On Birch and Swinnerton-Dyer’s cubic surfaces

There is an isomorphism of varieties given by

from the surface in $\mathbb{P}^{3}_{k}$ defined by (2), to the surface in $\mathbb{P}^{3}_{k}$ defined by

where $\theta^{\prime}=(-b+a\theta)(d-c\overline{\theta})$. A subsequent isomorphism given by scaling variables results in (3). ∎

There are 9 lines, $L_{i,j}$, defined by $0=x+\phi_{i}z+\psi_{i}w$ and

and 18 lines, $L_{(i,j,k),n}$ given by $z=Ax+By$ and $w=Cx+Dy$ such that $A,\ B,\ C,$ and $D$ satisfy the system of equations,

Here the value of $n$ in $L_{(i,j,k),n}$ corresponds to the three possible solutions of the system (5) for a fixed triple $(i,j,k)$.

As a result of the definition and intersection properties of lines on cubic surfaces, we can write a nine as a $3\times 3$ matrix

such that the intersection pairing has

Let $L^{\prime}/k$ be the field of definition for the 27 lines. A triple-nine for which the individual nines are fixed by $G_{k}$ is

The Galois group $G_{k}$ permutes the first nine, fixing no skew triple. The rows of the second two nines will be permuted via the permutation action on the roots $(\phi_{0},\phi_{1},\phi_{2})$. The action of $G_{k}$ on the columns of the second nines will determine whether or not any skew triple is fixed. If $3\mid[L^{\prime}\colon L]$, then the columns of the second two nines are permuted so by Lemma 3.1 $\operatorname{H^{1}}(G_{k},\operatorname{Pic}\overline{X})\simeq\mathbb{Z}/3\mathbb{Z}$. Otherwise $9\nmid[L^{\prime}\colon k]$ and the columns in the second two nines are fixed. As the first nine is the only $G_{k}$-stable nine containing these exceptional curves, there are no other possible triple nines and thus there are no elements of order 3 in $\operatorname{Br}X/\operatorname{Br}k$. Furthermore, any $G_{k}$-stable double six will have a set of 3 skew pairs which are $G_{k}$-stable. Therefore, Lemma 3.1 the only possibility is for $\operatorname{Br}X/\operatorname{Br}k\simeq\{1\}$. ∎

If $\operatorname{Br}X/\operatorname{Br}k$ is trivial, then the triple nine as in (6) will have enough fixed skew triples to build a set of six skew lines which is $G_{k}$-stable. In particular take two disjoint columns from the second two nines. We blow down $X$ along these six skew lines to obtain a degree 9 del Pezzo surface $X^{\prime}$ defined over $k$. It is well-known that degree 9 del Pezzo surfaces satisfy the Hasse principle. So by the Lang-Nishimura lemma $X$ must also satisfy the Hasse principle (cf. [13]). ∎

Take $D=L_{0,0}+L_{1,1}+L_{1,0}-\operatorname{div}(y)$. Clearly $D$ is not principal, as the intersection pairing between $D$ and a line, say $L_{1,1}$, which can be chosen as a generator of $\operatorname{Pic}\overline{X}$ over $\mathbb{Z}$ is non-zero, that is $(D,L_{1,1})=0-1+1-1\neq 0$. Then

In most cases, the scheme given by $X\cap V(x)$ will be a genus 1 curve and will subsequently have a $k_{\mathfrak{p}}$ point by the Hasse bound. This will be the case whenever $\mathfrak{p}\nmid 3d\mathcal{O}_{k}$.

Suppose $\mathfrak{p}\mid 3\mathcal{O}_{k}$ and $\mathfrak{p}\nmid d\mathcal{O}_{k}$. Then $X\cap V(x)\to\mathbb{P}^{1}$ defined by $[0\colon y\colon z\colon w]\mapsto[z\colon w]$ is one-to-one and surjective on $k_{\mathfrak{p}}$ points. Assumption 3 provides that at least one of these points is smooth.

For the primes $\mathfrak{p}$ of $k$ dividing $d$, to show $X(k_{\mathfrak{p}})\neq\emptyset$, it will suffice to find a $K_{{\cal P}}$ point for each prime ${\cal P}$ of $K$ dividing $\mathfrak{p}$. This is a result of the fact that on cubic surfaces the existence of $k_{\mathfrak{p}}$-rational points is equivalent to that for quadratic extensions of $k_{\mathfrak{p}}$ (cf. [6, Lem. 1.3.25]).

If ${\cal P}\mid d\mathcal{O}_{K}$ and ${\cal P}$ splits over $L$ then $X_{\cal P}$ is the union of 3 lines all defined over $K_{\cal P}/{\cal P}$ and has many $K_{\cal P}$ points.

Lastly, suppose ${\cal P}\mid d\mathcal{O}_{K}$ and ${\cal P}$ remains prime in $L$. Then we are in the case of $\operatorname{v}_{\cal P}(d)\leq\operatorname{v}_{\cal P}(\theta)$. Then consider the cubic surface $X^{\prime}$ in $\mathbb{P}^{3}_{k}$ defined by the equation,

which is isomorphic to $X$. Note that this equation has $\mathcal{O}_{{\cal P}_{1}}$ coefficients since $\operatorname{v}_{1}(\theta)\geq\operatorname{v}_{1}(d)$. Modulo ${\cal P}_{1}$, the defining equation for $X^{\prime}$ becomes

where $\overline{\theta}_{1}$ and $\left(\theta/d\right)_{1}$ are the restriction of the respective constants to the quotient $\mathcal{O}_{{\cal P}_{1}}/{\cal P}_{1}$. The surface $X^{\prime}_{1}$ has a smooth point $[\theta_{1}/d\colon-1\colon 1\colon 1]$ which will lift to a $K_{{\cal P}_{1}}$ point, $[x_{0}\colon y_{0}\colon z_{0}\colon w_{0}]\in X^{\prime}(K_{{\cal P}_{1}})$. Via the isomorphism, we have $[dx_{0}\colon y_{0}\colon dz_{0}\colon dw_{0}]\in X(K_{{\cal P}_{1}})$. ∎

(The structure of this proof follows that of [10, III.5.18].) In the infinite case, we must have $\operatorname{inv}_{\infty}(\mathcal{A}_{K}(P_{\infty}))=0$, as $[L\colon K]=3$ and $\operatorname{inv}_{\infty}(\mathcal{A}_{K}(P_{\infty}))=0$ or $1/2$.

Suppose that $v$ splits completely in $L$. Then $L_{v}=K_{v}$ and $(L_{v}/K_{v},f(P_{v}))$ is trivial. Thus we must have $\operatorname{inv}_{v}(\mathcal{A}_{K}(P_{v}))=0$.

If $v$ remains prime in $L$ then $[L_{v}\colon K_{v}]=3$. Take $P_{v}=[x_{0}\colon y_{0}\colon z_{0}\colon w_{0}]\in X(K_{v})$. Via scaling, assume that $x_{0},\ y_{0},\ z_{0}$ and $w_{0}$ are integral and at least one has valuation 0. Since $\prod_{i=0}^{2}(x_{0}+\phi_{i}z_{0}+\psi_{i}w_{0})$ is a norm from $L$ to $K$, $y_{0}=0$ would imply $x_{0}=z_{0}=w_{0}=0$, which is not possible. Thus $y_{0}\neq 0$. In particular, $f(P_{v})=\frac{x_{0}+\theta y_{0}}{y_{0}}$ is defined for all $P_{v}\in X(K_{v})$.

For simplicity, set $\operatorname{v}=\operatorname{v}_{v}$ and $N=\prod_{i=0}^{2}(x_{0}+\phi_{i}z_{0}+\psi_{i}w_{0})$. If $\operatorname{v}(N)=0$, then $\operatorname{v}(y_{0})=\operatorname{v}(x_{0}+\theta y_{0})=0$. Hence $\operatorname{inv}(\mathcal{A}_{K}(P_{v}))=0$. On the other hand, suppose $\operatorname{v}(N)>0$. Here the restriction of $N$ modulo $v$ is a norm on the residue class field of $L_{v}$ to that of $K_{v}$. Thus $N$ having positive valuation implies that the restriction $\overline{N}=0$. Hence $x_{0},z_{0},w_{0}\cong 0\pmod{v}$ so $\operatorname{v}(y_{0})=0$. In fact, $3\mid\operatorname{v}(N)$. Thus, $\operatorname{v}(d)+\operatorname{v}(x_{0}+\theta y_{0})+\operatorname{v}(x_{0}+\overline{\theta}y_{0})\equiv 0\mod 3$. However, $\operatorname{v}(x_{0}+\theta y_{0})=0$ or $\operatorname{v}(x_{0}+\overline{\theta}y_{0})=0$, since $v$ does not divide both $\theta$ and $\overline{\theta}$. In particular, $\operatorname{v}(x_{0}+\theta y_{0})\equiv 0\pmod{3}$ and

From the statement and proof of Theorem 4.2, we need only consider the primes ${\cal P}_{1}$ and ${\cal P}_{2}$ of $\mathcal{O}_{K}$ that lie above $\mathfrak{p}$. Via our assumption that no primes divide both $\theta$ and $\overline{\theta}$, we can assume that ${\cal P}_{1}\mid\theta$ and ${\cal P}_{2}\nmid\theta$. Take $\operatorname{v}_{i}=\operatorname{v}_{{\cal P}_{i}}$ to be the respective valuation maps. As $\operatorname{v}_{i}(dy(x+\theta y)(x+\overline{\theta}y))\ >\ 0$, we must be in the case that $\operatorname{v}_{i}(x_{0}),\ \operatorname{v}_{i}(w_{0}),\ \operatorname{v}_{i}(z_{0})\ >\ 0$ and $\operatorname{v}_{i}(y_{0})=0$. Then $\operatorname{v}_{2}(x+\theta y)=0$, so $\operatorname{inv}_{{\cal P}_{2}}(\mathcal{A}(P))=0$. On the other hand $\operatorname{v}_{1}(x+\theta y)=\operatorname{v}_{1}(dy(x+\theta y)(x+\overline{\theta}y))-\operatorname{v}_{1}(d)\equiv-\operatorname{v}_{1}(d)\mod 3$. In particular $\operatorname{inv}_{{\cal P}_{1}}(\mathcal{A}(P))=1/3$ or $2/3$. Thus

Lemma 4.1 guarantees that $X(\mathbf{A}_{k})\neq\emptyset$. Then Theorem 4.2 implies that the only possible non-zero summand in the invariant map is that corresponding to $p$ while Theorem 4.3 implies that this single invariant map is non-zero. Thus $\Phi_{\mathcal{A}}(P)\neq 0$ for every $P\in X(\mathbf{A}_{k})$ and $X$ has a Brauer–Manin obstruction to the Hasse Principle. ∎