Hilbert Symbols and the Quadratic Local Global Principle

By [L1], condition 1 means that $a{x}^2+b{y}^2=1$ has a solution, and then $(x,y,1)$ is a nontrivial zero of $z^2-a{x}^2-b{y}^2$. Conversely, let $(x,y,z)$ be a nontrivial zero of $z^2-a{x}^2-b{y}^2$. If $z\mathrm{\ne }0$, dividing by $z^2$ gives a solution of $a(x\mathrm{/}z{)}^2+b(y\mathrm{/}z{)}^2=1$. If $z=0$, then $y\mathrm{\ne }0$ and $r:=x\mathrm{/}y$ satisfies $b=-a{r}^2$. The explicit choice $X:=\frac{1+a}{2a},Y:=\frac{1-a}{2ar}$ then gives $a{X}^2+b{Y}^2=1$. Thus conditions 1 and 2 are equivalent.

If condition 2 holds and $y\mathrm{\ne }0$, then dividing the identity $z^2-a{x}^2-b{y}^2=0$ by $y^2$ gives $b=(z\mathrm{/}y{)}^2-a(x\mathrm{/}y{)}^2$. If instead $y=0$, then $a=(z\mathrm{/}x{)}^2$ is a square, say $a=r^2$, and $b={\left(\frac{b+1}{2}\right)}^2-a{\left(\frac{b-1}{2r}\right)}^2\mathrm{.}$ So condition 2 implies that $b$ has the form $u^2-a{v}^2$, which is exactly the norm condition in [L2]. Conversely, if $b=u^2-a{v}^2$, then $(v,1,u)$ is a nontrivial zero of $z^2-a{x}^2-b{y}^2$, so condition 2 holds and step 1.1 returns condition 1. Hence conditions 1 and 3 are equivalent.

Steps 1.1 and 2.1 prove the three formulations equivalent.

If $(a,b{)}_w=1$, choose $x,y$ with $a{x}^2+b{y}^2=1$ by [L1]. Then $(x{u}^{-1},y{v}^{-1})$ solves $(a{u}^2)X^2+(b{v}^2)Y^2=1$, so $(a{u}^2,b{v}^2{)}_w=1$.

The same argument with $u^{-1}$ and $v^{-1}$ shows that if $(a{u}^2,b{v}^2{)}_w=1$, then $(a,b{)}_w=1$. Hence the two symbols are equal, and only the square classes matter.

If $a>0$, then $x=1\mathrm{/}\sqrt{a}$ and $y=0$ solve $a{x}^2+b{y}^2=1$, so [L1] gives $(a,b{)}_{\mathrm{\infty }}=1$; the same argument works when $b>0$.

If $a<0$ and $b<0$, then $a{x}^2\le 0$ and $b{y}^2\le 0$ for all real $x,y$, so $a{x}^2+b{y}^2$ can never equal $1$. Therefore [L1] gives $(a,b{)}_{\mathrm{\infty }}=-1$. Combining with step 1.1 proves the formula.

By [L2], only the parities of $\alpha ,\beta$ and the unit square classes of $u,v$ matter, so it is enough to treat $\alpha ,\beta \in \{0,1\}$. We also use three consequences of [L1]. First, the defining equation is symmetric in $a$ and $b$, so $(a,b{)}_p=(b,a{)}_p$. Second, if $c$ is a square then $(a,c{)}_p=1$. Third, if $(a,c{)}_p=1$ then the norm subgroup from $Q_p(\sqrt{a})$ is multiplicative, so $(a,bc{)}_p=(a,b{)}_p$; by symmetry the same cancellation rule holds in the first argument.

If $\alpha =\beta =0$, both arguments are units. Consider the sets $U:=\{u{x}^2<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p:x\in {\mathbb{F}}_p\},V:=\{1-v{y}^2<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p:y\in {\mathbb{F}}_p\}\mathrm{.}$ Each has $(p+1)\mathrm{/}2$ elements, so they intersect. Hence there exist $x_0,y_0\in {\mathbb{F}}_p$ with $u{x}_0^2+v{y}_0^2\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$. At least one of $x_0,y_0$ is nonzero, so one partial derivative of $u{x}^2+v{y}^2-1$ is nonzero at $(x_0,y_0)$ modulo $p$; [L5] lifts this solution to ${\mathbb{Z}}_p$. Therefore $(u,v{)}_p=1$, agreeing with the displayed formula when $\alpha =\beta =0$.

Suppose $\alpha =1$ and $\beta =0$. Step 2.1 gives $(u,v{)}_p=1$, so the cancellation rule from step 1.1 yields $(pu,v{)}_p=(p,v{)}_p$. If $v$ is a square unit, then [L4] and step 1.1 give $(p,v{)}_p=1$. If $v$ is a nonsquare unit and $(p,v{)}_p=1$, then [L1] gives a primitive solution of $z^2-p{x}^2-v{y}^2=0$ over ${\mathbb{Z}}_p$. The congruence $z^2\equiv v{y}^2(\mathrm{m}\mathrm{o}\mathrm{d}p)$ forces $y$ to be divisible by $p$, for otherwise [L4] would make $v$ a square in ${\mathbb{Q}}_p$. Then $z^2=p{x}^2+p^2(\cdots )$, so primitivity forces $x$ to be a unit and therefore $v_p(z^2)=1$, impossible. Hence $(p,v{)}_p=-1$ in the nonsquare case. By [L3] and [L4], this is exactly $(v\mathrm{/}p)$, so $(pu,v{)}_p=\left(\frac{v}{p}\right)\mathrm{.}$

By symmetry, the case $\alpha =0,\beta =1$ gives $(u,pv{)}_p=\left(\frac{u}{p}\right)\mathrm{.}$

When $\alpha =\beta =1$, step 1.1 gives $(-pv,pv{)}_p=1$, so $(pu,pv{)}_p=(pu,pv{)}_p(-pv,pv{)}_p=(-p^{2}uv,pv{)}_p=(-uv,pv{)}_p=(pv,-uv{)}_p\mathrm{.}$ Now apply step 3.1 with $-uv$ in place of $v$: $(pv,-uv{)}_p=\left(\frac{-uv}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{u}{p}\right)\left(\frac{v}{p}\right)=(-1{)}^{(p-1)\mathrm{/}2}\left(\frac{u}{p}\right)\left(\frac{v}{p}\right),$ where the last identity is Euler's criterion from [L3]. This matches the displayed formula for $\alpha =\beta =1$.

Steps 2.1 through 4.2 settle all four parity cases, so the claimed formula holds for all $a=p^{\alpha }u$ and $b=p^{\beta }v$.

By [L2], only the parities of $\alpha ,\beta$ and the odd unit classes modulo $8$ matter, so it is enough to treat $\alpha ,\beta \in \{0,1\}$ and $u,v\in \{\pm 1,\pm 5\}$. As in the odd-prime proof, [L1] gives three useful identities: the symbol is symmetric; $(-c,c{)}_2=1$ for every $c$; and, if $(d,c{)}_2=1$, then $(b,c{)}_2=(bd,c{)}_2$. In particular, $(2u,2v{)}_2=(2u,2v{)}_2(-2v,2v{)}_2=(-4uv,2v{)}_2=(-uv,2v{)}_2=(2v,-uv{)}_2,$ so the case $\alpha =\beta =1$ reduces to the case $\alpha =1,\beta =0$.

First suppose $\alpha =\beta =0$, so both arguments are odd units. If one of $u,v$ is $1$, then the symbol is $1$. The remaining positive cases are $(5,-1{)}_2$, $(5,5{)}_2$, and $(5,-5{)}_2$ up to symmetry: they are witnessed respectively by $5\cdot 1^2-2^2=1,5\cdot 1^2+5\cdot 2^2=5^2,(-5,5{)}_2=1.$ For the negative cases $u,v\in \{-1,-5\}$, any primitive solution of $z^2=u{x}^2+v{y}^2$ would have at least one of $x,y$ odd. If exactly one of $x,y$ were odd, then the right-hand side would be congruent to $3$ or $7$ modulo $8$; if both were odd, it would be congruent to $2$ or $6$ modulo $8$. None of these is a $2$-adic square by [L3], so these pairs have symbol $-1$. Thus $(u,v{)}_2=(-1{)}^{ϵ(u)ϵ(v)}\mathrm{.}$

Next suppose $\alpha =1$ and $\beta =0$. If $v=1$, then $(2u,1{)}_2=1$. If $v=-1$, the formula predicts $(-1{)}^{ϵ(u)}$: for $u=1$ and $u=5$ the identities $2\cdot 1^2-1^2=1,10\cdot 1^2-3^2=1$ show that the symbol is $1$, while for $u=-1$ and $u=-5$ every primitive value of $-2x^2-y^2$ or $-10x^2-y^2$ is congruent to $2$, $5$, $6$, or $7$ modulo $8$, so the symbol is $-1$ by [L3]. If $v=5$, then for any primitive solution of $z^2=2u{x}^2+5y^2$ the right-hand side is congruent modulo $8$ to one of $5$, $2u$, or $2u+5$, namely to $5$, $2$, $6$, $7$, or $3$; none is a square, so $(2u,5{)}_2=-1$. If $v=-5$, the formula predicts $(-1{)}^{ϵ(u)+1}$: for $u=-1$ and $u=-5$ the choice $x=y=1$ gives right-hand sides $-7$ and $-15$, both congruent to $1$ modulo $8$ and therefore square by [L3], so the symbol is $1$; for $u=1$ and $u=5$, the same parity check as above shows that $2u{x}^2-5y^2$ is never a square modulo $8$, so the symbol is $-1$. Therefore $(2u,v{)}_2=(-1{)}^{ϵ(u)ϵ(v)+\omega (v)}$ for every odd-unit representative $v$.

Step 3.1 and symmetry give the case $\alpha =0,\beta =1$. For odd units modulo $8$, direct calculation gives $ϵ(xy)=ϵ(x)+ϵ(y),\omega (xy)=\omega (x)+\omega (y),ϵ(-1)=1,\omega (-1)=0$ in $\mathbb{Z}\mathrm{/}2\mathbb{Z}$. In the remaining case $\alpha =\beta =1$, step 1.1 and step 3.1 therefore give the exponent $ϵ(v)ϵ(-uv)+\omega (-uv)=ϵ(u)ϵ(v)+\omega (u)+\omega (v),$ which is exactly the displayed formula. Hence the formula holds for all $a=2^{\alpha }u$ and $b=2^{\beta }v$ in ${\mathbb{Q}}_2^{\times }$.

Step [L1] descends the symbol to square classes. Symmetry is immediate from the defining equation $a{x}^2+b{y}^2=1$. The explicit formulas of [L2] are multiplicative in each argument on the square-class group, so they give bilinearity at every rational place.

To prove nondegeneracy, fix a nonsquare class $c$. Over $\mathbb{R}$, [L2] shows that $(-1,c{)}_{\mathrm{\infty }}=-1$ when $c<0$. Over ${\mathbb{Q}}_p$ for odd $p$, write $c=p^{\gamma }w$: if $\gamma$ is odd, choose a nonsquare unit $u$ so that [L2] gives $(u,c{)}_p=-1$; if $\gamma$ is even, then $w$ is a nonsquare unit and [L2] gives $(p,c{)}_p=-1$. Over ${\mathbb{Q}}_2$, the classes of $-1,2,5$ generate the square-class group and [L2] shows that each nontrivial class is detected by one of them. Hence no nontrivial square class pairs trivially with every other one.

If $a{X}^2+b{Y}^2=t$ has a solution, divide by $t$ to obtain $(a\mathrm{/}t)X^2+(b\mathrm{/}t)Y^2=1$. By [L1], this means $(a\mathrm{/}t,b\mathrm{/}t{)}_v=1$. Since $(a\mathrm{/}t)\mathrm{/}(at)=t^{-2}$ and $(b\mathrm{/}t)\mathrm{/}(bt)=t^{-2}$ are squares, [L2] gives $(at,bt{)}_v=1$.

Conversely, if $(at,bt{)}_v=1$, then [L2] gives $(a\mathrm{/}t,b\mathrm{/}t{)}_v=1$, because the two pairs differ by multiplying each entry by the same square $t^{-2}$. By [L1], there exist $X,Y$ with $(a\mathrm{/}t)X^2+(b\mathrm{/}t)Y^2=1$, and multiplying by $t$ yields $a{X}^2+b{Y}^2=t$.

If the binary form $a{X}^2+b{Y}^2$ represents $-c$, then some $(u,v)$ satisfies $a{u}^2+b{v}^2=-c$, and therefore $(u,v,1)$ is a nontrivial isotropic vector of $a{X}^2+b{Y}^2+c{Z}^2$. Conversely, let $(x,y,z)$ be a nontrivial isotropic vector. If $z\mathrm{\ne }0$, then dividing by $z^2$ shows that $a{X}^2+b{Y}^2$ represents $-c$. If $z=0$, then $y\mathrm{\ne }0$ and $r:=x\mathrm{/}y$ satisfies $b=-a{r}^2$, so for any target $t$ the explicit choice $U:=(t+a)\mathrm{/}(2a)$ and $V:=(t-a)\mathrm{/}(2ar)$ gives $a{U}^2+b{V}^2=t$. In particular, the binary form represents $-c$. Thus ternary isotropy is equivalent to representation of $-c$ by $a{X}^2+b{Y}^2$.

Apply [L1] with $t=-c$. Then the representation condition of step 1.1 is exactly $(-ac,-bc{)}_v=1$.

By [L1], after choosing a basis we may write $q(x_1,\dots ,x_n)=a_1{x}_1^2+\cdots +a_n{x}_n^2$. If some $a_i=0$, then the corresponding basis vector is a nonzero isotropic vector. So we may assume $a_1,a_2,a_3\mathrm{\ne }0$ and restrict to the ternary subform $a_1{x}^2+a_2{y}^2+a_3{z}^2$.

Let $Q=\{u^2:u\in F\}$ be the set of square classes including $0$. By [L2] and [L3], $F^{\times }$ has even order, so the nonzero squares form an index-two subgroup and $\mathrm{\mid }Q\mathrm{\mid }=(\mathrm{\mid }F\mathrm{\mid }+1)\mathrm{/}2$. The sets $a_{1}Q$ and $-a_3-a_{2}Q$ therefore each have more than half the elements of $F$, so they intersect. Hence there exist $x,y\in F$ with $a_1{x}^2=-a_3-a_2{y}^2$, and then $(x,y,1)$ is a nonzero isotropic vector for the ternary subform and therefore for $q$.

By [L1], after scaling we may write $q=a_1{X}_1^2+\cdots +a_n{X}_n^2$ with integers $a_i$. If some $a_i=0$, then $q$ already has the rational isotropic vector with $X_i=1$ and every other coordinate $0$, hence it is isotropic over every ${\mathbb{Q}}_p$ and there is nothing more to prove. So assume from now on that every $a_i$ is nonzero. Exclude the finite set of primes dividing $2a_1\cdots a_n$. For any remaining odd prime $p$, all $a_i$ are units modulo $p$, so the reduction $\bar{q}$ over ${\mathbb{F}}_p$ still has dimension $n\ge 3$.

By [L2], the reduced form $\bar{q}$ has a nonzero isotropic vector $\bar{x}$. Since some coordinate ${\bar{x}}_j$ is nonzero and $a_j\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}p)$, the partial derivative $\mathrm{\partial }q\mathrm{/}\mathrm{\partial }{X}_j=2a_j{X}_j$ is nonzero at $\bar{x}$ modulo $p$. Fix lifts of the other coordinates and view $q$ as a polynomial in $X_j$ alone; then [L3] lifts the simple root ${\bar{x}}_j$ to a $p$-adic root. Thus $q$ is isotropic over ${\mathbb{Q}}_p$. Since only finitely many primes were excluded in step 1.1, the theorem follows.

By [L1], the map $(a,b)\mapsto {\prod }_v(a,b{)}_v$ is bilinear on ${\mathbb{Q}}^{\times }\mathrm{/}({\mathbb{Q}}^{\times }{)}^2$. This square-class group is generated by the classes of $-1$, $2$, and the odd primes. Hence it is enough to check the product formula on pairs of generators.

For the pair $(-1,-1)$, [L2] gives $(-1,-1{)}_{\mathrm{\infty }}=-1$ and $(-1,-1{)}_2=-1$, while $(-1,-1{)}_p=1$ for every odd prime $p$, so the global product is $1$. For the pair $(-1,2)$, the identity $-1\cdot 1^2+2\cdot 1^2=1$ shows $(-1,2{)}_v=1$ at every place $v$, so the global product is again $1$.

Let $p$ be an odd prime. Then [L2] gives $(-1,p{)}_v=1$ for $v\mathrm{\notin }\{2,p\}$ and $(-1,p{)}_2=(-1{)}^{(p-1)\mathrm{/}2},(-1,p{)}_p=\left(\frac{-1}{p}\right)=(-1{)}^{(p-1)\mathrm{/}2}$ by the first supplement from [L3], so the product for $(-1,p)$ is $1$. The same local formulas imply $(p,p{)}_v=(-1,p{)}_v$ for every place $v$, so the pair $(p,p)$ also has global product $1$.

If $p$ is an odd prime, then $(2,p{)}_v=1$ for $v\mathrm{\notin }\{2,p\}$ and $(2,p{)}_2=(-1{)}^{(p^2-1)\mathrm{/}8},(2,p{)}_p=\left(\frac{2}{p}\right)=(-1{)}^{(p^2-1)\mathrm{/}8}$ by the second supplement from [L3], so the pair $(2,p)$ has global product $1$.

If $p$ and $q$ are distinct odd primes, then $(p,q{)}_v=1$ for $v\mathrm{\notin }\{2,p,q\}$, while $(p,q{)}_2=(-1{)}^{(p-1)(q-1)\mathrm{/}4},(p,q{)}_p=\left(\frac{q}{p}\right),(p,q{)}_q=\left(\frac{p}{q}\right)\mathrm{.}$ Quadratic reciprocity [L3] says that the product of these three terms is $1$. Every generator pair therefore has global product $1$, and bilinearity from step 1.1 gives the reciprocity law for all $a,b\in {\mathbb{Q}}^{\times }$.

For each place $v\mathrm{\ne }w$, the assumed isotropy and [L1] give $(-ac,-bc{)}_v=1$. Multiplying these identities over all $v\mathrm{\ne }w$ and using [L2] with $x=-ac$, $y=-bc$ forces $(-ac,-bc{)}_w=1$ as well.

Applying [L1] again at the place $w$ shows that $q$ is isotropic over $Q_w$.

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 $m:=\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$.