A DICHOTOMY PRINCIPLE FOR REPRESENTATIONS OF PRIMES BY QUADRATIC FORMS

M. G. MAHMOUDI

Abstract. The quadratic value theorem provides a dichotomy principle for the representation of an irreducible polynomial p in the ring A = F [X_{1}, ⋯, X_{n}] where F is a field, by an anisotropic quadratic form q defined over F: either a scalar multiple of p is multiplicatively generated by the values of q over A, or q is anisotropic over the residue field of A at p. We examine the validity of this principle in wider contexts.

Mathematics Subject Classification: 11E04, 11E25.

1. Introduction

By the classical theorem in arithmetic on the sums of two squares, the positive integers which can be written as a sum of two squares are precisely those whose prime factors of the form 4k + 3 appear an even number of times in their prime factorization. A counterpart of this result for polynomial ring F [X] where F is a field of characteristic different from 2 is also known. It turns out that a polynomial g ∈ F [X] is a sum of squares of two polynomials, if the leading coefficient of g is a sum of squares of two elements of F and for every irreducible factor p of g for which −1 is not a square in the field F [X]/(p(X)), the exponent of p in the prime factorization of g is even (see [4, Thm 1, 2]). A multivariable version of the previous result is known as “quadratic value theorem” (see [2, 18.3]). This theorem states that if q is an arbitrary anisotropic quadratic form over a field F of arbitrary characteristic and g ∈ F [T] = F [t_{1}, ⋯, t_{n}] is a nonzero polynomial then a scalar multiple of g lies in the group generated by the nonzero values represented by q over the field F (T) = F (t_{1}, ⋯, t_{n}) if and only if q is isotropic over the residue field at p for each irreducible divisor p occurring to an odd power in the factorization of g. In the case where the set of all nonzero values represented by q form a group (e.g., if q is a Pfister form), this theorem indicates that a scalar multiple of a nonzero polynomial g ∈ F [T] is represented by q if and only if q is isotropic for each irreducible divisor p occurring to an odd power in the factorization of g. Note that the quadratic value theorem, together with Cassels-Pfister theorem (which guarantees a polynomial solution f_{1}, ⋯, f_{n} ∈ F [X] of a_{1}f_{1}(X)^{2} + ⋯ + a_{n}f_{n}(X)^{2} = g(X), a_{1}, ⋯, a_{n} ∈ F, if there exists a rational function solution f_{1}, ⋯, f_{n} ∈ F (X)), imply the above polynomial version of the two squares theorem.

Two main important steps in the proof of these types of theorems are, first, to determine the conditions under which a prime or irreducible element p can be represented by a certain quadratic form, secondly, to determine the conditions under which p divides a nonzero value represented by that form. A look at the above theorems reveals that there always exists a dichotomy between the primes which can be represented by the corresponding form (or a scalar multiple of them lies in the group generated by the nonzero values represented by that form) and those primes p for which the corresponding form is anisotropic over the residue field at that prime. For example, the classical theorem on sums of two squares in arithmetic shows that a prime p can be either written as a sum of two squares, i.e., p = x^{2} + y^{2} for some integers x and y, or p ∤ x^{2} + y^{2} unless p|x and p|y. In the language of quadratic forms, this means that the binary quadratic form q = X^{2} + Y^{2} is either anisotropic over 𝔽_{p} or p is represented by q over ℤ (see the precise definition of anisotropic forms in §2). We have a similar dichotomy for the polynomial version of the two squares theorem and the quadratic value theorem.

The quadratic value theorem provides a very helpful principle which can be used to give a unified treatment of many theorems in the theory of quadratic forms and some related results on sums of squares. For example, the well-known Artin-Springer theorem [13] which states that an anisotropic quadratic form remains anisotropic after a scalar extension of odd degree, is a quasi-immediate corollary of it. Also, another result which has a similar esprit (see [12]) stating that if F is an ordered field and f(X) is an irreducible polynomial on F which changes sign on F then the ordering of F can be extended to F [X]/(f(X)) and more generally a substitution principle, can be easily deduced from the quadratic value theorem (see [2, 18.9] and (6.3)).

Particular cases of the quadratic value theorem or very related results can be found in [8, Ch. X, 2.13], [10, Ch. 7, 1.7], [6], [7], [11, Ch. 2, 8.6] and [5, §5]. The proof of many of the above results is reminiscent of the proof of Artin-Springer theorem. Primes of the form 4k + 3 in the theorem on sums of two squares play a similar role as the polynomials of odd degree in the Artin-Springer theorem. According to [10, p. 99], the main idea of these proofs “comes from an old theorem of Legendre on ternary quadratic forms over ℤ.”

A first step, used in the proof of many of the above results, is to apply a suitable division algorithm. Roughly speaking, if in an integral domain A, a prime p of type 4k + 3 (or a polynomial of odd degree, depending on the context) divides an expression like a_{1}β_{1}^{2} + ⋯ + a_{n}β_{n}^{2}, then the technique consists in estimating the magnitude of a_{1}γ_{1}^{2} +⋯+a_{n}γ_{n}^{2} where γ_{1}, ⋯, γ_{n} are the remainders of the divisions of β_{1}, ⋯, β_{n} by p. The relation p|a_{1}γ_{1}^{2} + ⋯ + a_{n}γ_{n}^{2} leads to a contradiction, either by an infinite descent or by finding a smaller prime of type 4k + 3 dividing a_{1}γ_{1}^{2} + ⋯ + a_{n}γ_{n}^{2} and finishing by induction. This is actually what happens in the classical proof of Artin-Springer theorem where A is the polynomial ring F [X] over a field F and p is an irreducible polynomial odd degree.

Unfortunately the ordinary euclidean division algorithm is not always available. Already for the case where A = R[X] where R is an integral domain but not a field, one does not have such an algorithm with respect to the ordinary degree map. Nevertheless, we have a division algorithm in a weak sense: for every polynomial f(X), g(X) ∈ R[X] there exist a nonzero scalar α ∈ R and the polynomials q(X), r(X) ∈ R[X] such that αf(X) = g(X)q(X) + r(X) with deg(r(X)) + 1 ≤ deg(g(X)). In this situation, the constant polynomials, i.e., the elements of R such as α, are infinitesimal, in the sense that an inequality like deg(a(X)) ≤ deg(b(X)) does not change if one multiplies a(X) by the elements of R. Also, the degree function satisfies the ultrametric inequality deg(f(X)+g(X)) ≤ max{deg(f(X)), deg(g(X))}. It was perhaps Hurwitz, who for the first time, implicitly used such kinds of division algorithms in the context of number fields in [3]. Some of their number theoretics applications in connection with euclidean number fields appeared several decades later in [9]. For the polynomial ring F [X_{1}, ⋯, X_{n}] where F is a field, this division algorithm has been implicitly used in the proofs of a norm principle in [7] and in the quadratic value theorem in [2, 18.3].

In this work, we apply and further exploit this machinery for more general weak division algorithms. Inspired by the division algorithm mentioned above, we consider an (ϵ, δ, m, n)-division algorithm in an arbitrary integral domain A, with a degree like homomorphism d : A → (Γ, +) where Γ is a commutative positive totally ordered semigroup such that for every a, b ∈ A with b ≠ 0, there exist q, r ∈ A such that αa = bq + r and α comes from a subset S_{b} of the nonzero elements of A depending only on b such that d(α_{1} ⋯ α_{n}) + d(r) + δ ≤ d(b) for every α_{1}, ⋯, α_{n} ∈ S_{b} (this intuitively encodes the “infinitesimal” nature of the elements of S_{b}) and d(a_{1} + ⋯ + a_{m}) ≤ max{d(a_{1}), ⋯, d(a_{m})} + ϵ for every a_{1}, ⋯, a_{m} ∈ A (this interprets the “ultrametric” nature of d).

In such circumstances, we state and prove in (4.3), a general dichotomy principle which shows that under reasonable assumptions, for a quadratic form q(X_{1}, ⋯, X_{n}) with coefficients in a unique factorization domain A and a prime p ∈ A, one has two alternatives: either q is anisotropic over the residue field of A at p or there exists an invertible element c ∈ A such that cp belongs to the group generated by the nonzero values represented by the form q in the multiplicative group of the field of fractions of A. In order to obtain truly number theoretic applications of this result, we also need a corresponding Cassels-Pfister type theorem. This Cassels-Pfister theorem is proved in (5.1).

In the last section of this paper, by considering different weak division algorithms (e.g., the binary division algorithm in ℤ), we present some applications of the dichotomy principle given in this work.

2. Notation and Terminology

Unless otherwise stated, all fields considered in this work are of arbitrary characteristic. Let A be an integral domain. A quadratic form on A is a homogeneous polynomial q(X_{1}, ⋯, X_{n}) ∈ A[X_{1}, ⋯ X_{n}] of degree 2. The set of all nonzero values represented by q = q(X_{1}, ⋯, X_{n}) over A is denoted by D_{A}(q) or simply by D(q) if no confusion arises, in other words

D_{A}(q) = {q(α) ≠ 0 | α ∈ A^{n}}.

The group generated by D(q) in the field of fractions of A is denoted by ⟨D(q)⟩. The bilinear form associated to q is denoted by B, i.e.,

(1)  B(X, Y) = q(X + Y) − q(X) − q(Y), X, Y ∈ A^{n}.

The A-submodule rad(B) = {v ∈ A^{n} : B(v, w) = 0 for all w ∈ A^{n}} of A^{n} is called the radical of B. The standard Gram matrix of q = ∑_{i<j} a_{ij}X_{i}X_{j} is the matrix G = [b_{ij}] where b_{ii} = 2a_{ii} and b_{ij} = a_{ij} for 1 ≤ i ≠ j ≤ n. It follows that u^{t}Gu = 2q(u) for every column vector u ∈ A^{n}. A quadratic form q is called nondegenerate if the associated bilinear form B is nondegenerate. Alternatively q is nondegenerate if the standard Gram matrix of q is invertible (or equivalently rad(B) = 0). The form q is said to be isotropic, if there exists a vector u ∈ A^{n}\rad(B) such that q(u) = 0; such a vector u is called an isotropic vector for q. Hence, if q is nondegenerate, then q being isotropic simply means that there exists a nonzero vector u ∈ A^{n} such that q(u) = 0. The form q is called anisotropic if q is not isotropic, in other words q(u) = 0 implies that u ∈ rad(B). In this sense, the form X^{2} + Y^{2} is anisotropic over 𝔽_{2}.

For an integral domain A and a prime element p (resp. prime ideal P) of A, the residue field at p (resp. at P) is defined as the field of fractions of the integral domain A/(p) (resp. A/P) and is denoted by A(p) (resp. A(P)).

We recall that an ordered semigroup (S, ≤) is called positive if for every x, y ∈ S we have x ≤ xy and x ≤ yx. Let ℕ_{0} denote the set of nonnegative integers.

For a real number x, the largest integer less than or equal to x and the smallest integer greater than or equal to x are respectively denoted by ⌊x⌋ and ⌈x⌉. The fractional part of a positive real number x is denoted by {x}, i.e., {x} = x − ⌊x⌋.

3. A weak division algorithm

Consider a commutative totally ordered positive semigroup (Γ, +, ≤). We consider two extra symbols −∞, +∞ with the following formal properties ±∞ + γ = ±∞, ±∞+(±∞) = ±∞ and −∞ < γ < +∞ for all γ ∈ Γ. Let m, n be two positive integers and ϵ, δ ∈ Γ ∪ {±∞}. We say that an integral domain A has an (ϵ, δ, m, n)- division algorithm with respect to a degree-like function d : A → Γ ∪ {−∞} if the following conditions are satisfied:

(i) d(a) = −∞ if and only if a = 0. (ii) d(ab) = d(a) + d(b) for all a, b ∈ A. (iii) d(a_{1} + a_{2} + ⋯ + a_{m}) ≤ max{d(a_{1}), d(a_{2}), ⋯, d(a_{m})} + ϵ for all a_{i} ∈ A. (iv) For every b ∈ A with b ≠ 0, there exists a subset S_{b} of nonzero elements of A with 1 ∈ S_{b} such that for every a ∈ A there exists q, r ∈ A and α ∈ S_{b} with αa = qb + r and d(α_{1}α_{2} ⋯ α_{n}) + d(r) + δ ≤ d(b) for all α_{i} ∈ S_{b}, i = 1, ⋯, n.

We say that d is δ-strict, if in the condition (iv) above, d(α_{1}α_{2} ⋯ α_{n})+d(r)+δ < d(b). We say that d is ϵ-strict if in the condition (iii) above we have the strict inequality d(a_{1} + a_{2} + ⋯ + a_{m}) < max{d(a_{1}), d(a_{2}), ⋯, d(a_{m})} + ϵ if for some i, j we have d(a_{i}) ≠ d(a_{j}).

Lemma 3.1. (a) For every invertible element u of A, the element e = d(u) is the minimum element of the image of d.

(b) We have e + e = e.

(c) If Γ is a cancellative semigroup then e is the neutral element of Γ.

Proof. (a) Since Γ is a positive semigroup, we have d(u) ≤ d(u) + d(u^{−1}a) = d(a) for every a ∈ A.

(b) We have e ≤ e + e by the positiveness of Γ. On the other hand, if u is an invertible element of A we have e + e = d(u) + d(u) = d(u^{2}) ≤ e since u^{2} is invertible. Hence, e + e = e.

(c) This follows from the general fact that in a cancellative semigroup an element e is the neutral element if and only if e is idempotent. □

Example 3.2. (a) For every positive integers n, m, the absolute value function d : ℤ → (ℕ_{0}, +) ∪ {−∞}, d(x) = |x|, x ≠ 0 and d(0) = −∞ gives a (+∞, 1, m, n)- division algorithm. In this case for every nonzero integer b, we can take S_{b} = {1}. This is the classical euclidean division algorithm.

(b) If A[X] is the polynomial ring over an arbitrary integral domain A, then the ordinary degree function d : A[X] → (ℕ_{0}, +) ∪ {−∞} gives a (0, 1, m, n)-division algorithm for every positive integers n, m. In this case for every polynomial b ∈ A[X], one can take S_{b} = A\{0}. If A is a field, then we have the ordinary division algorithm of polynomials and one can take S_{b} = {1}.

Lemma 3.3. Let c be a real number with c > 1. Then, for every real number x, there exists an integer k ∈ {1, 2, ⋯, ⌈c − 1⌉} and an integer q such that |kx − q| ≤ 1/c. This inequality is strict unless there exists an integer k ∈ {1, 2, ⋯, ⌈c − 1⌉} such that \{kx\} = \frac{c−1}{c}.

Proof. First consider the case where 1 < c ≤ 2. We should prove that there exists an integer q such that |x − q| ≤ 1/c. In fact there exists an integer q such that |x − q| ≤ 1/2 ≤ 1/c and the proof is complete.

Now consider the case where c > 2. If for some k ∈ {1, ⋯, ⌈c − 1⌉}, the quantity {kx} belongs to one of the intervals [0, 1/c[ or [(c − 1)/c, 1[ then for q = 0 or q = 1 we obtain |kx − q| ≤ 1/c and the proof is complete (note that this inequality is strict unless \{kx\} = \frac{c−1}{c}). Otherwise, for every k ∈ \{1, ⋯, ⌈c − 1⌉\} we have {kx} ∈ [1/c, (c − 1)/c[. Since [1/c, (c − 1)/c[ is the union of the intervals I_{1} ∪ ⋯ I_{⌈c−1⌉−1} where

I_{i} = [\frac{1}{c} + (i − 1)\frac{c − 2}{c(⌈c − 1⌉ − 1)}, \frac{1}{c} + i\frac{c − 2}{c(⌈c − 1⌉ − 1)}[,

by pigeonhole principle there exists an interval I_{i} for some i with 1 ≤ i ≤ ⌈c−1⌉−1 such that {kx} ∈ I_{i}. It follows that for some m, n ∈ {1, ⋯, ⌈c−1⌉} with m > n we have \{mx\}, \{nx\} ∈ I_{i}. This implies that |\{mx\}−\{nx\}| < length(I_{i}) = \frac{c−2}{c(⌈c−1⌉−1)} ≤ 1/c. Now if we put q = ⌊mx⌋ − ⌊nx⌋ we obtain |(m − n)x − q| = |{mx} − {nx}| < 1/c. □

In the below lemma we use the convention log_{c}(0) = −∞ for a real number c > 1.

Lemma 3.4. Let c > 1 be a real number and m, n be positive integers. Then ℤ has a (log_{c} m, 1 − n log_{c}(⌈c − 1⌉), m, n)-division algorithm with respect to the function d : ℤ → ℝ_{≥0} ∪ {−∞} given by d(x) = log_{c} |x|.

Proof. We should check the validity of (i), (ii), (iii) and (iv). We obviously have (i) and (ii). To prove (iii), consider a_{1}, ⋯, a_{m} ∈ ℤ with a_{i} ≠ 0 for all i. Choose i with 1 ≤ i ≤ m such that

|a_{i}| = max{|a_{1}|, |a_{2}|, ⋯, |a_{m}|}.

We obtain d(a_{1}+⋯+a_{m}) = log_{c}(|a_{1}+a_{2}+⋯+a_{m}|) ≤ log_{c}(|a_{1}|+|a_{2}|+⋯+|a_{m}|) ≤ log_{c}(m|a_{i}|) = log_{c}(|a_{i}|) + log_{c} m and the proof of (iii) is finished.

To prove (iv), consider a nonzero b ∈ ℤ. Take the set S_{b} = {1, ⋯, ⌈c − 1⌉} (here S_{b} does not depend on b). Consider an element a ∈ ℤ. First note that since d(−1) = 0, we have d(z) = d(−z) for every z ∈ ℤ, hence proving the result for −a or −b instead of a or b are equivalent. Thus, without loss of generality we may assume that both a and b are positive. Also, note that if x = a/b is an integer then by taking α = 1 and r = 0, the validity of (iv) is established. Hence, we may assume that x is not integer. According to (3.3), there exists an integer q and an element α ∈ {1, ⋯, ⌈c − 1⌉} such that |αx − q| ≤ 1/c hence

(2)  d(αx − q) = log_{c}(|αx − q|) ≤ log_{c}(1/c) = −1

We have αx − q = (αa − qb)/b. Putting r := αa − bq ∈ ℤ, we obtain αa = qb + r. It remains to show that d(α_{1}α_{2} ⋯ α_{n}) + d(r) + δ ≤ d(b) or equivalently log_{c}(α_{1}α_{2} ⋯ α_{n}) + log_{c}(|r/b|) + δ ≤ 0 for every α_{1}, ⋯, α_{n} ∈ S_{b}. But this inequality can be proved as follows: log_{c}(α_{1} ⋯ α_{n}) + log_{c}(|r/b|) + δ = log_{c}(α_{1} ⋯ α_{n}) + log_{c}(|αx − q|) + δ ≤ n log_{c}(⌈c − 1⌉) − 1 + δ = 0 by (2) and the fact that α_{i} ≤ ⌈c − 1⌉ for i = 1, ⋯, n. □

4. A dichotomy principle

Definition 4.1. Let (Γ, +) be an abelian torsion-free group. We say that a quadratic form q(X_{1}, ⋯, X_{n}) = ∑_{i≤j} a_{ij}X_{i}X_{j} ∈ A[X_{1}, ⋯, X_{n}] is Γ-equilibrated if every solution (γ_{1}, ⋯, γ_{n}) ∈ Γ^{n} of the equations γ_{i}+γ_{j} = γ_{i′} +γ_{j′} = γ_{i′′} +γ_{j′′} = ⋯ = γ where γ_{1}, ⋯, γ_{n} ∈ Γ and γ ∈ G is a constant and (i, j), (i′, j′), (i′′, j′′), ⋯ are all indices for which a_{ij}, a_{i′j′}, a_{i′′j′′}, ⋯ are nonzero, is of the form γ_{1} = γ_{2} = ⋯ = γ_{n}.

Example 4.2. Every diagonal quadratic form q = a_{1}X_{1}^{2} + ⋯ + a_{n}X^{2}, a_{i} ∈ A is ℤequilibrated since the equations 2γ_{1} = ⋯ = 2γ_{n} = γ imply that γ_{1} = ⋯ = γ_{n}. The two dimensional form q(X_{1}, X_{2}) = X_{1}X_{2} is not ℤ-equilibrated since the equation γ_{1} + γ_{2} = γ does not imply that γ_{1} = γ_{2}.

Theorem 4.3. Let A be a unique factorization domain. Assume that A has an (ϵ, δ, m, n − 1)-division algorithm with respect to a degree-like function d : A → Γ∪{−∞} where Γ is a commutative positive totally ordered cancellative semigroup. Assume further that the image of d has the least element property and let e be the minimum element of the image of d and suppose that δ > e.

Let q(X_{1}, ⋯, X_{n}) = ∑_{i≤j} a_{ij}X_{i}X_{j} ∈ A[X_{1}, ⋯, X_{n}] be a quadratic form with m nonzero terms and let μ = max{d(a_{ij}) : 1 ≤ i, j ≤ n}. Assume that for all primes p for which q is degenerate over A(p), there exists an invertible element c ∈ A such that pc ∈ ⟨D(q)⟩. Consider the following situations:

(v) 2δ > μ + ϵ. (vi) 2δ ≥ μ + ϵ and d is δ-strict. (vii) μ = e, 2δ ≥ ϵ, q is Γ-equilibrated, d is ϵ-strict, and for every element a ∈ A with d(a) ≠ e there exists a prime element in A, depending only on d(a), which divides a. Assume that in the situations (v) and (vi) for all primes p of degree e (if any) and in the situation (vii) for all primes p of degree ≤ ϵ (if any) one of the following alternatives holds: either q is anisotropic over A(p) or there exists an invertible element c ∈ A such that pc ∈ ⟨D_{A}(q)⟩. Then the same alternatives hold for all primes p in all situations (v), (vi) and (vii).

Proof. By (3.1), e is necessarily the neutral element of Γ. Since the image of d is well-ordered, we can proceed by induction on the degree of p. Assume that the assertion is true for all primes of degree less than γ, for some γ in the image of d. It is enough to show that if p is a prime element of degree γ such that q is nondegenerate over A(p) and q is isotropic over A(p) then there exists an invertible element c ∈ A such that pc ∈ ⟨D(q)⟩. By the hypotheses, this assertion is true if q is degenerate over A(p). Hence, we may that q is nondegenerate over A(p). We may also assume that d(p) > e (resp. d(p) > ϵ) in the situations (v) and (vi) (resp. (vii)), since otherwise the conclusion is guaranteed by the hypotheses. Let ξ = (α_{1}, ⋯, α_{n}) ∈ A(p)^{n} be an isotropic vector for q. Hence, q(ξ) = 0 and ξ ∉ rad(B_{A(p)}) = {0}. It follows that there exist elements f_{1}, ⋯, f_{n}, h ∈ A such that at least one of f_{i}’s is nonzero modulo p and

(3)  q(f_{1}, ⋯, f_{n}) = ph.

Since A has an (ϵ, δ, m, n − 1)-division algorithm with respect to d, there exists a subset S_{p} of nonzero elements of A and an element α_{i} ∈ S_{p} such that

(4)  α_{i}f_{i} = q_{i}p + r_{i}, i = 1, ⋯, n,

where q_{i}, r_{i} ∈ A and for every β_{1}, ⋯, β_{n−1} ∈ S_{p}

(5)  d(β_{1} ⋯ β_{n−1}) + d(r_{i}) + δ ≤ d(p), i = 1, ⋯, n.

We claim that for every β ∈ S_{b} we have p ∤ β. In fact, if for some β ∈ S_{p} we have p|β, then (5) implies that d(p)+δ ≤ d(p), hence d(p)+δ = d(p) since Γ is a positive semigroup. This implies that δ = e, because Γ is cancellative, this contradicts our assumption that δ > e. In particular, we have p ∤ α_{i} for all i = 1, ⋯, n. We now claim that at least one of r_{i}’s is nonzero. In fact, if r_{i} = 0 for all i = 1, ⋯, n then the relation (4) together with the fact that p ∤ α_{i} imply that p|f_{i} for all i. This contradicts our assumption that at least one of f_{i}’s is nonzero modulo p. Now, by multiplying both sides of (3) by α_{1}^{2} ⋯ α_{n}^{2} and using (4), we obtain

(6)  q(f′_{1}, ⋯, f′_{n}) = ph′, for some h′ ∈ A,

where f′_{1} = α_{2}α_{3} ⋯ α_{n}r_{1}, ⋯, f′_{n} = α_{1}α_{2} ⋯ α_{n−1}r_{n}. Since

α_{1}, α_{2}, ⋯, α_{n} ∈ S_{p},

using (5), we obtain d(f′_{i}) + δ ≤ d(p) for every i = 1, ⋯, n. Hence, from the beginning we may assume that in the relation (3), the following additional condition holds

(7)  d(f_{i}) + δ ≤ d(p).

We may assume that the terms f_{1}, ⋯, f_{n} do not have a common prime divisor; in fact if a prime element p_{1} ∈ A divides all f_{i}’s then by (3) we obtain p_{1}^{2}|ph. As we have mentioned, at least one of f_{i}’s is nonzero modulo p, hence p and p_{1} are not associated primes, it follows that p_{1}^{2}|h and we may cancel p_{1}^{2} from both sides of (3).

Since we have an (ϵ, δ, m, n − 1)-division algorithm with respect to d, we obtain

(8)  d(q(f_{1}, ⋯, f_{n})) ≤ max{d(a_{ij}) + d(f_{i}f_{j})) : a_{ij} ≠ 0} + ϵ

By adding 2δ to both sides of (8) and using (7) and the assumption d(a_{ij}) ≤ μ for a_{ij} ≠ 0 we obtain:

(9)  d(q(f_{1}, ⋯, f_{n})) + 2δ ≤ μ + 2d(p) + ϵ

In the situation (v) or (vi), at least one of the inequalities 2δ ≥ μ + ϵ or (5) (hence (7)) is strict. Note that since (9) is a consequence of (7), the inequality (9) is strict too. Hence, in both cases, from (9) we obtain the strict inequality

(10)  d(q(f_{1}, ⋯, f_{n})) < 2d(p).

We claim that in the situation (vii), this inequality holds as well. First note that the condition μ = e implies that d(a_{ij}) = e for all nonzero a_{ij}’s. To prove (10), note that as mentioned at the beginning of the proof, we may assume that d(p) > ϵ in the situation (vii). If for some i < j we have d(f_{i}) ≠ d(f_{j}) then as q is Γ-equilibrated, there exist two pairs (r, s) and (r′, s′) with r ≤ s and r′ ≤ s′ such that a_{rs} ≠ 0 and a_{r′s′} ≠ 0 and d(f_{r}f_{s}) ≠ d(f_{r′}f_{s′}). This relation together with the hypothesis in (vii) that d is ϵ-strict, imply that the inequality (8) is strict too. Consequently,

(9) is strict and we obtain (10). If for all i < j, d(f_{i}) = d(f_{j}), then by the last hypothesis in (vii), either f_{1}, ⋯, f_{n} have a common prime divisor or d(f_{i}) = e for i = 1, ⋯, n. But since f_{i}’s do not have a common prime divisor, it follows that the first case cannot occur. Hence, d(f_{i}) = e, for all i. Now, using (3) and (8) we obtain d(p) ≤ d(q(f_{1}, ⋯, f_{n})) ≤ e + ϵ = ϵ which is a contradiction. Thus, we have shown the validity of (10) in all situations (v), (vi) and (vii).

The relation (10) together with (3) imply that

(11)  d(h) < d(p).

First consider the case where h is a non-invertible. Let h = p_{1} ⋯ p_{r} be a decomposition of h into prime elements p_{i}’s. By (11) we have d(p_{i}) < d(p) for i = 1, ⋯, r. We claim that there exists an invertible element c_{i} ∈ A such that p_{i}c_{i} ∈ ⟨D(q)⟩ for i = 1, ⋯, r. The correctness of this claim is guaranteed by the hypotheses if q is degenerate over A(p_{i}). If q is nondegenerate over A(p_{i}) then considering the relation (3) modulo p_{i} (and remarking that f_{1}, ⋯, f_{n} do not have a common prime divisor, hence at least one of them is nonzero modulo p_{i}) we obtain that q(X_{1}, ⋯, X_{n}) is isotropic over A(p_{i}). By induction, the second alternative holds and the proof of the claim is completed. Hence, for every i we have p_{i}c_{i} ∈ ⟨D(q)⟩. It follows that hc_{1} ⋯ c_{r} ∈ ⟨D(q)⟩ and the relation (3) implies that pc_{1} ⋯ c_{r} ∈ ⟨D(q)⟩ as desired.

Finally, consider the case where c = h is invertible. The relation (3) implies that pc ∈ ⟨D(q)⟩ and the proof is complete. □

Corollary 4.4. With the same hypotheses as in (4.3), assume further that q represents 1, then for a nonzero element f ∈ A the following conditions are equivalent:

(a) There exists an invertible element f_{0} ∈ A such that f_{0}f ∈ ⟨D(q)⟩.

(b) The form q is isotropic over A(p) for every prime divisor p of f occurring to an odd power in the factorization of f.

Proof. (a)⇒(b): We may assume that f is not invertible; Let p a prime divisor of f occurring to an odd power in the factorization of f. By (a) there exist anisotropic vectors v_{1}, ⋯, v_{r}, u_{1}, ⋯, u_{s} ∈ A^{n} such that

f_{0}fq(u_{1}) ⋯ q(u_{s}) = q(v_{1}) ⋯ q(v_{r}).

We can write u_{i} = p^{α_{i}}u′_{i} and v_{j} = p^{β_{j}}v′_{j} for some nonnegative α_{i}, β_{j} ∈ ℤ such that at least one entry of u′_{i} and v′_{j} is not divisible by p. From the above relation we obtain

p^{2α}f_{0}fq(u′_{1}) ⋯ q(u′_{s}) = p^{2β}q(v′_{1}) ⋯ q(v′_{r}),

where α = α_{1} + ⋯ + α_{s} and β = β_{1} + ⋯ + β_{r}. Since p occurs to an odd power in a factorization of f, the above relation implies that either there exists an index i such that p|q(v′_{i}) or there exists an index j such that p|q(u′_{j}). In both cases, q becomes isotropic over A(p). (b)⇒(a) If f is invertible, then by taking f_{0} = f^{−1}, (a) holds. Otherwise, it suffices to prove that for every prime divisor p of f occurring to an odd power in the factorization of f, there exists an invertible element c such that pc ∈ ⟨D(q)⟩, but this is exactly the conclusion of (4.3). □

5. A Cassels-Pfister’s representation theorem

Theorem 5.1. Let A be an integral domain. Assume that A has an (ϵ, δ, m, n)- division algorithm with respect to a degree-like function d : A → Γ ∪ {−∞} where Γ is a commutative positive totally ordered cancellative semigroup. Assume further that the image of d has the least element property and let e is the minimum element of the image of d and suppose that δ > e and d^{−1}(e) consists only of invertible elements. Let q(X_{1}, ⋯, X_{n}) = ∑_{i≤j} a_{ij}X_{i}X_{j} ∈ A[X_{1}, ⋯, X_{n}] be an anisotropic quadratic form with m nonzero terms and let μ = max{d(a_{ij}) : 1 ≤ i, j ≤ n}. Assume that one of the following conditions hold

(v) 2δ > μ + ϵ. (vi) 2δ ≥ μ + ϵ and d is δ-strict. In this case, if an element b ∈ A is represented by q over the field of fractions K of A then b is already represented by q over A.

Proof. As we have mentioned in (3.1), e is the neutral element of Γ. Since b is represented by q over the field of fractions of A, there exist f_{0}, f_{1}, ⋯, f_{n} ∈ A with f_{0} ≠ 0 such that

(12)  q(\frac{f_{1}}{f_{0}} , ⋯, \frac{f_{n}}{f_{0}} ) = b.

We proceed by induction on d(f_{0}). If d(f_{0}) = e, the hypothesis on d^{−1}(e) implies that f_{0} is invertible, hence f_{i}/f_{0} ∈ A for i = 1, ⋯, n and the proof is complete. We may assume that d(f_{0}) > e and d(f_{0}) is minimal (with f_{0} ≠ 0) among all representations of the form (12). The relation (12) leads to

(13)  q(f_{1}, ⋯, f_{n}) − bf_{0}^{2} = 0.

Thus, if we put Q(X_{1}, ⋯, X_{n+1}) = q(X_{1}, ⋯, X_{n})−bX_{n+1}^{2}, for u = (f_{1}, ⋯, f_{n}, f_{0}) we have Q(u) = 0. We may assume that f_{1}, ⋯, f_{n}, f_{0} do not have a non-invertible common divisor, otherwise we can cancel it from both sides of (13). As we have an (ϵ, δ, m, n)-division algorithm with respect to d, there exists a subset S_{f_{0}} of nonzero elements of A such that

(14)  α_{i}f_{i} = g_{i}f_{0} + r_{i},

where r_{i}, g_{i} ∈ A, α_{i} ∈ S_{f_{0}} and for every β_{1}, ⋯, β_{n} ∈ S_{f_{0}} we have

(15)  d(β_{1} ⋯ β_{n}) + d(r_{i}) + δ ≤ d(f_{0}).

In particular, we can take r_{0} = 0 and α_{0} = g_{0} = 1. Consider the vector

(16)  v = (\frac{g_{1}}{α_{1}} , ⋯, \frac{g_{n}}{α_{n}} , \frac{g_{0}}{α_{0}} ) = (\frac{g_{1}}{α_{1}} , ⋯, \frac{g_{n}}{α_{n}} , 1) ∈ K^{n+1}.

We have Q(v) ≠ 0, since we would otherwise obtain

q(\frac{α_{2} ⋯ α_{n}g_{1}}{α_{1} ⋯ α_{n}} , ⋯, \frac{α_{1} ⋯ α_{n−1}g_{n}}{α_{1} ⋯ α_{n}}) = b

which contradicts the minimality of the degree of f_{0} (note that by (15) we have d(α_{1} ⋯ α_{n}) < d(α_{1} ⋯ α_{n}) + δ ≤ d(f_{0}) since δ > e). The map τ_{v} : K^{n+1} → K^{n+1} defined by

(17)  τ_{v}(x) = x − \frac{B(x, v)}{Q(v)}v,

where B is the bilinear form associated to Q as defined in (1), is an isometry (c.f., [2, 7.2]). If char(K) ≠ 2 (resp. char(K) = 2 ), the map τ_{v} is usually called the reflection (resp. orthogonal transvection) along v. Since Q(u) = 0 we obtain Q(τ_{v}(u)) = 0, in other words

Q(u − \frac{B(u, v)}{Q(v)}v) = 0.

Multiplying this relation by Q(v)^{2} we obtain Q(Q(v)u − B(u, v)v) = 0. Put

(18)  w = Q(v)u − B(u, v)v.

Hence Q(w) = 0. We can write w = (w_{1}, ⋯, w_{n}, w_{0}) where

w_{i} = Q(v)f_{i} − B(u, v)\frac{g_{i}}{α_{i}} .

We try to simplify the last entry w_{0} of w. We have

(19)  \begin{aligned}w_{0} & = Q(v)f_{0} − B(u, v)\frac{g_{0}}{α_{0}} = Q(v)f_{0} − B(u, v) \\ & = f_{0}(Q(v) − B(u, v)\frac{1}{f_{0}} ) = f_{0}Q(v − \frac{1}{f_{0}} u) = \frac{1}{f_{0}} Q(f_{0}v − u) \end{aligned}

The i-th entry of f_{0}v − u is equal to f_{0} \frac{g_{i}}{α_{i}} − f_{i} = \frac{1}{α_{i}} (f_{0}g_{i} − α_{i}f_{i}) = −\frac{r_{i}}{α_{i}} by (14), hence

(20)  f_{0}v − u = (−\frac{r_{1}}{α_{1}} , ⋯, −\frac{r_{n}}{α_{n}} , −\frac{r_{0}}{α_{0}} ) = (−\frac{r_{1}}{α_{1}} , ⋯, −\frac{r_{n}}{α_{n}} , 0)

As Q(w) = 0, we obtain Q(w′) = 0 where w′ = α_{1} ⋯ α_{n}w ∈ A^{n+1}. We can write w′ = (w′_{1}, ⋯, w′_{n}, w′_{0}) where w′_{i} = α_{1} ⋯ α_{n}w_{i}. Thus,

(21)  q(w′_{1}, ⋯ w′_{n}) − bw′_{0}^{2} = 0

Using (19) and (20) we obtain

(22)  \begin{aligned}w′_{0} = α_{1} ⋯ α_{n}w_{0} & = (α_{1} ⋯ α_{n})\frac{1}{f_{0}} Q(f_{0}v − u) \\ & = (α_{1} ⋯ α_{n})\frac{1}{f_{0}} Q(−\frac{r_{1}}{α_{1}} , ⋯, −\frac{r_{n}}{α_{n}} , 0) \end{aligned}

Now by setting

(23)  (f′_{1}, ⋯, f′_{n}, f′_{0}) = (−α_{0}α_{2} ⋯ α_{n}r_{1}, ⋯, −α_{0}α_{1} ⋯ α_{n−1}r_{n}, 0)

and multiplying (22) by α_{1} ⋯ α_{n}f_{0} we obtain

(24)  α_{1} ⋯ α_{n}f_{0}w′_{0} = Q(f′_{1}, ⋯, f′_{n}, f′_{0}) = Q(f′_{1}, ⋯, f′_{n}, 0) = q(f′_{1}, ⋯, f′_{n})

Assume that d(a_{ij}f′_{i}f′_{j}) is maximum among all terms of

q(f′_{1}, ⋯, f′_{n}) = ∑_{i<j} a_{ij}f′_{i}f′_{j}

for some i, j. The relations (23), (24) and the fact that there exists an (ϵ, δ, m, n) division algorithm with respect to d imply that

(25)  \begin{aligned}d(w′_{0}) + d(f_{0}) + d(α_{1} ⋯ α_{n}) & = q(f′_{1}, ⋯, f′_{n}) \\ & ≤ max\{d(a_{ij}) + d(f′_{i}f′_{j})\} + ϵ \\ & = d(a_{ij}) + d(f′_{i}) + d(f′_{j}) + ϵ \\ & ≤ μ + 2d(α_{1} ⋯ α_{n}) + d(r_{i}) + d(r_{j}) + ϵ. \end{aligned}

Adding 2δ to all sides of (25) and thanks to (15) we obtain:

(26)  \begin{aligned}d(w′_{0}) + d(f_{0}) + 2δ & ≤ d(w′_{0}) + d(f_{0}) + 2δ + d(α_{1} ⋯ α_{n}) \\ & ≤ μ + 2d(α_{1} ⋯ α_{n}) + d(r_{i}) + d(r_{j}) + ϵ + 2δ \\ & ≤ μ + 2d(f_{0}) + ϵ, \end{aligned}

hence

(27)  d(w′_{0}) + 2δ ≤ μ + d(f_{0}) + ϵ.

In the case (v), we have 2δ > μ+ϵ, hence (27) leads to the inequality d(w′_{0}) < d(f_{0}). In the case (vi), the inequality (15) is strict, hence the second inequality in (26) is strict too. Hence, the inequality (27) which is an implication of (26) is strict. Thus, in this case we also have the inequality d(w′_{0}) + 2δ < μ + d(f_{0}) + ϵ. This together with the hypothesis 2δ ≥ μ + ϵ in (vi) imply that d(w′_{0}) < d(f_{0}). Thus, in both cases we obtain the inequality d(w′_{0}) < d(f_{0}). Also, note that w′_{0} is nonzero; since otherwise (21) and the fact that q is anisotropic implies that w′_{1} = ⋯ = w′_{n} = 0. We obtain then w_{1} = ⋯ = w_{n}. Hence, w = (w_{1}, ⋯, w_{n}, w_{0}) = 0. Now the definition of w given in (18) implies that u and v are linearly independent. Since Q(u) = 0, we obtain Q(v) = 0. This is a contradiction, because we have shown that Q(v) ≠ 0. Hence, we have proved that w′_{0} is nonzero and d(w′_{0}) < d(f_{0}). This together with (21) contradict our assumption of the minimality of d(f_{0}). □

6. Some applications

Example 6.1. (Cassels-Pfister’s Theorem, [10, 2.3]) Consider the polynomial ring F [t] over a field F. Let q(X_{1}, ⋯, X_{n}) = ∑ a_{ij}(t)X_{i}X_{j} be an anisotropic quadratic form where a_{ij}(t) ∈ F [t] is of degree at most 1 for every i and j. Let f(t) ∈ F [t] be a polynomial which is represented by q over the field of fractions F (t) of F [t]. Then f(t) is already represented by q over F [t].

In this situation, we use the ordinary degree function d : F [X] → ℕ_{0} mentioned in (3.2)(b). The function d induces a (0, 1, m, n)-division algorithm for every m, n. In particular, we can take n (resp. m) as the number of variables (resp. the number of nonzero terms) of the form q. With the notations of (5.1) we have μ ≤ 1. Thus, the condition 2δ > μ + ϵ is satisfied and the conclusion holds. Note that the hypothesis on d^{−1}(e) in (5.1) prevents to replace the field F with an arbitrary unique factorization domain.

Example 6.2. (Quadratic value theorem, [2, Thm. 18.3]) Let F be a field and let F [T] = F [t_{1}, ⋯, t_{r}]. Consider a nondegenerate anisotropic quadratic form q(X_{1}, ⋯, X_{n}) = ∑ a_{ij}X_{i}X_{j} where a_{ij} ∈ F. Let p be a prime element of F [T]. Then either q(X_{1}, ⋯, X_{n}) remains anisotropic over F [T](p) or there exists a nonzero element c ∈ F such that cp ∈ ⟨D_{F [T]}(q)⟩.

We prove this result by induction on r. For r = 0, the conclusion holds since F [T] = F does not contain any prime element. Hence, let assume that r ≥ 1. Put A = F [t_{1}, ⋯, t_{r−1}], hence F [T] = A[t]. According to (3.2)(b), we have a (0, 1, m, n − 1) division algorithm on A[t] with respect to the usual degree function d : A[t] → ℕ_{0}, where m denotes the number of nonzero terms of q. We have μ = 0, hence the condition 2δ ≥ μ+ϵ is satisfied. In order to apply (4.3), it suffices to check that for every prime p ∈ A[t] of order e = 0 (hence p ∈ A) either q is anisotropic over A[t](p) or there exists an invertible element c ∈ A[t] such that cp ∈ ⟨D_{A[t]}(q)⟩. We have a ring isomorphism A[t]/(p) ≃ A(p)[t] since p ∈ A, hence we have a field isomorphism A[t](p) ≃ A(p)(t). If q is isotropic over A[t](p), the relation A[t](p) ≃ A(p)(t) implies that q is isotropic over A(p) = F [t_{1}, ⋯, t_{r}](p). By induction there exists a nonzero c ∈ F such that cp ∈ ⟨D_{F [t_{1},··· ,t_{r−1}]}(q)⟩ ⊆ ⟨D_{F [t_{1},··· ,t_{r}]}(q)⟩ and the proof is complete. Thus, we have checked the validity of the result for all primes of degree ≤ e = 0. The conclusion now follows from (5.1).

Example 6.3. (Compare [2, 18.9]) Let F be a field and let F [T] = F [t_{1}, ⋯, t_{r}]. Consider a nondegenerate anisotropic quadratic form q(X_{1}, ⋯, X_{n}) = ∑ a_{ij}X_{i}X_{j} where a_{ij} ∈ F. Let p(T) be an irreducible polynomial in F [T]. Suppose that there exists an element c ∈ F^{r} such that p(c) ∉ ⟨D_{F [T]}(q)⟩. Then q is anisotropic over F [T](p(T)).

By (6.2) if q is not anisotropic over F [t](p(T)), there exists an invertible scalar λ ∈ F such that λp(T) ∈ ⟨D_{F [T]}(q)⟩. It follows that there exist polynomials f_{i}(T) ∈ D_{F [T]}(q) and g_{j}(T) ∈ D_{F [T]}(q) for i = 1, ⋯, n and j = 1, ⋯, m, such that f_{1}(T) ⋯ f_{n}(T)p(T) = g_{1}(T) ⋯ g_{m}(T). Putting T := c in this identity, implies that p(c) ∈ ⟨D_{F [T]}(q)⟩, which is a contradiction.

Example 6.4. (Multivariable version of Artin-Springer’s theorem) With the same notation as in (6.3), if the degree of p with respect to one of its variables t ∈ {t_{1}, ⋯, t_{r}} is odd then q remains anisotropic over F [T](p).

It is enough to compare the degrees of both sides of f_{1}(T) ⋯ f_{n}(T)p(T) = g_{1}(T) ⋯ g_{m}(T) with respect to the corresponding variable t.

Example 6.5. (Two squares theorem)

(a) If an integer is a sum of two rational squares, then it can be written as a sum of two integral squares.

(b) Every prime number of the form 4k + 1 can be written as a sum of squares of two integers.

(c) If p is a prime number of the form 4k + 3 and p|a^{2} + b^{2} for some integers a and b then p divides both a and b.

Consider the quadratic form q(X_{1}, X_{2}) = X_{1}^{2} + X_{2}^{2}. Here, we have n = m = 2. By (3.4) for c = 2, the ring ℤ has an (ϵ, δ, m, n) = (1, 1, 2, 2)-division algorithm with respect to the degree function d(x) = log_{2} |x|. Note that the image of d is a discrete, hence a well-ordered subset of nonnegative reals. Hence, for A = ℤ, Γ = (ℝ_{≥0}, +), we can apply (4.3) if the respective hypotheses are satisfied. We have μ = 0, hence 2δ > ϵ + μ is satisfied and the conclusion of (a) follows from (4.3).

Again using (3.4) for c = 2, the ring ℤ has an (ϵ, δ, m, n − 1) = (1, 1, 2, 1)- division algorithm with respect to the degree function d(x) = log_{2} |x|. Also, q is nondegenerate over all 𝔽_{p} for all primes p. In this situation we have μ = 0, hence 2δ > μ + ϵ holds. Consider a prime number p of the form 4k + 1. According to (4.3), either q remains anisotropic over 𝔽_{p} or pc is a sum of two squares in ℚ for some c ∈ \{1, −1\}. The former case is ruled out (for instance, one has p|(\frac{p−1}{2}!)^{2} + 1 by Wilson’s theorem). In the latter case, c = −1 is rules out since q is a positive definite form. Hence, p is a sum of two squares in ℚ. Now (6.5) (a) implies that p is a sum of two integral squares.

(2) If p does not divide a or b then q becomes isotropic over 𝔽_{p}. By (4.3) we deduce that p is represented by q over ℚ, hence over ℤ. Now a congruence argument modulo 4 yields a contradiction.

Example 6.6. (a) A nonnegative integer can be written, in the form x^{2} + xy + y^{2} for some x, y ∈ ℚ if and only if it can be written in the same form for some x, y ∈ ℤ.

(b) A nonnegative prime p ≠ 3 can be written in the form x^{2} + xy + y^{2} for some x, y ∈ ℤ if and only if p of the form 3k + 1.

Consider the quadratic form q(X_{1}, X_{2}) = X_{1}^{2} + X_{1}X_{2} + X_{2}^{2} over ℤ. Here we have n = 2 and m = 3. According to (3.4), for c = 2 we have an (ϵ, δ, m, n) = (log_{2} 3, 1, 3, 2) on ℤ with respect to the degree function d : ℤ → ℝ_{≥0} given by d(x) = log_{2} |x|. Here we have μ = 0, hence the condition 2δ > μ + ϵ is satisfied and the conclusion of (a) follows from (5.1).

To prove (b), it is enough by (a) to show that p ∈ D_{ℚ}(q) if and only if p is of the form 3k + 1. We first show that ⟨D_{ℤ}(q)⟩ = D_{ℚ}(q). To see this note that since the discriminant of q is 3, the form q is equivalent to the form q′(X_{1}, X_{2}) = X_{1}^{2} + 3X_{2}^{2} over ℚ. Hence, ⟨D_{ℤ}(q)⟩ = ⟨D_{ℤ}(q′)⟩ = D_{ℚ}(q′) = D_{ℚ}(q) (in fact, more generally this argument shows that if a quadratic form q defined over an integral domain A is equivalent to a multiplicative form over the field of fractions K of A then ⟨D_{A}(q)⟩ = D_{K}(q)). Hence, it is enough to prove that p ∈ ⟨D_{ℤ}(q)⟩ if and only if p is of the form 3k +1. Again using (3.4), for c = 2 we have an (ϵ, δ, m, n−1) = (log_{2} 3, 1, 3, 1) division algorithm with respect to the degree function d defined above. Note that p = 3 is the only prime number for which q is nondegenerate over 𝔽_{p}, by the way 3 ∈ D_{ℚ}(q). By (4.3), for every prime p, either q is anisotropic over 𝔽_{p} or p ∈ ⟨D_{ℤ}(q)⟩. But since the discriminant of q is 3, it is anisotropic precisely when the Legendre symbol (−3/p) is nontrivial or equivalently p is not of the form 3k +1.

Example 6.7. (L. Aubry, 1912, [1, p. 273]) Let n be an integer which is a sum of three rational squares. Then n is a sum of three integral squares.

Consider the quadratic form q(X_{1}, X_{2}, X_{3}) = X_{1}^{2}+X_{2}^{2}+X_{3}^{2}. We have m = n = 3. According to (3.4), for c = 2 there exists a (ϵ, δ, m, n) = (log_{2}(3), 1, 3, 3)-division algorithm with respect to the degree function d(x) = log_{2}(|x|) on ℤ. Here we have μ = 0 and the condition 2δ > μ + ϵ holds. According to (5.1), we conclude that n is a sum of three integral squares.

Example 6.8. (Weak version of the four squares theorem) Every prime number p can be written as a sum of four rational squares.

Consider the quadratic form q(X_{1}, X_{2}, X_{3}, X_{4}) = X_{1}^{2} + X_{2}^{2} + X_{3}^{2} + X_{4}^{2}. Here we have m = n = 4. We use the same notation as in the proof of (6.5). According to (3.4), for c = 2 the ring ℤ has a (2, 1, 4, 3)-division algorithm with respect to the degree-like function d(x) = log_{2} |x|. Here μ = 0 and 2δ = μ + ϵ and we apply (4.3) (vii). It follows that q remains anisotropic over 𝔽_{p} or p is represented by q over ℚ. The first case is ruled out since every form of dimension ≥ 3 over a finite field is isotropic. Hence, p is a sum of four squares in ℚ.

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M. G. Mahmoudi, mmahmoudi@sharif.ir, Department of Mathematical Sciences, Sharif University of Technology, P. O. Box 11155-9415, Tehran, Iran. Fax: (+98) (21) 6616-5117