Vanishing of the Brauer group of a del Pezzo surface of degree 4

TL;DR

This paper constructs a specific degree 4 del Pezzo surface over a field where the Brauer group behaves unexpectedly, showing limitations of existing computational algorithms.

Contribution

It provides an explicit example of a del Pezzo surface with a trivial Brauer group despite nontrivial Galois cohomology, highlighting the boundaries of current methods.

Findings

The constructed surface has trivial Brauer group over the base field.

The Galois cohomology group is nontrivial, isomorphic to Z/2Z.

The example demonstrates the algorithm's limitations in certain cases.

Abstract

We explicitly construct a del Pezzo surface $X$ of degree 4 over a field $k$ such that $\operatorname{H}^1(k,\operatorname{Pic}\overline X)$ is isomorphic to $\mathbb{ZZ}/2\mathbb{Z}$ while $\operatorname{Br} X/\operatorname{Br} k$ is trivial. This proves that the algorithm to compute the Brauer group in [VAV] cannot be generalized in some cases.

Equations

$$0\to \mathrm{Pic}X\to (\mathrm{Pic}\overline{X}{)}^{G_k}\to \mathrm{Br}k\to {\mathrm{Br}}_{1}X\to {\text{H}}^1(k,\mathrm{Pic}\overline{X})\to {\text{H}}^3(k,{\mathbb{G}}_m),$$

$$\frac{{\mathrm{Br}}_{1}X}{\mathrm{im}(\mathrm{Br}k\to \mathrm{Br}X)}\simeq {\text{H}}^1(k,\mathrm{Pic}\overline{X})$$

$$\begin{array}{rl}Q& :a{x}_0^2+b{x}_1^2+x_2^2+c{x}_4^2=0\\ Q^{{}^{\prime }}& :bc{x}_0^2+x_1^2+x_2^2+a{x}_3^2=0.\end{array}$$

$$C_{P_i}=W\cap Q_{t_i}\cap L_{P_i}=W\cap L_{P_i}\text{ and }{C}_{P_i}^{{}^{\prime }}=W\cap Q_{t_i}\cap L_{P_i}^{{}^{\prime }}=W\cap L_{P_i}^{{}^{\prime }}$$

$$\frac{1}{2}(H+C_0+C_1+C_2+C_3+C_4),C_0,C_1,C_2,C_3,C_4$$

$${\text{id}}_Y\times {\sigma }^{{}^{\prime }}:\overline{Y}=Y{\times }_K\mathrm{Spec}\overline{K}\to \overline{Y}.$$

$${\Pi }_{T\in \mathcal{S}}\mathrm{Aut}({\Gamma }_T)\cap \mathcal{O}(K_Y^{\perp })\subset \mathrm{Aut}(\Gamma ).$$

$$(\ast )\mathrm{deg}(\mathcal{T})=2,{\Pi }_{T\in \mathcal{T}}{N}_{K(T)/K}({ϵ}_T)\in K^{\times 2},\text{and }{ϵ}_T\notin K(T{)}^{\times 2}\text{ for every }T\in \mathcal{T}.$$

$$\sigma \mapsto \{\begin{array}{ll}-H+{\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i& \text{ if }\sigma \notin G_{K_{\mathcal{T}}}\\ 0& \text{ otherwise}\end{array}$$

$$0\to \mathrm{Pic}\overline{Y}\stackrel{\times 2}{\to }\mathrm{Pic}\overline{Y}\to (\mathrm{Pic}\overline{Y}/2\mathrm{Pic}\overline{Y})\to 0$$

$$\frac{(\mathrm{Pic}\overline{Y}/2\mathrm{Pic}\overline{Y}{)}^{G_K}}{((\mathrm{Pic}\overline{Y}{)}^{G_K}/2(\mathrm{Pic}\overline{Y}{)}^{G_K})}\stackrel{\sim }{\to }{\text{H}}^1(K,\mathrm{Pic}\overline{Y})[2];D\mapsto (\sigma \mapsto \frac{1}{2}(d-\sigma d))$$

$$\sigma \mapsto \{\begin{array}{ll}-H+{\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i& \text{ if }\sigma \notin G_{K_{\mathcal{T}}}\\ 0& \text{ otherwise}.\end{array}$$

$${\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i-\sigma ({\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i)={\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i-{\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i^{{}^{\prime }}=-2H+2{\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i.$$

$${\Sigma }_{t_i\in (\mathcal{S}-\mathcal{T})(\overline{K})}\pm C_i\in (\mathrm{Pic}\overline{Y}{)}^{G_K}.$$

$$\sigma (\pm C_j+{\Sigma }_{\{t_i\ne t_j,t_i\in (\mathcal{S}-\mathcal{T})(\overline{K})\}}\pm C_i)=\mp C_j+D$$

$${\Sigma }_{t_i\in (\mathcal{S}-\mathcal{T})(\overline{K})}\pm C_i\notin (\mathrm{Pic}\overline{Y}{)}^{G_K}.$$

$${\Sigma }_{t_i\in (\mathcal{S}-\mathcal{T})(\overline{K})}\pm C_i\in (\mathrm{Pic}\overline{Y}{)}^{G_K}.$$

$${\mathrm{Br}}_{cyc}(V,L)\stackrel{\sim }{\to }\frac{\ker ({\overline{N}}_{L/K})}{\mathrm{im}(1-\sigma )}\hookrightarrow {\text{H}}^1(K,\mathrm{Pic}\overline{V}).$$

$$\sigma \mapsto \{\begin{array}{ll}D& \text{ if }\sigma \notin G_L\\ 0& \text{ otherwise}\end{array},$$

$$N_{L/k}(D)=(f).$$

$$\frac{\ker ({\overline{N}}_{L/K})}{\mathrm{im}(1-\sigma )}\simeq {\text{H}}^1(\text{Gal}(L/K),\mathrm{Pic}{V}_L).$$

$${\text{H}}^1(\text{Gal}(L/K),\mathrm{Pic}{V}_L)\hookrightarrow {\text{H}}^1(K,\mathrm{Pic}\overline{V}).$$

$${\mathrm{Br}}_{cyc}(Y,k_{\mathcal{T}})\stackrel{\sim }{\to }\frac{\ker ({\overline{N}}_{K_{\mathcal{T}}/K})}{\mathrm{im}(1-\sigma )}\hookrightarrow {\text{H}}^1(K,\mathrm{Pic}\overline{Y}).$$

$${\mathcal{A}}_{\mathcal{T}}:=\left(K_{\mathcal{T}}/K,l^{-2}{\Pi }_{T\in \mathcal{T}}{\text{N}}_{K(T)/K}(l_T)\right),$$

$$\sigma \mapsto \{\begin{array}{ll}-H+{\Sigma }_{t_i\in \mathcal{T}(\overline{K})}{C}_i& \text{ if }\sigma \notin G_{K_{\mathcal{T}}}\\ 0& \text{ otherwise}\end{array}$$

$$\begin{array}{rl}Q& :a{x}_0^2+b{x}_1^2+x_2^2+c{x}_4^2=0\\ Q^{{}^{\prime }}& :bc{x}_0^2+x_1^2+x_2^2+a{x}_3^2=0.\end{array}$$

$$0\to \mathrm{Pic}X\to (\mathrm{Pic}\overline{X}{)}^{G_k}\to \mathrm{Br}k\to {\mathrm{Br}}_{1}X\to {\text{H}}^1(k,\mathrm{Pic}\overline{X})\to {\text{H}}^3(k,{\mathbb{G}}_m),$$

$$f(\lambda ,\mu )=\det (\lambda A+\mu A^{{}^{\prime }})=32\mu \lambda (\lambda +\mu )(b\lambda +\mu )(a\lambda +bc\mu ).$$

$$T_0=[1:0],T_1=[0:1],T_2=[1:-1],T_3=[1:-b],\text{and }{T}_4=[bc:-a]$$

$${ϵ}_{T_0}=16abc\sim abc\mathrm{mod}k(T_0{)}^{\times 2}=k^{\times 2}.$$

Full text

1. Introduction

The Brauer group of a field $k$, denoted by $\operatorname{Br}k$, was defined by Richard Brauer in 1929. It classifies central simple algebras over a field $k$ with respect to the Morita equivalence. Two central simple algebras $\mathscr{A}$ and $\mathscr{B}$ are said to be Morita equivalent over $k$ if there there exist two positive integers $n$ and $m$ such that $\mathscr{A}\otimes\textup{M}_{n}(k)\simeq\mathscr{B}\otimes\textup{M}_{m}(k)$ as $k$-algebras. Moreover, $\operatorname{Br}k$ is isomorphic to the Galois cohomology group $\textup{H}^{2}(k,k_{s}^{\times})$; we say this is the cohomological definition of the Brauer group. The cohomological definition of the Brauer group of a field can be generalized to define the Brauer group of a scheme $X$ denoted by $\operatorname{Br}X$ as the étale cohomology $\textup{H}_{\text{\'{e}t}}^{2}(X,\mathbb{G}_{m})$, where $\mathbb{G}_{m}$ is the sheaf of units on $X$. When $X$ is projective, geometrically integral and smooth, $\operatorname{Br}X$ is realized concretely as the equivalence classes of Azumaya algebras over $X$. This has opened the door to many arithmetic and geometric applications of the Brauer group.

Various theorems and algorithms have been implemented to compute the quotient of the Brauer group $\operatorname{Br}X/\operatorname{im}(\operatorname{Br}k\rightarrow\operatorname{Br}X)$ for certain schemes. For example, such algorithms have been implemented for most cubic surfaces; See [CTKS], [EJa] and [EJb]. Algorithms have also been implemented for some del Pezzo surfaces of degree 2 as in [Cor], and for most del Pezzo surfaces of degree 4 as in [BBFL], and [VAV].

In this paper we highlight a part of the algorithm to compute $\operatorname{Br}X/\operatorname{im}(\operatorname{Br}k\rightarrow\operatorname{Br}X)$ for a del Pezzo surface of degree 4 as in [VAV, Section 4.1]. Furthermore, we prove that it cannot be generalized in some cases.

Let $X$ be a del Pezzo surface of degree 4. By embedding $X$ anticanonically into $\mathbb{P}^{4}$, we view it as the intersection of two quadrics as in [Wit, Proposition 3.26]. The two quadrics $Q$ and $Q^{{}^{\prime}}$ define a pencil $\{\lambda Q+\mu Q^{{}^{\prime}}:[\lambda,\mu]\in\mathbb{P}^{1}\}$. By [Wit, Proposition 3.26], the pencil has five degenerate geometric fibers which are rank 4 quadrics. Let $\mathscr{S}$ be the degree 5 subscheme of $\mathbb{P}^{1}$ representing the degeneracy locus of the pencil. For every closed point $T\in\mathscr{S}$, denote by $k(T)$ the residue field of $T$ and by $Q_{T}$ the corresponding quadric in the pencil.

The proof of the algorithm to compute $\operatorname{Br}X/\operatorname{im}(\operatorname{Br}k\rightarrow\operatorname{Br}X)$ for a del Pezzo surface of degree 4 as in [VAV, Section 4.1] uses the exact sequence, resulting from the Hochschild-Serre spectral sequence,

[TABLE]

where $G_{k}$ is the absolute Galois group of $k$ and $\operatorname{Br}_{1}X=\ker(\operatorname{Br}X\rightarrow\operatorname{Br}\overline{X})$. The Hochschild-Serre spectral sequence yields the isomorphism

[TABLE]

under some arithmetic assumptions related to the solvability of the quadrics $Q_{T}$ over $k(T)$ for $T\in\mathscr{S}$. Depending on the field $k$, the map $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\rightarrow\textup{H}^{3}(k,\mathbb{G}_{m})$ might be non trivial as we prove for a certain del Pezzo surfaces of degree 4 in this paper. In other words, this proves that the arithmetic assumption in the algorithm in [VAV, Section 4.1] is necessary; changing the arithmetic input of the algorithm leads to a different outcome. In particular, we prove the following theorem.

Theorem 1.1**.**

Let $k=\mathbb{Q}^{cycl}(a,b,c)$ where $a,b$ and $c$ are independent transcendental elements. Let $X$ be the del Pezzo surface of degree 4 in $\mathbb{P}^{4}_{k}$ defined by the intersection of the following two quadrics

[TABLE]

Then $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/2\mathbb{Z}$ while $\frac{\operatorname{Br}X}{\operatorname{Br}k}$ is trivial.

Uematsu has done similar work for an affine diagonal quadric [Uema] and for a diagonal cubic surface [Uemb]. In contrast to his proofs, our proof does not rely on computing the boundary map $d_{2}^{1,1}:H^{1}(k,\operatorname{Pic}\overline{X})\rightarrow H^{3}(k,\mathbb{G}_{m})$ of the Hochschild-Serre spectral sequence. Instead it relies on the algorithm in [VAV, Section 4.1] and the work done by Harpaz [Hara]. In [Hara], Harpaz proves that for a specific example of an affine diagonal quadric $U$ over a field $F$, $\operatorname{Br}U/\operatorname{Br}F$ vanishes while $H^{1}(F,\operatorname{Pic}\overline{U})\simeq\mathbb{Z}/2\mathbb{Z}$. In his argument, he base extends $X$ to a field $L$; over this field he shows that $\operatorname{Br}X_{L}/\operatorname{Br}L\simeq\mathbb{Z}/2\mathbb{Z}$. Finally, he proves that the generating class $\mathscr{A}$ of $\operatorname{Br}X_{L}/\operatorname{Br}L$ is not in the image of $\operatorname{Br}X/\operatorname{Br}k$. We carry a similar proof to Harpaz’s argument; we base extend to the field $L=k(\sqrt{a})$. For the explicit computation of the class of the algebra $\mathscr{A}$ that generates $\operatorname{Br}X_{L}/\operatorname{Br}L$, we use the algorithm in [VAV, Section 4.1].

Outline: Section 2 provides some background about del Pezzo surfaces of degree 4. In particular, we highlight a part of the algorithm in [VAV, Section 4.1] to relate $H^{1}(k,\operatorname{Pic}\overline{X})$ to $\operatorname{Br}X/\operatorname{Br}k$. We prove the main theorem in Section 3.

Notation: Throughout this paper, we use $k$ to denote a field and $X$ to denote a scheme. We denote by $\overline{k}$ a fixed algebraic closure of $k$. We denote by $\overline{X}$ the base change of $X$ to $\overline{k}$. Let $G_{k}$ denote the absolute Galois group of $k$. Denote by $H^{i}(k,A)$ the $i$-th group cohomology of the Galois group $G_{k}$ and the $G_{k}$-module $A$.

Acknowlegements

I thank my advisor, Bianca Viray, for suggesting the problem and for the many helpful discussions. I also thank Yonatan Harpaz for emailing us his typed notes [Hara].

2. Background

2.1. Background on del Pezzo Surfaces of Degree 4

We summarize some facts that we need in this paper about del Pezzo surfaces of degree 4 and we fix some notation. For further details, this information can be found in [VAV, Section 2], [VAa, Section 1], and [Wit].

Let $Y$ be a del Pezzo surface of degree 4 over a field $K$. By embedding it anticanonically into $\mathbb{P}^{4}$, we view it as the intersection of two quadrics $Q$ and $Q^{{}^{\prime}}$ as in [Wit, Proposition 3.26]. The two quadrics $Q$ and $Q^{{}^{\prime}}$ define a pencil $\{\lambda Q+\mu Q^{{}^{\prime}}:[\lambda,\mu]\in\mathbb{P}^{1}\}$. By [Wit, Proposition 3.26], the pencil has five degenerate geometric fibers which are rank 4 quadrics.

Let $\mathscr{S}$ be the degree 5 subscheme of $\mathbb{P}^{1}$ representing the degeneracy locus of the pencil. For every closed point $T\in\mathscr{S}$, denote by $K(T)$ the residue field of $T$ and by $Q_{T}$ the corresponding quadric in the pencil. Let $\epsilon_{T}$ be the discriminant of a smooth rank 4 quadric obtained by restricting $Q_{T}$ to a hyperplane $H_{T}$ in $\mathbb{P}^{4}$ not containing the vertex of $Q_{T}$. By [Wit, Section 3.4.1], the square class of $\epsilon_{T}$ does not depend on the choice of $H_{T}$. So we consider $\epsilon_{T}$ as an element in $K(T)/K(T)^{\times 2}$. Over $\overline{K}$, let $\mathscr{S}(\overline{K})=\{t_{0},\ldots,t_{4}\}$ where $t_{0},\ldots,t_{4}\in\mathbb{P}^{1}(\overline{k})$. Denote by $K(t_{i})$ the smallest field contained in $\overline{k}$ and containing $t_{i}$, and by $Q_{t_{i}}$ the corresponding quadric. Let $\epsilon_{t_{i}}$ be the discriminant of the smooth rank 4 quadric obtained by restricting $Q_{t_{i}}$ to a hyperplane $H_{i}$ in $\mathbb{P}^{4}$ not containing the vertex of $Q_{t_{i}}$. By [Wit, Section 3.4.1], the square class of $\epsilon_{t_{i}}$ does not depend on the choice of $H_{i}$. So we consider $\epsilon_{t_{i}}$ as an element in $K(t_{i})/K(t_{i})^{\times 2}$

Now we turn our attention to the Picard group of $\overline{Y}$ and the Galois action on its classes. By [VAa, Theorem 1.6], $\overline{Y}$ is the blowup of $\mathbb{P}^{2}$ at 5 points. Let $\{e_{1},\ldots,e_{5}\}$ be the classes of the exceptional divisors associated to the blown up points. Let $l$ be the class of the pullback of a line in $\mathbb{P}^{2}$ that does not not pass through any of the blown up points. By [Harb, Proposition V.3.2], $\operatorname{Pic}\overline{Y}\simeq\mathbb{Z}^{6}$ and admits $\{e_{1},\ldots,e_{5},l\}$ as a basis. Furthermore, we will describe $\operatorname{Pic}\overline{Y}$ in terms of some conic classes that will be useful for the algorithm in [VAV, Section 4.1].

Theorem 2.1**.**

[BBFL*, Theorem 2]**

Let $Y$ be a del Pezzo surface of degree $4$ over a field $K$. There are $10$ families of conics on $Y$. Moreover, their classes $C_{0},\ldots,C_{4},C_{0}^{{}^{\prime}},\ldots,C_{4}^{{}^{\prime}}$ in $\operatorname{Pic}\overline{Y}$ can be written in terms of the basis of $\operatorname{Pic}\overline{Y}$ as $C_{i}=l-e_{i}$ and $C_{i}^{{}^{\prime}}=H-C_{i}$ where $H$ is the hyperplane class of $Y$. Over $\overline{K}$ the conics in each class form a pencil.∎*

As in [VAV, Section 2.3], we define the conic classes $C_{0},\ldots,C_{4},C_{0}^{{}^{\prime}},\ldots,C_{4}^{{}^{\prime}}$ in $\operatorname{Pic}\overline{Y}$ as follows. For $i=0,\ldots,4$, let $K_{i}$ be a finite extension of $K$ such that the rank 4 quadric $Q_{t_{i}}$ has a smooth $K_{i}$-point and such that $[K_{i}(\sqrt{\epsilon}_{t_{i}}):K_{i}]=[K(t_{i})(\sqrt{\epsilon}_{t_{i}}):K(t_{i})]$. Let $P_{i}$ be any smooth $K_{i}$-point on $Q_{t_{i}}$ and $H_{P_{i}}$ be the hyperplane tangent to $Q_{t_{i}}$ at $P_{i}$. By [VAV, Lemma 2.1], we have $Q_{t_{i}}\cap H_{P_{i}}=L_{P_{i}}\cup L_{P_{i}}^{{}^{\prime}}$ for some planes $L_{P_{i}}$ and $L_{P_{i}}^{{}^{\prime}}$ defined over $K_{i}(\sqrt{\epsilon}_{t_{i}})$. We define $C_{P_{i}}=Y\cap L_{P_{i}}$ and $C_{P_{i}}^{{}^{\prime}}=Y\cap L_{P_{i}}^{{}^{\prime}}$. We have,

[TABLE]

for some smooth quadric $W$ in the pencil associated to $Y$. Hence $C_{P_{i}}$ and $C_{P_{i}}^{{}^{\prime}}$ are conics on $Y$. A different choice of $K_{i}$-smooth point $\tilde{P}_{i}$ on $Q_{t_{i}}$ leads to two different planes $L_{\tilde{P}_{i}}$ and $L_{\tilde{P}_{i}}^{{}^{\prime}}$, and hence two different conics. Because $Q_{t_{i}}$ is a singular rank 4 quadric over $\overline{K}$, it is a cone over a smooth quadric $Q^{{}^{\prime\prime}}$ in $\mathbb{P}^{3}$. The two families of lines on $Q^{{}^{\prime\prime}}$ induce two families of planes on $Q_{t_{i}}$. So $L_{P_{i}},L_{P_{i}}^{{}^{\prime}},L_{\tilde{P}_{i}}$ and $L_{\tilde{P}_{i}}^{{}^{\prime}}$ belong to two pencils. So without loss of generality we may assume that $C_{P_{i}}\sim C_{\tilde{P_{i}}}$ and $C_{P_{i}}^{{}^{\prime}}\sim C_{\tilde{P_{i}}}^{{}^{\prime}}$. We denote the classes of these conics by $C_{i}$ and $C_{i}^{{}^{\prime}}$ which are independent of the chosen point on $Q_{t_{i}}$. By Theorem 2.1, the classes of these conics are the 10 possible classes of conics on $Y$ and $C_{i}^{{}^{\prime}}=H-C_{i}$ for every $i\in\{0,\ldots,4\}$ where $H$ is the hyperplane class of $Y$.

Proposition 2.2**.**

[VAV*, Proposition 2.2]**

After possibly interchanging the $C_{i}$ and the $C_{i}^{{}^{\prime}}$ for some indices $i$, we may assume that the Picard group $\operatorname{Pic}\overline{Y}\cong\mathbb{Z}^{6}$ is freely generated by the following classes*

[TABLE]

where $H$ is the hyperplane class of $Y$.∎

Consider $\sigma\in G_{K}$, and let $\sigma^{{}^{\prime}}\colon\operatorname{Spec}\overline{K}\rightarrow\operatorname{Spec}\overline{K}$ be the corresponding morphism on schemes. After base changing to $\overline{K}$ we get

[TABLE]

The morphism above induces an automorphism on $\operatorname{Pic}\overline{Y}$. This defines the action of $G_{K}$ on $\operatorname{Aut}(\operatorname{Pic}\overline{Y})$. This action fixes the canonical class and the intersection multiplicity [Man74, Theorem 23.8].

Let $\Gamma$ be the graph of ten vertices indexed by $C_{i}$ and $C_{i}^{{}^{\prime}}$ whose edges join $C_{i}$ and $C_{i}^{{}^{\prime}}$. The group $\operatorname{Aut}(\Gamma)$ is the semi-direct product $(\mathbb{Z}/2\mathbb{Z})^{5}\rtimes S_{5}$ where the $\mathbb{Z}/2\mathbb{Z}$ entries represent the exchanges for each $\{C_{i},C_{i}^{{}^{\prime}}\}$ and the $S_{5}$ entry represents a permutation of the sets $\{C_{i},C_{i}^{{}^{\prime}}\}$ for $i\in\{0,\ldots,4\}$. Let $\mathscr{O}(K_{Y}^{\bot})$ be the subgroup of $\operatorname{Aut}(\Gamma)$ that fixes the orthogonal complement of $K_{Y}$ in $\operatorname{Pic}\overline{Y}$. By the discussion in [KST, p:8-10], there is a natural embedding $\mathscr{O}(K_{Y}^{\bot})\hookrightarrow\operatorname{Aut}(\Gamma)$ of index 2. Moreover by [KST, p:8-10] the image of this embedding constitutes of all automorphisms that are the product of an even number of exchanges and an element of $S_{5}$.

Fix $T\in\mathscr{S}$. Let $\Gamma_{T}$ be subgraph with $2\deg(T)$ vertices indexed by $C_{i}$ and $C_{i}^{{}^{\prime}}$ for $t_{i}\in T(\overline{K})$.

Proposition 2.3**.**

[VAV*, Proposition 2.3]**

The action of $G_{K}$ on $\operatorname{Pic}\overline{Y}$ induces an action on $\Gamma$ that factors through*

[TABLE]

Moreover, $G_{K}$ acts transitively on $\{C_{i},C_{i}^{{}^{\prime}}:t_{i}\in T(\overline{K})\}$ if and only if $\epsilon_{T}\notin K(T)^{\times 2}$.

Proof.

By the discussion above, the action of $G_{K}$ factors through $\Pi_{T\in\mathscr{S}}\operatorname{Aut}(\Gamma_{T})\cap\mathscr{O}(K_{Y}^{\bot})$. Moreover, by definition the pair $\{C_{i},C_{i}^{{}^{\prime}}\}$ is defined over $K(t_{i})$ and each individual conic over $K(t_{i},\sqrt{\epsilon_{T}})$. So $\epsilon_{T}\notin K(T)^{\times 2}$ is exactly the condition for a transitive action of $G_{K}$ on $\{C_{i},C_{i}^{{}^{\prime}}\}$. ∎

2.2. An algorithm to relate $\textup{H}^{1}(K,\operatorname{Pic}\overline{Y})$ and $\operatorname{Br}Y/\operatorname{Br}K$

Throughout this section, we assume that there is a subscheme $\mathscr{T}\subset\mathscr{S}$ that satisfies the conditions

[TABLE]

Because $\mathscr{T}$ satisfies (*), Lemma [VAV, Lemma 3.1] allows us to assume that $\epsilon_{T}\in\operatorname{im}(K^{\times}/K^{\times 2}\rightarrow K(T)^{\times}/K(T)^{\times 2})$. Define $K_{\mathscr{T}}:=K(\sqrt{\epsilon_{T}})$ which is independent of $T$ because $\deg(\mathscr{T})=2$ and $\Pi_{T\in\mathscr{T}}N_{K(T)/K}(\epsilon_{T})\in K^{\times 2}$. By Lemma [VAV, Lemma 3.1] and since $\epsilon_{T}\notin K(T)^{\times 2}$, $K_{\mathscr{T}}$ is a quadratic extension of $K$.

Proposition 2.4**.**

The cocycle in $\textup{H}^{1}(K,\operatorname{Pic}\overline{Y})$ given by

[TABLE]

is non trivial if and only if there exists $T\in\mathscr{S}-\mathscr{T}$ such that $\epsilon_{T}\notin K(T)^{\times 2}$.

Proof.

This proof can be found in [VAV, Section 3.2, Proposition 3.3] assuming that $Q_{T}$ has a smooth $K(T)$-point for all $T\in\mathscr{T}$. However the proof works without the assumption that $Q_{T}$ has a smooth $K(T)$-point for all $T\in\mathscr{T}$. We repeat the proof for the reader’s convenience.

The long exact sequence on cohomology associated to the short exact sequence

[TABLE]

induces an isomorphism

[TABLE]

where $d$ is a lift of $D$ to $\operatorname{Pic}\overline{Y}$.

We prove that the divisor $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\in(\operatorname{Pic}\overline{Y}/2\operatorname{Pic}\overline{X})^{G_{K}}$ and that its image of under the isomorphism (2.1), is the cocycle $\alpha\colon G_{K}\rightarrow\operatorname{Pic}\overline{Y}$ defined by

[TABLE]

By definition, for every $t_{i}\in\mathscr{S}(\overline{K})$ the pair of conics $\{C_{i},C_{i}^{{}^{\prime}}\}$ is defined over $K(t_{i})$ and each individual conic is defined over $K(t_{i},\sqrt{\epsilon_{t_{i}}})$. Hence $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}$ is fixed by $\sigma$ for every $\sigma\in G_{K_{\mathscr{T}}}$. Furthermore, if $\sigma\notin G_{K_{\mathscr{T}}}$ then

[TABLE]

By the previous two statements, $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\in(\operatorname{Pic}\overline{Y}/2\operatorname{Pic}\overline{Y})^{G_{K}}$. Moreover, by the explicit description of ismorphism (3.2), $\alpha\in\textup{H}^{1}(K,\operatorname{Pic}\overline{Y})$ as defined above is the image of $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}$ by the isomorphism (2.1).

Therefore $\alpha$ is trivial if and only if $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\in 2\operatorname{Pic}\overline{Y}+(\operatorname{Pic}\overline{Y})^{G_{K}}$. We will prove that $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\notin 2\operatorname{Pic}\overline{Y}+(\operatorname{Pic}\overline{Y})^{G_{K}}$ if and only if there exists $T\in\mathscr{S}-\mathscr{T}$ such that $\epsilon_{T}\notin K(T)^{\times 2}$. We determine an equivalent criterion to $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\notin 2\operatorname{Pic}\overline{Y}+(\operatorname{Pic}\overline{Y})^{G_{K}}$ by using the generators of $\operatorname{Pic}\overline{Y}$. By Equation (2.2), $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\notin(\operatorname{Pic}\overline{Y})^{G_{K}}$. Moreover, by the same argument as in the proof of Lemma 3.3 any combination of the generators of $\operatorname{Pic}\overline{Y}$ involving an odd coefficient of $\frac{H+\Sigma_{i}C_{i}}{2}$ is not fixed by $G_{K}$. Therefore $\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}\in 2\operatorname{Pic}\overline{Y}+(\operatorname{Pic}\overline{Y})^{G_{K}}$ if and only if there exists a choice of signs such that

[TABLE]

If there exists $T\in\mathscr{S}-\mathscr{T}$ such that $\epsilon_{T}\notin K(T)^{\times 2}$ then by Proposition 2.3 $G_{K}$ acts transitively on each pair $\{C_{i},C_{i}^{{}^{\prime}}:t_{i}\in T(\overline{K})\}$. Then for some $t_{j}\in T(\overline{K})$ and by using the fact that $C_{j}^{{}^{\prime}}=H-C_{j}$, there exists $\sigma\in G_{K}$ such that

[TABLE]

where $D$ is a linear combination of the $C_{i}$’s excluding $C_{j}$. Then for any choice of signs

[TABLE]

This proves that $\alpha$ is a non trivial cocycle in $\textup{H}^{1}(K,\operatorname{Pic}\overline{Y})$.

Conversely, if for every $T\in\mathscr{S}-\mathscr{T}$ we have $\epsilon_{T}\in K(T)^{\times 2}$ then for any choice of signs

[TABLE]

So $\alpha$ is trivial in $\textup{H}^{1}(K,\operatorname{Pic}\overline{Y})$. ∎

Let $V$ be a nice scheme over a field $K$. Let $L$ be a cyclic extension of $K$; denote by $\sigma$ the generator of $\textup{Gal}(L/K)$. Let $f$ be any element in $K(V)^{\times}$. We denote by $\operatorname{Br}_{cyc}(V,L)$ the set of classes of algebras $[(L/K,f)]$ in the image of $\operatorname{Br}V/\operatorname{Br}_{0}V\rightarrow\operatorname{Br}K(V)/\operatorname{Br}_{0}V$ where $\operatorname{Br}_{0}V=\operatorname{im}(\operatorname{Br}K\rightarrow\operatorname{Br}V)$. We view $1-\sigma$ as endomorphisms of $\textup{Div}V_{L}$. We let $N_{L/K}\colon\textup{Div}V_{L}\rightarrow\textup{Div}V_{K}$ and $\overline{N}_{L/K}\colon\operatorname{Pic}V_{L}\rightarrow\operatorname{Pic}V_{K}$ be the usual norm maps.

Theorem 2.5**.**

There exists an injection $\operatorname{Br}_{cyc}(V,L)\hookrightarrow\textup{H}^{1}(K,\operatorname{Pic}\overline{V})$ given by the composition of the maps

[TABLE]

The image of the class $[(L/K,f)]$ under the above composition is the cocycle

[TABLE]

where $D$ is the divisor such that $(f)=N_{L/K}(D)$.∎

Proof.

The first isomorphism follows from [VAb, Theorem 3.3]. By [VAb, Theorem 3.3], the isomorphism maps the class of the algebra $[(L/K,f)]$ to the divisor $D$ such that

[TABLE]

The extension $L/K$ is cyclic; so by using the explicit resolution we compute

[TABLE]

The image of $D$ under this isomorphism is the cocycle $\alpha$ that maps $\sigma\mapsto D$.

Furthermore, the first part of the inflation-restriction exact sequence yields the injection

[TABLE]

The image of $\alpha$ under the above inflation map is the required cocycle. ∎

Applying the map in Theorem 2.5 to the del Pezzo surface $Y$ and the cyclic extension $K_{\mathscr{T}}/K$, we get the map

[TABLE]

For the remainder of this section, we assume that for every subscheme $\mathscr{T}\subset\mathscr{S}$ satisfying $(*)$, the quadric $Q_{T}$ has a smooth $K(T)$-point for every $T\in\mathscr{T}$. Let

[TABLE]

where $l_{T}$ is a $K(T)$-linear form such that the associated hyperplane is tangent to $Q_{T}$ at a smooth point for every $T\in\mathscr{T}$ and $l$ is any linear form.

Proposition 2.6**.**

Let $\mathscr{T}\subset\mathscr{S}$ satisfy $(*)$. The cyclic algebra $\mathscr{A}_{\mathscr{T}}$ is in the image of $\operatorname{Br}Y\rightarrow\operatorname{Br}K(Y)$. Further, if there exists $T\in\mathscr{S}-\mathscr{T}$ such that $\epsilon_{T}\notin K(T)^{\times 2}$ then the algebra $\mathscr{A}_{\mathscr{T}}$ is non trivial. In particular, it maps under the map 2.4 to the nontrivial cocycle

[TABLE]

in $\textup{H}^{1}(K,\operatorname{Pic}\overline{Y})$.

Proof.

To prove that $\mathscr{A}_{\mathscr{T}}\in\operatorname{im}(\operatorname{Br}X\rightarrow\operatorname{Br}K(Y))$, we prove that $\mathscr{A}_{\mathscr{T}}$ is unramified at every codimension 1 point $x\in Y^{(1)}$, i.e., $\partial_{x}(\mathscr{A})=0$ by the residue sequence, [Groa, Grob, Groc]. The prime divisors which correspond to a valuation such that the function $l^{-2}\Pi_{T\in\mathscr{T}}\textup{N}_{K(T)/K}(l_{T})$ has an odd valuation at are $C_{T}$ and $C_{T}^{{}^{\prime}}$ for every $T\in\mathscr{T}$. However, for every $T\in\mathscr{T}$, $K_{\mathscr{T}}\subset\kappa(C_{T}\cup C_{T}^{{}^{\prime}})$ because $C_{T}$ and $C_{T}^{{}^{\prime}}$ are conjugate over $K_{\mathscr{T}}$. So by [GS, 7.5.1], $\mathscr{A}$ is unramified at $C_{T}$ and $C_{T}^{{}^{\prime}}$ for every $T\in\mathscr{T}$. Hence $\mathscr{A}\in\operatorname{im}(\operatorname{Br}Y\rightarrow\operatorname{Br}K(Y))$ by the residue sequence, [Groa, Grob, Groc].

By definition of the maps defining 2.4, the class of the algebra $\mathscr{A}_{\mathscr{T}}$ gets mapped to a cocycle $\alpha$ that maps $G_{K_{\mathscr{T}}}$ to the identity and maps any element $\sigma\notin G_{K_{\mathscr{T}}}$ to a divisor $D$ such that $\textup{N}_{K_{\mathscr{T}}/K}(D)=\textup{div}(l^{-2}\Pi_{T\in\mathscr{T}}\textup{N}_{K(T)/K}(l_{T}))=-2H+\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}+\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}^{{}^{\prime}}$. So $D$ can be chosen as $-H+\Sigma_{t_{i}\in\mathscr{T}(\overline{K})}C_{i}$. The cocycle $\alpha$ is non trivial by Proposition 3.4. So $\mathscr{A}_{\mathscr{T}}$ is non trivial as well. ∎

3. An example of a del Pezzo surface of degree 4 with trivial Brauer Group and non trivial $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})$

In this section we prove the main theorem.

Theorem 3.1**.**

Let $k=\mathbb{Q}^{cycl}(a,b,c)$ where $a,b$ and $c$ are independent transcendental elements. Let $X$ be the del Pezzo surface of degree 4 in $\mathbb{P}^{4}_{k}$ defined by the intersection of the following two quadrics

[TABLE]

Then $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/2\mathbb{Z}$ while $\frac{\operatorname{Br}X}{\operatorname{Br}k}$ is trivial.

The following corollary follows from the algorithm in [VAV, Section 4.4].

Corollary 3.2**.**

The assumption that $Q_{T}$ has a smooth $k(T)$-point for every $T\in\mathscr{T}$ in [VAV, Section 4.4] cannot be omitted entirely to apply the algorithm for computing $\operatorname{Br}X/\operatorname{Br}k$.∎

The proof of Theorem 3.1 relies on the functoriality with respect to base extension from $k$ to $L=k(\sqrt{a})$ of the Hochschild-Serre spectral sequence:

[TABLE]

where $\operatorname{Br}_{1}X=\ker(\operatorname{Br}X\rightarrow\operatorname{Br}\overline{X})$. Since $X$ is rational then $\operatorname{Br}_{1}X=\operatorname{Br}X$. Later in this section, we prove that the algorithm in [VAV, Section 4.1] can be applied over $L$ and that the generating Brauer class of $\operatorname{Br}X_{L}/\operatorname{Br}L$ is not in the image of $\operatorname{Br}X/\operatorname{Br}k$.

3.1. Degeneracy locus of $X$

Let $A$ and $A^{{}^{\prime}}$ be the matrices associated to the quadratic forms of the quadrics $Q$ and $Q^{{}^{\prime}}$ respectively. The characteristic polynomial of the pencil of quadrics $\{\lambda Q+\mu Q^{{}^{\prime}}:[\lambda:\mu]\in\mathbb{P}^{1}\}$ is

[TABLE]

Since all the irreducible factors of $f(\lambda,\mu)$ are distinct, it is separable in $k[\lambda,\mu]$. Therefore [Wit, Proposition 3.26], $X$ is smooth. Moreover, the degeneracy locus $\mathscr{S}$ of this del Pezzo surface consists of the five degree 1 points

[TABLE]

corresponding to the linear factors of $f(\lambda,\mu)$.

For each point in the degeneracy locus $\mathscr{S}$, we compute the corresponding quadric and discriminant as explained in Chapter 3. We show the case corresponding to $T_{0}$ and summmarize the rest in an array below. The singular locus of the quadric $Q_{T_{0}}=V(ax_{0}^{2}+bx_{1}^{2}+x_{2}^{2}+cx_{4}^{2})$ is $[0:0:0:1:0]$. So we may choose $H_{T_{0}}$ to be $V(x_{3})$. By a direct computation,

[TABLE]

The quadrics and discriminants corresponding to all the points $\{T_{0},\ldots,T_{4}\}$ are summarized below.

[TABLE]

3.2. Computing $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})$ and $\textup{H}^{1}(L,\operatorname{Pic}\overline{X})$

The first part of the following Lemma is used to compute $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})$ and $\textup{H}^{1}(L,\operatorname{Pic}\overline{X})$ and the second part will be used later in the proof of Theorem 3.1 to replace $\operatorname{Br}_{0}X$ and $\operatorname{Br}_{0}X_{L}$ by $\operatorname{Br}k$ and $\operatorname{Br}L$ respectively.

Lemma 3.3**.**

Let $K=k$ or $L$, where $k$ and $L$ are the fields defined before. For $X$ as before we have,

1. (1)

$\frac{(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}}{((\operatorname{Pic}\overline{X})^{G_{K}}/2(\operatorname{Pic}\overline{X})^{G_{K}})}=\langle C_{0}+C_{1}\rangle$** 2. (2)

$(\operatorname{Pic}\overline{X})^{G_{K}}=\langle H\rangle$.

Proof.

We start by proving that $(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}=\langle H,C_{0}+C_{1}\rangle$. By Proposition 2.2, classes in $\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X}$ can be represented by divisors of the form $D=\beta(\frac{H+\Sigma_{i}C_{i}}{2})+\Sigma_{i}\alpha_{i}C_{i}$ where $\beta$ and $\alpha_{i}$ are either 0 or 1 for all $i\in\{0,\ldots,4\}$. Let $\sigma$ be any element in $G_{K}$. By Proposition 2.3 and since each $T\in\mathscr{S}$ has degree 1, $\sigma$ is the product of an even number of exchanges between $C_{i}$ and $C_{i}^{{}^{\prime}}=H-C_{i}$ for $i\in\{0,\ldots 4\}$. Let $I\subset\{0,\ldots,4\}$ be the set of indices of the exchanges that are factors of $\sigma$. Since $\sigma$ is the product of an even number of exchanges, $I$ has even cardinality. The Galois element $\sigma$ is determined by the set of indices $I$; so we denote it by $\sigma_{I}$. We are interested in characterizing $D\in(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}$ or equivalently $D\in\operatorname{Pic}\overline{X}$ such that $D-\sigma_{I}D\in 2\operatorname{Pic}\overline{X}$ for every $\sigma_{I}\in G_{K}$. First we compute $\sigma_{I}D$ for any $\sigma_{I}\in G_{K}$. Let $E:=\frac{H+\Sigma_{i}C_{i}}{2}$ and $\gamma:=\Sigma_{i\in I}\alpha_{i}$.

[TABLE]

We expand and arrange $D-\sigma_{I}D$ as

[TABLE]

By transitivity of the action of $G_{K}$, Proposition 2.3, we may assume $I$ to be non trivial. By Considering the coefficients of $D-\sigma_{I}D$ we get

[TABLE]

Hence $\beta=0$. So $\gamma$ is even.

Now we consider the possibilities of the even cardinality sets $I\subset\{0,\ldots,4\}$. Since $C_{0},C_{1}$ are defined over $K(\sqrt{\epsilon_{T_{0}}})=K(\sqrt{\epsilon_{T_{1}}})$, then either both exchanges between $\{C_{0},C_{0}^{{}^{\prime}}\}$ and $\{C_{1},C_{1}^{{}^{\prime}}\}$ are factors of $\sigma\in G_{K}$ or both are not. Hence $\#(I\cap\{0,1\})\neq 1$. By the computation of $D-\sigma_{I}D$ before when $I=\{0,1\}$, we deduce that $\alpha_{0}+\alpha_{1}\equiv 0\pmod{2}$. Since $I$ has even cardinality and $\#(I\cap\{0,1\})\neq 1$ and by transitivity of the action of $G_{K}$ on $\{C_{2},C_{2}^{{}^{\prime}}\}$, $\{C_{3},C_{3}^{{}^{\prime}}\}$, and $\{C_{4},C_{4}^{{}^{\prime}}\}$, the nontrivial possibilities of $I-\{0,1\}\cap I$ are $\{2,3\}$, $\{2,4\}$, and $\{3,4\}$. The class of the divisor $D\pmod{2}$ is fixed by $\sigma_{I}$ where $I-\{0,1\}\cap I$ is $\{2,3\}$, $\{2,4\}$, or $\{3,4\}$ if and only if $\alpha_{2}+\alpha_{3}$, $\alpha_{2}+\alpha_{4}$, or $\alpha_{3}+\alpha_{4}$ are even respectively. From the discussion before we deduce

[TABLE]

Since $\alpha_{i}\in\{0,1\}$, from the above congruences we deduce that $D$ is [math], $C_{0}+C_{1}$, $C_{2}+C_{3}+C_{4}$, or $\Sigma_{i}C_{i}$. Hence $(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}=\langle C_{0}+C_{1},C_{2}+C_{3}+C_{4}\rangle$. Further,

[TABLE]

So we may rewrite the generators as $(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}=\langle H,C_{0}+C_{1}\rangle$.

We prove that $\frac{(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}}{((\operatorname{Pic}\overline{X})^{G_{K}}/2(\operatorname{Pic}\overline{X})^{G_{K}})}=\langle C_{0}+C_{1}\rangle$ or equivalently that $(\operatorname{Pic}\overline{X})^{G_{K}}=\langle H\rangle$. Let $\Sigma_{i}a_{i}C_{i}$ be a nontrivial combination of the $C_{i}$’s. Let $j\in\{0,\ldots 4\}$ be an arbitrary element such that $a_{j}\neq 0$. By Proposition 2.3 and the fact that $\epsilon_{j}$ is not a square in $K(T_{j})^{\times}=K^{\times}$, there exists a $\tau\in G_{K}$ such that $\tau C_{j}=C_{j}^{{}^{\prime}}=H-C_{j}$. Therefore, $\tau(\Sigma_{i}a_{i}C_{i})=-\alpha_{j}C_{j}+D^{{}^{\prime}}$ where $D^{{}^{\prime}}$ is a linear combination of $C_{i}$ $i\neq j$. Hence $\Sigma_{i}a_{i}C_{i}$ is not fixed by $G_{K}$. Therefore $(\operatorname{Pic}\overline{X})^{G_{K}}=\langle H\rangle$ and this proves (1) and (2). ∎

Proposition 3.4**.**

For $X$ and $k$, and $L$ as before, we have $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\simeq\textup{H}^{1}(L,\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/2\mathbb{Z}$ and $\textup{Res}\colon\textup{H}^{1}(k\operatorname{Pic}\overline{X})\rightarrow\textup{H}^{1}(L,\operatorname{Pic}\overline{X})$ is an isomorphism.

Proof.

Consider the following short exact sequence

[TABLE]

The connecting morphism in the induced long exact sequence yields

[TABLE]

where $K=k$ or $L$, and $d$ is a lift of $D$ to $\operatorname{Pic}\overline{X}$. By Lemma 3.3 (1) and by the isomorphism (3.2), we deduce that $\textup{H}^{1}(K,\operatorname{Pic}\overline{X})[2]\simeq\mathbb{Z}/2\mathbb{Z}$. Further, $\textup{H}^{1}(K,\operatorname{Pic}\overline{X})$ is 2-torsion because the cardinality of $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})$ divides $4$ as verified in Magma [BCP] script in the arXiv distribution of [VAV]. So $\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\simeq\textup{H}^{1}(L,\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/2\mathbb{Z}$. Moreover, the image of the cocycle corresponding to $C_{0}+C_{1}$, that generates $\frac{(\operatorname{Pic}\overline{X}/2\operatorname{Pic}\overline{X})^{G_{K}}}{((\operatorname{Pic}\overline{X})^{G_{K}}/2(\operatorname{Pic}\overline{X})^{G_{K}})}$ by Lemma 3.3, under the restriction map $\textup{Res}\colon\textup{H}^{1}(k\operatorname{Pic}\overline{X})\rightarrow\textup{H}^{1}(L,\operatorname{Pic}\overline{X})$ is the cocycle corresponding to $C_{0}+C_{1}$. So $\textup{Res}\colon\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\rightarrow\textup{H}^{1}(L,\operatorname{Pic}\overline{X})$ is an isomorphism. ∎

3.3. The Brauer group of $X$ over $L=k(\sqrt{a})$

We use the algorithm in [VAV, Section 4.1] to prove that $\operatorname{Br}X_{L}/\operatorname{Br}L\simeq\mathbb{Z}/2\mathbb{Z}$ and to explicitly construct a non trivial algebra $\mathscr{A}$ whose class generates $\operatorname{Br}X_{L}/\operatorname{Br}L$. This algebra will be used in the proof of Theorem 3.1.

Proposition 3.5**.**

The Brauer group $\operatorname{Br}X_{L}/\operatorname{Br}L$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ and is generated by the class of the quaternion algebra:

[TABLE]

Moreover, $\mathscr{A}-\sigma\mathscr{A}\sim(c,b)\in\operatorname{Br}L$.

Proof.

The same computation of the degeneracy locus, associated quadrics, and discriminants of $X$ in Section 4.1 still work over $L$. Moreover $\mathscr{T}=\{T_{0},T_{1}\}$ is a degree two subscheme of $\mathscr{S}$ such that $\Pi_{T\in\mathscr{T}}N_{L(T)/L}=\epsilon_{T_{0}}\epsilon_{T_{1}}=a^{2}b^{2}c^{2}\in L(T)^{\times 2}=L^{\times 2}$, and the element $\epsilon_{T_{0}}=\epsilon_{T_{1}}=bc$ is non trivial in $L^{\times}/L^{\times 2}$. Hence $\mathscr{T}$ satisfies $(*)$ as defined in Section 2. By direct computations we show $\{T_{0},T_{1}\}$ is the only subscheme of $\mathscr{S}$ that satisfies $(*)$. Moreover, the quadrics $Q_{T_{0}}:ax_{0}^{2}+bx_{1}^{2}+x_{2}^{2}+cx_{4}^{2}=0$ and $Q_{T_{1}}:bcx_{0}^{2}+x_{1}^{2}+x_{2}^{2}+ax_{3}^{2}=0$ have smooth $L$-points $P_{T_{0}}=[i:0:\sqrt{a}:0:0]$ and $P_{T_{1}}=[0:i:1:0:0]$ respectively. Further, $\epsilon_{T_{2}}$ is not a square in $L^{\times}$. So by the algorithm in [VAV, Section 4.1], $\operatorname{Br}X_{L}/\operatorname{Br}L\simeq\textup{H}^{1}(L,\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/2\mathbb{Z}$ and is generated by the algebra

[TABLE]

where $l_{T}$ is the $L(T)$-linear form such that the associated hyperplane is tangent to $Q_{T}$ at a smooth point $P_{T}$ of $Q_{T}$ for $T\in\mathscr{T}$, and $l$ is any linear form.

Computing the linear forms we get, $l_{T_{0}}:2aix_{0}+2\sqrt{a}x_{2}$, and $l_{T_{1}}:2ix_{1}+2x_{2}$. Substituting these into $\mathscr{A}$ yields the required algebra.

We have

[TABLE]

By the defining equation of $Q^{{}^{\prime}}$, $ax_{0}^{2}+x_{2}^{2}=-bx_{1}^{2}-cx_{4}^{2}=-b(x_{1}^{2}-cb(i\frac{x_{4}}{b})^{2})$. So $\mathscr{A}-\sigma\mathscr{A}\sim(bc,b)\sim(c,b)$. ∎

3.4. Characterizing $\operatorname{im}(\operatorname{Br}X/\operatorname{Br}k\rightarrow\operatorname{Br}X_{L}/\operatorname{Br}L)$

Let $\textup{Gal}(L/k)=\langle\sigma\rangle$.

Lemma 3.6**.**

Let $\textup{Gal}(L/k)=\langle\sigma\rangle$. If $\mathscr{A}-\sigma\mathscr{A}\neq x-\sigma x$ for every $x\in\operatorname{Br}L$ then $[\mathscr{A}]\notin\operatorname{im}(\operatorname{Br}X/\operatorname{Br}k\rightarrow\operatorname{Br}X_{L}/\operatorname{Br}L)$.

Proof.

By [GS, Proposition 3.3.17], the generalized inflation-restriction sequence for the field extension $L(X)/k(X)$ is

[TABLE]

If the class of $\mathscr{A}$ is in the image of $\operatorname{Br}X/\operatorname{Br}k\rightarrow\operatorname{Br}X_{L}/\operatorname{Br}L$, then there exists $x\in\operatorname{Br}L$ such that $\mathscr{A}-x\in\operatorname{im}(\operatorname{Br}X\rightarrow\operatorname{Br}X_{L})$. Hence $\mathscr{A}-x\in\operatorname{im}(\operatorname{Br}k(X)\rightarrow\operatorname{Br}L(X))$. By the generalized inflation-restriction sequence above, $\mathscr{A}-x$ is fixed by $\textup{Gal}(L/k)=\langle\sigma\rangle$. Rearranging we get that $\mathscr{A}-\sigma\mathscr{A}=x-\sigma x$. ∎

3.5. Proof of Theorem 3.1

First we may replace $\operatorname{Br}_{0}X$ and $\operatorname{Br}_{0}X_{L}$ by $\operatorname{Br}k$ and $\operatorname{Br}L$ respectively because the map $(\operatorname{Pic}\overline{X})^{G_{K}}=\langle H\rangle\rightarrow\operatorname{Br}K$ is trivial by the exact sequence that follows from the Hochschild-Serre spectral sequence where $K=k$ or $L$.

For the sake of contradiction we assume that $\operatorname{Br}X/\operatorname{Br}k$ is non trivial. Since there is an injection $\operatorname{Br}X/\operatorname{Br}k\hookrightarrow\textup{H}^{1}(k,\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/2\mathbb{Z}$ by the exact sequence that follows from the Hochschild-Serre spectral sequence and $\operatorname{Br}X/\operatorname{Br}k$ is nontrivial, there is a unique nontrivial class in $\operatorname{Br}X/\operatorname{Br}k$; denote this class by $[\mathscr{B}]$. By Proposition 3.4, Proposition 3.5 and the functoriality of the Hochschild-Serre spectral sequence we have

[TABLE]

So $[\mathscr{B}]\in\operatorname{Br}X/\operatorname{Br}k$ gets mapped to $[\mathscr{A}]\in\operatorname{Br}X_{L}/\operatorname{Br}L$ as defined in Proposition 3.5 by the field extension map. We will show that any algebra in the class of $[\mathscr{A}]\in\operatorname{Br}X_{L}/\operatorname{Br}L$ is not in the image of the map $\operatorname{Br}X\rightarrow\operatorname{Br}X_{L}$, thus resulting in a contradiction. By Lemma 3.6 it is enough to prove that $\mathscr{A}-x$ is not fixed by $\sigma$ for all $x\in\operatorname{Br}L$.

Suppose that there exists $x\in\operatorname{Br}L$ such that $\mathscr{A}-x$ is fixed by $\sigma$, i.e., $\mathscr{A}-\sigma\mathscr{A}=x-\sigma x$. By Proposition 3.5, both sides of the equation $\mathscr{A}-\sigma A=x-\sigma x$ are in $\operatorname{Br}L$. Let $Y=\operatorname{Spec}{\mathbb{Q}^{cycl}(b,c)[\sqrt{a}]}$. Let $P:\sqrt{a}=0$ be a divisor on $Y$. Because $L$ is the function field of $Y$ and $P$ is a divisor on $Y$, we have the residue sequence, ([Groa], [Grob],[Groc]),

[TABLE]

If $x$ is cyclic of degree $n$ then $\partial_{P}(x)$ is determined by a degree $n$ cyclic extension of $\mathbb{Q}^{cycl}(b,c)$ and a choice of a generator of the Galois group of the extension. By Kummer Theory the cyclic extension determined by $\partial_{P}(x)$ is of the form $\mathbb{Q}^{cycl}(b,c)(\sqrt[n]{\alpha})/\mathbb{Q}^{cycl}(b,c)$ where $\alpha\in\mathbb{Q}^{cycl}(b,c)^{\times}$. Let $f=f(b,c)\in L$ be a function of $b$, and $c$. By [GS, 7.5.1], the residue $\partial_{P}$ of the cyclic algebra $(\sqrt{a},f(b,c))_{n}\in\operatorname{Br}L$ is determined by the cyclic extension

[TABLE]

and a choice of generator of the Galois group of this extension. Therefore we may choose $f$ such that $\partial_{P}((\sqrt{a},f(b,c))_{n})=\partial_{P}(x)$. Furthermore $\sigma((\sqrt{a},f(b,c))_{n})=(-\sqrt{a},f(b,c))_{n}=(\sqrt{a},f(b,c))_{n}$ because $-1$ is an $n$-th root of unity in $L$. Since $(\sqrt{a},f(a,b))_{n}$ is fixed by $\sigma$, subtracting $(\sqrt{a},f(b,c))$ from $x$ will not change $x-\sigma x$. Moreover, $(\sqrt{a},f(b,c))$ has the same residue as $x$ at $P$; so we may assume without loss of generality that $x$ is unramified at $P$. By Murkurjev-Suslin Theorem and the fact that $\partial_{P}$ is a homomorphism we extend this argument to any central simple algebra $x\in\operatorname{Br}L$. So we assume $x$ is unrafimied at $P$ in general. We claim that $\mathscr{A}-\sigma\mathscr{A}$ is also unramified at $P$ and we prove it later. Hence we may specialize the equation $\mathscr{A}-\sigma\mathscr{A}=x-\sigma x$ at the divisor $P$,

[TABLE]

The action of $\sigma$ on $x$ commutes with specialization. Furthermore, $P$ is invariant under $\sigma$. Hence $(\sigma x)_{|P}=\sigma x_{|P}=x_{|P}$. So $(\mathscr{A}-\sigma\mathscr{A})_{|P}$ is trivial in $\operatorname{Br}\kappa$ where $\kappa$ is the residue field of $Y$ at $P$. In our example $\kappa=\mathbb{Q}^{cycl}(b,c)$. Since $\mathscr{A}-\sigma\mathscr{A}\sim(b,c)$, by Proposition 3.5, is a constant Azumaya algebra in $\operatorname{Br}\mathscr{O}_{Y,\eta}$, $(\mathscr{A}-\sigma\mathscr{A})_{|P}=\mathscr{A}-\sigma\mathscr{A}\sim(b,c)$. Hence $(b,c)$ is trivial in $\operatorname{Br}\mathbb{Q}^{cycl}(b,c)$. This is a contradiction because the extension $\mathbb{Q}^{cycl}(b)(\sqrt{c^{-v_{D}(b)}b^{v_{D}(c)}})/\mathbb{Q}^{cycl}(b)$ associated to $\partial_{c=0}((c,b))$ by [GS, 7.5.1] is non trivial.∎

Now we prove the claim that the algebra $\mathscr{A}-\sigma\mathscr{A}$ is unramified at $P\colon\sqrt{a}=0$. We have the residue sequence, ([Groa], [Grob],[Groc])

[TABLE]

By [GS, 7.5.1], the residue $\partial_{P}$ of the cyclic algebra $\mathscr{A}-\sigma\mathscr{A}\sim(b,c)\in\operatorname{Br}L$ is determined by the cyclic extension $\mathbb{Q}^{cycl}(b,c)(\sqrt{b^{-v_{P}(c)}c^{v_{P}(b)}})/\mathbb{Q}^{cycl}(b,c)$ and a choice of a generator of the Galois group of this extension. The valuations $v_{P}(c)=v_{P}(b)=0$. By the last two sentences, we deduce that $\partial_{P}(\mathscr{A}-\sigma\mathscr{A})=0$. Therefore $\mathscr{A}-\sigma\mathscr{A}$ is unramified at $P$.∎