Integral points on twisted Markoff surfaces

TL;DR

This paper investigates the integral Hasse principle on twisted Markoff surfaces using Brauer-Manin obstruction, providing explicit examples with 4-torsion Brauer group elements and detailed local invariant computations.

Contribution

It introduces new examples of twisted Markoff surfaces with nontrivial Brauer groups and explicit representatives, advancing understanding of the Brauer-Manin obstruction in this context.

Findings

Identified examples with 4-torsion elements in the Brauer group

Constructed explicit representatives for these Brauer group elements

Computed local invariants at special places for these examples

Abstract

We study the integral Hasse principle for affine varieties of the form ax^2+y^2+z^2-xyz=m ,using Brauer-Manin obstruction, and we produce examples whose Brauer groups include 4-torsion elements .We will construct their explicit representatives and compute local invariants in special places.

Equations

$$a{x}^2+y^2+z^2-xyz=m$$

$$x^2+y^2+z^2-xyz=m.$$

$$m-4\mathrm{mod}{Q}^{{\times }^2}\in ⟨\pm 1,2,3,5⟩\subset Q^{\times }/Q^{{\times }^2}.$$

$$a{x}^2+y^2+z^2-xyz=m.$$

$$m-4a\mathrm{mod}{Q}^{{\times }^2}\in ⟨\pm 1,2,3,\{p_i{\}}_{1\le i\le s}⟩\subset Q^{\times }/Q^{{\times }^2}.$$

$$\mathrm{Br}(X)\to \mathrm{Br}(k(X))$$

$${\mathrm{Br}}_1(X)=\mathrm{Ker}[\mathrm{Br}(X)\to \mathrm{Br}(\overline{X})]$$

$$a{x}^{2}t+y^{2}t+z^{2}t-xyz=m{t}^3$$

$$H_1:x=t=0;H_2:y=t=0;H_3:z=t=0$$

$$\left\{\begin{array}[]{ccc}l_{1}(\varepsilon,\delta):&x=2\varepsilon t,&y-\varepsilon z=\delta\sqrt{m-4a}t\\ l_{2}(\varepsilon,\delta):&y=2\varepsilon\sqrt{a}t,&z-\varepsilon\sqrt{a}x=\delta\sqrt{m-4a}t\\ l_{3}(\varepsilon,\delta):&z=2\varepsilon\sqrt{a}t,&\sqrt{a}x-\varepsilon y=\delta\sqrt{m-4a}t\\ l_{4}(\varepsilon,\delta):&\sqrt{a}x=\varepsilon\sqrt{m}t,&\sqrt{a}y=\frac{1}{2}(\varepsilon\sqrt{m}+\delta\sqrt{m-4a})z\\ l_{5}(\varepsilon,\delta):&y=\varepsilon\sqrt{m}t,&z=\frac{1}{2}(\varepsilon\sqrt{m}+\delta\sqrt{m-4a})x\\ l_{6}(\varepsilon,\delta):&z=\varepsilon\sqrt{m}t,&x=\frac{1}{2a}(\varepsilon\sqrt{m}+\delta\sqrt{m-4a})y\\ \end{array}\right.$$

$$(l_i(1,1),l_j(1,1))=0whenever1\le i\ne j\le 6.$$

$$a{x}^{2}t+y^{2}t+z^{2}t-xyz=m{t}^3$$

$${\mathrm{Br}}_1(X)/{\mathrm{Br}}_0(X)\cong H^1(k,\mathrm{Pic}(\overline{X}))$$

$$(H_j,l_i(1,1))=\{\begin{array}{ll}1& \text{i−j≡0 or 3 mod 6}\\ 0& \text{otherwise}\end{array}$$

$$H_j=l-l_j(1,1)-l_{j+3}(1,1)$$

$$\{\begin{array}{l}\sigma (l_1)=l_1,\sigma (l_2)=l-l_3-l_4\\ \sigma (l_3)=l-l_2-l_4,\sigma (l_4)=l-l_2-l_3\\ \sigma (l_5)=l_5,\sigma (l_6)=l_6\\ \sigma (l)=2l-l_2-l_3-l_4\end{array}$$

$$\{\begin{array}{l}\sigma (l_1)=2l-l_1-l_2-l_3-l_5-l_6\\ \sigma (l_2)=l-l_1-l_6\\ \sigma (l_3)=l-l_1-l_5\\ \sigma (l_4)=l-l_5-l_6\\ \sigma (l_5)=2l-l_1-l_3-l_4-l_5-l_6\\ \sigma (l_6)=2l-l_1-l_2-l_4-l_5-l_6\\ \sigma (l)=4l-2l_1-2l_5-2l_6-l_2-l_3-l_4\end{array}$$

$$\{\begin{array}{l}\sigma (l_1)=l_1\\ \sigma (l_2)=l-l_3-l_4\\ \sigma (l_3)=l-l_2-l_4\\ \sigma (l_4)=2l-l_2-l_3-l_4-l_5-l_6\\ \sigma (l_5)=l-l_4-l_6\\ \sigma (l_6)=l-l_4-l_5\\ \sigma (l)=3l-l_2-l_3-2l_4-l_5-l_6\end{array}$$

$$\{\begin{array}{l}\sigma (l_1)=2l-l_1-l_2-l_3-l_5-l_6\\ \sigma (l_2)=l-l_1-l_6,\sigma (l_3)=l-l_1-l_5\\ \sigma (l_4)=l_4,\sigma (l_5)=l-l_1-l_3\\ \sigma (l_6)=l-l_1-l_2\\ \sigma (l)=3l-2l_1-l_2-l_3-l_5-l_6\end{array}$$

$$a{x}^2+y^2+z^2-xyz=m$$

$$B{r}_1(U)/B{r}_0(U)\cong \mathbb{Z}/2\oplus \mathbb{Z}/2$$

$$({\oplus }_{i=1}^6\mathbb{Z}{l}_i\oplus \mathbb{Z}l)/(l-l_j-l_{j+3}:1\le j\le 3)\cong {\oplus }_{i=1}^4\mathbb{Z}{\overline{l}}_i.$$

$$\{\begin{array}{l}\tau ({\overline{l}}_1)={\overline{l}}_1\\ \tau ({\overline{l}}_2-{\overline{l}}_4)={\overline{l}}_1-{\overline{l}}_3+l_4\\ \tau ({\overline{l}}_3-{\overline{l}}_4)={\overline{l}}_1-{\overline{l}}_2+l_4\end{array}$$

$$Ker(1+\tau )=({\overline{l}}_1-{\overline{l}}_3-{\overline{l}}_2+2{\overline{l}}_4),Im(\tau -1)=({\overline{l}}_1-{\overline{l}}_3-{\overline{l}}_2+2{\overline{l}}_4).$$

$$H^1(G,Pic(\overline{U}))\cong H^1(H,Pic(\overline{U}{)}^{⟨\sigma ⟩})\cong H^1(⟨\theta ⟩,Pic(\overline{U}{)}^{⟨\sigma ,\tau ⟩})$$

$$\{\begin{array}{l}\theta ({\overline{l}}_1)=-{\overline{l}}_1\\ \theta ({\overline{l}}_2-{\overline{l}}_3)={\overline{l}}_3-{\overline{l}}_2\end{array}$$

$$H^1(⟨\theta ⟩,Pic(\overline{U}{)}^{⟨\sigma ,\tau ⟩})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

$$H^1(G,Pic(\overline{U}))\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

$$(2z-xy{)}^2-4(m-4a)=(x^2-4)(y^2-4a)$$

$$\{x\pm 2=0\}\cap \{((x\mp 2)(y^2-4a)=0\}$$

Full text

[TABLE]

using Brauer-Manin obstruction, and we produce examples whose Brauer groups include $4$-torsion elements . We use methods of [5] to describe them and in some cases we show that there is no Brauer-Manin obstruction to the integral Hasse principle for them.

1. Introduction

In recent papers [4] and [10], Colliot-Thélène , Wei, Xu, D. Loughran and V. Mitankin, studied the integral Hasse principle and strong approximation for Markoff surfaces, using the Brauer-Manin obstruction. For Markoff surfaces , D. Loughran and V. Mitankin obtained the following beautiful result :

Assume that $m\in\mathbb{Z}$ is such that affine surface $\mathcal{U}_{m}$ defined by

[TABLE]

has a Brauer-Manin obstruction to the integral Hasse principle. Then

[TABLE]

As they pointed out , this can be seen as an analogue of the finiteness of exceptional spinor classes in the study of the representation of an integer by a ternary quadratic form (see [3 ,§7]).

Now, fix $m,a\in\mathbb{Z},m\neq 0,4a$. Let $U\subset{\mathbb{A}}^{3}_{\mathbb{Z}}$ be the affine scheme over $\mathbb{Z}$ defined by the equation

[TABLE]

We study the Brauer-Manin obstruction to the integral Hasse principle for U. In particular, we have similar results :

Theorem 1.1**.**

Assume that $[Q(\sqrt{a},\sqrt{m},\sqrt{m-4a}):Q]=8$, let $(a,m)=p_{1}^{n_{1}}p_{2}^{n_{2}}\cdots p_{s}^{n_{s}}$, where $(a,m)$ is the greatest common divisor of a and m, $p_{i}$ are prime for $1\leq i\leq s$ . If there is a Brauer-Manin obstruction to the integral Hasse principle for U, we have

[TABLE]

Moreover, we will give examples whose Brauer groups include $4$-torsion elements, and with some assumptions, we can show that there is no Brauer-Manin obstruction to the integral Hasse principle for them.

As noted in [11], an often used strategy for proving that a class $\mathcal{A}\in{\mathrm{Br}}(U)$ of order n gives no obstruction to the Hasse principle is to demonstrate the existence of a finite place v of k such that the evaluation map $X(k_{v})\to({\mathrm{Br}}k_{v})[n]$, sending a point $P\in Y(k_{v})$ to the evaluation $\mathcal{A}(P)\in({\mathrm{Br}}k_{v})[n]$, is surjective .However,the local invariant of non-cyclic algebra is difficult to compute in general. Based on ideas of [5], we will construct explicit representatives of non-cyclic Brauer classes on affine surfaces, and compute its local invariants in special places.

Notation Let $k$ be a field and $\overline{k}$ a separable closure of $k$. If $X$ is a k-variety,we write $\overline{X}=X\times_{k}\overline{k}$. If $X$ is an integral $k$-variety, we let $k(X)$ denote the function field of $X$. If $X$ is a geometrically integral $k$-variety, we let $\overline{k}(X)$ denote the function field of $\overline{X}$. We let $\mathrm{Pic}(X)=H^{1}_{{\operatorname{\acute{e}t}}}(X,{\mathbb{G}}_{m})$ denote the Picard group of a scheme $X$. We let ${\mathrm{Br}}(X)=H^{2}_{{\operatorname{\acute{e}t}}}(X,{\mathbb{G}}_{m})$ denote the Brauer group of a scheme $X$. If $X$ is a regular integral $k$-variety, the natural map

[TABLE]

is injective. We let

[TABLE]

denote the algebraic Brauer group of a k-variety.

2. Algebraic Brauer group of cubic surface

Follow J.-L.Colliot-Thélène,Dasheng Wei,and Fei Xu, we have

Lemma 2.1**.**

Let $X\subset\mathbb{P}_{k}^{3}$ defined by equation

[TABLE]

over a field k of characteristic zero with $a\in k^{\times},a\notin k^{2},$ then X is smooth if and only if $m\neq 0,4a$.In this case,the 27 lines in X are defined over k($\sqrt{a},\sqrt{m},\sqrt{m-4a}$) by the following equations:

[TABLE]

and

[TABLE]

*with $\varepsilon=\pm 1$, and $\delta=\pm 1$.Moreover the intersection number

[TABLE]

Proof.

The results follow from straight forward computation. ∎

.

Proposition 2.2**.**

Let $X\subset\mathbb{P}_{k}^{3}$ defined by equation

[TABLE]

over a field k of characteristic zero with $a\in k^{\times},a\notin k^{2}$,and $m\neq 0,4a$.Then ${\mathrm{Br}}(X)/{\mathrm{Br}}_{0}(X)=0$ or ${\mathrm{Br}}(X)/{\mathrm{Br}}_{0}(X)\cong\mathbb{Z}/2$ with generator $((\frac{x}{t})^{2}-4,m-4a)$.

Proof.

The proof is completely similar to [4,Proposition3.2] , for later application,we only give computations of ${\mathrm{Br}}(X)/{\mathrm{Br}}_{0}(X)$ in somes cases. Since X is geometrically rational,we have ${\mathrm{Br}}(X)={\mathrm{Br}}_{1}(X)$. Since$X(k)\neq\emptyset$, we have the following isomorphism

[TABLE]

by the Hochschild-Serre spectral sequence. By [7,Chapter V,Proposition4.8],there is $l\in\mathrm{Pic}(\overline{X})$ satisfying the following intersecton property

$(l,l)=1\quad(l,l_{i}(1,1))=0$ for $1\leq i\leq 6$.

such that $\{H_{i}(1,1):\ 1\leq i\leq 6\}\cup\{l\}$ forms a basis of $\mathrm{Pic}(\overline{X})$.

Since

[TABLE]

where $1\leq j\leq 3,1\leq i\leq 6$. One concludes that

[TABLE]

in $\mathrm{Pic}(\overline{X})$ for $1\leq j\leq 3$ by [7,Chapter V,Proposition 4.8(e)]. For simplicity, we write $l_{i}$ for $l_{i}(1,1)$ with $1\leq i\leq 6$ . If $[k(\sqrt{a},\sqrt{m},\sqrt{m-4a}):k]=2$, there exists $\sigma\in Gal(k(\sqrt{a})/k)$ such that $\sigma(\sqrt{a})=-\sqrt{a}$ .

1.$\sqrt{m}\in k,\sqrt{m-4a}\in k$, we have

[TABLE]

Since $Ker(1+\sigma)=(l-l_{2}-L_{3}-l_{4})$,   $Im(\sigma-1)=(l-l_{2}-L_{3}-l_{4})$, we have $H^{1}(k,Pic(\overline{X}))=0$ .

2. $\sqrt{m}\in k,\sqrt{m-4a}\notin k$, we have

[TABLE]

Since

$Ker(1+\sigma)=(l-l_{1}-l_{6}-l_{2},l-l_{1}-l_{5}-l_{3},l-l_{4}-l_{5}-l_{6}),$

$Im(\sigma-1)=(l-l_{1}-l_{6}-l_{2},l-l_{1}-l_{5}-l_{3},l-l_{4}-l_{5}-l_{6}),$

we have $H^{1}(k,Pic(\overline{X}))=0$ .

3. $\sqrt{m}\notin k,\sqrt{m-4a}\in k$, we have

[TABLE]

Since

$Ker(1+\sigma)=(l-l_{2}-l_{3}-l_{4},l-l_{4}-l_{5}-l_{6}),$

$Im(\sigma-1)=(l-l_{2}-l_{3}-l_{4},l-l_{4}-l_{5}-l_{6}),$

we have $H^{1}(k,Pic(\overline{X}))=0$.

4.$\sqrt{m}\notin k,\sqrt{m-4a}\notin k$ , we have

[TABLE]

Since

$Ker(1+\sigma)=(l-l_{1}-l_{2}-l_{6},l-l_{1}-l_{3}-l_{5}),$

$Im(\sigma-1)=(l-l_{1}-l_{2}-l_{6},l-l_{1}-l_{3}-l_{5}),$

we have $H^{1}(k,Pic(\overline{X}))=0$. ∎

Proposition 2.3**.**

Let U be the affine variety over a field of characteristic zero defined by the equation

[TABLE]

*where $a\in k^{\times},a\notin k^{2}$, $m\neq 0,4a$. If $[k(\sqrt{a},\sqrt{m},\sqrt{m-4a}):k]=8$, we have

[TABLE]

with generators $(x-2,m-4a)$,$(x+2,m-4a)$

Proof.

Let $G=Gal(k(\sqrt{a},\sqrt{m},\sqrt{m-4a})/k)$, there exist $\sigma$,$\tau$ and $\theta$ $\in$ G,such that

$\sigma(\sqrt{a})=-\sqrt{a},\quad\sigma(\sqrt{m})=\sqrt{m},\ \ \sigma(\sqrt{m-4a})=\sqrt{m-4a}.$

$\tau(\sqrt{a})=-\sqrt{a},\quad\tau(\sqrt{m})=-\sqrt{m},\ \ \tau(\sqrt{m-4a})=\sqrt{m-4a}.$

$\theta(\sqrt{a})=-\sqrt{a},\quad\theta(\sqrt{m})=\sqrt{m},\ \ \theta(\sqrt{m-4a})=-\sqrt{m-4a}.$

By [4,Proposition 2.2], $Pic(\overline{U})$ is given by the following quotient group

[TABLE]

By computations in proposition 2.2, we have

1. (i)

$\begin{cases}\sigma(\overline{l}_{1})=\overline{l}_{1}\\ \sigma(\overline{l}_{2})=\overline{l}_{1}-\overline{l}_{3}\\ \sigma(\overline{l}_{3})=\overline{l}_{1}-\overline{l}_{2}\\ \sigma(\overline{l}_{4})=\overline{l}_{1}+\overline{l}_{4}-\overline{l}_{2}-\overline{l}_{3}\\ \end{cases}$

Since $Ker(1+\sigma)=(\overline{l}_{1}-\overline{l}_{3}-\overline{l}_{2})$ , $Im(\sigma-1)=(\overline{l}_{1}-\overline{l}_{3}-\overline{l}_{2})$,

we have $H^{1}(\langle\sigma\rangle,Pic(\overline{U}))=0$. 2. (ii)

$\begin{cases}\theta(\overline{l}_{1})=-\overline{l_{1}}\\ \theta(\overline{l}_{2})=\overline{l}_{3}-\overline{l}_{1}\\ \theta(\overline{l}_{3})=\overline{l}_{2}-\overline{l}_{1}\\ \theta(\overline{l}_{4})=\overline{l}_{2}+\overline{l}_{3}-\overline{l}_{1}-\overline{l}_{4}\\ \end{cases}$ 3. (iii)

$\begin{cases}\tau(\overline{l}_{1})=\overline{l}_{1}\\ \tau(\overline{l}_{2})=\overline{l}_{1}-\overline{l}_{3}\\ \tau(\overline{l}_{3})=\overline{l}_{1}-\overline{l}_{2}\\ \tau(\overline{l}_{4})=-\overline{l}_{4}\\ \end{cases}$

Let $H={\langle\tau,\theta\rangle}$, we have the following exact sequence

$0\to H^{1}(H,Pic(\overline{U})^{\langle\sigma\rangle})\to H^{1}(G,Pic(\overline{U}))\to H^{1}(\langle\sigma\rangle,Pic(\overline{U}))=0$.

where $Pic(\overline{U})^{\langle\sigma\rangle}=(\overline{l}_{1},\overline{l}_{2}-\overline{l}_{4},\overline{l}_{3}-\overline{l}_{4})$. Let us compute $H^{1}(H,Pic(\overline{U})^{\langle\sigma\rangle})$, we have the following exact sequence

$0\to H^{1}(\langle\theta\rangle,Pic(\overline{U})^{\langle\sigma,\tau\rangle})\to H^{1}(H,Pic(\overline{U})^{\langle\sigma\rangle})\to H^{1}(\langle\tau\rangle,Pic(\overline{U})^{\langle\sigma\rangle})$.

Since

[TABLE]

we have

[TABLE]

One concludes that $H^{1}(\langle\tau\rangle,Pic(\overline{U})^{\langle\sigma\rangle})=0$, hence

[TABLE]

where $Pic(\overline{U})^{\langle\sigma,\tau\rangle}=(\overline{l}_{1},\overline{l}_{2}-\overline{l}_{3})$. Since

[TABLE]

one has

[TABLE]

We obtain

[TABLE]

Note that $ax^{2}+y^{2}+z^{2}-xyz=m$ is equivalent to

[TABLE]

Arguing in the same way as in the proof of [4,Theorem 3.4], one obtains the generators $(x-2,m-4a)$ and $(x+2,m-4a)$. Indeed since

[TABLE]

is a closed subset of codimension $\geq 2$ on $U$, one obtains that $(x\pm 2,m-4a)\in{\mathrm{Br}}_{1}(U)$. This implies that

[TABLE]

Now we show that $B$ is not constant.

[TABLE]

The generic fibre $U_{\eta}\xrightarrow{\pi_{\eta}}\eta$ induces

[TABLE]

by [6,Theorem 5.4.1]. Since $[k(\sqrt{a},\sqrt{m},\sqrt{m-4a}):k]=8$, the residue of $(x^{2}-4,m-4a)$ at $(m-ax^{2})$ is different from that of $(x^{2}-4,m-ax^{2})$. This implies that $\pi_{\eta}^{*}(x^{2}-4,m-4a)$ is not constant by the Faddeev exact sequence. Since $\pi_{\eta}^{*}(x^{2}-4,m-4a)$ is the pull-back of $B$ by the projection map $U_{\eta}\rightarrow U$, one concludes that $B$ is not constant. ∎

3. Examples of Brauer-Manin obstruction

We now give examples of Brauer-Manin obstruction to the integral Hasse principle. Here the results are inspired by the results in [10,§5.3,§5.4]

Lemma 3.1**.**

If $p$ is an odd prime with $(p,m-4a)=1$, then the following elements

[TABLE]

vanish over $U(\mathbb{Z}_{p})$. If $m-4a>0$,these elements vanish over $U(\mathbb{R})$. In particular, if $a<0$, $(x^{2}-4,m-4a)_{\infty}=(z^{2}-4a,m-4a)_{\infty}=(y^{2}-4a,m-4a)_{\infty}=0$.

Proof.

Arguing in the same way as in the proof of [4,Lemma 5.1 ], one can easily verify this. ∎

Lemma 3.2**.**

Let $p\mid(m-4a)$ be odd , if $p\nmid a$, any singular point T$(x,y,z)\in U(\mathbb{F}_{p})$ satisfies

[TABLE]

Proof.

Since $T\in U(\mathbb{F}_{p})$ is singular, we have

[TABLE]

We obtain $x^{2}=4,\quad y^{2}=4a$. ∎

Lemma 3.3**.**

*Let $p\geq 3$ such that $p\mid(m-4a)$ and $p\nmid a$, with $ord_{p}(m-4a)$ even but $m-4a\notin{Q_{p}^{\times}}^{2}$. Let $\mathcal{B}_{1}=(x^{2}-4,m-4a),\ \mathcal{B}_{2}=(x+2,m-4a).$  For all $T\in U(\mathbb{Z}_{p})$, we have

If $(\frac{a}{p})=-1$,*

[TABLE]

If $(\frac{a}{p})=1$,

[TABLE]

Proof.

Note that $ax^{2}+y^{2}+z^{2}-xyz=m$ is equivalent to

[TABLE]

As $(m-4a)\notin{Q_{p}^{\times}}^{2}$ and $ord_{p}(m-4a)$ is even, it follows that $ord_{p}((x^{2}-4)(y^{2}-4a))$ is even.

If $ord_{p}(x^{2}-4)$ is even ,one obtains $ord_{p}(x+2)$ is even, hence $\{inv_{p}\mathcal{B}_{1}(T),inv_{p}\mathcal{B}_{2}(T)\}=\{\{0,0\}\}$. Assume that $ord_{p}(x^{2}-4)$ is odd, then $ord_{p}(y^{2}-4a)$ is odd, thus $y^{2}\equiv 4a\bmod p$. If $(\frac{a}{p})=-1$, this is a contradiction. Now let $(\frac{a}{p})=1$, if $ord_{p}(x+2)$ is odd, we obtain

[TABLE]

If $ord_{p}(x-2)$ is odd, we obtain

[TABLE]

We now show that these possibilities can be realised. We first consider $(\frac{a}{p})=-1$, we have smooth point $(2,0,0)\in U(\mathbb{F}_{p})$ by lemma 3.2, hence $U(\mathbb{Z}_{p})\neq\emptyset$. Now we consider $(\frac{a}{p})=1$, we have smooth point $(0,2\sqrt{a},0)\in U(\mathbb{F}_{p})$ by lemma 3.2, hence there exists $T\in U(\mathbb{Z}_{p})$ such that

[TABLE]

Suppose $ord_{p}(m-4a)=2$, let $s\in\mathbb{F}_{p}^{\times}$ such that $(\frac{\frac{m-4a}{p^{2}}-s}{p})=1$. Let $s^{\prime}\in\mathbb{Z}_{p}$ such that $s^{\prime}\equiv s\bmod p$ . There is $y_{0}\in\mathbb{Z}_{p}$ such that ${y_{0}}^{2}-4a=ps^{\prime}$ by Hensel lemma. Let $x_{0}=p-2$, we consider the following equation

[TABLE]

over $\mathbb{Z}_{p}$. That is

[TABLE]

By Hensel lemma, one can see that the equation has solutions. Let $t_{0}$ denote one of the solutions and let $z_{0}\in\mathbb{Z}_{p}$ such that $2z_{0}-x_{0}y_{0}=pt_{0}$, then $T_{0}=(x_{0},y_{0},z_{0})\in U(\mathbb{Z}_{p})$, we have

[TABLE]

If we let $x_{1}=p+2$, one can see that there exists $T_{1}=(x_{1},y_{1},z_{1})\in U(\mathbb{Z}_{p})$, we have

[TABLE]

For $ord_{p}(m-4a)>2$, the proof is similar .

∎

Proposition 3.4**.**

Suppose conditions of Proposition 2.3 are satisfied, if there exists $p\geq 5$ such that $p\nmid a$ and $ord_{p}(m-4a)$ is odd, there is no Brauer–Manin obstruction to integral Hasse principle.

Proof.

We can assume that $U(A_{\mathbb{Z}})\neq\emptyset$, where $A_{\mathbb{Z}}={\mathrm{R}}\times\prod_{p}\mathbb{Z}_{p}$ ,otherwise there is nothing to prove. Let

[TABLE]

to prove the proposition , it suffices to show for all $(\varepsilon_{1},\varepsilon_{2})\in(\mathbb{Z}/2\mathbb{Z})^{2}$, there exists $\zeta\in U(\mathbb{Z}_{p})$ such that

[TABLE]

We first consider the case $(\frac{a}{p})=1$.

Since $p\geq 5$,there exist $t,s\in\mathbb{F}_{p}$, such that the Legendre symbol $(\frac{t^{2}-4a}{p})=1$, $(\frac{s^{2}-4a}{p})=-1$, let

[TABLE]

One can see that $\upsilon_{1}$ and $\upsilon_{2}$ are smooth points of $U(\mathbb{F}_{p})$ by lemma 3.2, hence there exist $\mu_{1},\mu_{2}\in U(\mathbb{Z}_{p})$ such that $\mu_{i}\equiv\upsilon_{i}\bmod p$ for $1\leq i\leq 2$, we obtain

[TABLE]

Since $p\geq 5$ ,we can choose $e,f\in\mathbb{F}_{p}$ such that the Legendre symbol $(\frac{e}{p})=-1$,$(\frac{e^{2}-4e}{p})=1$, $(\frac{f}{p})=-1$, $(\frac{f^{2}-4f}{p})=-1$. Let

[TABLE]

One can see that $\upsilon_{3}$ and $\upsilon_{4}$ are smooth points of $U(\mathbb{F}_{p})$ by lemma 3.2, hence there exist $\mu_{3},\mu_{4}\in U(\mathbb{Z}_{p})$ such that $\mu_{i}\equiv\upsilon_{i}\bmod p$ for $3\leq i\leq 4$, we obtain

[TABLE]

Now assuing $(\frac{a}{p})=-1$. Let $\eta_{1}=\mu_{1},\eta_{2}=\mu_{2}$, we have

[TABLE]

Let

[TABLE]

where $\alpha^{2}=ae,\beta^{2}=af$,one can see that $\xi_{3}$ and $\xi_{4}$ are smooth points of $U(\mathbb{F}_{p})$ by lemma 3.2, hence there exist $\eta_{3},\eta_{4}\in U(\mathbb{Z}_{p})$ such that $\eta_{i}\equiv\xi_{i}\bmod p$ for $3\leq i\leq 4$, we obtain

[TABLE]

The proposition is established. ∎

As corollary, we obtain Theorem 1.1 .

Lemma 3.5**.**

Given $a,m\in\mathbb{Z}$, the equation $ax^{2}+y^{2}+z^{2}-xyz=m$ has solutions in $(\mathbb{Z}_{p})^{3}$ for all primes p except for the following two cases:

1. (i)

$a\equiv 1\ \mathrm{mod}\ 4,m\equiv 3\ \mathrm{mod}\ 4$. 2. (ii)

$a\equiv 1\ \mathrm{mod}\ 3,m\equiv\pm 3\ \mathrm{mod}\ 9$.

Proof.

We break up the proof into several cases. Let

[TABLE]

For $p>3$,we have

1. (i)

$p\nmid a,p\mid m$. If $p\equiv 1\ \mathrm{mod}\ 4$, there exists $(0,t,s)\in(\mathbb{F}_{p})^{3}$, such that $f(0,t,s)\equiv 0\ \mathrm{mod}\ p,f_{y}(0,t,s)\not\equiv 0\ \mathrm{mod}\ p$, hence f has a zero in $(\mathbb{Z}_{p})^{3}$.

If $p\equiv 3\ \mathrm{mod}\ 4$,we first consider$(\frac{a}{p})=-1$,there exists $(1,0,t)\in(\mathbb{F}_{p})^{3}$,such that $f(1,0,t)\equiv 0\ \mathrm{mod}\ p,f_{z}(1,0,t)\not\equiv 0\ \mathrm{mod}\ p$, hence f has a zero in $(\mathbb{Z}_{p})^{3}$. Now suppose $(\frac{a}{p})=1$, there exists $(3,t,2t)\in(\mathbb{F}_{p})^{3}$,such that $f(3,t,2t)\equiv 0\ \mathrm{mod}\ p,f_{y}(3,t,2t)\not\equiv 0\ \mathrm{mod}\ p$, hence f has a zero in $(\mathbb{Z}_{p})^{3}$. 2. (ii)

$p\nmid a,p\nmid m$. If $(\frac{m}{p})=-1$, since $p>3$, there exists $(0,t,s)\in(\mathbb{F}_{p})^{3}$,such that $f(0,t,s)\equiv 0\ \mathrm{mod}\ p,f_{y}(0,t,s)\not\equiv 0\ \mathrm{mod}\ p$, hence f has a zero in $(\mathbb{Z}_{p})^{3}$.

If $(\frac{m}{p})=1$,there exists $(0,t,0)\in(\mathbb{F}_{p})^{3}$,such that $f(0,t,0)\equiv 0\ \mathrm{mod}\ p,f_{y}(0,t,0)\not\equiv 0\ \mathrm{mod}\ p$, hence f has a zero in $(\mathbb{Z}_{p})^{3}$. 3. (iii)

$p\mid a,p\mid m$. There exists $(2,1,1)\in(\mathbb{F}_{p})^{3}$, such that $f(2,1,1)\equiv 0\ \mathrm{mod}\ p,f_{x}(2,1,1)\not\equiv 0\ \mathrm{mod}\ p$, hence f has a zero in $(\mathbb{Z}_{p})^{3}$. 4. (iv)

$p\mid a,p\nmid m$. An argument similar to (ii), one can prove f has a zero in $(\mathbb{Z}_{p})^{3}$.

For $p=3$, we have

1. (i)

$3\nmid a,3\mid m$. If $a\equiv 1\ \mathrm{mod}\ 3$, f has only zero $(0,0,0)\in(\mathbb{F}_{3})^{3}$. So f has no zeros in $(\mathbb{Z}_{3})^{3}$ when $m\equiv\pm 3\ \mathrm{mod}\ 9$. Now assume $9\mid m$,one can easily check the equation $ax^{2}+y^{2}+z^{2}-3xyz=\frac{m}{9}$ has a solution $(x_{0},y_{0},z_{0})\in(\mathbb{Z}_{3})^{3}$ using Hensel lemma. Hence $f(3x_{0},3y_{0},3z_{0})=0$.

If $a\equiv 2\ \mathrm{mod}\ 3$, since $f(1,1,0)\equiv 0\ \mathrm{mod}\ 3,f_{y}(1,1,0)\not\equiv 0\ \mathrm{mod}\ 3$, f has a zero in $(\mathbb{Z}_{3})^{3}$. 2. (ii)

$3\nmid a,3\nmid m$. If $m\equiv 1\ \mathrm{mod}\ 3$, since $f(0,1,0)\equiv 0\ \mathrm{mod}\ 3,f_{y}(0,1,0)\not\equiv 0\ \mathrm{mod}\ 3$, f has a zero in $(\mathbb{Z}_{3})^{3}$.

If $m\equiv 2\ \mathrm{mod}\ 3$, since $f(0,1,1)\equiv 0\ \mathrm{mod}\ 3,f_{y}(0,1,1)\not\equiv 0\ \mathrm{mod}\ 3$,f has a zero in $(\mathbb{Z}_{3})^{3}$. 3. (iii)

$3\mid a,3\mid m$. Since $f(2,1,1)\equiv 0\ \mathrm{mod}\ 3,f_{x}(2,1,1)\not\equiv 0\ \mathrm{mod}\ 3$,f has a zero in $(\mathbb{Z}_{3})^{3}$. 4. (iv)

$3\mid a,3\nmid m$. If $m\equiv 1\ \mathrm{mod}\ 3$, since $f(0,1,0)\equiv 0\ \mathrm{mod}\ 3,f_{y}(0,1,0)\not\equiv 0\ \mathrm{mod}\ 3$, f has a zero in $(\mathbb{Z}_{3})^{3}$.

If $m\equiv 2\ \mathrm{mod}\ 3$, since $f(0,1,1)\equiv 0\ \mathrm{mod}\ 3,f_{y}(0,1,1)\not\equiv 0\ \mathrm{mod}\ 3$, f has a zero in $(\mathbb{Z}_{3})^{3}$.

For $p=2$, we have

1. (i)

$2\nmid a,2\mid m$. Since $f(1,1,1)\equiv 0\ \mathrm{mod}\ 2,f_{y}(1,1,1)\not\equiv 0\ \mathrm{mod}\ 2$, f has a zero in $(\mathbb{Z}_{2})^{3}$. 2. (ii)

$2\nmid a,2\nmid m$. If $a\equiv m\ \mathrm{mod}\ 8$, since $f(1,0,0)\equiv 0\ \mathrm{mod}\ 8,ord_{2}(f_{x}(1,0,0))=1$, f has a zero in $(\mathbb{Z}_{2})^{3}$. If $a\equiv m-4\ \mathrm{mod}\ 8$, since $f(1,2,0)\equiv 0\ \mathrm{mod}\ 8,ord_{2}(f_{x}(1,2,0))=1$, f has a zero in $(\mathbb{Z}_{2})^{3}$.

If $a\equiv 3\ \mathrm{mod}\ 4,m\equiv 1\ \mathrm{mod}\ 4$, we first consider $m\equiv 1\ \mathrm{mod}\ 8$, since $f(0,1,0)\equiv 0\ \mathrm{mod}\ 8,ord_{2}(f_{y}(0,1,0))=1$, f has a zero in $(\mathbb{Z}_{2})^{3}$. Now suppose $m\equiv 5\ \mathrm{mod}\ 8$, since $f(0,1,2)\equiv 0\ \mathrm{mod}\ 8,ord_{2}(f_{y}(0,1,2))=1$, f has a zero in $(\mathbb{Z}_{2})^{3}$.

If $a\equiv 1\ \mathrm{mod}\ 4,m\equiv 3\ \mathrm{mod}\ 4$, note that all solutions of $f\equiv 0\ \mathrm{mod}\ 2$ are $(1,0,0),(0,1,0)$ and $(0,0,1)$. If we take for $(1,0,0)$, this implies $a\equiv m\ \mathrm{mod}\ 4$,a contradiction. If we take for $(0,1,0)$ or $(0,0,1)$, this implies $m\equiv 1\ \mathrm{mod}\ 4$, a contradiction. So f has no zero in $(\mathbb{Z}_{2})^{3}$ in this case. 3. (iii)

$2\mid a,2\mid m$. Since $f(0,1,1)\equiv 0\ \mathrm{mod}\ 2,f_{x}(1,1,1)\not\equiv 0\ \mathrm{mod}\ 2$, f has a zero in $(\mathbb{Z}_{2})^{3}$. 4. (iv)

$2\mid a,2\nmid m$. Since $f(1,1,1)\equiv 0\ \mathrm{mod}\ 2,f_{x}(1,1,1)\not\equiv 0\ \mathrm{mod}\ 2$, f has a zero in $(\mathbb{Z}_{2})^{3}$.

∎

Proposition 3.6**.**

Let $U$ be the scheme over $\mathbb{Z}$ given by

[TABLE]

where $a,d$ are odd integers such that $(a,d)=1$, $3\nmid(a-1)$, $\sqrt{a}\notin Q$, $p\equiv\pm 1\ \mathrm{mod}\ 8$ or $(\frac{a}{p})=-1$ for $p\mid d$. Then there is a Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

By lemma 3.5, we have $U(A_{\mathbb{Z}})\neq\emptyset$ . Let

[TABLE]

we will show that for each point $T\in U(\mathbb{Z}_{p})$, we have

[TABLE]

If $p\nmid 2d^{2}$ the claim follows from Lemma 3.1 . If $p\mid d$ and $p\equiv\pm 1\bmod 8$ ,we have $2\in{Q_{p}^{\times}}^{2}$. Thus $2d^{2}\in{Q_{p}^{\times}}^{2}$.If $p\mid d$ and $(\frac{a}{p})=-1$, the claim follows from lemma 3.3. Finally, since $m-4a>0$, the claim is trivial for $p=\infty$ . It remains to examine $p=2$.

Assme now $p=2$. Let $T\in U(\mathbb{Z}_{2})$, one easily see that there is at least one coordinate of T belonging to ${\mathbb{Z}_{2}^{\times}}$. A simple Hilbert symbol calculation implies the claim for $p=2$.

∎

Proposition 3.7**.**

Let $U$ be the scheme over $\mathbb{Z}$ given by

[TABLE]

where a is an even integer such that $a\equiv 1\ \mathrm{mod}\ 3$ and $\sqrt{a}\notin Q$, $p\equiv\pm 1\ \mathrm{mod}\ 12$ or $(\frac{a}{p})=-1$ for $p\mid d$. When $\sqrt{4a+3d^{2}}\notin Q$, there is a Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

One can see $U(A_{\mathbb{Z}})\neq\emptyset$ by lemma 3.5.

Let $\mathcal{B}=(x^{2}-4,3)=(z^{2}-4a,3)=(y^{2}-4a,3)$, we will show that for each point $T\in U(\mathbb{Z}_{p})$, we have

[TABLE]

so that $\mathcal{B}$ gives an obstruction to the Hasse principle.

If $p\nmid 6d^{2}$ the claim follows from Lemma 3.1 . If $p\mid d$ and $p\equiv\pm 1\bmod 12$ ,we have $3\in{Q_{p}^{\times}}^{2}$. Thus $3d^{2}\in{Q_{p}^{\times}}^{2}$.If $p\mid d$ and $(\frac{a}{p})=-1$, the claim follows from lemma 3.3. Finally, since $m-4a>0$, the claim is trivial for $p=\infty$ . It remains to examine $p=2,3$.

Assme now $p=2$. Let $T\in U(\mathbb{Z}_{2})$, one easily see that there is at least one coordinate of T belonging to ${\mathbb{Z}_{2}^{\times}}$. A simple Hilbert symbol calculation implies the claim for $p=2$.

For $p=3$, note that $ax^{2}+y^{2}+z^{2}-xyz=4a+3d^{2}$ is equivalent to the following equations

[TABLE]

Then for any $P\in U(\mathbb{Z}_{3})$, there are two coordinates of $P$ belonging to $3\mathbb{Z}_{3}$. We can assume $x,y\in 3\mathbb{Z}_{3}$, since $(x^{2}-4,3)_{3}=(y^{2}-4a,3)_{3}=\frac{1}{2}$, one concludes that ${\mathrm{inv}}_{3}\mathcal{B}(P)=\frac{1}{2}$. The proposition is established. ∎

Proposition 3.8**.**

Let U be the scheme over $\mathbb{Z}$ given by

[TABLE]

where $4\mid a,a\equiv 1\bmod 3$ and $\sqrt{a}\notin Q$, $d\in\mathbb{Z}$ whose prime divisors are congruent to $\pm 1\ \mathrm{mod}\ 12$ and $\pm 1\ \mathrm{mod}\ 8$ or $\pm 5\ \mathrm{mod}\ 12$ and $\pm 3\ \mathrm{mod}\ 8$. Then there is a Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

Note that if $U(\mathbb{Z}_{2})\neq\emptyset$, since $4\mid a$,for any local solution $T(x,y,z)\in U(\mathbb{Z}_{2})$, y or z is in ${\mathbb{Z}_{2}^{\times}}$. We assume z is in ${\mathbb{Z}_{2}^{\times}}$,hence $z^{2}-4a\equiv 1\bmod 8$. Thus $z^{2}-4a\in{Q_{2}^{\times}}^{2}$. Let $\mathcal{B}=(x^{2}-4,6)=(z^{2}-4a,6)=(y^{2}-4a,6)$, we obtain ${\mathrm{inv}}_{2}\mathcal{B}(T)=0$.

By lemma 3.5, we have $U(A_{\mathbb{Z}})\neq\emptyset$. A similar argument in the proof of Proposition 3.6, we obtain

[TABLE]

so that $\mathcal{B}$ gives an obstruction to the Hasse principle.

∎

Proposition 3.9**.**

Let U be the scheme over $\mathbb{Z}$ given by

[TABLE]

where $a,d$ are odd integers such that $(a,d)=1$, $ord_{5}(a)\geq 2$ ,and $\sqrt{a}\notin Q$, the prime divisors of $d$ are congruent to $\pm 1\ \mathrm{mod}\ 8$ and $\pm 1\ \mathrm{mod}\ 5$ or $\pm 3\ \mathrm{mod}\ 8$ and $\pm 2\ \mathrm{mod}\ 5$. Then there is a Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

By lemma 3.5, one can prove $U(A_{\mathbb{Z}})\neq\emptyset$. Let $\mathcal{B}=(x^{2}-4,10)=(z^{2}-4a,10)=(y^{2}-4a,10)$, we will show that for each point $T\in U(\mathbb{Z}_{p})$, we have

[TABLE]

so that $\mathcal{B}$ gives an obstruction to the Hasse principle.

If $p\nmid 10d^{2}$ the claim follows from Lemma 3.1 . If $p\mid d$, then $10\in{Q_{p}^{\times}}^{2}$. Thus $10d^{2}\in{Q_{p}^{\times}}^{2}$. Finally, since $m-4a>0$, the claim is trivial for $p=\infty$ . It remains to examine $p=2,5$.

Assme now $p=5$. Since $25\mid a$,for any local solution $T(x,y,z)\in U(\mathbb{Z}_{5})$, y or z is in ${\mathbb{Z}_{5}^{\times}}$. We assume $z$ is in $\mathbb{Z}_{5}^{\times}$,hence $z^{2}-4a\in Q_{5}^{{\times}^{2}}$. we obtain ${\mathrm{inv}}_{5}$ $\mathcal{B}(T)=0$.

For $p=2$,for any local solution $T(x,y,z)\in U(\mathbb{Z}_{2})$, there is at least one coordinate of T belonging to ${\mathbb{Z}_{2}^{\times}}$. We assume z is in ${\mathbb{Z}_{2}^{\times}}$,hence $z^{2}-4a\equiv 5\bmod 8$. we obtain ${\mathrm{inv}}_{2}\mathcal{B}(T)=1/2$.

so that $\mathcal{B}$ gives a obstruction to the Hasse principle.

∎

Remark 3.10**.**

We can take $m=4a+2qd^{2}$, where $q$ is an odd prime , one easily obtains similar conclusions.

Proposition 3.11**.**

Let U be the scheme over $\mathbb{Z}$ given by

[TABLE]

where $q$ is an odd prime, t is an odd integer such that $3\nmid(t-1)$ ,$\sqrt{t}\notin Q$, $(t,d)=1$ and the prime divisors of d are congruent to $\pm 1\ \mathrm{mod}\ 8$. Then there is a Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

By lemma 3.5, one can prove $U(A_{\mathbb{Z}})\neq\emptyset$.Let $\mathcal{B}=(x^{2}-4,2)=(z^{2}-4a,2)=(y^{2}-4a,2)$, we will show that for each point $T\in U(\mathbb{Z}_{p})$, we have

[TABLE]

so that $\mathcal{B}$ gives an obstruction to the Hasse principle.

If $p\nmid 2q^{2}d^{2}$ the claim follows from lemma 3.1 . If $p\mid d$, then $2\in{Q_{p}^{\times}}^{2}$. Thus $2q^{2}d^{2}\in{Q_{p}^{\times}}^{2}$. Finally, since $m-4a>0$, the claim is trivial for $p=\infty$ . Note that for any local solution $T(x,y,z)\in U(\mathbb{Z}_{2})$, there is at least one coordinate of T belonging to ${\mathbb{Z}_{2}^{\times}}$. we obtain ${\mathrm{inv}}_{2}\mathcal{B}(T)=1/2$. It remains to examine $p=q$.

If y or z is in ${\mathbb{Z}_{q}^{\times}}$, we have $y^{2}-4a\in{Q_{q}^{\times}}^{2}$ or $z^{2}-4a\in{Q_{q}^{\times}}^{2}$. If not, for any point $M(x,y,z)\in U(\mathbb{Z}_{q})$,let $y=qy^{\prime},z=qz^{\prime}$, then $(x,y^{\prime},z^{\prime})$ is the solution of the following equation

[TABLE]

We obtain ${\mathrm{inv}}_{q}\mathcal{B}(M)=0$ by lemma 3.1 . ∎

Proposition 3.12**.**

Let U be the scheme over $\mathbb{Z}$ given by

[TABLE]

where $q$ is an odd prime such that $q\equiv\pm 3\bmod 8$ , then there is a Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

One can easily check $U(A_{\mathbb{Z}})\neq\emptyset$ by lemma 3.5. Let $\mathcal{B}=(x^{2}-4,2q)=(z^{2}+4q,2q)=(y^{2}+4q,2q)$, we will show that for each point $T\in U(\mathbb{Z}_{p})$, we have

[TABLE]

We only need to consider $p=2,q,\infty.$ Note that for any local solution $T(x,y,z)\in U(\mathbb{Z}_{2})$, there is at least one coordinate of T belonging to ${\mathbb{Z}_{2}^{\times}}$. we obtain ${\mathrm{inv}}_{2}\mathcal{B}(T)=1/2$. Since $y^{2}+4q>0$, the claim is trivial for $p=\infty$ . It remains to examine $p=q$.

If y or z is in ${\mathbb{Z}_{q}^{\times}}$, we have $y^{2}+4q\in{Q_{q}^{\times}}^{2}$ or $z^{2}+4q\in{Q_{q}^{\times}}^{2}$.If not, for any point $M(x,y,z)\in U(\mathbb{Z}_{q})$,let $y=qy^{\prime},z=qz^{\prime}$, then $(x,y^{\prime},z^{\prime})$ is the solution of the following equation

[TABLE]

Thus $x^{2}\equiv 2\bmod q$, a contradiction. We obtain ${\mathrm{inv}}_{q}\mathcal{B}(M)=0$. ∎

4. Review of bicyclic group cohomolgy

Let $G=\mathbb{Z}/n\oplus\mathbb{Z}/m$, with generators $t$ and $s$.Put $N_{t}:=1+t+\cdots+t^{n-1}$ and $\Delta_{t}:=1-t$ in $\mathbb{Z}[G]$, similar put $N_{s}:=1+s+\cdots+s^{m-1}$ and $\Delta_{s}:=1-s$ in $\mathbb{Z}[G]$. For trivial G-module $\mathbb{Z}$ ,we have the following resolution

[TABLE]

where

[TABLE]

If we are given a G-module M,then applying $Hom_{G}(-,M)$ to the above complex,the groups $H^{i}(G,M)$ are homology groups of the following complex:

[TABLE]

where

[TABLE]

We introduce the notations: $Z^{1}(G,M):=ker(d^{1})$,and $Z^{2}(G,M):=ker(d^{2})$,then we have

[TABLE]

For subgroup $\langle t\rangle$, we have the following resolution

[TABLE]

The injection from $\mathbb{Z}[t]$ to the first factor $\mathbb{Z}[G]$ of $\mathbb{Z}[G]^{i+1}$ induces the restriction

$H^{i}(G,M)\to H^{i}(\langle t\rangle,M)$

$(a_{0},...a_{i})\to a_{0}$

Similar for subgroup $\langle s\rangle$, the injection from $\mathbb{Z}[s]$ to the last factor $\mathbb{Z}[G]$ of $\mathbb{Z}[G]^{i+1}$ induces the restriction

$H^{i}(G,M)\to H^{i}(\langle s\rangle,M)$

$(a_{0},...a_{i})\to a_{i}$

5. Special examples

Example 1**.**

Let U be an affine variety over a field of characteristic zero defined by the equation

[TABLE]

where $a\in k^{\times},a\notin k^{2},m\neq 0,4a$. By[4,Proposition 2.2],$Pic(\overline{U})$ is given by the following quotient group

[TABLE]

Here we give explicit condition which $H^{1}(k,Pic(\overline{U}))\cong\mathbb{Z}/2\oplus\mathbb{Z}/4$, and use methods of Colliot-Thélène, D. Kanevsky, J.-J. Sansuc [5] to describe the 4 torsion elements .

Lemma 5.1**.**

When $[k(\sqrt{a},\sqrt{m},\sqrt{m-4a}):k]=4$ and $\frac{\sqrt{m-4a}}{\sqrt{ma}}\in k$, $H^{1}(k,Pic(\overline{U}))\cong\mathbb{Z}/2\oplus\mathbb{Z}/4$

Proof.

Let $G=Gal(k(\sqrt{a},\sqrt{m})/k)$,there exist $\sigma,\tau\in G$,such that

[TABLE]

By computation of proposition 2.3, we have

1. (i)

\left\{\begin{array}[]{l}\sigma(\overline{l}_{1})=-\overline{l_{1}}\\ \sigma(\overline{l}_{2})=\overline{l}_{3}-\overline{l}_{1}\\ \sigma(\overline{l}_{3})=\overline{l}_{2}-\overline{l}_{1}\\ \sigma(\overline{l}_{4})=\overline{l}_{2}+\overline{l}_{3}-\overline{l}_{1}-\overline{l}_{4}\\ \end{array}\right. 2. (ii)

\left\{\begin{array}[]{l}\tau(\overline{l}_{1})=\overline{l}_{1}\\ \tau(\overline{l}_{2})=\overline{l}_{1}-\overline{l}_{3}\\ \tau(\overline{l}_{3})=\overline{l}_{1}-\overline{l}_{2}\\ \tau(\overline{l}_{4})=-\overline{l}_{4}\\ \end{array}\right.

Since

[TABLE]

we have $H^{1}(\langle\tau\rangle,Pic(\overline{U}))\cong\mathbb{Z}/2$. By computation,we have $Pic(\overline{U})^{\langle\tau\rangle}=(\overline{l}_{1},\overline{l}_{2}-\overline{l}_{3})$, and since

[TABLE]

one concludes that

[TABLE]

Hence,we have the following sequence

[TABLE]

by [6.proposition 3.3.14]. To show $H^{1}(G,Pic(\overline{U}))$ has 4-torsion elements, we use bicyclic group cohomology . Now we identify classes in $H^{1}(G,Pic(\overline{U}))$ with pairs $(a,b)\in Z^{1}(G,Pic(\overline{U}))$ modulo those of the form $(\Delta_{\sigma}(v),\Delta_{\tau}(v))$, where

[TABLE]

Then any element of $H^{1}(G,Pic(\overline{U}))$ is the class of

[TABLE]

where $x_{1},x_{2},y_{1}$ and $y_{2}$ $\in\mathbb{Z}$. If we let $y_{2}$ be odd, it’s easy to prove it’s 4-torsion element. ∎

Remark 5.2**.**

Using methods of Colliot-Thélène,Dasheng Wei,and Fei Xu,we can obtain all 2-torsion elements: $(x+2,m-4a)$,$\quad(x-2,m-4a)$, $(x^{2}-4,m-4a)$

We take $x_{1}=2,x_{2}=1,y_{1}=0,y_{2}=1$,we obtain this class $(\overline{l}_{1}+\overline{l}_{4}-\overline{l}_{2},\overline{l}_{4})$, since $\overline{l}_{1}+\overline{l}_{4}=\overline{l}_{2}+\overline{l}_{5}$ in $Pic(\overline{U})$, $(\overline{l}_{5},\overline{l}_{4})$ is a $4$-torsion element in $H^{1}(G,Pic(\overline{U}))$.Let $K=k(\sqrt{a},\sqrt{m})$, one has the following commutative diagram of exact sequences

[TABLE]

where the morphism $\partial$ is the connecting homomorphism of the following exact sequence

[TABLE]

$d^{1}(l_{5},l_{4})=(l_{5}(1,1)+l_{5}(1,-1),l_{5}(1,1)+l_{4}(-1,1)-l_{5}(-1,1)-l_{4}(1,1),l_{4}(1,1)+l_{4}(1,-1))\in Z^{2}(G,Div(U_{K}))$, let

[TABLE]

By [1,proposition 7.1(b) and proposition 8.4],we have

[TABLE]

This implies that $div(\frac{g}{f})=l_{5}(1,1)+l_{4}(-1,1)-l_{5}(-1,1)-l_{4}(1,1)$

Since $div(y-\sqrt{m})=l_{5}(1,1)+l_{5}(1,-1)$,  $div(x-\frac{\sqrt{m}}{\sqrt{a}})=l_{4}(1,1)+l_{4}(1,-1))$, we obtain

[TABLE]

Now we claim $(\sqrt{m}y-m,\frac{g(2\sqrt{a}-\sqrt{m}+\sqrt{m-4a})}{f(2\sqrt{a}+\sqrt{m}-\sqrt{m-4a})},x-\frac{\sqrt{m}}{\sqrt{a}})$ $\in Z^{2}(G,K(U)^{\times})$, it suffices to show

[TABLE]

in $K(U)^{\times}$, one can directly check this by using rational point $(0,0,\sqrt{m})$ of $U_{K}$.The cocycle determines a non-cyclic Azumaya algebras $\mathcal{A}$ on U.

.

Proposition 5.3**.**

Let U be the affine scheme defined by

[TABLE]

*where $a,m\in\mathbb{Z}$. Let $K=Q(\sqrt{m},\sqrt{a})$, when

1. (i)

$[K:Q]=4$*, $\frac{\sqrt{m-4a}}{\sqrt{ma}}\in Q$ * 2. (ii)

for any prime q,its decomposition group in $Gal(K/Q)$ is cyclic 3. (iii)

*There exists a prime $p\geq 5$,such that p splits in $Q(\sqrt{m})$ and has ramification index 2 in *$Q(\sqrt{a})$

then there is no Brauer-Manin obstruction to the integral Hasse principle for U.

Proof.

We can assume that $U(A_{\mathbb{Z}})\neq\emptyset$, where $A_{\mathbb{Z}}={\mathrm{R}}\times\prod_{p}\mathbb{Z}_{p}$ , otherwise there is nothing to prove. Note that since p splits in $Q(\sqrt{m})$, its decomposition group is $\langle\sigma\rangle$. Hence for any $T\in U(\mathbb{Z}_{p})$, $\mathcal{A}(T)=(\sqrt{m}y-m,a)\in{\mathrm{Br}}(Q_{p})$.By (iii), we can assume $ord_{p}(a)=1$. Note that

[TABLE]

Let $\mathcal{B}_{1}=(x+2,a)$, $\mathcal{B}_{2}=(y-\sqrt{m},a)$,to prove the proposition , it suffices to show for all $(\varepsilon_{1},\varepsilon_{2})\in(\mathbb{Z}/2\mathbb{Z})^{2}$, there exists $\zeta\in U(\mathbb{Z}_{p})$ such that

[TABLE]

Since $ord_{p}(a)$ is odd, we have p $\mid m$ by (i) ,in fact $ord_{p}(m)$ is even by (iii). Let $s,t\in\mathbb{F}_{p}^{\times}$, such that the Legendre symbol $(\frac{s}{p})=1$, $(\frac{t}{p})=-1$, let

[TABLE]

One can see that $\upsilon_{1}$ and $\upsilon_{2}$ are smooth points of $U(\mathbb{F}_{p})$, hence there exist $\mu_{1},\mu_{2}\in U(\mathbb{Z}_{p})$ such that $\mu_{i}\equiv\upsilon_{i}\bmod p$ for $1\leq i\leq 2$, we obtain

[TABLE]

By Dirichlet theorem,there exist a prime $l$,such that Legendre symbol $(\frac{l}{p})=-1$ and $p\nmid(l+1)$. Let

[TABLE]

One can check that $\upsilon_{3}$ and $\upsilon_{4}$ are smooth points of $U(\mathbb{F}_{p})$, hence there exist $\mu_{3},\mu_{4}\in U(\mathbb{Z}_{p})$ such that $\mu_{i}\equiv\upsilon_{i}\bmod p$ for $3\leq i\leq 4$, we obtain

[TABLE]

The proposition is established. ∎

Acknowledgements

I thank my advisor Fei Xu for countless helpful conversations and suggestions, and also for his patience. I also thank Yang Zhang for many useful discussions.