9
$\begingroup$

Let $X$ be a smooth variety over a perfect field $k$ with $X(k) \neq \emptyset$. Then is the natural map \begin{equation} \mathrm{Pic}(X) \to (\mathrm{Pic}(X_{\bar{k}}))^{\mathrm{Gal}(\bar{k}/k)} \qquad (1) \end{equation} surjective?

Remarks:

  1. If $X$ is a proper, then the map (1) is in fact an isomorphism. This is usually proved using the Hochshild--Serre spectral sequence.
  2. The map (1) need not be injective in general. Take $X$ to be the complement of a closed point of degree two in $\mathbb{P}^1_k$. Then $\mathrm{Pic}(X) = \mathbb{Z}/2\mathbb{Z}$ but $\mathrm{Pic}(X_{\bar{k}}) = 0$.
$\endgroup$
  • 2
    $\begingroup$ Not sure what "variety" means for you, but $X$ has to at least be assumed to be connected (hence geometrically connected since $X(k)$ is non-empty). $\endgroup$ – nfdc23 Dec 11 '17 at 7:01
  • 1
    $\begingroup$ For a smooth, geometrically irreducible variety that has a zero-cycle of degree $1$, that map is surjective. Surjectivity of that map is one consequence of vanishing of the elementary obstruction of Colliot-Th'el`ene and Sansuc. $\endgroup$ – Jason Starr Dec 11 '17 at 10:33
  • 2
    $\begingroup$ First: Yes I'm assuming that $X$ is irreducible, hence geometrically integral. @Jason Starr: If $X_{\bar{k}}$ has no non-constant invertible regular functions (e.g. $X$ proper) then I can see how this works, but I don't see how to use the elementary obstruction in the general case. Can you please provide more details? $\endgroup$ – Daniel Loughran Dec 11 '17 at 18:25
  • 1
    $\begingroup$ @DanielLoughran. That is a good point. Most of the references on the elementary obstruction assume that $H^0(X,\mathcal{O}_X^\times)$ equals $k^\times$. However, I believe that is not necessary for surjectivity of the natural map (1). I will try to write more shortly. $\endgroup$ – Jason Starr Dec 11 '17 at 18:35
  • 2
    $\begingroup$ matwbn.icm.edu.pl/ksiazki/aa/aa76/aa7626.pdf by Coray and Manoil calls the surjectivity of (1) the property BigPic and discusses its relation to the Hasse principle on curves. $\endgroup$ – Chris Wuthrich Dec 12 '17 at 13:09
6
$\begingroup$

Updated. The example by @Lucifer is completely correct. Thanks to @Count Dracula who explained the example proposed by @Lucifer. That example is fine. I am keeping the counterexamples below, since they arise in a different way: as Severi-Brauer schemes over multiplicative group schemes over a field (and schemes mapping to such Severi-Brauer schemes). The counterexamples are given after a general statement that can (sometimes) be used to prove surjectivity of the morphism (1).

Setup. Denote by $S$ the scheme $\text{Spec}\ k$ (to save typing). As in Borovoi-- Colliot-Thélène -- Skorobogatov, denote by $\mathfrak{g}$ the Galois group $\text{Gal}(\overline{k}/k)$. Let $\pi:X\to S$ be a smooth, separated, quasi-compact morphism with geometrically irreducible fiber. Let $\sigma:S\to X$ be a section of $\pi$. Denote by $G_X$, resp. $G_{X,\sigma}$, the group $S$-scheme that represents the functor sending every smooth $S$-scheme $T$ to $\mathbb{G}_{m,X}(X\times_S T)$, resp. to the kernel of $$\sigma^*(T):\mathbb{G}_{m,X}(X\times_S T) \to \mathbb{G}_{m,S}(T).$$ Pullback by $\sigma$ defines a group homomorphism, $$\sigma^*:\text{Br}(X)\to \text{Br}(S).$$ Denote the kernel by $\text{Br}(X)_\sigma$. Pullback by $\pi$ defines a group homomorphism, $$\pi^*:H^2_{\mathfrak{g}}(G_{X,\sigma}(\overline{k})) \to \text{Br}(X)_\sigma.$$

Lemma. The cokernel of the natural restriction morphism (1) equals the kernel of $H^2_{\mathfrak{g}}(G_{X,\sigma}(\overline{k})) \to \text{Br}(X)_\sigma$. The group $G_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group. In particular, the cokernel of the natural restriction morphism is a torsion group.

Proof. The relevant terms in the Hochschild-Serre spectral sequence give the following long exact sequence, $$0 \to H^1_{\mathfrak{g}}(G_X(\overline{k})) \to \text{Pic}(X)\xrightarrow{\text{res}} \text{Pic}(X_{\overline{k}})^{\mathfrak{g}} \xrightarrow{\delta} H^2_{\mathfrak{g}}(G_X(\overline{k})) \to \text{Br}(X).$$ By Hilbert's Theorem 90, the first term is $H^1_{\mathfrak{g}}(G_{X,\sigma}(\overline{k}))$. Pullback by the section $\sigma$ defines a splitting of the map $$H^2_{\mathfrak{g}}(\mathbb{G}_{m,S}(\overline{k}))\to \text{Br}(X).$$ Thus, the cokernel of $\text{res}$ equals the kernel of $\pi^*$.

The group $G_{X,\sigma}(\overline{k})$ is finitely generated by Rosenlicht. Every torsion-element satisfies an integral equation over $\overline{k}$. Since $X_{\overline{k}}$ is integral, $\overline{k}$ is integrally closed in $H^0(X_{\overline{k}},\mathcal{O}_X)$. Thus, $G_{X,\sigma}(\overline{k})$ is a finitely generated, torsion-free Abelian group. The profinite Galois cohomology group $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is torsion. Thus, $H^2_{\mathfrak{g}}(G_\sigma(\overline{k}))$ is a torsion group. QED

The contravariant functor $G$ that sends a pointed $S$-scheme, $(X,\sigma)$, to the group scheme $G_{X,\sigma}$ has an adjoint. For every smooth group $S$-scheme $M$ with $M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ isomorphic to $\mathbb{G}_{m,\overline{k}}^\rho$, the group scheme $G_{M,e}$ is the Cartier dual group scheme $M^D$. For every morphism of group $S$-schemes, $$\phi:M^D\to G_{X,\sigma},$$ there is a unique $S$-morphism, $$f:X\to M,$$ sending $\sigma$ to the identity section $e$ and such that the pullback map on unit group $S$-schemes is $\phi$.

Proposition. The Galois cohomology group $H^2_{\mathfrak{g}}(M^D)$ is canonically isomorphic to the kernel, $\text{Br}_1(M)$, of the pullback map $$\text{pr}_M^*:\text{Br}(M)\to \text{Br}(M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}).$$

Proof. Since $\text{Pic}(\mathbb{G}_{m,\overline{k}}^\rho)$ is the trivial group, this follows from the previous lemma. QED

The Cartier dual group $S$-scheme to $G_\sigma$ is a smooth group $S$-scheme $M$ of multiplicative type. There is a unique $S$-morphism $$f:X\to M,$$ mapping $\sigma$ to the identity section $e$ and such that the induced homomorphism of groups of units is an isomorphism of $S$-group schemes.

Proposition. The cokernel of $\text{res}$ is identified with the kernel of the group homomorphism $$f^*:\text{Br}_1(M) \to \text{Br}(X).$$ In particular, if there exists a degree $1$ zero-cycle in the generic fiber of $f$, then $\text{res}$ is surjective.

Proof. This follows from the previous proposition and functoriality of the Hochschild-Serre spectral sequence with respect to $f$. If the fiber of $f$ over $\eta=\text{Spec}\ k(M)$ has a degree $1$ zero-cycle, then the pullback map $\text{Br}(k(M))\to \text{Br}(X_\eta)$ is injective by the lemma. Since $M$ is smooth, it follows that also $f^*$ is injective. QED

Counterexamples. This suggests how to construct a counterexample, namely as a Severi-Brauer scheme over a multiplicative group scheme $M$ over $k$.

Let $k$ be a finite field $\mathbb{F}_q$. Let $M$ be $\mathbb{G}_{m,k}= \text{Spec}\ k[t,t^{-1}]$. The Cartier dual group scheme $M^D$ has $M^D(k) = M^D(\overline{k}) \cong \mathbb{Z}$, generated by the element $t$.

For every integer $\ell$, let $k_\ell=\mathbb{F}_{q^\ell}$ be a degree $\ell$ extension of $k$ with cyclic Galois group generated by a Frobenius map. This extension together with the generator of the Galois group is unique up to unique isomorphism. Associated to the cyclic extension $k_\ell/k$ and the element $t\in M^D(k)$, there is a cyclic algebra $A_{\ell,t}$ that is a locally free $\mathcal{O}_M$-module of rank $\ell^2$ and that is an Azumaya algebra over $\mathcal{O}_M$. The Brauer class of $A_{\ell,t}$ in $\text{Br}(M)$ is an element of order $\ell$ generating the full $\ell$-torsion subgroup of $\text{Br}(M)$. By Tsen's Theorem, the Brauer group of $M\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is trivial. Thus, the class of $A_{\ell,t}$ generates a cyclic subgroup of order $\ell$ in $\text{Br}(M)_k$.

Now let $f:X\to M$ be the Severi-Brauer scheme parameterizing left ideals in $A_{\ell,t}$ of rank $\ell$. Since the Brauer group of $k$ is trivial (by Wedderburn or Chevalley-Warning), the restriction of $f$ over the $k$-rational point $e\in M(k)$ is a trivial Severi-Brauer scheme. Thus, there exists a section $$\sigma:\text{Spec}\ k \to X,$$ with image in the fiber of $f$ over $e$. The Picard group of $X\times_{\text{Spec}\ k} \text{Spec}\ \overline{k}$ is $\mathbb{Z}$, and the generator restricts on each projective space (geometric) fiber of $f$ to a generator $\mathcal{O}(1)$ of the projective space. However, there cannot be such an invertible sheaf on $X$, or else $\mathcal{A}_{\ell,t}$ would have zero Brauer class.

$\endgroup$
  • $\begingroup$ I'm just trying to make all this very explicit. A special case of your construction is the following. Let $a$ be a non-square in $k$. Then take $$X: x^2 -ay^2 = tz^2 \quad \subset \mathbb{P}^2 \times \mathbb{G}_m.$$ This is conic bundle over $\mathbb{G}_m$ with a rational point. There is no section over $k$, but are you claiming that the the class in the Picard group of a section over $k(\sqrt{a})$ is Galois invariant? $\endgroup$ – Daniel Loughran Dec 13 '17 at 5:15
  • 1
    $\begingroup$ @DanielLoughran. Yes, that is what I am claiming. For the change of variables, $[u,v,z]=[x-y\sqrt{a},x+y\sqrt{a},z]$, the closed subscheme is $\text{Zero}(uv-tz^2)\subset \mathbb{P}^2\times \mathbb{G}_m$. There is a (non-Galois-invariant) section of the projection to $\mathbb{G}_m$ given by $\text{Zero}(v,z)$. The invertible sheaf of that section is the unique ample generator of the Picard group of the Picard group of $X\times_{\text{Spec}\ k}\text{Spec}\ k(\sqrt{a})$. Thus, the invertible sheaf is Galois invariant, even though the divisor is not Galois invariant. $\endgroup$ – Jason Starr Dec 13 '17 at 10:14
4
$\begingroup$

This is not true. Let $k$ be a field of characteristic not $2$ and $E$ is an elliptic curve with origin $0$ and $k$-rational $2$-torsion point $P$. Assume $E$ has a third rational point. Then $X = E \setminus \{0, P\}$ has at least one rational point. We have $$ Pic(X) = E(k)/\langle P \rangle $$ But the Galois invariants in $Pic(X_{\overline{k}})$ are equal to $E'(k)$ where $E \to E'$ is the isogeny whose kernel is the order $2$ subgroup scheme generated by $P$. Thus all you have to do is find an example of $k, E, P$ where $E(k) \to E'(k)$ is not surjective. For example if the ground field $k$ is finite this will be the case because $|E(k)| = |E'(k)|$ in that case.

$\endgroup$
  • 1
    $\begingroup$ Why do you claim that the Galois invariants of $\text{Pic}(X_{\overline{k}})$ equal $E'(k)$? Consider the case that $k$ is already algebraically closed. Then $\text{Pic}(X_{\overline{k}})$ equals $\text{Pic}(X)$. $\endgroup$ – Jason Starr Dec 12 '17 at 8:48
  • 1
    $\begingroup$ No Jason, it would be $E(k)/\langle P \rangle = E'(k)$ if $k$ is algebraically closed. $\endgroup$ – Count Dracula Dec 12 '17 at 14:34
  • $\begingroup$ @CountDracula. Now I understand. $\endgroup$ – Jason Starr Dec 12 '17 at 18:22
  • $\begingroup$ You are the most fairminded person I know. $\endgroup$ – Count Dracula Dec 12 '17 at 20:27
  • $\begingroup$ @CountDracula. I know that is not true. $\endgroup$ – Jason Starr Dec 12 '17 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.