[proofplan] We prove the easy direction by extending a rational isotropic vector to each completion. For the converse, we first show that any nonzero radical already gives a rational isotropic vector, so the remaining case is nonsingular. We then diagonalize the nonsingular rational form and invoke the nonsingular diagonal Hasse-Minkowski local-global theorem as the external arithmetic input; applying that theorem to the diagonal representative gives a rational zero. [/proofplan]

[step:Pass rational isotropy to every completion] Let $q: \mathbb{Q}^n \to \mathbb{Q}$ be a nonzero quadratic form, and suppose that $q$ is isotropic over $\mathbb{Q}$. Thus there exists $x = (x_1, \dots, x_n) \in \mathbb{Q}^n \setminus \{0\}$ such that $q(x) = 0$. For each field $K \in \{\mathbb{R}\} \cup \{\mathbb{Q}_p : p \text{ is prime}\}$, let \begin{align*} \iota_K: \mathbb{Q} &\to K \end{align*} be the standard embedding of $\mathbb{Q}$ into $K$, and let \begin{align*} q_K: K^n &\to K \end{align*} be the scalar extension of $q$, obtained by applying $\iota_K$ to the coefficients of $q$. The vector $\iota_K(x) = (\iota_K(x_1), \dots, \iota_K(x_n))$ is nonzero in $K^n$, because $\iota_K$ is injective. Since polynomial evaluation commutes with scalar extension, \begin{align*} q_K(\iota_K(x)) = \iota_K(q(x)) = \iota_K(0) = 0. \end{align*} Hence $q$ is isotropic over $\mathbb{R}$ and over $\mathbb{Q}_p$ for every prime $p$. [/step]

[step:Remove the radical and diagonalize over $\mathbb{Q}$]Assume conversely that $q$ is isotropic over $\mathbb{R}$ and over $\mathbb{Q}_p$ for every prime $p$. Let $B_q: \mathbb{Q}^n \times \mathbb{Q}^n \to \mathbb{Q}$ denote the symmetric [bilinear form](/page/Bilinear%20Form) associated to $q$, defined by \begin{align*} B_q(x,y) = \frac{1}{2}\bigl(q(x+y)-q(x)-q(y)\bigr). \end{align*} Let $R \subset \mathbb{Q}^n$ be the radical \begin{align*} R = \{x \in \mathbb{Q}^n : B_q(x,y)=0 \text{ for every } y \in \mathbb{Q}^n\}. \end{align*} For every $r \in R$, the defining identity for the polar form gives \begin{align*} q(r)=B_q(r,r)=0, \end{align*} because $r$ is orthogonal to every vector of $\mathbb{Q}^n$, in particular to itself. Hence, if $R \ne \{0\}$, any nonzero $r \in R$ is already a rational isotropic vector and the proof is complete. We may therefore assume $R=\{0\}$, so $q$ is nonsingular. For notational compatibility with the final step, set $W:=\mathbb{Q}^n$ and $q_W:=q$. By the diagonalization theorem for quadratic forms over fields of characteristic not equal to $2$, since $\operatorname{char}(\mathbb{Q})=0$, there are a rational basis $e_1, \dots, e_m$ of $W$ and nonzero coefficients $a_1, \dots, a_m \in \mathbb{Q}^{\times}$ such that the diagonal representative \begin{align*} Q: \mathbb{Q}^m &\to \mathbb{Q} \\ X=(X_1,\dots,X_m) &\mapsto a_1X_1^2 + \cdots + a_mX_m^2 \end{align*} is isometric to $q_W$. Isometry preserves isotropy over $\mathbb{Q}$ and after scalar extension to every completion. Thus it remains to show that this nonsingular diagonal form $Q$ has a nonzero rational zero.[/step]

[guided]The first technical point is that a quadratic form may have a radical. We name it explicitly: the associated symmetric bilinear form is \begin{align*} B_q: \mathbb{Q}^n \times \mathbb{Q}^n &\to \mathbb{Q}, \\ (x,y) &\mapsto \frac{1}{2}\bigl(q(x+y)-q(x)-q(y)\bigr), \end{align*} and its radical is \begin{align*} R = \{x \in \mathbb{Q}^n : B_q(x,y)=0 \text{ for every } y \in \mathbb{Q}^n\}. \end{align*} For any $r \in R$, the polar identity evaluated at $(r,r)$ gives \begin{align*} B_q(r,r)=\frac{1}{2}\bigl(q(2r)-2q(r)\bigr)=\frac{1}{2}\bigl(4q(r)-2q(r)\bigr)=q(r), \end{align*} where the equality $q(2r)=4q(r)$ is the homogeneity of a quadratic form. Since $r \in R$ implies $B_q(r,r)=0$, every vector in the radical satisfies $q(r)=0$. Therefore, if $R$ contains a nonzero vector, the theorem is already proved: that vector lies in $\mathbb{Q}^n \setminus \{0\}$ and is a rational zero of $q$. Thus the only remaining case is $R=\{0\}$. This means precisely that the polar form $B_q$ has zero radical, so $q$ is nonsingular. We set $W:=\mathbb{Q}^n$ and $q_W:=q$; this keeps the notation compatible with the final transportation step while making clear that no local isotropic vector has to be projected away from a radical component. Now we use the standard diagonalization theorem for quadratic forms over fields of characteristic not equal to $2$. Since $\operatorname{char}(\mathbb{Q})=0$, the theorem applies and gives a basis $e_1,\dots,e_m$ of $W$ and coefficients $a_1,\dots,a_m \in \mathbb{Q}^{\times}$ such that the map \begin{align*} Q: \mathbb{Q}^m &\to \mathbb{Q} \\ X=(X_1,\dots,X_m) &\mapsto a_1X_1^2 + \cdots + a_mX_m^2 \end{align*} is isometric to $q_W$. An isometry is an invertible rational linear change of variables, so it carries nonzero zeros to nonzero zeros over $\mathbb{Q}$ and, after extending scalars, over $\mathbb{R}$ and every $\mathbb{Q}_p$. Therefore the remaining problem is exactly the nonsingular diagonal case.[/guided]

[step:Apply the local-global criterion for nonsingular diagonal forms]We invoke the nonsingular diagonal Hasse-Minkowski local-global theorem as an external arithmetic prerequisite: if \begin{align*} Q: \mathbb{Q}^m &\to \mathbb{Q} \\ X=(X_1,\dots,X_m) &\mapsto a_1X_1^2 + \cdots + a_mX_m^2 \end{align*} is a nonsingular diagonal quadratic form with $a_i \in \mathbb{Q}^{\times}$ for every $1 \le i \le m$, and if $Q$ is isotropic over $\mathbb{R}$ and over $\mathbb{Q}_p$ for every prime $p$, then $Q$ is isotropic over $\mathbb{Q}$. This theorem is the standard arithmetic input proved separately from Hilbert symbols, the classification of quadratic forms over local fields, Hilbert reciprocity, and weak approximation; those ingredients are not reproved in this entry. The form $Q$ constructed in the previous step is nonsingular and diagonal with coefficients in $\mathbb{Q}^{\times}$. The local isotropy hypothesis for $q$ transfers to $Q$ because scalar extension and rational isometry preserve the equation defining isotropy. Therefore the claim applies and gives a vector $X \in \mathbb{Q}^m \setminus \{0\}$ such that \begin{align*} Q(X)=0. \end{align*} Transporting $X$ through the inverse of the rational diagonalizing isometry gives a vector $w \in W \setminus \{0\}$ with $q_W(w)=0$.[/step]

[guided]At this point the proof has been reduced to the exact form in which the arithmetic theorem is normally stated. The diagonal form is \begin{align*} Q: \mathbb{Q}^m &\to \mathbb{Q} \\ X=(X_1,\dots,X_m) &\mapsto a_1X_1^2 + \cdots + a_mX_m^2, \end{align*} with every $a_i$ nonzero, so $Q$ is nonsingular. For every completion $K \in \{\mathbb{R}\} \cup \{\mathbb{Q}_p : p \text{ prime}\}$, the scalar extension $Q_K:K^m \to K$ is isometric to the scalar extension of $q_W$. Since isotropy is exactly the existence of a nonzero solution to the polynomial equation $Q_K(X)=0$, this isometry transfers the assumed local zeros of $q$ to local zeros of $Q$. We now invoke the nonsingular diagonal Hasse-Minkowski local-global theorem stated at the beginning of this step as an external arithmetic prerequisite. Its hypotheses are precisely the facts just verified: $Q$ is nonsingular, diagonal, rational, and isotropic over $\mathbb{R}$ and over every $\mathbb{Q}_p$. The conclusion is the existence of a rational vector $X \in \mathbb{Q}^m \setminus \{0\}$ such that \begin{align*} Q(X)=0. \end{align*} The diagonalizing isometry is a rational invertible [linear map](/page/Linear%20Map) from $\mathbb{Q}^m$ onto $W$. Applying this map to $X$ gives $w \in W \setminus \{0\}$, and preservation of the quadratic form gives \begin{align*} q_W(w)=Q(X)=0. \end{align*} Thus the nonsingular part has a rational isotropic vector.[/guided]

[step:Return the isotropic vector to the original form] The inclusion map \begin{align*} \jmath: W &\to \mathbb{Q}^n \end{align*} is the rational linear inclusion determined by the direct-sum decomposition $\mathbb{Q}^n = R \oplus W$. From the previous step there exists $w \in W \setminus \{0\}$ with $q_W(w)=0$. Since $q_W$ is the restriction of $q$ to $W$, we have \begin{align*} q(\jmath(w)) = q_W(w) = 0. \end{align*} Also $\jmath(w) \ne 0$ because $\jmath$ is injective. Hence $q$ has a nonzero zero in $\mathbb{Q}^n$, so $q$ is isotropic over $\mathbb{Q}$. Combining this converse with the first step proves that $q$ is isotropic over $\mathbb{Q}$ if and only if it is isotropic over $\mathbb{R}$ and over $\mathbb{Q}_p$ for every prime $p$. [/step]