p-adic analysis 3

Shu Sasaki

26th February 2011

1 Hasse-Minkowski theorem

Hasse-Minkowski: Let f(X_{1}, …, X_{n}) ∈ Q[X_{1}, …, X_{n}] be a quadratic form, i.e., a homogeneous polynomial of degree 2 in X_{1}, …, X_{n}. Then f(X_{1}, …, X_{n}) has a solution in Q if and only if f has a solution in Q_{p} for every p, including p = ∞. Prove this by following Serre’s Cours d’arithmétique (1995).

1.1 Quadratic forms over a field F

1.1.1 Quadratic forms in the new sense (after Serre)

Let A be a commutative ring. Let V be a module over A. A function Q : V → A is called a quadratic form on V (over A) if (1) Q(ax) = a^{2}Q(x) for a ∈ A and x ∈ V, (2) The function (x, y) ↦ Q(x+y)−Q(x)−Q(y) is a bilinear form on V. In what follows, we shall take A to be a field F of characteristic ≠ 2, e.g. Q_{p}, R, Q. Furthermore we assume dim_{F} V < ∞.

Define q to be (x, y) ↦ q(x, y) = \frac{1}{2}(Q(x + y) − Q(x) − Q(y)) which “makes sense” since the characteristic of F is not 2. This is a symmetric (q(x, y) = q(y, x) since V is additively abelian as well as F is commutative) bilinear (by (2)) form on V and call it the scalar product (“le produit scalaire”) associated to Q. Clearly q(x, x) = \frac{1}{2}(Q(2x) − 2Q(x)) = \frac{1}{2}(4Q(x) − 2Q(x)) = Q(x). The correspondence Q ↦ q defines a bijection between quadratic forms and symmetric bilinear forms on V over a field F of characteristic ≠ 2.

A morphism of (V, Q) into (V′, Q′) is a liner map f : V → V′ such that Q′ ∘ f = Q. In which case, q′(f(x), f(y)) = q(x, y).

Choose a basis e = (e_{1}, …, e_{n}) of V. The matrix A = (a_{ij}) associated to Q with respect to e is defined by a_{ij} = q(e_{i}, e_{j}). Since q is symmetric, a_{ij} = a_{ji} and A is a symmetric n-by-n matrix with entries in F. By definition, if x = ∑_{i} x_{i}e_{i}, Q(x) = ∑_{i,j} a_{ij}x_{i}x_{j}.

If we change e by an invertible matrix M, the matrix associated to Q is conjugated as MAM^{t}. This multiply the determinant by (detM)^{2}. We define discQ, the discriminant of Q, to be detA in F^{×}/F^{×2}, which is independent of the choices.

Let (V, Q) be a quadratic form over F. Two elements x, y of V are called orthogonal if q(x, y) = 0. The set of elements in V orthogonal to a subset W ⊆ V is denoted by W^{⊥}. If V_{1} and V_{2} are two vector subspaces of V, they are said to be orthogonal if V_{1} ⊂ V_{2}^{⊥}. The codimension of the orthogonal complement V^{⊥} in V, i.e., dimV − dimV^{⊥}, is called the rank of Q.

If V^{⊥} = {0}, say Q is non-degenerate. Equivalently, discQ ≠ 0 in F^{×}/F^{×2}.

Let W_{1}, …, W_{m} be the vector subspaces of V. Say V is the orthogonal direct sum of the W_{i} if they are pairwise orthogonal and V is the direct sum of the W_{i}.

A basis e = (e_{1}, …, e_{n}) of a quadratic form (V, Q) is called orthogonal if V is the orthogonal direct sum of the F e_{i}. In which case, its elements are pairwise orthogonal. Furthermore, the matrix of Q with respect to the basis is the diagonal matrix diag(a_{1}, …, a_{n}), and if x = ∑_{i} x_{i}e_{i} then Q(x) = a_{1}x_{1}^{2} + ⋯ a_{n}x_{n}^{2}.

THEOREM Every quadratic form (V, Q) has an orthogonal basis. Proof. By induction on the dimension of V. □

An element x of (V, Q) is isotropic if Q(x) = 0. A subspace U of V is called isotropic if all its elements are isotropic.

A (two-dimensional) quadratic form having a basis consisting of two isotropic elements x, y such that q(x, y) ≠ 0 (i.e., not orthogonal) is called a hyperbolic plane.

PROPOSITION. Let (V, Q) be a non-degenerate quadratic form over F. Let x be a non-zero isotropic element of (V, Q). Then there exists a subspace U of V which contains x and which is a hyperbolic plane. Proof. Since (V, Q) is non-degenerate, there exists v ∈ V such that q(x, v) ≠ 0 (otherwise every element of V was orthogonal to x, which would imply that F x ⊆ V^{⊥} ≠ {0} and V^{⊥} ≠ {0} contradicting the assumption that (V, Q) is non-degenerate). Let z = v/q(x, v). Then q(x, z) = 1. Let y = 2z − q(z, z)x. Then Q(y) = Q(2z − q(z, z)x) = q(2z − q(z, z)x, 2z − q(z, z)x) = 4q(z, z) − 2q(z, z)q(x, z) − 2q(z, z)q(z, x) + q(z, z)^{2}q(x, x) = 0) and therefore y is isotropic. On the other hand q(x, y) = q(x, 2z − q(z, z)x) = 2q(x, z) − q(z, z)q(x, x) = 2q(x, z) = 2 ≠ 0 and hence U = F x + F y is a hyperbolic plane containing x, as desired. □

COROLLARY. Let (V, Q) be a non-degenerate quadratic form over F and contains a non-zero isotropic element. Then Q(V) = F. Proof. By the proposition, we may assume V is a hyperbolic plane. Let x, y be a basis and WLOG assume q(x, y) = 1. If a ∈ F, then a = Q(x + (a/2)y) (since Q(x + (1/2)y) = q(x + (a/2)y, x + (a/2)y) = 2(a/2)q(x, x) = a) and therefore Q(V) = F. □

1.1.2 Quadratic forms in the classical sense

Let f(X_{1}, …, X_{n}) = ∑_{i=1}^{n} a_{ii}X_{i} + 2 ∑_{i<j} a_{ij}X_{i}X_{j} be a quadratic form (in the classical sense) in n variables over F. Define a_{ji} = a_{ij} if i > j and thus the symmetric matrix A = (a_{ij}). Then (F^{n}, f) is a quadratic form in the sense of Serre as above (“Q” on V = F^{n} is a function which sends (x_{1}, …, x_{n}) to f(x_{1}, …, x_{n}) ∈ F)with its “representation matrix” (with respect to the standard basis A.

Say two quadratic forms f and f′ in the classical sense are equivalent if (F^{n}, f) and (F^{n′}, f′) (necessarily n = n′) are isomorphic. In which case write f ∼ f′. Clearly this is equivalent to saying that A and A′ are conjugate.

Let f(X_{1}, …, X_{n}) and g(X_{1}, …, X_{m}) are two quadratic forms over F. Then by f + g we mean the quadratic form (X_{1}, …, X_{n}, X_{n+1}, …, X_{n+m}) ↦ f(X_{1}, …, X_{n})+g(X_{n+1}, …, X_{n+m}) in n+m variables. In the same vein, by f − g we mean f + (−g).

A form f = f(X_{1}, X_{2}) in n = 2 variables is called hyperbolic if f ∼ X_{1}X_{2} ∼ X_{1}^{2} − X_{2}^{2}.

Say a form f(X_{1}, …, X_{n}) represents an element a ∈ F if there exists (x_{1}, …, x_{n}) ∈ F^{n}, (x_{1}, …, x_{n}) ≠ (0, …, 0), such that f(x_{1}, …, x_{n}) = a.

PROPOSITION If f represents 0 and is non-degenerate, f ∼ f_{2} +g with f_{2} hyperbolic. In particular, f represents all elements of F. Proof. This follow from the proposition and corollary in the preceding section. □

COROLLARY Let g = g(X_{1}, …, X_{n−1}) be a non-degenerate form and a ∈ F^{×}. TFAE:

(1) g represents a;

(2) g ∼ h + aZ^{2} where h is a quadratic form in n − 2 variables;

(3) the form f = g − aZ^{2} represents 0.

Proof. (2) ⇒ (1): immediate. (1) ⇒ (2): if g represents a, there exists non-zero x ∈ F^{n−1} such that g(x) = a, in other words, the quadratic form (F^{n}, g) has an element x such that Q(x) = a. Then V is the direct sum of F x and its orthogonal complement (F x)^{⊥}, and hence g ∼ h + aZ^{2} where h is the quadratic form g restricted to (F x)^{⊥}. (2) ⇒ (3): immediate. (3) ⇒ (2): if f = g − aZ^{2} represents 0, there exists x_{1}, …, x_{n−1}, z ∈ F such that g(x_{1}, …, x_{n−1}) − az^{2} = 0. If z = 0 then g represents 0, in which case g, since it is non-degenerate, represent any element in F, in particular, a ∈ F. If z ≠ 0, then dividing through by z, we get g(x_{1}/z, …, x_{n−1}/z) − a = 0 and g represents a. □

COROLLARY g and h are two non-degenerate quadratic forms of rank ≥ 1, and let f = g − h. TFAE:

(1) f represents 0;

(2) there exists a ∈ F^{×} such that g and h both represent;

(3) there exists a ∈ F^{×} such that g − aZ^{2} and h − aZ^{2} represent 0.

Proof. (2) ⇔ (3): immediate from the corollary above ((1) ⇔ (3)). (2) ⇒ (1): immediate. (1) ⇒ (2); a non-trivial zero of f can be written in the form (x, y) such that g(x) = h(y). Let a denote g(x)(= h(y)) If a ≠ 0, then we are done. If a = 0 then one of the forms, say, g represents 0. Since g is non-degenerate, g represents any element of F, in particular, any non-zero value h takes. □

THEOREM Let f be a quadratic form in n variables over F. There exist a_{1}, …, a_{n} ∈ F such that f ∼ a_{1}X_{1}^{2} + ⋯ + a_{n}X_{n}^{2}. Proof. This follows from a theorem about quadratic basis stated earlier. □

Define the rank of f to be #{i|a_{i} ≠ 0}. Clearly rkf = n if and only if discf = a_{1} … a_{n} ≠ 0 if and only if f is non-degenerate.

1.2 Quadratic forms over Q_{p}

Let I_{Q} denote the set of places in Q, i.e, the prime numbers and “∞”.

1.2.1 Hilbert symbols

In this section, by F, we mean either Q_{p} or R. For a, b ∈ F^{×}, define the Hilbert symbol (a, b) ∈ {±1} of a, b (relative to F) by (a, b) = 1 if z^{2} − ax^{2} − by^{2} has a non-trivial solution (z, x, y) ≠ (0, 0, 0) in F; (a, b) = −1 otherwise. Clearly it defines a map (−, −) : F^{×}/F^{×2} × F^{×}/F^{×2} → {±1}.

THEOREM. Let F be either Q_{p} or R. For a, b ∈ F^{×}, (a, b) = 1 if and only if a ∈ N_{F (\sqrt{b})/F} F (\sqrt{b})^{×}. Proof. Suppose that b is a square: b = c^{2}. In which case, (c, 0, 1) is a solution for Z^{2} −aX^{2} −bY^{2} = 0.

Hence (a, b) = 1.

Suppose that b is not a square. Then F (\sqrt{b}) is quadratic over F. Every element of F (\sqrt{b}) is of the form z + \sqrt{b}y and N(z + \sqrt{b}y) = z^{2} − by^{2}. If a ∈ NF (\sqrt{b})^{×}, then there exists z, y ∈ F such that a = z^{2} = by^{2}, i.e., z^{2}−a−by^{2} = 0, and hence (z, 1, y) is a solution for Z^{2} = aX^{2}−bY^{2} = 0. In which case (a, b) = 1. Conversely, if (a, b) = 1, then there exist (z, x, y) ≠ (0, 0, 0) in F^{3} such that z^{2} −ax^{2} −by^{2} = 0. Note that x ≠ 0 since otherwise z^{2} = by^{2} and b would be a square. Hence (z/x)^{2} − a − b(y.x)^{2} = 0, i.e., a = (z/x)^{2} − b(y/x)^{2}, i.e., a is a norm of z/x − \sqrt{b}y/x. □

The identical proof works for proving:

“COROLLARY”. Let F = Q. Let a, b ∈ Q. There exists a solution Z^{2} − aX^{2} − bY^{2} = 0 has a solution ≠ (0, 0, 0) in Q if and only if a ∈ NF (\sqrt{b})^{×}.

EXERCISES (1) (a, b) = (b, a) and (a, c^{2}) = 1, (2) (a, −a) = 1 and (a, 1−a) = 1 (3) (a, b) = 1 implies (aa′, b) = (a′, b) (4) (a, b) = (a, −ab) = (a, (1 − a)b).

THEOREM (“Hilbert symbols in terms of Legendre symbols”)

If F = R, (a, b) = 1 if a or b is > 0 while (a, b) = −1 if a and b are < 0.

If F = Q_{p}, writing a = p^{v(a)}a′, b = p^{v(b)}b′ where a′, b′ ∈ Z_{p}^{×}, we have, if p ≠ 2

(a, b) = (−1)^{v(a)v(b) \frac{1}{2} (p−1)}(a′/p)^{v(b)}(b′/p)^{v(a)},

where by (−/p) we mean the “reduction mod p” Z_{p}^{×} → F_{p}^{×} followed by the Legendre symbol, and, if p = 2,

(a, b) = (−1)^{ϵ(a′)ϵ(b′)+v(a)ω(b′)+v(b)ω(a′)}.

where, following Serre’s Cours d’arithmetique and Corps locaux, ϵ : Z_{2}^{×} → (Z/2Z) is a homomorphism which sends x to ((x − 1)/2 mod 2) (EXERCISE: ϵ(x) = 0 if x ≡ 1 mod 4; ϵ(x) = 1 if x ≡ −1 mod 4) while ω : Z_{2}^{×} → (Z/2Z) is a homomorphism which sends x to ((x^{2} − 1)/8 mod 2) (EXERCISE: ω(x) = 0 if x ≡ ±1 mod 8; ω(x) = 1 if x ≡ ±5 mod 8). Proof. See Chapitre III, 1.2, Théorème 1 in Serre’s Cours d’arithmétique. □

LEMMA. Suppose p ≠ 2. Let a, b ∈ Z_{p}^{×}. Then (a, b) = 1.

SUBLEMMA 1 (Warning). Let k be a finite field and #k is a power of p. Let f(X_{1}, …, X_{n}) be a polynomial in k[X_{1}, …, X_{n}] of total degree d > n. Then #{x = (x_{1}, …, x_{n}) ∈ k^{n}|f(x) = 0} is divisible by p.

Proof. STP the complement {x = (x_{1}, …, x_{n}) ∈ k^{n}|f(x) ≠ 0}, which we shall denote by N, is divisible by p. Observe that for x ∈ k, x^{#−1} = 0 if x = 0 while x^{#−1} = 1 otherwise. Observe furthermore that ∑_{x∈k} x^{i} = 0 for every 0 ≤ i < #k − 1. Note that (N mod p) = ∑_{x=(x_{1},...,x_{n})∈k^{n}} f(x)^{#k−1} because if f(x) ≠ 0, then f(x)^{#k−1} = 1 as observed above. On expanding f(X_{1}, …, X_{n})^{#k−1} and subsequently evaluating at x = (x_{1}, …, x_{n}), we see sums of the form ∑_{x_{1},...,x_{n}∈k} x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}} with i_{1} + ⋯ + i_{n} ≤ (#k − 1)d < (#k − 1)n. From the inequality, there exists i such that i < (#k − 1) and therefore ∑_{x_{i}} x_{i}^{i} = 0. □

SUBLEMMA 2 (Chevalley). Suppose that a polynomial f as above is homogeneous. Then there exists x = (x_{1}, …, x_{n}) ∈ k^{n} such that f(x) = 0.

Proof. Since f is homogeneous, f(0) = 0. By SUBLEMMA 1, this means that there are at least p − 1 non-zero roots in k^{n} of f. □

Proof of the lemma. By SUBLEMMA 2, there exists z_{0}, x, y ∈ Z not all of them are divisible by p (“primitive” in the sense of Serre) such that z_{0}^{2} − ax^{2} − by^{2} ≡ 0 mod p. By symmetry, WLOG, p does not divide z_{0}. Then the polynomial f(X) = X^{2} − ax^{2} − by^{2} ∈ Z_{p}[X] satisfies f(z_{0}) ≡ 0 mod p, i.e., ||f(z_{0})|| ≤ 1/p while, since f′(z_{0}) = 2z_{0}, p ≠ 2, p |̸z_{0}, ||f′(z_{0})|| = 1. So ||f(z_{0})|| < ||f′(z_{0})||^{2} and it follows from Hensel’s lemma that there exists z ∈ Z_{p} such that f(z) = 0, i.e., z^{2} −ax^{2} −by^{2} = 0. (z, x, y) is a solution for Z^{2} − aX^{2} − bY^{2} and thus (a, b) = 1. □

COROLLARY (aa′b) = (a, b)(a′, b).

In particular,

COROLLARY The Hilbert symbol defines a non-degenerate bilinear forms on F^{×}/F^{×2}.

1.2.2 Hilbert symbols II

THEOREM. (Hilbert) If a, b ∈ Q^{×}, for all but finitely many v ∈ I_{Q}, (a, b)_{v} = 1 and

∏_{v∈I_{Q}}(a, b)_{v} = 1.

Proof See Chapitre III, 2.1, Théorm̀e 3 of Serre’s Cd’A. □

USEFUL LEMMAS (without proofs)

LEMMA (“Approximation theorem”). Let S be a finite subset of I_{Q}. Then the image of Q in ∏_{v∈S} Q_{v} (which comes equipped with the product topology) is dense. Proof. Follows from Chinese Remainder Theorem. See Chapitre III, 2.2, Lemme 2 of Serre’s Cd’A.□

LEMMA (“Dirichlet”) If m and n are relatively prime integers ≥ 1, then there exists infinitely many prime numbers p such that p ≡ m mod n. Proof. See Chapitre VI (in “Deuxième partie”), 4.1, Théorm̀e 2 of Serre’s Cd’A. □

THEOREM. Let {a_{i}}_{i∈I} be a finite set of elements in Q^{×} and let {ϵ_{i,v}}_{i∈I,v∈I_{Q}} be a set of elements ϵ_{i,v} ∈ {±1}. Then there exists x ∈ Q^{×} such that (a_{i}, x)_{v} = ϵ_{i,v} if and only if (1) almost all the ϵ_{i,v} are 1, (2) for every i ∈ I, ∏_{v∈I_{Q}} ϵ_{i,v} = 1, and (3) for every v ∈ I_{Q}, there exists x_{v} ∈ Q_{v}^{×} such that (a_{i}, x_{v})_{v} = ϵ_{i,v} for all i ∈ I.

Proof. If there exists x ∈ Q^{×} as in the theorem, then (1) and (2) follow from Hilbert’s theorem. (3) is trivial. Conversely, suppose (1), (2) and (3). After multiplying the a_{i} (by squares of integers), we may suppose the a_{i} are integers. Define S = {∞, 2}∪{v ∈ I_{Q}|∃i ∈ I, v|a_{i}} and T = {v ∈ I_{Q}|∃i ∈ I, ϵ_{i,v} = −1}. Clearly #S < ∞, while #T < ∞ because of (1).

(I) \underline{S ∩ T = ∅}: suppose S∩T = ∅. Define two positive integers m = ∏_{t∈T,t≠∞} t and n = 8 ∏_{s∈S,s≠2,∞} s. Since S ∩ T = ∅, m and n are relatively prime. By Dirichlet’s theorem, there exists a prime number p such that p ≡ m mod n with p ∉ (S ∪ T). Let x = mp. This is what we are looking for, i.e., it satisfies (a_{i}, x)_{v} = ϵ_{i,v} for all i ∈ I and v ∈ I_{Q}.

(I-1) \underline{v ∈ S}: if v ∈ S, it follows from S ∩ T = ∅ that v ∉ T and therefore ϵ_{i,v} = 1 for every i ∈ I. We shall show that, for every i ∈ I, (a_{i}, x)_{v} = 1 for v ∈ S.

(I-1-1) \underline{v = ∞}: in which case, since x > 0, (a_{i}, x)_{∞} = 1.

(I-1-2) \underline{v ≠ ∞}: since p ≡ m mod n, x = mp ≡ m^{2} mod n. Thus x ≡ m^{2} mod 8, and x ≡ m^{2} mod v ≠ 2. Since v ∉ T and v ≠ p, v divides neither x nor m, i.e., x and m are v-adic units. It then follows that x is a square y^{2}, say, in Q_{v}^{×}. Then (a_{i}, x)_{v} = (a_{i}, y^{2})_{v} = (a_{i}, y)_{v}(a_{i}, y)_{v} = 1.

(I-2) \underline{v ∉ S}: if v ∉ S, then every a_{i} is a v-adic unit, i.e., in Z_{v}^{×}. Since v ≠ 2, (a_{i}, b)_{v} = (a_{i}/v)^{v(b)} for all b ∈ Q_{v}^{×} (see the theorem in the previous (sub)section, about “Hilbert symbols in terms of Legendre symbols”).

(I-2-1) \underline{v ∉ (T ∪ \{p\})}: since v ∉ T, v does not divide m, and v ≠ p so v does not divide x = mp; the v-valuation v(x) = 0. Then, by the formula, (a_{i}, x)_{v} = (a_{i}/v)^{0} = 1. On the other hand, since v ∉ T, ϵ_{i,v} = 1 (by the definition of T) and hence (a_{i}, x)_{v} = 1 = ϵ_{i,v}.

(I-2-2) \underline{v ∈ T}: by the definition of x, every v ∈ T divides x exactly once; v(x) = 1. By the assumption (3), there exists x_{v} ∈ Q_{v}^{×} such that (a_{i}, x_{v})_{v} = ϵ_{i,v} for all i ∈ I. Since v ∈ T, for at least one i ∈ I, ϵ_{i,v} = −1. Then v(x_{v}) = 1 mod 2 (otherwise v(x_{v}) = 0 mod 2, and −1 = ϵ_{i,v} = (a_{i}, x_{v})_{v} = (a_{i}/v)^{v(x_{v})} = 0 and contradiction!). Then (a_{i}, x)_{v} = (a_{i}/v)^{v(x)} = (a_{i}/v)^{1} = (a_{i}/v)^{v(x_{v})} = (a_{i}, x_{v})_{v} = ϵ_{i,v}.

(I-2-3) \underline{v = p}: one deduces this case from what we know already, using Hilbert’s theorem. For every i ∈ I, (a_{i}, x)_{p} = ∏_{v≠p}(a_{i}, x)_{v} = ∏_{v≠p} ϵ_{i,v} = ϵ_{i,p} (the first equality follows from Hilbert’s theorem, the second follows from (I-1-1) through to (II-2-2), and the third follows from from the assumption (2).

(II) \underline{S ∩ T ≠ ∅}: suppose S ∩ T ≠ ∅. Since Q_{v}^{×2} is an open subgroup of Q_{v}^{×}, ∏_{v∈S} x_{v}Q_{v}^{×2} ⊂ ∏_{v∈S} Q_{v}^{×} is an open neighbourhood of (x_{v})_{v∈S} (where the x_{v} are those assumed in (3)). By the approximation theorem, there exists y ∈ Q^{×} which is in the neighbourhood, i.e., y ∈ ∏_{v∈S} x_{v}Q_{v}^{×2}, i.e., y/x_{v} ∈ Q_{v}^{×2}. Then (a_{i}, y)_{v} = (a_{i}, x_{v})_{v} = ϵ_{i,v} for every v ∈ S. Set ϵ′_{i,v} = ϵ_{i,v}(a_{i}, y)_{v}. Then {ϵ′_{i,v}}_{i∈I,v∈I_{Q}} satisfies (1),

(2) and (3), and ϵ′_{i,v} = ϵ_{i,v}(a_{i}, y)_{v} = ϵ_{i,v}^{2} = 1 for every v ∈ S. From the case above (with empty “S′”), there exists z ∈ Q^{×} such that (a_{i}, z)_{v} = ϵ′_{i,v} for every i ∈ I and v ∈ I_{Q}. Then let x = yz. Check this x works. □

1.2.3 Invariants

Given a quadratic form (V, Q) over Q_{p}, we may define two invariants.

(1) the discriminant d(Q) ∈ Q_{p}^{×}/Q_{p}^{×2}: if e = (e_{1}, …, e_{n}) is an orthogonal basis of V, d(Q) = a_{1} … a_{n} ∈ Q_{p}^{×} where a_{i} = q(e_{i}, e_{i}) for i = 1, …, n.

(2) ϵ(Q) ∈ {±1}: if e is an orthogonal basis and a_{i} = q(e_{i}, e_{i}), define ϵ(e) = ∏_{i<j}(a_{i}, a_{j}) where (−, −) ∈ {±1} is the Hilbert symbol.

THEOREM The number ϵ(e) does not depend on the choice of the orthogonal basis e. Proof. See Chapitre IV, 2.1, Théorm̀e 5 of Serre’s Cd’A. It is a piece of linear algebra. □

Thus we write ϵ(Q) for ϵ(e).

1.2.4 Representations of p-adic numbers

Let F denote Q_{p} for brevity.

LEMMA. (1) The number of elements in F^{×}/F^{×2} is 2^{n} with n = 2 (resp. n = 3) if p ≠ 2 (resp. p = 2).

(2) For a ∈ F^{×}/F^{×2} and ϵ = ±1, let H_{a,ϵ} denote the “hyperplane” {x ∈ F^{×}/F^{×2}|(x, a) = ϵ}. When a = 1, #H_{1,1} = 2^{n} and #H_{1,−1} = 0; when a ≠ 1, #H_{a,ϵ} = 2^{n−1}.

(3) Let a, a′ ∈ F^{×}/F^{×2} and ϵ, ϵ′ ∈ {±1}. Assume H_{a,ϵ} ≠ ∅ and H_{a′,ϵ′} ≠ ∅. Then H_{a,ϵ} ∩ H_{a′,ϵ′} = ∅ if and only if a = a′ and ϵ = −ϵ′.

Proof. (1) Proved already. (2) If a = 1, any x ∈ F^{×}/F^{×2} satisfies (x, 1) = 1. Hence H_{1,1} = F^{×}/F^{×2} and H_{1,−1} = ∅. If a ≠ 1, then the map F^{×}/F^{×2} → {±1} sending b to (a, b) is surjective and its kernel precisely H_{a,1}. Therefore #(F^{×}/F^{×2}) = #H_{a,1}#{±1}, i.e., #H_{a,1} = 2^{n}/2 = 2^{n−1}. Since H_{a,−1} is the complement of H_{a,1} in F^{×}/F^{×2}, #H_{a,−1} = 2^{n−1}. (3) Suppose H_{a,ϵ} ∩ H_{a′,ϵ′} = ∅. Since they are non-empty sets, according to (2), H_{a,ϵ} and H_{a′,ϵ′} have both size 2^{n−1} and they are complement to one another. Furthermore, H_{a,1} = H_{a′,1} and H_{a,−1} = H_{a′,−1}. Hence (x, a) = (x, a′) for every x ∈ F^{×}/F^{×2}. Since the Hilbert symbol is non-degenerate bilinear form on F^{×}/F^{×2}, this implies that a = a′. In which case ϵ = −ϵ′. □

THEOREM. Let f be a quadratic form of rank n over Q_{p}. Let d = d(f) and ϵ = ϵ(f). Then f represents 0 if and only if (1) n = 2 and d = −1, (2) n = 3 and (−1, −d) = ϵ, (3) n = 4 and either d ≠ 1 or d = 1 and ϵ = (−1, −1), (4) n ≥ 5.

Proof. f ∼ a_{1}X_{1}^{2} + ⋯ a_{n}X_{n}^{2}.

\underline{(1)}: f(X_{1}, X_{2}) = a_{1}X_{1}^{2} + a_{2}X_{2}^{2} represents 0 if and only if −a_{1}/a_{2} is a square. But in F^{×}/F^{×2}, −a_{1}/a_{2} = −a_{1}a_{2} = −d(f). So −a_{1}/a_{2} is a square if and only if d(f) = 1 in F^{×}/F^{×,2}.

\underline{(2)}: f(X_{1}, X_{2}, X_{3}) = a_{1}X_{1}^{2} + a_{2}X_{2}^{2} + a_{3}X_{3}^{2} represents 0 if and only if −a_{3}f = −a_{3}a_{1}X_{1}^{2} − a_{3}a_{2}X_{2}^{2} − a_{3}^{2}X_{3}^{2} represents 0 if and only if −a_{3}f = −a_{3}a_{1}X_{1}^{2}−a_{3}a_{2}X_{2}^{2}−X_{3}^{2} represents 0 if and only if (−a_{3}a_{1}, −a_{3}a_{2}) = 1. Expanding it, this holds if and only if (−1, −1)(−1, a_{1}a_{2}a_{3})(a_{1}, a_{2})(a_{2}, a_{3})(a_{3}, a_{1}) = 1 if and only if (−1, −d(f))ϵ(f) = 1 if and only if (−1, −d(f)) = ϵ(f).

\underline{(3)}: f(X_{1}, X_{2}, X_{3}, X_{4}) = a_{1}X_{1}^{2} + a_{2}X_{2}^{2} + a_{3}X_{3}^{2} + a_{4}X_{4}^{2} represents 0 if and only if there exists x ∈ F^{×}/F^{×2} which is represented both by g(X_{1}, X_{2}) = a_{1}X_{1}^{2}+a_{2}X_{2}^{2} and h(X_{3}, X_{4}) = −a_{3}X_{3}^{2}−a_{4}X_{4}^{2}. By (2) of the corollary below, which follows from (2) of the theorem (that we have just proved!), this holds if and only if (x, −d(g)) = ϵ(g) and (x, −d(h)) = ϵ(h), i.e., (x, −a_{1}a_{2}) = (a_{1}, a_{2}) and (x, −a_{3}a_{4}) = (−a_{3}, −a_{4}).

Define A (resp. B) to the the set of x ∈ F^{×}/F^{×,2} such that (x, −a_{1}a_{2}) = (a_{1}, a_{2}) (resp. (x, −a_{3}a_{4}) = (−a_{3}, −a_{4}). Clearly A ≠ ∅ (e.g. a_{1} ∈ A) and B ≠ ∅ (e.g. −a_{3} ∈ B). f does not represents 0 over F if and only if A ∩ B = ∅. By the preceding lemma, this holds if and only if a_{1}a_{2} = a_{3}a_{4} and (a_{1}, a_{2}) = −(−a_{3}, −a_{4}) if and only if a_{1}a_{2}a_{3}a_{4} = (a_{3}a_{4})^{2} = 1, i.e., d(f) = 1 and ϵ(f) = −(−1, −1). Indeed, unraveling the definition, ϵ(f) = (a_{1}, a_{2})(a_{1}a_{2}, a_{3}a_{4})(a_{3}, a_{4}). Which equals (a_{1}, a_{2})(a_{3}, a_{4})(a_{3}a_{4}, a_{3}a_{4}) (since a_{1}a_{2} = a_{3}a_{4}) and thus (a_{1}, a_{2})(a_{3}, a_{4})(−1, a_{3}a_{4}) (since (A, A) = (−1, A)). This is (a_{1}, a_{2})(−a_{3}, −a_{4})(−1, −1))

(Springer-Verlag (1970) edition has a typo here!) since (a_{3}, a_{4})(−1, a_{3}a_{4}) = (a_{3}, a_{4})(−1, a_{3})(−1, a_{4}) = (−a_{3}, a_{4})(−1, a_{3}) = (−a_{3}, a_{4})(−1, −1)(−1, a_{3}) = (−a_{3}, −a_{4})(−1, −1).

\underline{(4)}: STP n = 5. By the preceding lemma and (2) of the corollary below, a form of rank 2 represents as least 2^{r−1} elements in F^{×}/F^{×2}. So does any form of rank ≥ 2. Since 2^{r−1} ≥ 2, f represents at least one element of F^{×}/F^{×2} distinct from d. In which case, f ∼ aX^{2} + g where g is of rank 4 and d(g) = d/a ≠ 1. By the n = 4 case above, g then represents 0, and therefore f represents 0. □

COROLLARY. Let f be as in the theorem. Let a ∈ Q_{p}^{×}/Q_{p}^{×2}. Then f represents a if and only if (1) n = 1 and a = d (2) n = 2 and (a, −d) = ϵ, (3) n = 4 and either a ≠ −d or a = −d and ϵ = (−1, −d),

(4) n ≥ 4.

Proof. Given a quadratic form f in n ≥ 1 variables as above and a ∈ Q^{×}, define a quadratic form g in n + 1 variables by g = f − aZ^{2}. By definition d(g) = −ad(f) and ϵ(g) = (−a, d(f))ϵ(f). Corollary then follows. For example, to prove (2) in COROLLARY, use (3) in THEOREM: (a, −d(g)) = ϵ(g), i.e., (−1, ad(f)) = (−a, d(f))ϵ(f). Since (−1, ad(f)) = (−1, a)(−1, d(f)) = (a, −1)(−1, d(f)) = (a, −1)(a, d(f)^{2})(−1, d(f)) = (a, −d(f))(a, d(f))(−1, d(f)) = (a, −d(f))(−a, d(f)) and (−a, d(f)) ∈ {±1}, we have (a, −d(f)) = ϵ(f).□

1.3 Quadratic forms over Q

THEOREM(Hasse-Minkowski) A quadratic form f over Q represents 0 if and only if f_{v} (which is f thought of as a quadratic form over Q_{v} by Q ↪ Q_{v}) represents 0 for every v ∈ I_{Q}.

Proof. If f over Q represents 0, it clearly represents 0 over Q_{v} for every v ∈ I_{Q}. The essence of the theorem is in its converse. WLOG, assume f = a_{1}X_{1}^{2} + ⋯ + a_{n}X_{n}^{2} where a_{i} ∈ Q. We may furthermore assume WLOG a_{1} = 1.

\underline{n = 2}: in which case we may write f = X_{1}^{2} −aX_{2}^{2}. Since f_{∞} represents 0, a > 0. Write a = ∏_{p} p^{v_{p}(a)}. For every p, since f_{p} represents 0, a is a square in Q_{p}; and hence v_{p}(a) is even. So a is a square in Q and f represents 0.

\underline{n = 3}: we may write f(X_{1}, X_{2}, X_{3}) = X_{1}^{2} − aX_{2}^{2} − bX_{3}^{2}. Multiply a and b by squares if necessary, assume a and b are square-free integers. Furthermore, by symmetry, assume WLOG |a|_{∞} ≤ |b|_{∞}. We shall prove the case n = 3 by induction on m = |a| + |b|.

\underline{n = 3, m = 2}: in which case |a| = |b| = 1, i.e., f = X_{1}^{2} ± X_{2}^{2} ± X_{3}^{2}. Since f_{∞} represents 0, one can eliminate the possibility X_{1}^{2} + X_{2}^{2} + X_{3}^{2}. Then we are done. For example, X_{1}^{2} + X_{2}^{2} − X_{3}^{2} represents 0 as (X_{1}, X_{2}, X_{3}) = (0, 1, 1) works.

\underline{n = 3, m > 2}: fix m > 2 and suppose that the the case holds for “less than m”. We shall prove the case for m. Since m > 2, |b| > 2. Write b as the product of distinct primes. Let p be one of the prime factors.

\underline{(a/p) = 1}: in other words, a is a square mod p. Proof. By hypothesis, there exist (X, Y, Z) ∈ Q_{p}^{3} such that Z^{2} − aX^{2} − bY^{2} = 0. Let v = inf{v_{p}(X), v_{p}(Y), v_{p}(Z)}. Let (x, y, z) = p^{−v}(X, Y, Z); so (x, y, z) ∈ Z_{p}^{3}, z^{2} − ax^{2} − by^{2} = 0 and not all of {x, y, z} are divisible by p (“primitive” in the sense of Serre). Since b ≡ 0 mod p (by the definition of p), z^{2} − ax^{2} ≡ 0 mod p. If x ≡ 0 mod p, then z^{2} ≡ 0 mod p and thus z ≡ 0 mod p. Since p^{2}|x^{2} and p^{2}|z^{2}, p^{2}|by^{2}. But as p divides b exactly once, p^{2}|y^{2} and thus p|y. This contradicts \{x, y, z\} being not simultaneously divisible by p. So \underline{x ≢ 0 mod p}. Since p does not divide x, x ∈ Z_{p}^{×} and the (multiplicative) inverse x^{−1} ∈ Z_{p}^{×}. Therefore a ≡ (zx^{−1})^{2} mod p, i.e., a is a square mod p.

Since a is a square mod p for every p|b, a is a square mod b. Say a + bb′ = c^{2} for some b′ ∈ Z and c ∈ Z. One can choose b′ so that |c| ≥ |b|/2. Since bb′ = c^{2} − a, bb′ is a norm of the quadratic extension Q(\sqrt{a})/Q.

Let f = X_{1}^{2} − aX_{2}^{2} − bX_{3}^{2} and f′ = X_{1}^{2} − aX_{2}^{2} − b′X_{3}^{2}. Then

LEMMA. f has a solution ≠ (0, 0, 0) in F if and only if f′ has a solution ≠ (0, 0, 0) in F for F = Q or Q_{v}.

Proof. f represents 0 if and only if b ∈ N_{F (\sqrt{a})/F} F (\sqrt{a})^{×} if and only if b′ ∈ NF (\sqrt{a})^{×} (as NF (\sqrt{a})^{×} is a group and bb′ ∈ NF (\sqrt{a})^{×}) if and only if f′ represents 0.

For F = Q_{v}, we may also argue: bb′ ∈ N_{F (\sqrt{a})/F} F (\sqrt{a})^{×} implies (a, bb′) = 1. By the bilinearity of the local Hilbert symbol, (a, b′) = (a, b). □

Observe that |b′| = |(c^{2} − a)/b| ≤ |b|/4 + 1 < |b| as |b| ≥ 2. But b′ is not necessarily square-free and the “inductive hypothesis” does not apply just yet. Write b′ = b′′d^{2} and let f′′(X_{1}, X_{2}, X_{3}) = X_{1}^{2} − aX_{2}^{2} − b′′X_{3}^{2}. By assumption, f represents 0 in Q_{v} for every v. Which holds if and only if f′ represents 0 over Q_{v} for every v (by the lemma) if and only if f′′ represents 0 over Q_{v}. By the inductive hypothesis, since b′′ is square-free and |a| + |b′′| < |a| + |b| = m, this holds if and only if f′′ represents 0 over Q if and only if f′ represents 0 over Q if and only if f represents 0 over Q (by the lemma). □

\underline{n = 4}: we may write f(X_{1}, X_{2}, X_{3}, X_{4}) = aX_{1}^{2} + bX_{2}^{2} − (cX_{3}^{2} + dX_{4}^{2}). For v ∈ I_{Q}, f_{v} represents 0 if and only if there exists x_{v} ∈ Q_{v}^{×} which is represented both by aX_{1}^{2} + bX_{2}^{2} and cX_{3}^{2} + dX_{4}^{2}. This holds if and only if (x_{v}, −ab)_{v} = (a, b)_{v} and (x_{v}, −cd)_{v} = (c, d)_{v}. By Hilbert’s theorem ∏_{v}(a, b)_{v} = ∏_{v}(c, d)_{v} = 1, it follows from a theorem above that there exists x ∈ Q^{×} such that (x, −ab)_{v} = (a, b)_{v} and (x, −cd)_{v} = (c, d)_{v} for every v. In which case, aX_{1}^{2} + bX_{2}^{2} − xZ^{2} (resp. cX_{1}^{2} + dX_{2}^{2} − xZ^{2}0 represents 0 over Q_{v} and, by the case n = 2 above, aX_{1}^{2} + bX_{2}^{2} − xZ^{2} (resp. cX_{1}^{2} + dX_{2}^{2} − xZ^{2}) represents 0 over Q. It then follows that x is represented by aX_{1}^{2} + bX_{2}^{2} (resp. cX_{1}^{2} + dX_{2}^{2}) over Q and thus f represents 0 over Q.

\underline{n ≥ 5}: we shall prove this by induction on n. Write f = h − g where h = h(X_{1}, X_{2}) = a_{1}X_{1}^{2} + a_{2}X_{2}^{2} and g = g(X_{3}, …, X_{n}) = −(a_{3}X_{3}^{2} + ⋯ + a_{n}X_{n}^{2}).

Define S to be a finite subset of I_{Q} consisting of ∞, 2 and all p such that v_{p}(a_{i}) ≠ 0 for at least one i ≥ 3. In other words, the complement of S is the set of finite odd p such that the coefficients of h are p-adic units, i.e., ∈ Z_{p}^{×}.

Let v ∈ S. f_{v} represents 0 implies that there exists a_{v} ∈ Q_{v}^{×} represented by g and h, i.e., there exists x_{i,v} (i = 1, …, n) such that a_{v} = h(x_{v,1}, x_{v,2}) = g(x_{v,3}, …, x_{v,n}).

LEMMA. There exist x_{1}, x_{2} ∈ Q such that if a = h(x_{1}, x_{2}) then (a/a_{v}) ∈ Q_{v}^{×2}.

Proof. h defines h^{S} : ∏_{v∈S}(Q_{v}^{×} × Q_{v}^{×}) → ∏_{v∈S} Q_{v}^{×} which sends (x_{v})_{v∈S} where x_{v} = (x_{v,1}, x_{v,2}) to (h(x_{v}))_{v∈S}. This is clearly continuous. For (a_{v})_{v∈S} ∈ ∏_{v∈S} Q_{v}^{×} defined above, since Q_{v}^{×2} is an open subgroup of Q_{v}^{×}, a_{v}Q_{v}^{×2} is an open neighbourhood of a_{v} in Q_{v}^{×} and ∏_{v∈S} a_{v}Q_{v}^{×2} is an open neighbourhood of (a_{v})_{v∈S}; therefore (h^{S})^{−1}(∏_{v∈S} a_{v}Q_{v}^{×2}) is an open neighbourhood of (x_{v})_{v∈S}. By the approximation theorem, there exists x = (x_{1}, x_{2}) ∈ Q in the neighbourhood and it maps by h^{S} into the neighbourhood ∏_{v∈S} a_{v}Q_{v}^{×2}, i.e., h(x) ∈ a_{v}Q_{v}^{×2} for every v ∈ S as required. □

Let f_{1} = f_{1}(Z, X_{3}, …, X_{n}) = aZ^{2} − g(X_{3}, …, X_{n}) in n − 1 variables.

\underline{v ∈ S}: by definition, for v ∈ S, g represents a_{v} over Q_{v}. It then follows from the lemma that g represents a over Q_{v} (because if g(x_{3}, …, x_{n}) = a and a = a_{v}b_{v}^{2} for some b_{v} ∈ Q_{v}^{×}, g(b_{v}x_{3}, …, b_{v}x_{n}) = b_{v}^{2}g(x_{3}, …, x_{n}) = b_{v}^{2}a_{v} = a). So f_{1} represents 0 over Q_{v}.

\underline{v ∉ S}: if v ∉ S, the coefficients −a_{3}, …, −a_{n} of g are v-adic units. Then d_{v}(g) = (−1)^{n−2}a_{3} ⋯ a_{n} ∈ Z_{v}^{×}. Furthermore, ϵ_{v}(g) = 1 as v ≠ 2 and therefore it is the product of 1. It then follows that g represents a over Q_{v}, therefore f_{1} represents 0 over Q_{v}.

So for every v, f_{1} represents 0 over Q_{v}. By the inductive hypothesis, f_{1} then represents 0 over Q and therefore g represents a over Q. Finally, since h represents a over Q, f represents 0 over Q. □

COROLLARY Let a ∈ Q^{×}. Then f represents a if and only of f_{v} represents a for every v ∈ I_{Q}. Proof. Apply the theorem for aZ^{2} − f. □