimension $$1$$). Thus, the pullback homomorphism from $$\text{Pic}(Y)\oplus \text{Pic}(Z)$$ to $$\text{Pic}(X)$$ is an isomorphism.

Now let $$D'$$ be the divisor in $$X'$$ parameterizing pairs $$(q,L)$$ such that $$q$$ equals $$p$$, and let $$D$$ be $$D'\times W$$. The normal bundle of $$D'$$ in $$X'$$ is anti-ample. Thus, the normal bundle of $$D$$ in $$X$$ is not nef. For the reason mentioned above, the divisors $$D_Y$$ and $$D_Z$$ are not both effective.

Original post. Here are the details. Fix a vector space $$V$$ of dimension $$4$$ together with a linear subspace $$U$$ of dimension $$1$$. Denote by $$\text{Flag}(1,3;V)$$ the partial flag variety parameterizing ordered pairs $$(A,B)$$ of a $$1$$-dimensional linear subspace $$A$$ contained in a $$3$$-dimensional linear subspace $$B$$ contained in $$V$$. Denote by $$X$$ the closed subvariety of $$X$$ parameterizing ordered pairs such that $$U$$ is contained in $$B$$.

Let $$Y$$ be $$\mathbb{P}V$$, the parameter space of $$1$$-dimensional linear subspaces $$A$$ of $$V$$. Let $$Z$$ be the linear $$2$$-plane in $$\text{Grass}(3,V)$$ that parameterizes $$3$$-dimensional subspaces $$B$$ of $$V$$ that contain $$U$$, i.e., the dual projective space of $$V/U$$. Denote by $$f$$ and $$g$$ the forgetful morphisms from $$X$$ that remember only $$A$$, respectively $$B$$.

The fiber of $$f$$ over every point of $$Y$$ other than $$[U]$$ is a projective space $$\mathbb{P}^1$$. The fiber of $$f$$ over $$[U]$$ is the full projective space $$Z = \mathbb{P}(V/U)^\vee \cong \mathbb{P}^2$$. In all cases, the fiber is irreducible of positive dimension. The fiber of $$g$$ over every point of $$Z$$ is a P^1 $$\mathbb{P}^2$$.

In fact, the embedding of $$X$$ in $$Y\times Z$$ realizes $$X$$ as a projective space subbundle (of relative dimension 1 $$2$$)over $$Z$$ inside the (constant) projective space bundle $$Y\times Z$$ over $$Z$$ (of relative dimension $$2$$). From this and the formula for the Picard group of a projective space bundle, it is straightforward to see that the Picard group of $$X$$ is the isomorphic image under pullback of $$\text{Pic}(Y)\oplus \text{Pic}(Z)$$.

Now consider the divisor $$D$$ in $$X$$ parameterizing pairs $$(A,B)$$ such that $$A$$ equals $$U$$. The divisor $$D_Y$$ is effective, linearly equivalent to a hyperplane class in the projective space $$Y$$. However, the divisor $$D_Z$$ is not effective. In fact, it is linearly equivalent to the negative of the hyperplane class in the projective space $$Z$$.

• You are correct that I wrote the wrong fiber dimension for $g$. The correct fiber dimension is $2$. Commented Jul 16, 2021 at 14:35
• The reason my example is more complicated than the blowing up of the plane is the requirement by the OP that each fiber dimension is positive. Of course I started with the example by @Pop and modified it to satisfy the requirement. Commented Jul 16, 2021 at 14:46