Arithmetic of Châtelet surfaces under extensions of base fields

Abstract

For Châtelet surfaces defined over number fields, we study two arithmetic properties, the Hasse principle and weak approximation, when passing to an extension of the base field. Generalizing a construction of Y. Liang, we show that for an arbitrary extension of number fields L/K,  there is a Châtelet surface over K which does not satisfy weak approximation over any intermediate field of L/K,  and a Châtelet surface over K which satisfies the Hasse principle over an intermediate field $L'$ if and only if $[L': K]$ is even.

1 Introduction

Throughout this paper, let K be a number field, and let $\Omega _K$ be the set of all nontrivial places of K. For each $v\in \Omega _K,$ let $K_v$ denote the completion of K at v. Let $S\subset \Omega _K$ be a finite subset. Let $\text{\AA }_K$ (respectively, $\text{\AA }_K^S$) be the ring of adèles (adèles without S components) of K. We always assume that a field L is a finite extension of K. Let $S_L\subset \Omega _L$ denote the set of places of L lying over places in S.

Let X be a proper algebraic variety over K. The set X(K) of K-rational points of X can be viewed as a subset of the set $X(\text{\AA }_K)$ of adelic points via the diagonal embedding. We say that X is a counterexample to the Hasse principle if $X(\text{\AA }_K)\ne \emptyset$ whereas $X(K)=\emptyset .$ By the properness of X,  the set of adelic points $X(\text{\AA }_K^S)$ can be identified with the product $\prod _{v\in \Omega _K\backslash S}X(K_v),$ and hence equipped with the product topology of v-adic topologies. We say that X satisfies weak approximation (respectively, weak approximation off S ) if X(K) is dense in $X(\text{\AA }_K)$ (respectively,in $X(\text{\AA }_K^S)$), cf. [14, Chapter 5.1].

In this paper, we focus on the case where X is a Châtelet surface over K,  i.e., a smooth projective model of affine surface in $\text{\AA }_K^3$ defined by the equation

$$\begin{aligned} y^2-az^2=P(x), \end{aligned}$$

(1)

where $a\in K^\times ,$ and P(x) is a separable degree-4 polynomial in K[x]. Over a fixed number field K,  the Hasse principle and weak approximation for Châtelet surfaces (and many other varieties) have been studied in a lot of earlier papers (e.g., [4,5,6, 12], etc.). In [8], Liang pioneered the study of non-invariance of arithmetic properties under extensions of base fields and proved, among others, that for any number field K,  there is a Châtelet surface V over K and a quadratic extension L/K such that $V(K)\ne \emptyset ,$ V satisfies weak approximation, but the base extension $V_L$ does not satisfies weak approximation off all archimedean places of L. In this paper, we generalize Liang’s construction and obtain further results that apply to an arbitrary extension L/K.

Our main results are the following:

Theorem 1.1

(Theorem 4.1) Let L/K be any extension of number fields, and let $S \subset \Omega _K\backslash \{$all complex and 2-adic places$\}$ be a finite nonempty subset such that every place in S splits completely in L.

Then, there exists a Châtelet surface V defined over K with $V(K)\ne \emptyset ,$ such that for every intermediate field $K\subset L'\subset L$ and every finite subset $T'\subset \Omega _{L'},$ the base extension $V_{L'}$ satisfies weak approximation off $T'$ if and only if $T'\cap S_{L'}\ne \emptyset .$ In particular, the surface $V_{L'}$ does not satisfy weak approximation for every $K\subset L'\subset L.$

Theorem 1.2

(Theorem 5.1) For any extension of number fields L/K,  there exists a Châtelet surface V over K with $V(\text{\AA }_K)\ne \emptyset ,$ such that for every intermediate field $K\subset L'\subset L,$

• If the degree $[L':K]$ is odd, then the surface $V_{L'}$ is a counterexample to the Hasse principle, i.e.,$V(L')=\emptyset .$ In particular, the surface V is a counterexample to the Hasse principle.

• If the degree $[L':K]$ is even, then the surface $V_{L'}$ satisfies weak approximation. In particular, in this case, $V(L')\ne \emptyset .$

To construct the Châtelet surfaces in these theorems, the parameters in the equation (1), i.e.,the element $a\in K$ and the coefficients of the polynomial P(x),  need to be chosen carefully using approximation theorems for the affine line and Čebotarev’s density theorem. To verify the statements about weak approximation and the Hasse principle, we shall analyze the Brauer-Manin obstruction and use the well-known theorem that for Châtelet surfaces this obstruction (to weak approximation or the Hasse principle) is the only one ([4, 5]).

2 Notation and preliminaries

2.1 Notation

Given a number field K,  let ${{\mathcal {O}}}_K$ be the ring of its integers. Let $\infty _K\subset \Omega _K$ be the subset of all archimedean places, and let $2_K\subset \Omega _K$ be the subset of all 2-adic places. Let $\infty _K^r\subset \infty _K$ be the subset of all real places, and let $\infty _K^c\subset \infty _K$ be the subset of all complex places. Let $\Omega _K^f=\Omega _K\backslash \infty _K$ be the set of all finite places of K. For $v\in \infty _K,$ let $\tau _v:K\hookrightarrow K_v$ be the embedding of K into its completion. For $v\in \Omega _K^f,$ let ${{\mathcal {O}}}_{K_v}$ be its valuation ring. Let $K^2$ denote the set of square elements of K. Let ${{\mathcal {O}}}_S=\bigcap _{v\in \Omega ^f_K\backslash S}(K\cap {{\mathcal {O}}}_{K_v})$ be the ring of S-integers. A strong approximation theorem [1, Chapter II §15] states that K is dense in $\text{\AA }_K^S$ for any nonempty S. In this paper, we only use the following special case:

Lemma 2.1

The set K is dense in $\text{\AA }_K^{2_K}.$

We will use the following version of Čebotarev’s density theorem, cf. [10, Chapter VII §13].

Theorem 2.2

(Čebotarev) The set of places of K splitting completely in L has positive density.

2.2 Hilbert symbol

We use the Hilbert symbol $(a,b)_v\in \{\pm 1\},$ for $a,b\in K_v^\times$ and $v\in \Omega _K.$ By definition, $(a,b)_v=1$ if and only if $x_0^2-ax_1^2-bx_2^2=0$ has a $K_v$-solution in ${\mathbb {P}}^2$ with homogeneous coordinates $(x_0:x_1:x_2),$ which equivalently means that the curve defined over $K_v$ by the equation $x_0^2-ax_1^2-bx_2^2=0$ in ${\mathbb {P}}^2$ is isomorphic to ${\mathbb {P}}^1.$ The Hilbert symbol gives a symmetric bilinear form on $K_v^\times /K_v^{\times 2}$ with value in ${\mathbb {Z}}/2{\mathbb {Z}},$ cf. [13, Chapter XIV, Proposition 7]. And this bilinear form is nondegenerate, cf. [13, Chapter XIV, Corollary 7].

2.3 Preparation lemmas

We state the following lemmas for later use. Lemmas 2.3–2.8 have already been given in the paper [15]. We give their statements here for the convenience of the reader.

Lemma 2.3

Let v be an odd place of K. Let $a, b\in K_v^\times$ such that v(a), v(b) are even. Then $(a,b)_v=1.$

Lemma 2.4

Let v be an odd place of K. Let $a,b,c\in K_v^\times$ such that $v(b)<v(c).$ Then $(a,b+c)_v=(a,b)_v.$

Lemma 2.5

The set $K_v^{\times 2}$ is an open subgroup of $K_v^\times .$ If $v\in \Omega _K^f,$ then ${{\mathcal {O}}}_{K_v}^\times$ is also an open subgroup of $K_v^\times .$ So, they are nonempty open subset of $K_v.$

Lemma 2.6

Let $v\in \Omega _K^f.$ For any $n\in {\mathbb {Z}},$ the set $\{x\in K_v|v(x)=n\}$ is a nonempty open subset of $K_v.$

Lemma 2.7

Let $v\in \Omega _K^f.$ For any $a\in K_v^\times ,$ the sets $\{x\in K_v^\times |(a,x)_v=1\},$ $\{x\in K_v^\times |(a,x)_v=1\}\cap {{\mathcal {O}}}_{K_v}$ and $\{x\in {{\mathcal {O}}}_{K_v}^\times |(a,x)_v=1\}$ are nonempty open subsets of $K_v.$

Lemma 2.8

Let $v\in \Omega _K^f.$ For any $a\in K_v^\times ,$ the sets $\{x\in K_v^\times |(a,x)_v=-1\}$ and $\{x\in K_v^\times |(a,x)_v=-1\}\cap {{\mathcal {O}}}_{K_v}$ are open subsets of $K_v.$ Furthermore, if $a\notin K_v^{\times 2},$ then they are nonempty.

Lemma 2.9

Let $v\in \Omega _K^f.$ For any $a\in K_v^\times$ with v(a) odd, the set $\{x\in {{\mathcal {O}}}_{K_v}^\times |(a,x)_v=-1\}$ is a nonempty open subset of $K_v.$

Proof of Lemma 2.9

By Lemmas 2.5 and 2.8, the set is open in $K_v.$ We need to show that it is nonempty. Since $a\notin K_v^2,$ by the nondegeneracy of the bilinear form given by the Hilbert symbol, there exists an element $b\in K_v^\times$ such that $(a,b)_v=-1.$ If v(b) is odd, let $b'=-ab.$ Then $(a,b')_v=(a,-ab)_v=(a,-a)_v(a,b)_v=-1.$ Replacing b by $b'$ if necessary, we can assume that v(b) is even. Choose a prime element $\pi _v\in K_v.$ Then $\pi _v^{-v(b)}\in K_v^{\times 2},$ so the element $b\pi _v^{-v(b)}$ is in this set.

2.4 Brauer-Manin obstruction

Cohomological obstructions have been used to explain failures of the Hasse principle and nondensity of X(K) in $X(\text{\AA }_K^S).$ Let $Br (X)=H^2_{et}(X,{\mathbb {G}}_m)$ be the Brauer group of X. Let ${{\,\textrm{inv}\,}}_v:Br (K_v)\rightarrow {\mathbb {Q}}/{\mathbb {Z}}$ be the local invariant map. The Brauer-Manin pairing

$$\begin{aligned} X(\text{\AA }_K)\times Br (X)\rightarrow {\mathbb {Q}}/{\mathbb {Z}}, \end{aligned}$$

suggested by Manin [9], between $X(\text{\AA }_K)$ and $Br (X),$ is provided by local class field theory. The left kernel of this pairing is denoted by $X(\text{\AA }_K)^{Br },$ which is a closed subset of $X(\text{\AA }_K).$ By the global reciprocity in class field theory, there is an exact sequence:

$$\begin{aligned} 0\rightarrow Br (K)\rightarrow \bigoplus \limits _{v\in \Omega _K} Br (K_v)\rightarrow {\mathbb {Q}}/{\mathbb {Z}}\rightarrow 0, \end{aligned}$$

which induces an inclusion: $X(K)\subset pr^S(X(\text{\AA }_K)^Br ).$

Remark 2.10

For any smooth, proper and rationally connected variety X defined over a number field K,  it is conjectured by Colliot-Thélène [2] that the K-rational points set X(K) is dense in $X(\text{\AA }_K)^Br .$ Colliot-Thélène’s conjecture holds for Châtelet surfaces, cf. [4, 5].

3 Châtelet surfaces

Let K be a number field. Given an equation (1), let $V^0$ be the affine surface in $\text{\AA }^3_K$ defined by this equation. Let V be the natural smooth compactification of $V^0$ given in [14, Section 7.1], which is called the Châtelet surface given by this equation, cf. [12, Section 5]. By the Lang-Nishimura theorem ([7] and [11]) and the implicit function theorem, all smooth projective models of a given equation (1) are the same as to the discussion of the Hasse principle and weak approximation.

Remark 3.1

For any local field $K_v,$ if $a\in K_v^{\times 2},$ then V is birationally equivalent to ${\mathbb {P}}^2$ over $K_v.$ By the implicit function theorem, there exists a $K_v$-point on V.

Remark 3.2

For any local field $K_v,$ by smoothness of V,  the implicit function theorem implies that the nonemptiness of $V^0(K_v)$ is equivalent to the nonemptiness of $V(K_v),$ and that $V^0(K_v)$ is open dense in $V(K_v)$ with the v-adic topology. Given an element $A\in Br (V),$ since $Br (V)$ is torsion, the set of all possible values of the evaluation of A on $V(K_v)$ is finite. Indeed, by [14, Proposition 7.1.2], there exist only two possible values. They are determined by the evaluation of A on $V^0(K_v).$ In particular, if the evaluation of A on $V^0(K_v)$ is constant, then it is constant on $V(K_v).$

In the following two sections, we will construct two kinds of Châtelet surfaces.

3.1 Choice of the parameter a for the equation (1)

Given an extension of number fields L/K,  and a finite subset $S\subset \Omega _K\backslash (\infty _K^c\cup 2_K),$ we always use the following way to choose an element $a\in {{\mathcal {O}}}_K\backslash K^2$ for the parameter a in the equation (1).

By Theorem 2.2, we can take a place $v_0\in \Omega _K^f\backslash 2_K$ splitting completely in L. If $S\ne \emptyset ,$ let $S_0=S,$ otherwise, let $S_0= \{v_0\}.$ Then $S_0\ne \emptyset .$

For $v\in \Omega _K,$ by Lemma 2.5, the set $K_v^{\times 2}$ is a nonempty open subset of $K_v.$ For $v\in \Omega _K^f,$ by Lemma 2.6, the set $\{a\in K_v| v(a)$ is odd$\}$ is a nonempty open subset of $K_v.$ Using weak approximation for the affine line $\text{\AA }^1,$ we can choose an element $a\in K^\times$ satisfying the following conditions:

• $\tau _v(a)<0$ for all $v\in S_0\cap \infty _K,$

• $a\in K_v^{\times 2}$ for all $v\in 2_K,$

• v(a) is odd for all $v\in S_0\backslash \infty _K.$

These conditions do not change by multiplying an element in $K^{\times 2},$ so we can assume $a\in {{\mathcal {O}}}_K.$ The conditions that v(a) is odd for all $v\in S_0\backslash \infty _K,$ and that $\tau _v(a)<0$ for all $v\in S_0\cap \infty _K,$ imply $a\in {{\mathcal {O}}}_K\backslash K_v^2$ for all $v\in S_0.$ So $a\in {{\mathcal {O}}}_K\backslash K^2.$

Remark 3.3

Let $S'=\{v\in \infty _K^r | \tau _v(a)<0\}\cup \{v\in \Omega _K^f\backslash 2_K | v(a)\mathrm{~is ~odd} \},$ then $S'$ is a finite set. By the conditions that $\tau _v(a)<0$ for all $v\in S\cap \infty _K,$ and that v(a) is odd for all $v\in S\backslash \infty _K,$ we have $S'\supset S.$ Then $S'\ne \emptyset .$

Lemma 3.4

Given an extension of number fields L/K,  and a finite subset $S\subset \Omega _K\backslash (\infty _K^c\cup 2_K),$ we choose an element $a\in {{\mathcal {O}}}_K$ as above. If there exists one place in S splitting completely in L or $S=\emptyset ,$ then $a\in {{\mathcal {O}}}_K\backslash L^2.$

Proof

We use the notion $S_0$ as above. By our assumption, we can take a place $v\in S_0$ splitting completely in L. Take a place $w\in \Omega _L$ such that w|v. Then $K_v= L_w.$ By the choice of a,  if $v\in S_0\backslash \infty _K,$ then v(a) is odd; if $v\in S_0\cap \infty _K,$ then $\tau _v(a)<0$. In both cases, we have $a\in {{\mathcal {O}}}_K\backslash K_v^2={{\mathcal {O}}}_K\backslash L_w^2$. Hence $a\in {{\mathcal {O}}}_K\backslash L^2.$ $\square$

4 Weak approximation under extensions of base fields

In the paper [8], Liang studied the non-invariance of weak approximation under extensions of base fields. More precisely, for any number field K,  Liang [8, Proposition 3.4] proved that there exist a Châtelet surface V over K and a quadratic extension L/K such that $V(K)\ne \emptyset ,$ V satisfies weak approximation, but the base extension $V_L$ does not satisfy weak approximation, even off $\infty _L.$ In this section, we generalize Liang’s construction and obtain further results that apply to an arbitrary extension L/K.

4.1 Choice of parameters for the equation (1)

With $a\in {{\mathcal {O}}}_K\backslash K^2$ chosen as in Section 3, we choose an element $b\in K^\times$ in the following way.

Let $S'=\{v\in \infty _K^r | \tau _v(a)<0\}\cup \{v\in \Omega _K^f\backslash 2_K | v(a)\mathrm{~is ~odd} \}$ be as in Remark 3.3, then $S'\supset S$ is a finite set. By Lemma 2.6, for $v\in S\backslash \infty _K,$ the set $\{b\in K_v|v(b)=-v(a)\}$ is a nonempty open subset of $K_v;$ for $v\in S'\backslash (S\cup \infty _K),$ the set $\{b\in K_v|v(b)=v(a)\}$ is a nonempty open subset of ${{\mathcal {O}}}_{K_v}.$ By Lemma 2.1, we can choose a nonzero element $b\in {{\mathcal {O}}}_S[1/2]$ satisfying the following conditions:

• $v(b)=-v(a)$ for all $v\in S\backslash \infty _K,$

• $v(b)=v(a)$ for all $v\in S'\backslash (S\cup \infty _K).$

We choose an element $c\in K^\times$ with respect to the chosen a, b in the following way.

Let $S''=\{v\in \Omega _K^f\backslash 2_K| v(b)\ne 0 \},$ then $S''$ is a finite set and $S'\backslash \infty _K \subset S''.$ By Theorem 2.2, we can take two different finite places $v_1,v_2\in \Omega _K^f\backslash S''$ splitting completely in L. If $v \in S\backslash \infty _K,$ then v(a) is odd. In this case, by Lemma 2.9, the set $\{c\in {{\mathcal {O}}}_{K_v}^\times |(a,c)_v=-1\}$ is a nonempty open subset of ${{\mathcal {O}}}_{K_v}.$ If $v\in \{v_1,v_2\},$ then $b\in {{\mathcal {O}}}_{K_v}^\times .$ In this case, by Lemma 2.6, the sets $\{c\in K_v|v(c)=1\}$ and $\{c\in K_v|v(1+cb^2)=1\}$ are nonempty open subsets of ${{\mathcal {O}}}_{K_v}.$ Also by Lemma 2.1, we can choose a nonzero element $c\in {{\mathcal {O}}}_K[1/2]$ satisfying the following conditions:

• $\tau _v(1+cb^2)<0$ for all $v\in S\cap \infty _K,$

• $\tau _v(c)>0$ for all $v\in (S'\backslash S)\cap \infty _K,$

• $(a,c)_v=-1$ and $v(c)=0$ for all $v\in S\backslash \infty _K,$

• $v_1(c)=1$ and $v_2(1+cb^2)=1$ for the chosen $v_1,v_2$ above.

Let $P(x)=(cx^2+1)((1+cb^2)x^2+b^2),$ and let $V_1$ be the Châtelet surface given by $y^2-az^2=(cx^2+1)((1+cb^2)x^2+b^2).$

Proposition 4.1

For any extension of number fields L/K,  and any finite subset $S \subset \Omega _K\backslash (\infty _K^c\cup 2_K)$ splitting completely in L,  there exists a Châtelet surface $V_1$ defined over K,  which has the following properties.

• The Brauer group $Br (V_1)/Br (K)\cong Br (V_{1L})/Br (L)\cong {\mathbb {Z}}/2{\mathbb {Z}},$ is generated by an element $A\in Br (V_1).$ The subset $V_1(K)\subset V_1(L)$ is nonempty.

• For any $v\in S,$ there exist $P_v$ and $Q_v$ in $V_1(K_v)$ such that the local invariants ${{\,\textrm{inv}\,}}_v(A(P_v))=0$ and ${{\,\textrm{inv}\,}}_v(A(Q_v))=1/2.$ For any $v\notin S,$ and any $P_v\in V_1(K_v),$ the local invariant ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$

• For any $v'\in S_L,$ there exist $P_{v'}$ and $Q_{v'}$ in $V_1(L_{v'})$ such that the local invariants ${{\,\textrm{inv}\,}}_{v'}(A(P_{v'}))=0$ and ${{\,\textrm{inv}\,}}_{v'}(A(Q_{v'}))=1/2.$ For any $v'\notin S_L,$ and any $P_{v'}\in V_1(L_{v'}),$ the local invariant ${{\,\textrm{inv}\,}}_{v'}(A(P_{v'})) =0.$

Proof

For the extension L/K,  and the finite set S,  we check that the Châtelet surface $V_1$ chosen as in Subsection 4.1 has the required properties.

By the choice of the places $v_1,$ the polynomial $x^2+c$ is an Eisenstein polynomial, so it is irreducible over $K_{v_1}.$ Since $v_1(a)$ is even, we have $K(\sqrt{a})K_{v_1}\ncong K_{v_1}[x]/(cx^2+1).$ So $K(\sqrt{a})\ncong K[x]/(cx^2+1).$ The same argument holds for the place $v_2$ and the polynomial $(1+cb^2)x^2+b^2.$ Since all places of S split completely in L,  by Lemma 3.4, we have $a\in {{\mathcal {O}}}_K\backslash L^2.$ By the splitting condition of $v_1, v_2,$ we have $L(\sqrt{a})\ncong L[x]/(cx^2+1)$ and $L(\sqrt{a})\ncong L[x]/((1+cb^2)x^2+b^2).$ So $P(x)=(cx^2+1)((1+cb^2)x^2+b^2)$ is separable and a product of two degree-2 irreducible factors over K and L. According to [14, Proposition 7.1.1], the Brauer group $Br (V_1)/Br (K)\cong Br (V_{1L})/Br (L)\cong {\mathbb {Z}}/2{\mathbb {Z}}.$ Furthermore, by Proposition 7.1.2 in loc. cit, we take the quaternion algebra $A=(a,cx^2+1)\in Br (V_1)$ as a generator element of this group. Then we have the equality $A=(a,cx^2+1)=(a,(1+cb^2)x^2+b^2)$ in $Br (V_1).$

Since $(x,y,z)=(0,b,0)$ is a rational point on $V_1^0,$ the set $V_1(K)$ is nonempty. We denote this rational point by $Q_0.$

We need to compute the evaluation of A on $V_1(K_v)$ for all $v\in \Omega _K.$

For any $v\in \Omega _K,$ the local invariant ${{\,\textrm{inv}\,}}_v(A(Q_0))=0.$ By Remark 3.2, it suffices to compute the local invariant ${{\,\textrm{inv}\,}}_v(A(P_v))$ for all $P_v\in V_1^0(K_v).$

• (1) Suppose that $v\in (\infty _K\backslash S')\cup 2_K.$ Then $a\in K_v^{\times 2},$ so ${{\,\textrm{inv}\,}}_v(A(P_v))=0$ for all $P_v\in V_1(K_v).$

• (2) Suppose that $v\in (S'\backslash S)\cap \infty _K.$ For any $x\in K,$ by the choice of c,  we have $\tau _v(cx^2+1)>0.$ Then $(a,cx^2+1)_v=1,$ so ${{\,\textrm{inv}\,}}_v(A(P_v))=0$ for all $P_v\in V_1(K_v).$

• (3) Suppose that $v\in S'\backslash (S\cup \infty _K).$ Take an arbitrary $P_v\in V_1^0(K_v).$ If ${{\,\textrm{inv}\,}}_v(A(P_v))=1/2,$ then $(a,cx^2+1)_v=-1=(a,(1+cb^2)x^2+b^2)_v$ at $P_v.$ By Lemma 2.4, the first equality implies $v(x)\le 0.$ Since $v(a)=v(b)>0$ and $v(c)\ge 0,$ by Lemma 2.4, we have $(a,(1+cb^2)x^2+b^2)_v=(a,x^2)_v=1,$ which is a contradiction. So ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$

• (4) Suppose that $v\in \Omega _K^f\backslash (S'\cup 2_K ).$ Take an arbitrary $P_v\in V_1^0(K_v).$ If ${{\,\textrm{inv}\,}}_v(A(P_v))=1/2,$ then $(a,cx^2+1)_v=-1=(a,(1+cb^2)x^2+b^2)_v$ at $P_v.$ Since v(a) is even, by Lemma 2.3, the first equality implies that $v(cx^2+1)$ is odd. Since $c\in {{\mathcal {O}}}_K[1/2],$ we have $v(x)\le 0.$ So $v(c+x^{-2})$ is odd and positive. Since $v(b)\ge 0,$ by Hensel’s lemma, we have $1+b^2(c+x^{-2})\in K_v^{\times 2}.$ So $(a,(1+cb^2)x^2+b^2)_v=(a,x^2)_v(a,1+b^2(c+x^{-2}))_v=1,$ which is a contradiction. So ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$

• (5) Suppose that $v\in S\cap \infty _K.$ Take $P_v=Q_0,$ then ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$ By the choice of b, c,  we have $\tau _v(\frac{b^2}{-cb^2-1})>\tau _v(\frac{1}{-c})>0.$ Take $x_0\in K$ such that $\tau _v(x_0)>\sqrt{\tau _v(\frac{b^2}{-cb^2-1})},$ then $\tau _v((cx_0^2+1)((1+cb^2)x_0^2+b^2))>0$ and $\tau _v(cx_0^2+1)<0.$ So there exists a $Q_v\in V_1^0(K_v)$ with $x=x_0.$ Then ${{\,\textrm{inv}\,}}_v(A(Q_v))=1/2.$

• (6) Suppose that $v\in S\backslash \infty _K.$ Take $P_v=Q_0,$ then ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$ Take $x_0\in K_v$ such that $v(x_0)< 0.$ Since $v(b)=-v(a)<0$ and $v(c)=0,$ by Lemma 2.4, we have $(a,cx_0^2+1)_v=(a,cx_0^2)_v=(a,c)_v$ and $(a,(1+cb^2)x_0^2+b^2)_v=(a,cb^2x_0^2)_v=(a,c)_v.$ So $(a,(cx_0^2+1)((1+cb^2)x_0^2+b^2))_v=(a,c)_v(a,c)_v=1.$ Hence, there exists a $Q_v\in V_1^0(K_v)$ with $x=x_0.$ Since $(a,c)_v=-1,$ we have ${{\,\textrm{inv}\,}}_v(A(Q_v))=1/2.$

Finally, we need to compute the evaluation of A on $V_1(L_{v'})$ for all $v'\in \Omega _L.$

For any $v'\in \Omega _L,$ the local invariant ${{\,\textrm{inv}\,}}_{v'}(A(Q_0))=0.$

• (1) Suppose that $v' \in S_L.$ Let $v\in \Omega _K$ be the restriction of $v'$ on K. By the assumption that v splits completely in L,  we have $K_v=L_{v'}.$ So $V_1(K_v)=V_1(L_{v'}).$ By the argument already shown, there exist $P_v, Q_v\in V_1(K_v)$ such that ${{\,\textrm{inv}\,}}_v(A(P_v))=0$ and ${{\,\textrm{inv}\,}}_v(A(Q_v))=1/2.$ View $P_v, Q_v$ as elements in $V_1(L_{v'}),$ and let $P_{v'}=P_v$ and $Q_{v'}=Q_v.$ Then ${{\,\textrm{inv}\,}}_{v'}(A(P_{v'}))={{\,\textrm{inv}\,}}_v(A(P_v))=0$ and ${{\,\textrm{inv}\,}}_{v'}(A(Q_{v'}))={{\,\textrm{inv}\,}}_v(A(Q_v))=1/2.$

• (2) Suppose that $v'\in \Omega _L\backslash S_L.$ This local computation is the same as the case $v\in \Omega _K\backslash S.$ $\square$

Remark 4.2

For any $v\in S,$ and any $P_v\in V_1(K_v),$ the local invariant of the evaluation of A on $P_v$ is 0 or $1/2.$ Let $U_1=\{P_v\in V_1(K_v)| {{\,\textrm{inv}\,}}_v(A(P_v))=0\}$ and $U_2=\{P_v\in V_1(K_v)| {{\,\textrm{inv}\,}}_v(A(P_v))=1/2\}.$ Then $U_1$ and $U_2$ are nonempty disjoint open subsets of $V_1(K_v),$ and $V_1(K_v)=U_1\bigsqcup U_2.$

Applying the global reciprocity law, the surface $V_1$ in Proposition 4.1 has the following weak approximation properties.

Proposition 4.3

Given an extension of number fields L/K,  and a finite subset $S \subset \Omega _K\backslash (\infty _K^c\cup 2_K)$ splitting completely in L,  let $V_1$ be a Châtelet surface satisfying those properties of Proposition 4.1.

• (1) If $S=\emptyset ,$ then $V_1$ and $V_{1\,L}$ satisfy weak approximation.

• (2) If $S\ne \emptyset ,$ then $V_1$ satisfies weak approximation off $S'$ for a finite subset $S'\subset \Omega _K$ if and only if $S'\cap S\ne \emptyset .$

• (3) If $S\ne \emptyset ,$ the surface $V_{1L}$ satisfies weak approximation off T for a finite subset $T\subset \Omega _L$ if and only if $T\cap S_L\ne \emptyset .$

Proof

According to [4, 5, Theorem B], the Brauer-Manin obstruction to the Hasse principle and weak approximation is the only one for Châtelet surfaces, so $V_1(K)$ is dense in $V_1(\text{\AA }_K)^{Br }.$

• (1) Suppose that $S=\emptyset ,$ then for any $(P_v)_{v\in \Omega _K}\in V_1(\text{\AA }_K),$ by Proposition 4.1, the sum $\sum _{v\in \Omega _K}{{\,\textrm{inv}\,}}_v(A(P_v))=0.$ Since $Br (V_1)/Br (K)$ is generated by the element A,  we have $V_1(\text{\AA }_K)^{Br }= V_1(\text{\AA }_K).$ So $V_1(K)$ is dense in $V_1(\text{\AA }_K)^{Br }= V_1(\text{\AA }_K),$ i.e. the surface $V_1$ satisfies weak approximation.

• (2)

  • (a) Suppose that $S'\cap S\ne \emptyset .$ Take $v_0\in S'\cap S.$ For any finite subset $R\subset \Omega _K\backslash \{v_0\},$ take a nonempty open subset $M=V_1(K_{v_0})\times \prod _{v\in R}U_v\times \prod _{v\notin R \cup \{v_0\}}V_1(K_v)\subset V_1(\text{\AA }_K).$ Take an element $(P_v)_{v\in \Omega _K}\in M$ with ${{\,\textrm{inv}\,}}_{v_0} A(P_{v_0})=0.$ By Proposition 4.1 and $v_0\in S,$ we can take an element $P_{v_0}'\in V_1(K_{v_0})$ such that ${{\,\textrm{inv}\,}}_{v_0} A(P_{v_0}')=1/2.$ By Proposition 4.1, the sum $\sum _{v\in \Omega _K\backslash \{v_0\}}{{\,\textrm{inv}\,}}_v(A(P_v))$ is 0 or $1/2$ in ${\mathbb {Q}}/{\mathbb {Z}}.$ If it is $1/2,$ then we replace $P_{v_0}$ by $P_{v_0}'.$ In this way, we get a new element $(P_v)_{v\in \Omega _K}\in M.$ And the sum $\sum _{v\in \Omega _K}{{\,\textrm{inv}\,}}_v(A(P_v))=0$ in ${\mathbb {Q}}/{\mathbb {Z}}.$ So $(P_v)_{v\in \Omega _K}\in V_1(\text{\AA }_K)^{Br }\cap M.$ Since $V_1(K)$ is dense in $V_1(\text{\AA }_K)^{Br },$ the set $V_1(K)\cap M\ne \emptyset ,$ which implies that $V_1$ satisfies weak approximation off $\{v_0\}.$ So $V_1$ satisfies weak approximation off $S'.$

  • (b) Suppose that $S\ne \emptyset$ and $S'\cap S= \emptyset .$ Take $v_0\in S,$ and let $U_{v_0}=\{P_{v_0}\in V_1(K_{v_0})| {{\,\textrm{inv}\,}}_{v_0}(A(P_{v_0}))=1/2\}.$ For $v\in S\backslash \{v_0\},$ let $U_v=\{P_v\in V_1(K_v)| {{\,\textrm{inv}\,}}_v(A(P_v))=0\}.$ For any $v\in S,$ by Remark 4.2, the set $U_v$ is a nonempty open subset of $V_1(K_v).$ Let $M=\prod _{v\in S}U_v\times \prod _{v\notin S }V_1(K_v).$ It is a nonempty open subset of $V_1(\text{\AA }_K).$ For any $(P_v)_{v\in \Omega _K}\in M,$ by Proposition 4.1 and the choice of $U_v,$ the sum $\sum _{v\in \Omega _K}{{\,\textrm{inv}\,}}_v(A(P_v))=1/2$ is nonzero in ${\mathbb {Q}}/{\mathbb {Z}}.$ So $V_1(\text{\AA }_K)^{Br }\cap M=\emptyset ,$ which implies $V_1(K)\cap M= \emptyset .$ Hence $V_1$ does not satisfy weak approximation off $S'.$

• (3) The same argument applies to $V_{1L}.$ $\square$

From the discussion in the proof of Proposition 4.3, we have the following weak approximation properties for Châtelet surfaces.

Theorem 4.1

For any extension of number fields L/K,  and any finite nonempty subset $S \subset \Omega _K\backslash (\infty _K^c\cup 2_K)$ splitting completely in L,  there exists a Châtelet surface V over K with $V(K)\ne \emptyset ,$ such that for every intermediate field $K\subset L'\subset L,$ $Br (V)/Br (K)\cong Br (V_{L'})/Br (L')\cong {\mathbb {Z}}/2{\mathbb {Z}},$ and for every finite subset $T'\subset \Omega _{L'},$ the base extension $V_{L'}$ satisfies weak approximation off $T'$ if and only if $T'\cap S_{L'}\ne \emptyset .$ In particular, the surface $V_{L'}$ does not satisfy weak approximation for every $K\subset L'\subset L.$

Proof

For the extension L/K and the set S,  let V be the Châtelet surface chosen as in Sect. 4.1. Applying the same argument about the field L to its subfield $L',$ the properties that we list are just what we have explained in Proposition 4.1 and Proposition 4.3. $\square$

Using the construction method in Sect. 4.1, we have the following example, which is a special case of Proposition 4.1.

Example 4.4

For $K={\mathbb {Q}}$ and $L={\mathbb {Q}}(\sqrt{3}),$ and let $S=\{73\}\subset \Omega _K.$ The prime numbers 11, 23, 73 split completely in L. Using the construction method in Sect. 4.1, we choose data: $S=S'=S''={73},~v_1=11,~v_2=23,~a=73,~b=1/73,~c=99$ and $P(x)=(99x^2+1)(5428x^2/5329+1/5329).$ Then the Châtelet surface given by $y^2-73z^2=P(x),$ has the properties of Propositions 4.1 and 4.3.

5 The Hasse principle under extensions of base fields

Iskovskikh [6] showed that the Châtelet surface over ${\mathbb {Q}}$ given by $y^2+z^2=(x^2-2)(-x^2+3)$ is a counterexample to the Hasse principle. A family of Châtelet surfaces over ${\mathbb {Q}}$ extending Iskovskikh’s example was given in [3], (see also [14, Pages 145-146]). Poonen [12, Proposition 5.1] generalized their arguments to any number field. For an arbitrary number field K,  he constructed a Châtelet surface defined over K,  which is a counterexample to the Hasse principle. He used Čebotarev’s density theorem for some ray class fields to choose the parameters for the equation (1). The Châtelet surface he constructed has the properties of [12, Lemma 5.5] (a special situation of the following Proposition 5.1: the case when $S=\{v_0\}$ for some place $v_0$ associated to some large prime element in ${{\mathcal {O}}}_K$), which is the main ingredient in the proof of [12, Proposition 5.1]. In this section, we generalize Poonen’s construction and consider the Hasse principle of Châtelet surfaces under extensions of base fields.

5.1 Choice of parameters for the equation (1)

Given an extension of number fields L/K,  and a finite subset $S\subset \Omega _K\backslash (\infty _K^c\cup 2_K),$ we choose $a\in {{\mathcal {O}}}_K\backslash K^2$ as in Section 3. Then we choose an element $b\in K^\times$ in the following way.

Let $S'=\{v\in \infty _K^r | \tau _v(a)<0\}\cup \{v\in \Omega _K^f\backslash 2_K | v(a)\mathrm{~is ~odd} \}$ be as in Remark 3.3, then $S'\supset S$ is a finite set. If $v \in S\backslash \infty _K,$ then v(a) is odd. Then by Lemma 2.9, the set $\{b\in {{\mathcal {O}}}_{K_v}^\times |(a,b)_v=-1\}$ is a nonempty open subset of ${{\mathcal {O}}}_{K_v}.$ If $v \in S'\backslash (S\cup \infty _K),$ then by Lemma 2.7, the set $\{b\in {{\mathcal {O}}}_{K_v}^\times |(a,b)_v=1\}$ is a nonempty open subset of ${{\mathcal {O}}}_{K_v}.$ By Lemma 2.1, we can choose a nonzero element $b\in {{\mathcal {O}}}_K[1/2]$ satisfying the following conditions:

• $\tau _v(b)<0$ for all $v\in S\cap \infty _K,$

• $\tau _v(b)>0$ for all $v\in (S'\backslash S)\cap \infty _K,$

• $(a,b)_v=-1$ and $v(b)=0$ for all $v\in S\backslash \infty _K,$

• $(a,b)_v=1$ and $v(b)=0$ for all $v\in S'\backslash (S\cup \infty _K).$

We choose an element $c\in K^\times$ with respect to the chosen a, b in the following way.

Let $S''=\{v\in \Omega _K^f\backslash 2_K | v(b)\ne 0 \},$ then $S''$ is a finite set and $S'\cap S''=\emptyset .$ By Theorem 2.2, we can take two different finite places $v_1,v_2\in \Omega _K^f\backslash (S'\cup S''\cup 2_K )$ splitting completely in L. If $v\in (S'\backslash \infty _K)\cup \{v_1,v_2\},$ then $b\in {{\mathcal {O}}}_{K_v}^\times .$ In this case, by Lemma 2.6, the sets $\{c\in K_v|v(bc+1)=v(a)+2\},$ $\{c\in K_v|v(c)=1\}$ and $\{c\in K_v|v(bc+1)=1\}$ are nonempty open subsets of ${{\mathcal {O}}}_{K_v}.$

If $v\in S'',$ by Lemma 2.7, the set $\{c\in {{\mathcal {O}}}_{K_v}^\times |(a,c)_v=1\}$ is a nonempty open subset of ${{\mathcal {O}}}_{K_v}.$ Also by Lemma 2.1, we can choose a nonzero element $c\in {{\mathcal {O}}}_K[1/2]$ satisfying the following conditions:

• $0<\tau _v(c)<-1/\tau _v(b)$ for all $v\in S\cap \infty _K,$

• $\tau _v(bc+1)<0$ for all $v\in (S'\backslash S)\cap \infty _K,$

• $v(bc+1)=v(a)+2$ for all $v\in S'\backslash \infty _K,$

• $(a,c)_v=1$ for all $v\in S'',$

• $v_1(c)=1$ and $v_2(bc+1)=1$ for the chosen $v_1,v_2$ above.

Let $P(x)=(x^2-c)(bx^2-bc-1),$ and let $V_2$ be the Châtelet surface given by $y^2-az^2=(x^2-c)(bx^2-bc-1).$

Proposition 5.1

For any extension of number fields L/K,  and any finite subset $S \subset \Omega _K\backslash (\infty _K^c\cup 2_K)$ splitting completely in L,  there exists a Châtelet surface $V_2$ defined over K,  which has the following properties.

• The Brauer group $Br (V_2)/Br (K)\cong Br (V_{2L})/Br (L)\cong {\mathbb {Z}}/2{\mathbb {Z}},$ is generated by an element $A\in Br (V_2).$ The subset $V_2(\text{\AA }_K)\subset V_2(\text{\AA }_L)$ is nonempty.

• For any $v\in \Omega _K,$ and any $P_v\in V_2(K_v),$

  $$\begin{aligned} {{\,\textrm{inv}\,}}_v(A(P_v))={\left\{ \begin{array}{ll} 0&{} if\quad v\notin S,\\ 1/2 &{} if \quad v\in S. \end{array}\right. } \end{aligned}$$

• For any $v'\in \Omega _L,$ and any $P_{v'}\in V_2(L_{v'}),$

  $$\begin{aligned} {{\,\textrm{inv}\,}}_{v'}(A(P_{v'}))={\left\{ \begin{array}{ll} 0&{} if\quad v'\notin S_L,\\ 1/2 &{} if \quad v'\in S_L. \end{array}\right. } \end{aligned}$$

Proof

For the extension L/K and the finite set S,  we check that the Châtelet surface $V_2$ chosen as in Sect. 5.1 has the required properties.

Firstly, we need to check that $V_2$ has an $\text{\AA }_K$-point.

• (1) Suppose that $v\in (\infty _K\backslash S')\cup 2_K.$ Then $a\in K_v^{\times 2}.$ By Remark 3.1, the surface $V_2$ admits a $K_v$-point.

• (2) Suppose that $v\in (S'\backslash S)\cap \infty _K.$ Let $x_0=0.$ Since $\tau _v(b)>0$ and $\tau _v(bc+1)<0,$ we have $\tau _v(c)<0$ and $\tau _v((x_0^2-c)(bx_0^2-bc-1))=\tau _v(c(bc+1))>0,$ which implies that $V_2^0$ admits a $K_v$-point with $x=0.$

• (3) Suppose that $v\in S'\backslash (S\cup \infty _K).$ Take $x_0\in K_v$ such that the valuation $v(x_0)<0.$ Since $b\in {{\mathcal {O}}}_{K_v}^\times$ and $c\in {{\mathcal {O}}}_K[1/2],$ by Lemma 2.4, we have $(a,x_0^2-c)_v=(a,x_0^2)_v=1$ and $(a,bx_0^2-bc-1)_v=(a,bx_0^2)_v=(a,b)_v.$ By the choice of b,  we have $(a,b)_v=1.$ Hence $(a,(x_0^2-c)(bx_0^2-bc-1))_v=(a,b)_v=1,$ which implies that $V_2^0$ admits a $K_v$-point with $x=x_0.$

• (4) Suppose that $v\in S''.$ By the choice of a, b, c,  we have $(a,c)_v=1,$ $bc+1\in {{\mathcal {O}}}_{K_v}^\times ,$ and the valuation v(a) is even. By Lemma 2.3, we have $(a,bc+1)_v=1.$ Let $x_0=0.$ Then $(a,(x_0^2-c)(bx_0^2-bc-1))_v=(a,c(bc+1))_v=(a,c)_v(a,bc+1)_v=1,$ which implies that $V_2^0$ admits a $K_v$-point with $x=0.$

• (5) Suppose that $v\in \Omega _K^f\backslash (S'\cup S''\cup 2_K ).$ Then $v(b)=0.$ Take $x_0\in K_v$ such that the valuation $v(x_0)<0.$ Since $b\in {{\mathcal {O}}}_{K_v}^\times$ and $c\in {{\mathcal {O}}}_K[1/2],$ by Lemma 2.4, we have $(a,x_0^2-c)_v=(a,x_0^2)_v=1$ and $(a,bx_0^2-bc-1)_v=(a,bx_0^2)_v=(a,b)_v.$ Since v(a) and v(b) are both even, by Lemma 2.3, we have $(a,b)_v=1.$ So $(a,(x_0^2-c)(bx_0^2-bc-1))_v=(a,b)_v=1,$ which implies that $V_2^0$ admits a $K_v$-point with $x=x_0.$

• (6) Suppose that $v\in S\cap \infty _K.$ Let $x_0=0.$ Then by the choice of a, b, c,  we have $\tau _v(a)<0,$ $\tau _v(c)>0$ and $\tau _v(bc+1)>0.$ So $(a,(x_0^2-c)(bx_0^2-bc-1))_v=(a,c(bc+1))_v=1,$ which implies that $V_2^0$ admits a $K_v$-point with $x=0.$

• (7) Suppose that $v\in S\backslash \infty _K.$ Choose a prime element $\pi _v$ and take $x_0=\pi _v.$ By the choice of a, b, c,  we have $b,c\in {{\mathcal {O}}}_{K_v}^\times ,~v(bx_0^2)=2,$ and $v(bc+1)=v(a)+2\ge 3.$ By Lemma 2.4, we have $(a,x_0^2-c)_v=(a,-c)_v$ and $(a,bx_0^2-bc-1)_v=(a,bx_0^2)_v.$ By Hensel’s lemma, we have $-bc=1-(bc+1)\in K_v^{\times 2}.$ So $(a,(x_0^2-c)(bx_0^2-bc-1))_v=(a,-bcx_0^2)_v=1,$ which implies that $V_2^0$ admits a $K_v$-point with $x=\pi _v.$

Secondly, we need to prove the statement about the Brauer group, and find the element A in this proposition.

By the choice of the places $v_1,$ the polynomial $x^2-c$ is an Eisenstein polynomial, so it is irreducible over $K_{v_1}.$ Since $v_1(a)$ is even, we have $K(\sqrt{a})K_{v_1}\ncong K_{v_1}[x]/(x^2-c).$ So $K(\sqrt{a})\ncong K[x]/(x^2-c).$ The same argument holds for the place $v_2$ and the polynomial $bx^2-bc-1.$ Since all places of S split completely in L,  by lemma 3.4, we have $a\in {{\mathcal {O}}}_K\backslash L^2.$ By the splitting condition of $v_1, v_2,$ we have $L(\sqrt{a})\ncong L[x]/(x^2-c)$ and $L(\sqrt{a})\ncong L[x]/(bx^2-bc-1).$ So $P(x)=(x^2-c)(bx^2-bc-1)$ is separable and a product of two degree-2 irreducible factors over K and L. According to [14, Proposition 7.1.1], the Brauer group $Br (V_2)/Br (K)\cong Br (V_{2L})/Br (L)\cong {\mathbb {Z}}/2{\mathbb {Z}}.$ Furthermore, by Proposition 7.1.2 in loc. cit, we take the quaternion algebra $A=(a,x^2-c)\in Br (V_2)$ as a generator element of this group. Then we have the equality $A=(a,x^2-c)=(a,bx^2-bc-1)$ in $Br (V_2).$

Thirdly, We need to compute the evaluation of A on $V_2(K_v)$ for all $v\in \Omega _K.$

By Remark 3.2, it suffices to compute the local invariant ${{\,\textrm{inv}\,}}_v(A(P_v))$ for all $P_v\in V_2^0(K_v)$ and all $v\in \Omega _K.$

• (1) Suppose that $v\in (\infty _K\backslash S')\cup 2_K.$ Then $a\in K_v^{\times 2},$ so ${{\,\textrm{inv}\,}}_v(A(P_v))=0$ for all $P_v\in V_2(K_v).$

• (2) Suppose that $v\in (S'\backslash S)\cap \infty _K.$ By the choice of b, c,  we have $\tau _v(b)>0$ and $\tau _v(bc+1)<0.$ So, for any $x\in K,$ we have $(a,bx^2-bc-1)_v=1.$ Hence ${{\,\textrm{inv}\,}}_v(A(P_v))=0$ for all $P_v\in V_2^0(K_v).$

• (3) Suppose that $v\in S'\backslash (S\cup \infty _K).$ By the choice of b,  we have $(a,b)_v=1.$ Take an arbitrary $P_v\in V_2^0(K_v).$

  If $v(x)<0$ at $P_v,$ by Lemma 2.4, we have $(a,x^2-c)_v=(a,x^2)_v=1.$ If $v(x)> 0$ at $P_v,$ since $b,c\in {{\mathcal {O}}}_{K_v}^\times$ and $v(bc+1)=v(a)+2\ge 3,$ by Lemma 2.4, we have $(a,x^2-c)_v=(a,-c)_v.$ By Hensel’s lemma, we have $-bc=1-(bc+1)\in K_v^{\times 2}.$ So $(a,x^2-c)_v=(a,-c)_v=(a,-bc)_v=1.$ If $v(x)=0$ at $P_v,$ since $b\in {{\mathcal {O}}}_{K_v}^\times$ and $v(bc+1)=v(a)+2\ge 3,$ by Lemma 2.4, we have $(a,bx^2-bc-1)_v=(a,bx^2)_v=1.$ So ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$

• (4) Suppose that $v\in \Omega _K^f\backslash ( S'\cup 2_K ).$ Take an arbitrary $P_v\in V_2^0(K_v).$ If ${{\,\textrm{inv}\,}}_v(A(P_v))=1/2,$ then $(a,bx^2-bc-1)_v=(a,x^2-c)_v=-1$ at $P_v.$ Since v(a) is even, by Lemma 2.3, the last equality implies that $v(x^2-c)$ is odd, so it is positive. So $v(bx^2-bc-1)=0.$ By Lemma 2.3, we have $(a,bx^2-bc-1)_v=1,$ which is a contradiction. So ${{\,\textrm{inv}\,}}_v(A(P_v))=0.$

• (5) Suppose that $v\in S\cap \infty _K.$ Take an arbitrary $P_v\in V_2^0(K_v).$ If $A(P_v)=0,$ then $(a,bx^2-bc-1)_v=(a,x^2-c)_v=1$ at $P_v.$ The last equality implies that $\tau _v(x^2-c)>0.$ By the choice of b,  we have $\tau _v(b)<0,$ so $\tau _v(bx^2-bc-1)<0,$ which contradicts $(a,bx^2-bc-1)_v=1.$ So ${{\,\textrm{inv}\,}}_v(A(P_v))=1/2.$

• (6) Suppose that $v\in S\backslash \infty _K.$ By the choice of b,  we have $(a,b)_v=-1.$ Take an arbitrary $P_v\in V_2^0(K_v).$ If $v(x)\le 0$ at $P_v,$ because of $b\in {{\mathcal {O}}}_{K_v}^\times$ and $v(bc+1)=v(a)+2\ge 3,$ by Lemma 2.4, we have $(a,bx^2-bc-1)_v=(a,bx^2)_v=-1.$ If $v(x)> 0$ at $P_v,$ since $b,c\in {{\mathcal {O}}}_{K_v}^\times$ and $v(bc+1)=v(a)+2\ge 3,$ by Lemma 2.4, we have $(a,x^2-c)_v=(a,-c)_v.$ By Hensel’s lemma, we have $-bc=1-(bc+1)\in K_v^{\times 2}.$

  So $(a,x^2-c)_v=(a,-c)_v=-(a,-bc)_v=-1.$ So ${{\,\textrm{inv}\,}}_v(A(P_v))=1/2.$

Finally, we need to compute the evaluation of A on $V_2(L_{v'})$ for all $v'\in \Omega _L.$

• (1) Suppose that $v' \in S_L.$ Let $v\in \Omega _K$ be the restriction of $v'$ on K. By the assumption that v splits completely in L,  we have $K_v=L_{v'}.$ So $V_2(K_v)=V_2(L_{v'}).$ Then for any $P_{v'}\in V_2(L_{v'}),$ denote $P_{v'}$ in $V_2(K_v)$ by $P_v.$ Then by the argument already shown, the local invariant ${{\,\textrm{inv}\,}}_{v'}(A(P_{v'}))={{\,\textrm{inv}\,}}_v(A(P_v))=1/2.$

• (2) Suppose that $v'\in \Omega _L\backslash S_L.$ This local computation is the same as the case $v\in \Omega _K\backslash S.$ $\square$

Remark 5.2

If the surface $V_2$ has a K-rational point Q,  then by the global reciprocity law, the sum $\sum _{v\in \Omega _K}{{\,\textrm{inv}\,}}_v(A(Q))=0$ in ${\mathbb {Q}}/{\mathbb {Z}}.$ If the number $\sharp S$ is odd, then from Proposition 5.1, this sum is ${\sharp S}/2,$ which is nonzero in ${\mathbb {Q}}/{\mathbb {Z}}.$ So, in this case, the surface $V_2$ has no K-rational point, which implies that the surface $V_2$ is a counterexample to the Hasse principle. If the number $\sharp S$ is even, then for any $(P_v)_{v\in \Omega _K}\in V_2(\text{\AA }_K),$ by Proposition 5.1, the sum $\sum _{v\in \Omega _K}{{\,\textrm{inv}\,}}_v(A(P_v))={\sharp S}/2$ is 0 in ${\mathbb {Q}}/{\mathbb {Z}}.$ Since $Br (V_2)/Br (K)$ is generated by the element A,  we have $V_2(\text{\AA }_K)^{Br }= V_2(\text{\AA }_K)\ne \emptyset .$ According to [4, 5, Theorem B], the Brauer-Manin obstruction to the Hasse principle and weak approximation is the only one for Châtelet surfaces. So, in this case, the set $V_2(K)\ne \emptyset ,$ and it is dense in $V_2(\text{\AA }_K)^{Br }= V_2(\text{\AA }_K),$ i.e. the surface $V_2$ has a K-rational point and satisfies weak approximation. In particular, if the number $\sharp S=0,$ i.e. $S=\emptyset ,$ though the Brauer group $Br (V_2)/Br (K)$ is nontrivial, it gives no obstruction to weak approximation for $V_2.$

Combining the construction method in Sect. 5.1 with the global reciprocity law, we can relate the properties in Proposition 4.1 to the Hasse principle and weak approximation.

Theorem 5.1

For any extension of number fields L/K,  there exists a Châtelet surface V over K with $V(\text{\AA }_K)\ne \emptyset ,$ such that for every intermediate field $K\subset L'\subset L,$ $Br (V)/Br (K)\cong Br (V_{L'})/Br (L')\cong {\mathbb {Z}}/2{\mathbb {Z}}$, and that the surface $V_{L'}$ has the following properties.

• If the degree $[L':K]$ is odd, then the surface $V_{L'}$ is a counterexample to the Hasse principle, i.e. $V(L')=\emptyset .$ In particular, the surface V is a counterexample to the Hasse principle.

• If the degree $[L':K]$ is even, then the surface $V_{L'}$ satisfies weak approximation. In particular, in this case, $V(L')\ne \emptyset .$

Proof

By Theorem 2.2, we can take a place $v_0\in \Omega _K\backslash (\infty _K^c\cup 2_K)$ splitting completely in L. Let $S=\{v_0\}.$ Using the construction method in Sect. 5.1, there exists a Châtelet surface V defined over K having the properties of Proposition 5.1. For any subfield $L'\subset L$ over K,  by the same argument as in the proof of Proposition 5.1, we have $Br (V)/Br (K)\cong Br (V_{L'})/Br (L')\cong {\mathbb {Z}}/2{\mathbb {Z}}.$ Since $v_0$ splits completely in L,  it also does in $L'.$ Since $\sharp S$ is odd, $[L':K]$ is odd if and only if $\sharp S_{L'}$ is odd. Applying the same argument about the field L to its subfield $L',$ the properties that we list are just what we have explained in Remark 5.2. $\square$

Remark 5.3

The Brauer group $Br (V)/Br (K)\cong Br (V_{L'})/Br (L')\cong {\mathbb {Z}}/2{\mathbb {Z}}$ in Theorem 5.1 is nontrivial, and it gives an obstruction to the Hasse principle for V,  and also for $V_{L'}$ if $[L':K]$ is odd. But it no longer gives an obstruction to weak approximation for $V_{L'}$ if $[L':K]$ is even.

Using the construction method in Sect. 5.1, we have the following example, which is a special case of Proposition 5.1.

Example 5.4

Let $K={\mathbb {Q}},$ and let $\zeta _7$ be a primitive 7-th root of unity. Let $\alpha =\zeta _7+\zeta _7^{-1}$ with the minimal polynomial $x^3+x^2-2x-1.$ Let $L={\mathbb {Q}}(\alpha ).$ Then $[L:K]=3.$ Let $S=\{13\}.$ Since $13^2\equiv 1\mod 7,41^2\equiv 1\mod 7,$ and $43\equiv 1\mod 7,$ the places 13, 41, 43 split completely in L. Using the construction method in Sect. 5.1, we choose data: $S=\{13\},~S'=\{13,29\},~S''=\{5\},~v_1=43,~v_2=41,~a=377,~b=5,~c=878755181$ and $P(x)=(x^2-878755181)(5x^2-4393775906).$ Then the Châtelet surface given by $y^2-377z^2=P(x)$ has the properties of Proposition 5.1.