# Galois invariant Picard group elements

Let $X$ be a smooth variety over a perfect field $k$ with $X(k) \neq \emptyset$. Then is the natural map $$\mathrm{Pic}(X) \to (\mathrm{Pic}(X_{\bar{k}}))^{\mathrm{Gal}(\bar{k}/k)} \qquad (1)$$ 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 $\mathrm{Pic}(X_{\bar{k}}) = 0$.
• 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). – nfdc23 Dec 11 '17 at 7:01
• 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. – Jason Starr Dec 11 '17 at 10:33
• 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? – Daniel Loughran Dec 11 '17 at 18:25
• @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. – Jason Starr Dec 11 '17 at 18:35
• 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. – Chris Wuthrich Dec 12 '17 at 13:09

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.

• 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? – Daniel Loughran Dec 13 '17 at 5:15
• @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. – Jason Starr Dec 13 '17 at 10:14

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.

• 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 $\text{Pic}(X)$. – Jason Starr Dec 12 '17 at 8:48
• No Jason, it would be $E(k)/\langle P \rangle = E'(k)$ if $k$ is algebraically closed. – Count Dracula Dec 12 '17 at 14:34
• @CountDracula. Now I understand. – Jason Starr Dec 12 '17 at 18:22
• You are the most fairminded person I know. – Count Dracula Dec 12 '17 at 20:27
• @CountDracula. I know that is not true. – Jason Starr Dec 12 '17 at 22:59