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What are some examples of smooth projective varieties $X$ over a finite field for which the Tate conjecture for divisors is known, and which admit a smooth morphism to a smooth projective curve? I am not really interested in the case where $X$ is an abelian variety and the map to the curve factors through an elliptic curve. Unfortunately I do not think K3 surfaces can be smooth over a curve.

EDIT: removed redundant flatness assumption and incorrect assumption that it reduces to the case where the base is $\mathbb{P}^1$

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    $\begingroup$ Smooth implies flat so that condition isn't needed. Wlog the fiber is a curve. The key invariants are the genus of the base and the genus of the fiber. If either is 0 the statement is known for boring reasons. If the fibration splits as a product after a finite cover of the base then it's OK by the Tate conjecture for divisors. Otherwise you can maybe use progress on function field BSD. $\endgroup$
    – Will Sawin
    Commented Nov 16 at 18:11
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    $\begingroup$ @willsawin thanks for the response. Could you please expand on how you conclude when the fibration splits after a finite cover? $\endgroup$
    – Vik78
    Commented Nov 16 at 18:32
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    $\begingroup$ @Vik78 Why do you write "equivalently to $\mathbb{P}^1$" in your question? There are surfaces of general type which do not admit a smooth morphism to $\mathbb{P}^1$, but which do admit a smooth morphism to some higher genus curve. These are sometimes called Kodaira surfaces. $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 16 at 20:34
  • $\begingroup$ @AriyanJavanpeykar Every curve admits a nonconstant morphism to $\mathbb{P}^1$, so that one can always compose a map to a curve with such a morphism. I see now that this alone does not guarantee smoothness in positive characteristic as the map of curves may be inseparable. Is that what goes wrong in your example? $\endgroup$
    – Vik78
    Commented Nov 16 at 21:16
  • $\begingroup$ @Vik78 A smooth projective morphism from a curve $B$ (the general case where your base curve $B$ has arbitrary genus) to $\mathbb{P}^1$ is a finite etale morphism. Since the etale fundamental group of $\mathbb{P}^1$ is trivial (over a separably closed field), this forces $B$ to be a genus $0$ curve. So there is no reduction in your question from the case of a base curve of arbitrary genus to a curve of genus $0$. $\endgroup$
    – Jason Starr
    Commented Nov 16 at 22:33

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Assume first that $X$ is a product of curves $C_1 \times C_2$. Then the Künneth formula expresses $H^1(X,\mathbb Q_\ell)$ as a sum of three pieces. Two of the pieces are generated by the classes of divisors of the form $C_1 \times y$ and $x \times C_2$, respectively, so the Tate conjecture is automatic for these. The thir piece is

$$ H^1(C_1,\mathbb Q_\ell) \otimes H^1(C_2,\mathbb Q_\ell) = \operatorname{Hom} ( H^1(C_1,\mathbb Q_\ell), H^1(C_2,\mathbb Q_\ell) (-1)$$ $$ = \operatorname{Hom} ( H^1(J(C_1),\mathbb Q_\ell), H^1(J(C_2),\mathbb Q_\ell) (-1) $$ by Poincaré duality and the isomorphism between the $H^1$ of curves and their Jacobians. So any class which is invariant (up to the cyclotomic character) comes from a map $J(C_1) \to J(C_2)$ by the Tate conjecture for abelian varieties over finite fields. Restricting, we get a map from $C_1$ to $J(C_2)$. Shifting, we get a map from $C_1$ to the space of divisor classes of degree $d$ on $C_2$ for some large $d$. Now the map from the space of divisors of degree $d$ to divisor classes of degree $d$ is a projective space bundle. Using this, we may lift the map to a map from $C_1$ to the space of divisors of degree $d$ on $C_2$

But a map from $C_1$ to the space of divisors of degree $d$ on $C_2$ is nothing but a divisor on $C_1 \times C_2$ (with no components of the form $x \times C_2$, and intersection with $C_1 \times y$ in degree $d$). The class of this divisor is the desired class, verifying the Tate conjecture for this $X$.

Now if $X$ has a finite cover of the form $C_1\times C_2$, given an invariant cohomology class on $X$ we can pull the class back to $C_1 \times C_2$, express it as a linear combination of divisor classes, and push the divisors forward to $X$, obtaining an expression for the original class after dividing by the degree of the map. Thus we also verify the Tate conjecture in this case.

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