Hasse-Minkowski theorem over Q

By [L1] and clearing denominators, write $q=a_1{X}_1^2+\cdots +a_n{X}_n^2$ with every $a_i\in {\mathbb{Z}}^{\times }$. Dimension $1$ is vacuous. In dimension $2$, local isotropy says that $-a_1{a}_2$ has even valuation at every finite prime and is positive at the real place, hence it is a rational square and $q$ is rationally isotropic. Dimension $3$ is [L2]. We proceed by induction on $n\ge 4$.

Suppose $n=4$. Write $q=⟨a_1,a_2⟩\perp ⟨a_3,a_4⟩$ and let $S$ consist of $\mathrm{\infty }$ and the primes dividing $2a_1{a}_2{a}_3{a}_4$. At every $v\in S$, local isotropy gives a common value $t_v$ represented by the first binary form whose negative is represented by the second. If that value is $0$, at least one of the two nondegenerate binary forms has a nontrivial zero and therefore represents every local element by the elementary parametrization used in Ternary isotropy via the Hilbert symbol; choose a nonzero value represented by the other binary form and use universality of the isotropic one to obtain a nonzero common value. Apply [L3] to obtain $t\in {\mathbb{Q}}^{\times }$ in the square class of $t_v$ at every $v\in S$, with even valuation outside $S$ except possibly at one extra odd prime $p_0$. If no exceptional prime occurs, choose any odd $p_0\mathrm{\notin }S$ as a harmless placeholder. Only finitely many of those even valuations are nonzero. Multiplying $t$ by the global square ${\left({\prod }_{p\mathrm{\notin }S\cup \{p_0\}}{p}^{-v_p(t)\mathrm{/}2}\right)}^2$ preserves every local square class and makes $v_p(t)=0$ for all $p\mathrm{\notin }S\cup \{p_0\}$.

Put $f_1=⟨a_1,a_2,-t⟩,f_2=⟨a_3,a_4,t⟩\mathrm{.}$ Both forms are isotropic at every place of $S$ by the choice of the local square classes. If $p\mathrm{\notin }S\cup \{p_0\}$, then $p$ is odd and all three coefficients of each form are units. Fact [L6], followed by [L5], makes both forms isotropic over ${\mathbb{Q}}_p$. Thus each is isotropic everywhere except possibly at $p_0$, and [L7] supplies isotropy there too. Applying the ternary theorem [L2] makes both forms rationally isotropic. An isotropic vector of $f_1$ with nonzero third coordinate scales to a representation of $t$ by $⟨a_1,a_2⟩$; if that coordinate is zero, the binary form is itself isotropic and the same elementary parametrization represents $t$. Applying the identical argument to $f_2$ gives rational identities $t=a_1{x}_1^2+a_2{x}_2^2,-t=a_3{x}_3^2+a_4{x}_4^2\mathrm{.}$ Their sum is a nontrivial rational zero of $q$.

Now suppose $n\ge 5$ and the result is known in smaller dimensions. Set $q^{\mathrm{\prime }}=⟨a_3,\dots ,a_n⟩$. By [L4], the set $S$ of places where $q^{\mathrm{\prime }}$ is not isotropic is finite. If it is empty, induction applied to $q^{\mathrm{\prime }}$ already gives a rational zero of $q$. Otherwise, fix $v\in S$ and choose a local zero of $q$. If its value $t_v:=a_1{x}_v^2+a_2{y}_v^2$ is nonzero, retain it. If $t_v=0$, anisotropy of $q^{\mathrm{\prime }}$ forces all the $q^{\mathrm{\prime }}$-coordinates of this zero to vanish, so $(x_v,y_v)$ is a nontrivial zero of the first binary form. Both coordinates are nonzero; put $r:=x_v\mathrm{/}{y}_v$, so $a_2=-a_1{r}^2$. For any $u\in Q_v$, the choices $X:=\frac{u+a_1}{2a_1},Y:=\frac{u-a_1}{2a_{1}r}$ satisfy $a_1{X}^2+a_2{Y}^2=u$, which is the elementary parametrization underlying [L5]. Choose any nonzero vector $z_v$ for $q^{\mathrm{\prime }}$; anisotropy gives $q^{\mathrm{\prime }}(z_v)\mathrm{\ne }0$, and use the displayed formula with $u=-q^{\mathrm{\prime }}(z_v)$ to obtain new $x_v,y_v$. This gives a local zero of $q$ for which $t_v=a_1{x}_v^2+a_2{y}_v^2\mathrm{\ne }0,$ and $q^{\mathrm{\prime }}$ represents $-t_v$. Make this choice at every $v\in S$.

By [L8], choose rational $x,y$ close enough to every $(x_v,y_v)$ that $t:=a_1{x}^2+a_2{y}^2$ lies in $t_v{Q}_v^{\times 2}$ for all $v\in S$. Then $⟨t⟩\perp q^{\mathrm{\prime }}$ is isotropic at every place in $S$ because $q^{\mathrm{\prime }}$ represents $-t$ there, and it is isotropic outside $S$ because $q^{\mathrm{\prime }}$ already is. This form has dimension $n-1$, so induction gives a rational zero of it. Substituting $t=a_1{x}^2+a_2{y}^2$ turns that zero into a rational zero of $q$.

Conversely, a rational isotropic vector remains isotropic after embedding $\mathbb{Q}$ into any completion.