Hilbert Symbols and the Quadratic Local Global Principle -- Examples

The theorem The real Hilbert symbol formula gives value $1$ whenever at least one argument is positive, so $(1,1{)}_{\mathrm{\infty }}=(1,-1{)}_{\mathrm{\infty }}=(-1,1{)}_{\mathrm{\infty }}=1$.

The same theorem gives $(-1,-1{)}_{\mathrm{\infty }}=-1$. Equivalently, the norm form from $\mathbb{C}=\mathbb{R}(\sqrt{-1})$ is $u^2+v^2$, which never represents $-1$ over $\mathbb{R}$, matching Equivalent formulations of the Hilbert symbol.

Write $3=3^1\cdot 1$ and $2=3^0\cdot 2$. The formula from The odd-prime Hilbert symbol formula gives $(3,2{)}_3=(-1{)}^0{\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^1=(2\mathrm{/}3)$.

The nonzero square class modulo $3$ is only $1$, so $2$ is a nonsquare and $(2\mathrm{/}3)=-1$. Hence $(3,2{)}_3=-1$.

For $a=b=-1$, one has $\alpha =\beta =0$, $u=v=-1$, so $ϵ(u)=ϵ(v)=1$ and $\omega (u)=\omega (v)=0$. Hence The two-adic Hilbert symbol formula gives $(-1,-1{)}_2=(-1{)}^{1\cdot 1}=-1$.

For $a=b=5$, one has $\alpha =\beta =0$, $u=v=5$, so $ϵ(5)=0$ and $\omega (5)=1$. The exponent in the theorem is therefore $0$, and $(5,5{)}_2=1$.

The equation is the isotropy problem for $X^2+Y^2-3Z^2$, so Ternary isotropy via the Hilbert symbol says it is locally soluble over ${\mathbb{Q}}_3$ exactly when $(3,3{)}_3=1$.

Write $3=3^1\cdot 1$ twice in the odd-prime formula. Then $(3,3{)}_3=(-1{)}^{(3-1)\mathrm{/}2}=-1$, so the form is not isotropic over ${\mathbb{Q}}_3$. Therefore the conic has no ${\mathbb{Q}}_3$-point and hence no rational point.

The diagonal coefficients of $q$ are $1,2,-15$, so the only primes dividing $2abc$ are $2,3,5$. The theorem Almost all local completions are isotropic in dimension at least three guarantees that only finitely many primes can be bad; the exact finite list for this form is determined in the next step.

The criterion Ternary isotropy via the Hilbert symbol reduces the local question to the single symbol $(15,30{)}_p$. For $p\mathrm{\notin }\{2,3,5\}$ both arguments are $p$-adic units, so the odd-prime formula gives value $1$. Thus every prime outside $\{2,3,5\}$ is good, and only $2,3,5$ need separate local computation.

The criterion for $q$ is the symbol $(2,2{)}_v$. At the real place this is $1$ because both arguments are positive. At every odd prime $p$, both arguments are units, so the odd-prime formula gives $(2,2{)}_p=1$.

Since every place except possibly $2$ gives value $1$, One local place is determined by the others for ternary forms forces the remaining $2$-adic value also to be $1$. Hence $q$ is isotropic over ${\mathbb{Q}}_2$ without a separate direct computation there; globally, $(1,1,1)$ is the corresponding visible isotropic vector.

The two binary subforms $X^2+2Y^2$ and $3Z^2+6W^2$ both represent the common value $3$: indeed $1^2+2\cdot 1^2=3$ and $3\cdot 1^2+6\cdot 0^2=3$. No approximation lemma is needed in this concrete instance because the common rational value is already explicit.

Substituting these representations gives $q(1,1,1,0)=3-3=0$, so $(1,1,1,0)$ is a rational isotropic vector. This concrete calculation is exactly what Hasse-Minkowski theorem over Q guarantees once the local square classes have been matched.

Consider Selmer's cubic $3X^3+4Y^3+5Z^3=0$. It has a real point because the one-variable equation $3x^3+5=0$ has a real root, giving $(x,0,1)$. It has a $2$-adic point because $f(X)=3X^3+9$ satisfies $f(1)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and $f^{\mathrm{\prime }}(1)=9\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}2)$, so [L1] lifts the mod-$2$ root and yields a point $(x,1,1)$ in ${\mathbb{Z}}_2^3$. It has a $3$-adic point because $80\equiv -1(\mathrm{m}\mathrm{o}\mathrm{d}27)$, so for $g(Y)=Y^3-80$ one has $\mathrm{\mid }g(-1){\mathrm{\mid }}_3=3^{-4}<3^{-2}=\mathrm{\mid }{g}^{\mathrm{\prime }}(-1){\mathrm{\mid }}_3^2$; [L2] therefore gives a $3$-adic root $y$ of $g$, and then $(0,y,-4)$ lies on the cubic. It has a $5$-adic point because $h(X)=3X^3+32$ satisfies $h(1)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}5)$ and $h^{\mathrm{\prime }}(1)=9\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}5)$, so [L1] yields a $5$-adic root $x$, giving the point $(x,2,0)$.

Conrad's cited note proves that Selmer's cubic has local points over every remaining ${\mathbb{Q}}_p$ but no nontrivial rational point over $\mathbb{Q}$. Thus the curve is locally soluble at every completion while globally insoluble, refuting any naive extension of Hasse-Minkowski from quadratic forms to cubic curves.

The ternary form $X^2-Y^2-2Z^2$ is rationally isotropic, for instance at the integer vector $(3,1,2)$. This is consistent with Hasse-Minkowski theorem over Q, which is only a rational statement.

The related integral representation problem $x^2-y^2=2$ has no integer solution, because $(x-y)(x+y)=2$ and the two factors must have the same parity, while every factorization of $2$ uses one odd factor and one even factor. So rational isotropy does not automatically produce an integral representation.