Classification of Quadratic Forms over \(\mathbb Q\)

Author

Liyve

Published

September 11, 2025

Modified

July 29, 2025

Abstract
A reading guidance of the first part of J. P. Serre’s A Course in Arithmetic

Notations

  • We denote by \(K\) an arbitrary field of characteristic \(\neq 2\).

  • \(\left(\dfrac a p \right)\) being the Legendre symbol.

  • We will often denote by the same letter an element and its class modulo.

  • Assume familiarity with quadratic residues and basic knowledge of \(p\)-adic numbers and \(p\)-adic field.

Motivation and Introduction

Quadratic Forms and Quadratic Spaces over Field.

What is a Quadratic Form/Quadratic Space?

Definition 1 (Quadratic Space) Let \(V\) be a vector space (finite-dimensional) over a field \(K\) of characteristic \(\neq2\). A function \(Q: V \to K\) is called quadratic form on \(V\) satisfying:

  • \(Q(\lambda v) = \lambda^2 Q(v)\) for all \(\lambda \in K\), \(v \in V\),
  • The function \(B_Q(u, v) = Q(u + v) - Q(u) - Q(v)\) is a symmetric bilinear form on \(V\).

A quadratic space is such a pair \((V, Q)\).

  • Put \(x\cdot y=\frac{1}{2}B_Q(x,y)\). One has \(Q(x)=x\cdot x\).
  • Given a basis \(\{e_1, \ldots, e_n\}\) of \(V\), the quadratic form \(Q\) can be associated with a symmetric matrix \(A = (a_{ij})\) where \(a_{ij}=e_i\cdot e_j\).

\[ \text{If }x=\sum_{i=1}^n x_i e_i\in V\text{, then }Q(x) = \sum_{i,j=1}^n a_{ij} x_i x_j. \]

Translations

Let us consider quadratic forms in a more familiar form:

  • \(f(X)=\sum_{i,j=1}^n a_{ij} X_i X_j\) is a quadratic form in \(n\) variables over \(K\), where \(a_{ij}=a_{ji}\).

  • The pair \((K^n, f)\) is a quadratic space.

  • The matrix \(A=(a_{ij})\) is associated with \(f\).

  • Let \(f(X_1,\cdots,X_n)\) and \(g(X_1,\cdots,X_m)\) be two quadratic forms, we denote \(f\oplus g\) the quadratic form \[ f(X_1,\cdots,X_n)+g(X_{n+1},\cdots,X_{n+m}) \] in \(n+m\) variables.

Invariant: Discriminant

Change the basis \(\{e_i\}\) to another basis \(\{e'_i\}\); the associated symmetric matrix \(A\) transforms as \(A' = PAP^T\).

  • Two quadratic forms are equivalent if their matrices are congruent under such a transformation.

  • We know that any symmetric matrix can always be diagonalized by a congruence transformation.

  • Without loss of generality, assume quadratic forms are of the shape \[ f \sim \sum_{i=1}^n a_i X_i^2 \] And \(A' = PAP^T\) give us: \(\det(A)=\det(A')\det(P)^2.\)

  • This means the image of \(\det(A)\) in \(K^\times/{K^\times}^2\) is a invariant, it’s called discriminant of \(Q\), and denoted by \(d(Q)\) or simply \(d\).

Review: Classification of Quadratic Forms over \(\mathbb R\) and \(\mathbb F_q\)

The case over \(\mathbb R\)

Theorem 1 (Sylvester’s law of inertia) Let \(f=\sum_{i,j=1}^na_{ij}X_iX_j\) be a quadratic form of rank \(n\) over \(\mathbb R\). Then \[ f\sim X_1^2+X_2^2+\cdots+X_r^2-X_{r+1}^2-\cdots-X_{r+s}^2. \] where \(r\) and \(s\) are non-negative integers, and \(r+s=n\), the pair \((r,s)\) is called signature of \(f\).

By diagonalizing via congruence and factoring out squares on the diagonal, we see that the only invariants for classifying real quadratic forms are:

  • the rank \(\operatorname{rank}f = n\).

  • the signature \((r,s):=(\#\text{positive eigenvalues},\#\text{negative eigenvalues})\).

The rank and signature are invariants.

General Ideas

On an arbitrary field \(K\):

  • The rank is always an invariants. Hence we may (and we shall always) reduce to classify the non-degenerate quadratic forms of \(\operatorname{rank} n\).

  • Two quadratic forms \(f=\sum_{i\neq j}a_{ij}X_iX_j\) and \(f'=\sum_{i\neq j}a'_{ij}X_iX_j\) satisfy: there exist \(t_{ij}\in {K^\times}^2\) s.t. \(a_{ij}=t_{ij}a'_{ij}\), then \(f\sim f'\).

  • The distribution of diagonal elements in \(K^\times/{K^\times}^2\) suffices to show the equivalence

    • \(\mathbb C^\times / (\mathbb C^\times)^2 \cong \{1\}\), suffices to classify by the rank.

    • \(\mathbb R^\times / (\mathbb R^\times)^2 \cong \{1,-1\}\), signature is also needed.

    • \(\mathbb F_q^\times / (\mathbb F_q^\times)^2 \cong \{1,a\}\), where \(a \in \mathbb F_q\) isn’t a square.

    • For \(\mathbb Q_p\) and \(\mathbb Q\)?

The case over \(\mathbb F_q\)

  • Following the above discussion, we might hope that the number of squares appearing on the diagonal would serve as a sufficient criterion for equivalence. However, this is not the case.

  • Consider the non-degenerate quadratic form of rank \(2\) in \(2\) variables over \(\mathbb F_q\) with a quadratic nonresidue \(a\), \(aX^2+aY^2\sim X^2+Y^2\).

    • Do a change of basis: \(X=sX'+tY'\) and \(Y=tX'-sY'\). If we requier \(aX'^2+aY'^2=X^2+Y^2\), then \(s^2+t^2=a\). Then we must focus on the existence of solution of eqation \(s^2+t^2=a\).

    • \(s^2\) and \(a−t^2\) have both \((q+1)/2\) possible values, the pigeonhole principle implies the eqation has a nonzero solution.

  • The discriminant \(\det(A)\in \mathbb F_q^\times/{\mathbb F_q^\times}^2\) is an invariant for classifying quadratic forms.

Hilbert Symbol

The existence of nonzero solutions to the equation \(aX^2+bY^2=Z^2\) in \(K^3\) seems to be of great importance.

Definition 2 (Hilbert symbol) Let \(a, b\in K^\times\): \[ (a, b) _K= \begin{cases} 1 & \text{if $Z^2-aX^2-bY^2=0$ has a nontrivial solution in $K^3$}, \\ -1 & \text{if $Z^2-aX^2-bY^2=0$ has no nontrivial solution in $K^3$}. \end{cases} \] The number \((a,b)_K\) is called the Hilbert symbol of \(a\) and \(b\) relative to \(K\).

(When there is no ambiguity, the subscript is often omitted.)

  • The symbol may also be viewed in \(K^\times / (K^\times)^2\) when working with non-degenerate forms.

The Hilbert Symbol over \(\mathbb Q_p\)

From now on, we always assume \(K = \mathbb{Q}_p\) for a prime \(p\)：

Theorem 2 (([1] p. 20, chap. 3, sec. 1.2, theorem 1)) Say \(a = p^\alpha u\) and \(b = p^\beta v\) are \(p\)-adic numbers where \(u,v \in \mathbb Z_p^\times\), then \[ \begin{aligned} &(a, b)=(-1)^{\alpha \cdot \beta \cdot \frac{p-1}{2}} \left(\dfrac{u}{p}\right)^\beta \left(\dfrac{v}{p}\right)^\alpha &\text{ if } p \neq 2\\ &(a, b)=(-1)^{\frac{u-1}{2}\frac{v-1}{2}+\alpha\frac{v^2-1}{8}+\beta\frac{u^2-1}{8}} &\text{ if } p=2 \end{aligned} \]

  • This means that Hilbert symbol is a symmetric non-degenerate bilinear form.

The Multiplicative Formula of Hilbert Symbol

Let \(\mathbb{V}=\mathbb{P} \cup \{\infty\}\), and \(\mathbb Q_\infty=\mathbb R\). If \(a,b\in \mathbb Q^\times\), \((a,b)_v\) denotes the Hilbert symbol of their images in \(\mathbb Q_v\) for all \(v\in\mathbb{V}\).

Proposition 1 (Multiplicative formula) If \(a,b\in \mathbb Q^\times\), we have \((a,b)_v=1\) for almost all \(v\in \mathbb{V}\) and \[ \prod_{v\in\mathbb{V}}(a,b)_v=1. \]

  • We have seen that the Hilbert symbol works for rank 2, but how do we generalize to rank \(>2\)?

  • Let \(\varepsilon(f)=\prod_{i<j}(a_i,a_j)\), which is called the Hasse invariant of \(f\).

The Two Invariants

We have reduced to work with non-degenerate diagonalized quadratic forms of rank \(n\).

  • Recall that the discriminant \[ d(f) = a_1 a_2 \dots a_n \in \mathbb Q_p^\times / (\mathbb Q_p^\times)^2 \] is an invariant.

  • Recall that the Hasse invariant \[ \varepsilon(f) := \prod_{1 \leq i < j \leq n} ( a_i, a_j ) \] is also an invariant.

  • If \(f=a_1X_1^2\oplus f_1\) where \(f_1=a_2X_2^2+\cdots+a_nX_n^2\), then we have: \[ \begin{aligned} d(f) &= \prod_{i=1}^na_i = a_1 \prod_{i=2}^na_i= a_1 d(f_1).\\ \varepsilon(f) &= \prod_{1 \leq i < j \leq n} ( a_i, a_j ) = \varepsilon(f_1) \cdot ( a_1, a_2\cdots a_n) = \varepsilon(f_1) \cdot ( a_1, a_1 d(f)) \end{aligned} \]

Representation Numbers of Quadratic Forms

Decomposition of Quadratic Forms

On an arbitrary field \(K\), we say that a quadratic form \(f\) represents \(a \in K\) if there exists a nonzero \(v \in V\) such that \(f(v) = a\).

  • It may be viewed in \(\{0\} \cup K^\times / (K^{\times})^2\).

Proposition 2 (([1] p. 33, chap. 4, sec. 1.6, corollary 1)) Let \(a \in K^\times\). TFAE:

  • \(f\) represents \(a\)
  • \(f \sim g \oplus aZ^2\) where \(g\) is of rank \(\operatorname{rank} f - 1\).
  • \(f \oplus -aZ^2\) represents \(0\).
  • To check if \(a\) can be represented by \(f\), it suffices to examine when a quadratic form represents \(0\).
  • Suppose \(f_1\) and \(f_2\) can both represent some \(a\in K^\times\), then we hope to reduce their rank and use induction in subsequent proofs.

Conditions for decomposing quadratic forms

We mention some results here without details.

Theorem 3 (Witt ([1] p. 31, chap. 4, sec. 1.5, theorem 3)) Every injective metric-preserving map from a subspace \(U\) of a quadratic space \(V\) to another quadratic space \(W\) may be extended to a metric-preserving map from \(V\) to \(W\).

  • If a quadratic space \((V, Q)\) has two isometric subspaces \(U\) and \(W\), then by Witt’s theorem, the isometry can be extended to an automorphism of \(V\). By restricting this automorphism to \(U^\perp\), we see that \(U^\perp\) and \(W^\perp\) are also isometric. The results about quadratic spaces can be translated into results about quadratic forms：

Theorem 4 (Witt’s cancellation ([1] p. 34, chap. 4, sec. 1.6, theorem 4)) \(f_1 \oplus g_1 \sim f_2 \oplus g_2\) and \(g_1 \sim g_2\) implies \(f_1 \sim f_2\).

When does a quadratic form represent \(0\), \(a\) (\(a \in K^\times\))?

Theorem 5 (([1] p. 36, chap. 4, sec. 2.2, theorem 6)) \(f\) represents \(0\) iff:

  • For \(n=2\): \(d=-1\);
  • For \(n=3\): \((-1,-d)= \varepsilon\);
  • For \(n=4\): \(d \neq 1\) or \(d=1\) and \(\varepsilon = (-1,-1)\);
  • For \(n\geq5\): no conditions.

By applying Theorem to \(f_a=f \oplus -aZ^2\), we obtain:

Corollary 1 (([1] p. 37, chap. 4, sec. 2.2, corollary to themrem 6)) \(f\) represents \(a\in K^\times/K^\times\) iff:

  • For \(n=1\): \(a=d\);
  • For \(n=2\): \(( a,-d ) = \varepsilon\);
  • For \(n=3\): \(a \neq d\) or \(a=d\) and \(\varepsilon = ( -1,-d )\);
  • For \(n\geq4\): no conditions.

Quadratic Forms over \(\mathbb{Q}_p\) and \(\mathbb{Q}\)

Quadratic Forms over \(\mathbb{Q}_p\)

Quadratic Forms \(f \sim g\) over \(\mathbb{Q}_p\)

Theorem 6 (([1] p. 39, chap. 4, sec. 2.3, theorem 7)) Two non-degenerate quadratic forms of rank \(n\) over \(\mathbb Q_p\) are equivalent iff they have the same discriminant \(d\) and Hasse invariant \(\varepsilon\).

  • \(f,g\) have same \(d\) and \(\varepsilon\), thus there exists \(a\in \mathbb Q_p^\times\) which both represented by \(f\) and \(g\).

  • Then \(f \sim f_1 \oplus aZ^2\), where \(f_1\) is of rank \(n-1\).

  • \(d\) and \(\varepsilon\) of \(f_1\) can be determined:

    • \(d(f_1) = a d(f) = a d(g) = d(g_1)\)
    • \(\varepsilon(f_1) = \varepsilon(f) \cdot (a, a d(f)) = \varepsilon(g) \cdot (a, a d(g)) = \varepsilon(g_1)\)

  • Thus \(f_1, g_1\) share the same \(d\) and \(\varepsilon\). QED by induction.

Classification of Quadratic Forms over \(\mathbb{Q}_p\)

The invariants \(d\) and \(\varepsilon\) are not independent; they satisfy the following relations:

  • For \(n=1\): \(\varepsilon=1\);
  • For \(n=2\): \(d\neq-1\) or \(\varepsilon=1\);
  • For \(n\geq3\): no conditions.

Skeleton of Proof:

  • \(n=1\): \(f=aX^2\) has \(\varepsilon=1\) and \(d=a\) is arbitrary.
  • \(n=2\): \(f=aX^2+Y^2\) has \(\varepsilon=(a,b)=(a,-ab)\). If \(d=ab=-1\), then \(\varepsilon=1\). Conversely:
    • if \(d=-1\), \(\varepsilon=1\): take \(f=X^2-Y^2\)
    • if \(d\neq-1\), since the Hilbert symbol is non-degenerate, there exists \(a\in \mathbb Q_p^\times\) such that \((a,-d)=\varepsilon\). Take \(f=aX^2+adY^2\).
    • (when \(d=-1\), \(f=X^2-Y^2=aX^2+adY^2\))
  • \(n=3\): Choose \(a\in \mathbb Q_p^\times/{\mathbb Q_p^\times}^2\) and \(a\neq d\). There exsits form \(g\) of rank 2 s.t. \(d(g)=ad, \varepsilon(g)=\varepsilon(a,-d)\). The form \(f=g\oplus aZ^2\) works.
  • \(n>3\): Take \(f=g(X_1,X_2,X_3)\oplus a_4X^2\oplus\cdots\oplus a_nX_n^2\).

Quadratic Forms over \(\mathbb{Q}\)

Quadratic Forms \(f \sim g\) over \(\mathbb{Q}\)

Theorem 7 (Hasse-Minkowski) \(f\) represents \(0\) over \(\mathbb Q\) iff it represents \(0\) over \(\mathbb R\) and all \(\mathbb Q_p\).

Theorem 8 (([1] p. 44, chap. 4, sec. 3.3, theorem 9)) Two non-degenerate quadratic forms of rank \(n\) over \(\mathbb Q\) are equivalent iff they are equivalent over each \(\mathbb{Q}_v\).

  • Suppose \(f \sim g\) over \(\mathbb{Q}_v\) for all \(v\), then there exists \(a \in \mathbb{Q}\) represented by both \(f\) and \(g\).

  • Thus \(f \sim aZ^2 \oplus f_1\), \(g \sim aZ^2 \oplus g_1\), where \(\operatorname{rank} f_1 = \operatorname{rank} g_1 = n-1\).

  • By Witt’s cancellation theorem, we have \(f_1 \sim g_1\) over \(\mathbb{Q}_v\) for all \(v \in \mathbb{V}\).

  • By induction on rank \(n\), \(f_1 \sim g_1\) over \(\mathbb{Q}\), thus \(f \sim g\) over \(\mathbb{Q}\).

Classification of Quadratic Forms over \(\mathbb{Q}\)

Proposition 3 (Conclusion over \(\mathbb Q\)) The invariants \(d\) and \(\varepsilon\) are not independent; they satisfy the following relations:

  • \(\varepsilon_v=1\) for almost \(v\in\mathbb V\), and \(\prod_{v\in\mathbb{V}}\varepsilon_v=1\).
  • \(\varepsilon_v=1\) if \(n=1\) and if \(n=2\) and if the image \(d_v\) of \(d\) in \(\mathbb Q_p^\times/{\mathbb Q_p^\times}^2\) is equal to \(-1\).
  • \(r,s\geq 0\) and \(r+s=\operatorname{rank}\).
  • \(d_\infty=(-1)^s\)
  • \(\varepsilon_\infty=(-1)^{s(s-1)/2}\)

Let \(d\), \((\varepsilon_v)_{v\in \mathbb V}\), and \((r,s)\) vertify the relations above, then there exists a quadratic form of rank \(n\) over \(\mathbb Q\) having for invariants \(d\), \((\varepsilon_v)_{v\in \mathbb V}\), and \((r,s)\).

Appendix

Lemmas Required for the Proof

Lemmas

Lemma 1 Let \(f_i\in\mathbb Z_p[X_1,\cdots,X_m]\) be homogeneous polynomials with \(p\)-adic integer coefficents. TFAE:

  • The \(f_i\) have a non trivial common zero in \((\mathbb Q_p)^m\)
  • The \(f_i\) have acommon primitive zero(i.e. solution \((z,x,y)\not\equiv(0,0,0) (\operatorname{mod}p)\)) in \((\mathbb Z_p)^m\)
  • For all \(n>1\), the \(f_i\) have a common primitive zero in \((\mathbb{Z}/p^n\mathbb{Z})^m\).

Lemma 2 Let \(a,b\in K^\times\) and let \(K_b=K(\sqrt{b})\). For \((a,b)=1 \iff a\in N(K_b^\times)\) of norms of elements of \(K_b^\times\).

Lemmas

Lemma 3 Let \(f=g\oplus -h\), TFAE:

  • \(f\) represents \(0\)
  • There exists \(a\in K^\times\) which is represented by \(g\) and \(h\).

Theorem 9 Let \((a_i)_{i\in I}\) be a finite family of elements in \(\mathbb Q^\times\) and let \((\varepsilon_{i,v})_{i\in I, v\in \mathbb V}\) be a family of numbers equal to \(\pm1\). In order that there exists \(x\in \mathbb Q\times\) such that \((a_i,x)_v=\varepsilon_{i,v}\) for all \(i\in I\) and \(v\in \mathbb V\) iff the following conditions be satisfied:

  • Almost all the \(\varepsilon_{i,v}=1\)
  • \(\prod_{v\in \mathbb{V}}\varepsilon_{i,v}=1\) for all \(i\in I\)
  • For all \(v\in \mathbb{V}\) there exists \(x_v\in \mathbb{Q}_p^\times\) such that \((a_i,x_v)_v=\varepsilon_{i,v}\) for all \(i\in I\).

Lemmas

Theorem 10 (Approximation Theorem) Let \(S\subseteq\mathbb V\) be a finite set. The image of \(\mathbb Q\) in \(\prod_{v\in S}\mathbb Q_v\) is dense.

Lemma 4 All quadratic forms in at least 3 variavles over \(\mathbb{F}_q\) have a non trivial zero.

Lemma 5 Suppose \(p\neq2\). Let \(f\) be a quadratic form with coefficients in \(\mathbb{Z}_p\) whose discriminant \(\operatorname{det(a_{ij})}\) is invertible. Let \(a\in \mathbb{Z}_p\), every primitive solution of the equation \(f(x)\equiv a(\operatorname{mod}p)\) lifts to a true solutons.

Proof of Hasse-Minkowski Theorem

Proof

Theorem 11 (Hasse-Minkowski) \(f\) represents \(0\) over \(\mathbb Q\) iff it represents \(0\) over \(\mathbb R\) and all \(\mathbb Q_p\).

  • The necessity is trivial. W.L.O.G., \(f=\sum_{i=1}^na_iX_i^2\), \(a_i\in\mathbb Q^\times\). By replacing \(f\) by \(a_1f\), we can soppose \(a_i=1\)

  • \(n=2\): Suppose \(f=X_1^2-aX_2^2\)

    • \(f_\infty\) represents \(0\) implies \(a>0\). Let \(a=\prod_{p \text{ prime}}p^{\nu_p(a)}\).
    • \(f_v\) represents \(0\) implies that \(\nu_p(a)\) is even. Then \(a\) is a square, \(f\) represents \(0\) over \(\mathbb Q\).

Proof

  • \(n=3\): Suppose \(f=X_1^2-aX_2^2-bX_3^2\), we can assume \(a,b\) are square-free and \(|a|\leq|b|\). Proceed by induction on \(m=|a|+|b|\).

  • If \(m=2\), then \(f=X_1^2\pm X_2^2\pm X_3^2\).

    • \(f_\infty\) represents \(0\) implies \(f\neq X_1^2+X_2^2+X_3^2\).
    • In other cases, \(f\) represents \(0\) by \(f(1,1,0)\).

  • If \(m>2\), then \(b\geq2\), let \(b=\pm p_1\cdots p_k\).

  • We need to show \(a\) is a square modulo \(p_i\) for all \(i\).

Proof

  • It is obvious if \(a\equiv0 \pmod{p_i}\).
  • Otherwise, \(a\) is a \(p_i\)-adic unit.
  • By hypothesis, \(f=X_1^2-aX_2^2-bX_3^2\) represents \(0\), i.e. \(z^2-ax^2-by^2\) has a nontrivial zero in \((\mathbb Q_{p_i})^3\).
  • By the lemma, \(z^2-ax^2-by^2\) has a primitive zero \((z,x,y)\) in \((\mathbb Z_{p_i})^3\).
  • We have \(z^2-ax^2\equiv0\pmod{p_i}\).
  • If \(x\equiv0\pmod{p_i}\), then \(z\equiv0\pmod{p_i}\).
  • Then \(p_i^2\mid by^2=z^2-ax^2\), but \(\nu_{p_i}(b)=1\) implies \(y\equiv0\pmod{p_i}\).
  • Thus \((z,x,y)\equiv(0,0,0) \pmod{p_i}\), which is a contradiction, hence \(x\not\equiv0\pmod{p_i}\).
  • Moreover, \(a=\left(\frac{z}{x}\right)^2\) is a square modulo \(p_i\).
  • Since \(\mathbb{Z}/b\mathbb{Z}\cong\prod_{i=1}^k\mathbb{Z}/p_i\mathbb{Z}\), \(a\) is a square modulo \(b\).

Proof

  • There exist \(t, b'\) integers such that \(t^2=a+bb'\).
  • We can choose \(t\) such that \(|t|\leq|\frac{b}{2}|\). \(bb'=t^2-a\) is a norm from \(K(\sqrt{a})\) where \(K=\mathbb Q\) or \(\mathbb{Q}_p\).
  • By above lemma \((a,bb')=1\), hence \((a,b)=1\iff(a,b')=1\).
  • That means \(f=X_1^2-aX_2^2-bX_3^2\) represents \(0\) iff \(f'=X_1^2-aX_2^2-b'X_3^2\) represents \(0\).
  • \(|b'|=|\frac{t^2-a}{b}|\leq|\frac{t^2}{b}|+|\frac{a}{b}|\leq\frac{|b|}{4}+1\leq|b|\).
  • Write \(b'=u^2b''\), where \(b''\) is square-free. We have \(|b''|\leq|b|\).
  • The inductive hypothesis applies to \(f''=X_1^2-aX_2^2-b''X_3^2\), so it represents \(0\), and the same is true for \(f\).

Proof

  • \(n=4\): Suppose \(f=aX_1^2+bX_2^2-(cX_3^2+dX_4^2)\). There exists \(a\in K^\times\) which is represented by \(aX_1^2+bX_2^2\) and \(cX_3^2+dX_4^2\).
    • \((x_v,-ab)_v=(a,b)_v\) and \((x_v,-cd)_v=(c,d)_v\) for all \(v\in \mathbb{V}\)
    • By above theorem there exists \(x\in \mathbb{Q}^\times\) s.t. \((x,-ab)_v=(a,b)_v\) and \((x,-cd)_v=(c,d)_v\) for all \(v\in \mathbb{V}\)
    • This means \(aX_1^2+bX_2^2\) and \(cX_3^2+dX_4^2\) represents \(x\) in \(\mathbb{Q}_p\)
    • i.e. \(aX^2+bY^2-xZ^2\) represents \(0\) in all \(\mathbb{Q}_v\) also \(\mathbb{Q}\), and the same argument applied to \(cX_3^2+dX_4^2\), the fact that \(f\) represents \(0\) in \(\mathbb{Q}\) follows from this.

Proof

  • \(n\geq5\): we use induction on \(n\). Suppose \(f=h\oplus -g\) with \(h=a_1X_1^2+a_2X_2^2\), \(g=-(a_3X_3^2+\cdots+a_nX_n^2)\).
  • Let \(S=\{p\in \mathbb{V}\mid\nu_{p}(a_i)\neq0, i\geq3\}\cup\{2,\infty\}\), it is a finite set.
  • Let \(v \in S\), \(f_v\) represents \(0\), so there exists \(a_v \in \mathbb{Q}_v^\times\) which is represented by both \(h\) and \(g\) in \(\mathbb{Q}_v\).
  • That is, there exist \(x_1^{(v)}, x_2^{(v)} \in \mathbb{Q}_v\) such that \(h(x_1^{(v)}, x_2^{(v)}) = a_v\), and \(x_3^{(v)}, \ldots, x_n^{(v)} \in \mathbb{Q}_v\) such that \(g(x_3^{(v)}, \ldots, x_n^{(v)}) = a_v\).
  • The set \(\mathbb Q_v^{\times 2}\) is open in \(\mathbb{Q}_v^\times\), so \(\prod a_v Q_v^{\times 2}\) is also open in \(\prod_{v \in S} \mathbb{Q}_v^\times\), and \(h\) is a continuous map.

Proof

  • By the Approximation Theorem, there exists \(a \in \mathbb{Q}^\times\) such that \(a \in a_v Q_v^{\times 2}\) for all \(v \in S\).

  • Thus, \((x_1,x_2)\in(\mathbb{Q})^2\) s.t. \(h(x_1,x_2)=a\), and \(a/a_v\in {\mathbb{Q}^\times}^2\) for all \(v\in S\).

  • Consider \(f_1=aZ^2\oplus -g\).

    • if \(v\in S\), \(g\) represents \(a_v\), also \(a\) since \(a/a_v\in {\mathbb{Q}^\times}^2\).
    • if \(v\not\in S\), the coeffients are \(v\)-adic units, the \(d(g)\) is also a unit. And because \(v\neq2\), we have \(\varepsilon(g)=1\).

  • By above lemma, there exist a solution, and it lifts a true solution.

  • In all case we see \(f_1\) represents \(0\) in \(\mathbb{Q}_v\), and \(\operatorname{rank}f_1=n-1\).

  • By inductive hypothesis: \(f_1\) represents \(0\) in \(\mathbb{Q}\). i.e. \(g\) represents \(a\), and \(h\) represents \(a\).

  • The proof is complete.

References

[1]
J.-P. Serre, A course in arithmetic, vol. 7. in Graduate texts in mathematics, vol. 7. New York, NY: Springer, 1973. doi: 10.1007/978-1-4684-9884-4.