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.

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    $\begingroup$ 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 $Z$. $\endgroup$ Commented Jul 16, 2021 at 8:43
  • $\begingroup$ 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 $Z$? $\endgroup$
    – Puzzled
    Commented Jul 16, 2021 at 8:58
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    $\begingroup$ 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. $\endgroup$
    – Puzzled
    Commented Jul 16, 2021 at 9:59
  • $\begingroup$ @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. $\endgroup$ Commented Jul 16, 2021 at 10:44

1 Answer 1


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 $Z$.

  • $\begingroup$ You are correct that I wrote the wrong fiber dimension for $g$. The correct fiber dimension is $2$. $\endgroup$ Commented Jul 16, 2021 at 14:35
  • $\begingroup$ 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. $\endgroup$ Commented Jul 16, 2021 at 14:46

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