Hasse-Minkowski for ternary forms over Q

By [L1], diagonalize $q$ and multiply the whole form by a nonzero rational scalar so that one coefficient is $1$. Multiplying a form by a nonzero scalar does not change its isotropic vectors. Every nonzero rational square class has a squarefree integer representative, so independent nonzero rational rescalings of the other two coordinates, followed by a permutation of them, put the form in the shape $q(X,Y,Z)=Z^2-a{X}^2-b{Y}^2$ with nonzero squarefree integers $a,b$ and $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }b\mathrm{\mid }$. The local isotropy hypothesis and [L2] then say that $(a,b{)}_v=1$ for every place $v$ of $\mathbb{Q}$. We prove rational isotropy by induction on $m:=\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$.

If $m=2$, then $a,b\in \{\pm 1\}$. The real isotropy hypothesis rules out the positive-definite form $Z^2+X^2+Y^2$, so at least one of $a,b$ is $1$. Then $q(1,0,1)=0$ or $q(0,1,1)=0$, giving a rational isotropic vector.

Assume $m>2$ and that every smaller value of the squarefree-coefficient measure $\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$ satisfies the theorem. If $a$ is a rational square, then squarefreeness gives $a=1$ and $q(1,0,1)=0$, so assume that $a$ is not a rational square. Let $p$ be a prime dividing $b$. Because $q$ is isotropic over ${\mathbb{Q}}_p$, after scaling a nontrivial local solution we obtain a primitive triple $(x_p,y_p,z_p)\in {\mathbb{Z}}_p^3$ with $z_p^2-a{x}_p^2-b{y}_p^2=0.$ If $p$ divided $x_p$, then the equation would also force $p\mid z_p$, and because $p\mid b$ is squarefree it would then force $p\mid y_p$, contradicting primitivity. Therefore $x_p$ is a unit and $a\equiv (z_p\mathrm{/}{x}_p{)}^2(\mathrm{m}\mathrm{o}\mathrm{d}p)\mathrm{.}$ So $a$ is a square modulo every prime dividing $b$. By [L5], choose an integer $t$ with $t^2\equiv a(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{\mid }b\mathrm{\mid })$ and then take its least absolute residue, so $\mathrm{\mid }t\mathrm{\mid }\le \mathrm{\mid }b\mathrm{\mid }\mathrm{/}2$. Define $b_0:=\frac{t^2-a}{b}\in \mathbb{Z}\mathrm{.}$ Then $\mathrm{\mid }{b}_0\mathrm{\mid }=\frac{\mathrm{\mid }{t}^2-a\mathrm{\mid }}{\mathrm{\mid }b\mathrm{\mid }}\le \frac{\mathrm{\mid }b\mathrm{\mid }}{4}+1<\mathrm{\mid }b\mathrm{\mid },$ because $m>2$ and $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }b\mathrm{\mid }$.

If $b_0=0$, then $a=t^2$ and $(1,0,t)$ is already a nontrivial rational zero of $q$, so the induction closes immediately. Assume henceforth that $b_0\mathrm{\ne }0$, and write $b_0=d{c}^2$ with $d$ a nonzero squarefree integer and $c\in {\mathbb{Z}}_{>0}$. Then $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }{b}_0\mathrm{\mid }<\mathrm{\mid }b\mathrm{\mid }$. Fix a place $v$ and work in the quadratic algebra $A_v:=Q_v[T]\mathrm{/}(T^2-a)$ with norm $N(u+wT)=u^2-a{w}^2$. Step 1.1 and [L3] give an element ${\beta }_v\in A_v$ with $N({\beta }_v)=b$. Since this norm is nonzero, ${\beta }_v$ is a unit, with inverse $\overline{{\beta }_v}\mathrm{/}b$. Also $N(t+T)=t^2-a=b{b}_0=bd{c}^2\mathrm{.}$ The norm formula is multiplicative by direct expansion, so ${\gamma }_v:=\frac{t+T}{c{\beta }_v}$ has norm $d$. By [L3], $(a,d{)}_v=1$, and [L2] says that the squarefree smaller form $q_1(X,Y,Z):=Z^2-a{X}^2-d{Y}^2$ is isotropic over $Q_v$. This holds at every place.

Because $\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }d\mathrm{\mid }<\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }=m$, the inductive hypothesis applies to $q_1$, so it has a nontrivial rational zero $(x,y,z)$. The assumption in step 2.2 that $a$ is not a rational square forces $y\mathrm{\ne }0$: otherwise $z^2=a{x}^2$ with $(x,z)\mathrm{\ne }(0,0)$. Thus $z^2-a{x}^2=d{y}^2\mathrm{.}$ Set $w:=\frac{z+x\sqrt{a}}{y}\in \mathbb{Q}(\sqrt{a})\mathrm{.}$ By [L4], $N(w)=d$ and $N(cw)=c^{2}d=b_0$, while $N(t+\sqrt{a})=b{b}_0$. Therefore $\eta :=\frac{t+\sqrt{a}}{cw}\in \mathbb{Q}(\sqrt{a})$ has norm $b$. Writing $\eta =u+v\sqrt{a}$ gives $u^2-a{v}^2=b,$ so $(v,1,u)$ is a nontrivial rational zero of $Z^2-a{X}^2-b{Y}^2=q$. This closes the induction.

The converse implication is immediate because a rational isotropic vector remains isotropic after embedding $\mathbb{Q}$ into any completion.