# Picard groups of Fano varieties in positive characteristic

Let $k$ be an algebraically closed field of characteristic $p \geq 0$. Let $X$ be a smooth Fano variety over $k$ and let $\ell \neq p$ be a prime.

Is the natural morphism $\mathrm{Pic}(X) \otimes \mathbb{Z}_\ell \to \mathrm{H}^2(X, \mathbb{Z}_\ell(1))$ an isomorphism?

When $p = 0$, this is an easy consequence of Kodaira vanishing and the exponential sequence, together various comparison theorems for étale cohomology. So my question is really about what happens in positive characteristic. For Fano varieties which lift to characteristic zero this is also probably quite easy.

In fact, a positive answer to the following weaker version might be sufficient for my purposes.

Is $\mathrm{rank}(\mathrm{Pic}(X) \otimes \mathbb{Z}_\ell) = \mathrm{rank}( \mathrm{H}^2(X, \mathbb{Z}_\ell(1)))$?

• It seems to me that the cokernel of your cycle class map has no torsion (to see this use the long exact sequence associated to the Kummer exact sequence for $\mu_{\ell^n}$ and take a projective limit over $n$). As a consequence, the second question implies the first, so that it is sufficient to consider $\mathbb{Q}_{\ell}$ coefficients. In fact, this shows : Tate for divisorson $X$ implies integral Tate for divisors $X$. – Olivier Benoist Jun 4 '15 at 9:14
• @OlivierBenoist: No, I disagree. – Jason Starr Jun 4 '15 at 21:54
• I was wrong about that: I had the wrong transition maps for the sequence of $\ell$-components of the Brauer group. – Jason Starr Jun 5 '15 at 11:17

First of all, that cycle class homomorphism is certainly surjective for all Fano manifolds that lift to characteristic $0$. Secondly, using the usual Kummer sequence, to prove that the cycle class map is surjective, it suffices to prove that the Brauer group is finite. I think this should not be too hard to prove via the usual argument: for a general point $x_0$ in $X$, and for a dominant, generically finite morphism $u:Y\times \mathbb{P}^1\to X$ that maps $Y\times\{0\}$ to $x_0$, we should be able to prove that every Severi-Brauer variety over $X$ pulls back to a trivial Severi-Brauer on $Y\times \mathbb{P}^1$. Thus, using restriction / corestriction for $u$, we should get finiteness of the Brauer group. Of course there is the issue of "rational chain connectedness" instead of "rational connectedness", but I think the extension of this argument should be okay.
Edit. I should point out, this argument only works if $\ell$ is "sufficiently large". It would also work for all $\ell$ prime to the characteristic if we tensor the $\mathbb{Z}_{\ell}$-modules with $\mathbb{Q}_{\ell}$.
Second edit. The argument does work integrally for all $\ell$ prime to the characteristic, as Olivier Benoist points out. I was using the wrong transition maps in computing the inverse limit.
• One can also use the triviality of the Chow group of zero cycles and the "Bloch-Srinivas" method to get the same conclusion. This method also works integrally, but again only for sufficiently large $\ell$. – ulrich Jun 4 '15 at 3:36
• @Jason: I don't quite follow, what is the problem with small $\ell$? Is the issue with possible $\ell^\infty$-torsion in the Brauer group? Won't this problem "go away" as we are taking a projective limit over all $\mathrm{Br}[\ell^n]$? – Daniel Loughran Jun 4 '15 at 8:05
• @Jason Starr : This problem (torsion in cokernel) occurs for fixed $r$, but disappears after taking the projective limit over $r$. There the cokernel is the $\ell$-adic Tate module of $Br(X)$, that has no torsion. The argument appears for instance (in the case of K3 surfaces, but it is general) in the proof of arXiv:1107.1221 Lemma 2.2.2. – Olivier Benoist Jun 5 '15 at 8:01
• @OlivierBenoist: I agree that reference says that always the cycle class map is surjective if the Brauer group is finite. I had the wrong transition maps. The transition maps between the constant $\ell$-parts of the Brauer group are multiplication by $\ell$, not the identity maps. So the map is surjective integrally. – Jason Starr Jun 5 '15 at 11:16