The Hasse Minkowski Theorem Lee Dicker University of Minnesota, REU Summer 2001

The Hasse Minkowski Theorem Lee Dicker University of Minnesota, REU Summer 2001 The Hasse-Minkowski Theorem rovides a characterization of the rational quadratic forms. What follows is a roof of the Hasse-Minkowski Theorem arahrased from the book, Number Theory by Z.I. Borevich and I.R. Shafarevich [1]. Throughout this aer, some familiarity with the -adic numbers and the Hilbert symbol is assumed and some basic facts about quadratic forms are stated without roof. Also, the roof of the Hasse Minkowski theorem given here uses the Dirichlet theorem on rimes in arithmetic rogressions. A roof of Dirichlet s theorem will not be given here (see [1], for a roof of the theorem) due to its length, but the result is stated resently. Theorem 0 (Dirichlet s theorem). Every residue class modulo m which consists of numbers relatively rime to m contains an infinite number of rime numbers. To begin, we state the definition of a quadratic form over a field K: Definition 1. A quadratic form, f, over the field K is a homogeneous olynomial of degree 2 with coefficients in K where f can be written n f = a ij x i x j, and a ij = a ji. i,j=1 Using the same notation as above, the symmetric matrix A = (a ij ) comletely determines the quadratic form f. We call A the matrix of f and we sometimes refer to det A as the determinant of f. Let X be the column vector of the variables x 1,..., x n, then we can write f = X t AX. The quadratic form g is equivalent to f = X t AX if the matrix of g, A 1, can be written as A 1 = C t AC, where C is an n n invertible matrix with entries in K. The quadratic form f is said to reresent zero in Kif there exist values α i K, not all zero such that f(α 1,..., α n ) = 0. Likewise, f reresents γ K if there are values α i K such that f(α 1,..., α n ) = γ. It is easily seen that two equivalent quadratic reresent the same elements in K. Furthermore, if a nonsingular quadratic form f (ı.e. the matrix of f is invertible) reresents zero in a field K, then f reresents all elements of K. Additionally, we note that every quadratic form is equivalent to a diagonal quadratic form. That is, if f is a quadratic form in n variables and A is the matrix of f, there exists an invertible matrix C such that the matrix C t AC is diagonal. This follows immediately from the Gram-Schmidt rocess from linear algebra. Also, it can be shown that if a diagonal quadratic form reresents zero in a field K and the number of elements of K is greater than five then there is a reresentation of zero where all the variables take on nonzero values in K. We are now reared to state the Hasse Minkowski Theorem: 1

Theorem 1 (Hasse Minkowski). A quadratic form with rational coefficients reresents zero in the field of rational numbers if and only if it reresents zero in the field of real numbers and in all fields of -adic numbers, Q (for all rimes ). The necessity of the condition is clear so we must show its sufficiency. In light of the facts that every quadratic form is equivalent to a diagonal quadratic form, and that reresenting zero is reserved when assing between equivalent quadratic forms, only diagonal quadratic forms need be considered in our roof. Essentially, the roof deends on the number of variables, n, of the quadratic form. For n = 1 the theorem is trivial. We divide the roof into four remaining cases. We consider searately when n = 2, n = 3, n = 4, or n 5. When n = 2, we first need a lemma ertaining to binary quadratic forms: Lemma 1. A binary quadratic form f(x, y) = ax 2 + 2bxy + cy 2 with determinant d = ac b 2 reresents zero in a field K if and only if d = α 2 for some α K. Sketch of roof. For the necessity of the condition, when d = 0 the roof is trivial. When d 0 we note the following two facts: A nonsingular binary quadratic form that reresents zero is equivalent to the form g = y 1 y 2 (see, for examle, [1]). Also, the determinants of equivalent quadratic forms differ by a nonzero factor which is a square in K. For the sufficiency, if f = ax 2 + by 2 and d = ab = α 2, then f(α, a) = 0. If the binary quadratic form f reresents zero in R and d is the determinant of f, it follows from Lemma 1 that d > 0, hence, d = k1 1 k2 2 ks s where k i is a rational integer. If f also reresents zero in Q for all rimes, d must a square in Q i (i = 1,..., s). Thus, k i must be even for i = 1,..., s and d is a square in Q. By Lemma 1, f reresents zero in Q. Before roving the cases where n 3 some things should be noted. The first of these observations follow from the fact that a quadratic form with rational integer coefficients f reresents zero if and only if cf reresents zero, where c is a nonzero constant in Q: throughout the roof Theorem 1, we may assume that the coefficients of the quadratic form f(x 1,..., x n ) are rational integers (if necessary, multily f by the least common multile of the denominators of the coefficients). Also, it is clear the equation f(x 1,..., x n ) = 0 (1) is solvable in Q (or in Q ) if and only if it is solvable in the rational integers, Z (resectively, in the -adic integers, Z ). We note that (1) is solvable in R if and only if the form f is indefinite. Moving on to the case n = 3, let f = a 1 x 2 + a 2 y 2 + a 3 z 2. Since f reresents zero in R, the coefficients a 1, a 2, a 3 do not all have the same sign. We may assume that two coefficients of f are ositive and one is negative (if necessary, we may multily f by 1). We may also assume that a 1, a 2, a 3 are rational integers, square-free (making a change of variables if necessary), and relatively rime. Suose a 1 and a 2 have a common rime factor, then multilying f by and setting x/ and y/ as new variables, we get a form with coefficients a 1 /, a 2 /, a 3. By reeating this rocess we obtain a quadratic form whose coefficients are ositive rational integers, airwise relatively rime and square-free: ax 2 + by 2 cz 2. (2) We now rove a theorem that is relevant to the case at hand and will also be needed again, later, in our roof of the Hasse Minkowski theorem. We begin with a lemma. 2

Lemma 2 (Hensel s lemma). Let F (x 1,..., x n ) be a olynomial whose coefficients are -adic integers. Let γ 1,..., γ n be -adic integers such that for some i (1 i n) we have then there exist -adic integers θ 1,..., θ n such that F (γ 1,..., γ n ) 0 (mod ), F x i (γ 1,..., γ n ) 0 (mod ), F (θ 1,..., θ n ) = 0 and θ i γ i (mod ) (i = 1,..., n). Proof. Consider the olynomial in x, f(x) = F (γ 1,..., γ i 1, x, γ i+1,..., γ n ). We will find α Z such that f(α) = 0 and α γ i (mod ). Then set θ j = γ j for i j, and θ i = α. It is easily seen that this suffices to rove the lemma. Let γ i = γ. We construct a sequence of -adic integers α 0, α 1,..., α m,... (3) congruent to γ modulo, such that f(α m ) 0 (mod m+1 ) for all m 0. For m = 0 take α 0 = γ. We roceed by induction. Suose for some m 1, α 0,..., α m 1 have been determined. Then, α m 1 γ (mod ) and f(α m 1 ) 0 (mod m ). We exand the olynomial f(x) in owers of x α m 1 : f(x) = β 0 + β 1 (x α m 1 ) + β 2 (x α m 1 ) 2 +... (β i Z ). By the induction hyothesis, β 0 = f(α m 1 ) = m A where A Z. Also, since α m 1 γ (mod ) and F x i (γ 1,..., γ n ) 0 (mod ), β 1 = f (α m 1 ) = B, where B Z and B is not divisible by. Set x = α m 1 + ξ m and we get f(α m 1 + ξ m ) = m (A + Bξ) + β 2 2m ξ 2 +... Since B 0 (mod ) we may choose ξ = ξ 0 Z such that A + Bξ 0 0 (mod ). Furthermore, since km 1 + m for k 2, we have f(α m 1 + ξ 0 m ) 0 (mod m+1 ). In addition m 1, so α m 1 + ξ 0 m γ (mod ) and we see that we may take α m = α m 1 + ξ 0 m. We now check that the sequence (3) converges -adically. By construction, v (α m α m 1 ) m (here v (ξ), ξ Q is the -adic value of ξ) and since the -adic numbers are comlete, (3) converges to a -adic integer, call it α. Clearly, α γ (mod ). Since f(α m ) 0 (mod m+1 ), lim m f(α m ) = 0. By the continuity of the olynomial f, lim m f(α m ) = f(α). It follows f(α) = 0. Now, for the theorem mentioned above: Theorem 2. Let 2 and 0 < r < n. The quadratic form F = F 0 + F 1 = ɛ 1 x 2 1 + + ɛ r x 2 r + (ɛ r+1 x 2 r+1 + + ɛ n x 2 n), where ɛ i (1 i n), is a -adic unit, reresents zero in Q if and only if at least one of the forms F 0 or F 1 reresents zero. Proof. The sufficiency of the condition is clear, we rove the necessity. Suose F reresents zero: ɛ 1 ξ 2 1 + + ɛ r ξ 2 r + (ɛ r+1 ξ 2 r+1 + + ɛ n ξ 2 n) = 0. (4) 3

Without loss of generality, assume that ξ i Z (1 i n) and that at least one ξ i is not divisible by. Suose ξ 1, or some ξ i among i = 1,..., r, is not divisble by. Consider equation (4) modulo, then and F 0 (ξ 1,..., ξ r ) 0 (mod ), F 0 x 1 (ξ 1,..., ξ r ) = 2ɛ 1 ξ 1 0 (mod ). By Hensel s lemma, F 0 reresents zero. Now assume ξ 1,..., ξ r are all divisible by, then some ξ i (r + 1 i n) is not divisible by and ɛ 1 ξ 2 1 + + ɛ r ξ 2 r 0 (mod 2 ). When looking at equation (4) modulo 2 we have F 1 (ξ r+1,..., ξ n ) 0 (mod 2 ), or after dividing by F 1 (ξ r+1,..., ξ n ) 0 (mod ). We roceed as above, alying Hensel s lemma, and conclude that F 1 reresents zero. Two useful corollaries follow: Corollary 1. If ɛ 1,..., ɛ r are -adic units and 2, then the quadratic form f = ɛ 1 x 2 1 + + ɛ r x 2 r reresents zero in Q if and only if the congruence f(x 1,..., x r ) 0 (mod ) has a nontrivial solution in Z. Corollary 2. Let f be the quadratic form from Corollary 1. If r 3, then f always reresents zero in Q. Proof. Corollary 1 is an immediate result of Theorem 2. Corollary 2 follows from Chevaley s theorem, which states that if F (x 1,..., x n ) is a form of degree less than n, then the congruence F (x 1,..., x n ) 0 (mod ), a rime, has a nontrivial solution. Getting back to the Hasse Minkowski theorem, we are considering the quadratic form (2) where a, b, c are ositive, square-free, rational integers that are airwise relatively rime. Let 2 be a rime divisor of c. Since (2) reresents zero in Q by assumtion, we may aly Theorem 2 and Corollary 1 and conclude that the congruence ax 2 + by 2 0 (mod ) has a nontrivial solution, (x 0, y 0 ). It follows that the form ax 2 + by 2 factors linearly, modulo : if we assume y 0 is not divisible by, we have ax 2 + by 2 ay 2 0 (xy 0 + yx 0 )(xy 0 yx 0 ) (mod ). Thus, since divides c, (2) factors into linear factors modulo : ax 2 + by 2 cz 2 L () (x, y, z)m () (x, y, z) (mod ), where L () and M () are integral linear forms. Similarly, for,, an odd rime divisor of a or b, the quadratic form (2) factors into linear integral forms modulo. Also, when = 2 we note that the quadratic form (2) factors linearly modulo 2 since ax 2 + by 2 cz 2 (ax + by cz) 2 (mod 2). By the Chinese Remainder theorem we may find integral linear forms L(x, y, z) and M(x, y, z) such that L(x, y, z) L () (x, y, z) (mod ), M(x, y, z) M () (x, y, z) (mod ) 4

for all rime divisors of a, b, and c. Since a, b, c are square-free and airwise relatively rime, we have the congruence ax 2 + by 2 cz 2 L(x, y, z)m(x, y, z) (mod abc). (5) Now give rational integer values to the variables x, y, z satisfying the inequalities 0 x < bc, 0 y < ac, 0 z < ab. (6) Without loss of generality, we may exclude the case a = b = c = 1, then, since a, b, c are airwise relatively rime and square-free, bc, ac, and ab are not integers. It follows that the number of triles (x, y, z) satisfying the inequalities (6) will be strictly greater than bc ac ab = abc. Since the number of triles is greater than the number of resiue classes modulo abc, there exist distinct triles (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) such that L(x 1, y 1, z 1 ) L(x 2, y 2, z 2 ) (mod abc). If we set it follows from the linearity of L that Looking at the congruence (5) we see that x 0 = x 1 x 2, y 0 = y 1 y 2, z 0 = z 1 z 2, L(x 0, y 0, z 0 ) 0 (mod abc). ax 2 0 + by 2 0 cz 2 0 0 (mod abc). (7) Since the triles (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) satisfy the inequalities (6), we have Thus, Combining (7) and (8) we see that either or x 0 < bc, y 0 < ac, z 0 < ab. abc < ax 2 0 + by 2 0 cz 2 0 < 2abc. (8) ax 2 0 + by 2 0 cz 2 0 = 0, (9) ax 2 0 + by 2 0 cz 2 0 = abc. (10) In the first case, (x 0, y 0, z 0 ) is a nontrivial reresentation of zero in Q. If the second case holds, we aeal to the following. Since b and c are square-free and relatively rime, bc is not a square. We show that (2) reresents zero if and ony if ac is the norm of some element from the field Q( bc). Suosing (9) holds, we assume without loss of generality that x 0 0, it follows that ( cz0 ) 2 ( y0 ) 2 ( cz0 ac = bc = N + y ) 0 bc. x 0 x 0 x 0 x 0 For the converse, if ac = N(u + v bc), then ac 2 + b(cv) 2 cu 2 = 0. Now suose (10) holds. Multilying by c, we get ac(x 2 0 bc) = (cz 0 ) 2 bcy 2 0. Setting α = x 0 + bc, β = cz 0 + y 0 bc, it follows that and acn(α) = N(β), ( β ac = N. α) Thus ac is the norm of β α Q( bc) and (2) reresents zero in Q. In rearation for the last two cases to be roved, that is n = 4 and n 5, we need another lemma. Before that, we review some roerties of the Hilbert symbol. 5

Definition 2. For any air α 0, β 0 of -adic numbers, the Hilbert symbol (α, β) is equal to +1 or 1 if the form αx 2 + βy 2 z 2 reresents zero in the field Q or not, accordingly. We extend the Hilbert symbol to the real numbers by calling R the field Q. Whereas the field Q is the comletion of Q with resect to a finite rime, we say the real numbers are the comletion of Q with resect to the infinite rime. What follows are some basic roerties of the Hilbert symbol (see, for examle, [1]): (α, β 1 β 2 ) = (α, β 1 ) (α, β 2 ), (α, β) = (β, α), (α, α) = (α, 1). For -adic units ɛ and η, and real numbers a and b, ( ɛ (, ɛ) =, (ɛ, η) = 1 for 2,, ) where ( ɛ ) is the Legendre symbol. (2, ɛ) 2 = ( 1) (ɛ2 1)/8, (ɛ, η) 2 = ( 1) [(ɛ 1)/2][(η 1)/2], (a, b) = 1, if a > 0 or b > 0, (a, b) = 1, if a < 0 and b < 0, Lemma 3. If a rational quadratic form in three variables reresents zero in all fields Q, where runs through all rimes and, excet ossibly for Q q, then it reresents zero in Q q. Proof. Consider the rational quadratic form ax 2 + by 2 z 2. It follows from Corollary 2 that (a, b) = 1 for all, excet ossibly when = 2, or is in the rime factorization of a or b. Thus, there are only finitely many values of for which (a, b) = 1 and the roduct (a, b), where takes on the value of all rimes and the symbol, makes sense. To rove the lemma it suffices to show that (a, b) = 1, (11) that is, the number of for which (a, b) = 1 is even. The basic roerties of the Hilbert symbol allow us to reduce the roof of (11) to three cases: (1) a = 1, b = 1, (2) a = q, b = 1 (q a rime), (3) a = q, b = q (q and q rimes). All of the following comutations follow from the basic roerties of the Hilbert symbol and, in one lace, the law of quadratic recirocity. Case (1): ( 1, 1) = ( 1, 1) 2 ( 1, 1) = ( 1) ( 1) = 1. Case (2): (q, 1) = (q, 1) q (q, 1) 2 = (2, 1) = (2, 1) 2 (2, 1) = 1 1 = 1, ( 1 ) ( 1) [(q 1)/2][( 1 1)/2] = ( 1) (q 1)/2 ( 1) [(q 1)/2][( 1 1)/2] = 1. q 6

Case (3): (2, q) = (2, q) q (2, q) 2 = ( 2 q ) ( 1) (q2 1)/8 = ( 1) (q2 1)/8 ( 1) (q2 1)/8 = 1, ( q (q, q ) = (q, q ) q (q, q ) q (q, q )( q ) ) 2 = q q ( 1) [(q 1)/2][(q 1)/2] = ( 1) [(q 1)/2][(q 1)/2] ( 1) [(q 1)/2][(q 1)/2] = 1. This roves (11), hence, the lemma. For the case n = 4 of the Hasse Minkowski theorem we will consider the quadratic form a 1 x 2 1 + a 2 x 2 2 + a 3 x 2 3 + a 4 x 2 4 (12) where a i (1 i 4) is a square-free integer. The form (12) reresents zero in R, thus, we may assume a 1 > 0, a 4 < 0. Furthermore, let g = a 1 x 2 1 + a 2 x 2 2 and h = a 3 x 2 3 a 4 x 2 4. To rove the theorem for n = 4, we show that there exists a rational number a, such that both g and h reresent a in Q. Let 1,..., s be all the distinct odd rimes that divide a 1, a 2, a 3, a 4. For each of these rimes and the rime = 2 choose a reresentation of zero in Q : a 1 ξ 2 1 + a 2 ξ 2 2 + a 3 ξ 2 3 + a 4 ξ 2 4 = 0. We ick the ξ i so that ξ i 0 (1 i 4) (see the review of quadratic forms, above). Set b = a 1 ξ 2 1 + a 2 ξ 2 2 = a 3 ξ 2 3 a 4 ξ 2 4. Choose our ξ i so that b is divisible by at most the first ower of (if b = 0, then g, h reresent zero and thus reresent all elements of Q ). The congruences a b 2 (mod 16) a b 1 (mod 2 1) (13). a b s (mod 2 s) determine a rational integer a unique modulo m = 16 2 1 2 s. Since b i is divisble by at most the first ower of i, following congruence holds: b i a 1 1 (mod i ). Any -adic unit congruent to 1 modulo is a square in Q (see [1]), thus b i a 1 is a square in Q i. Similarly, b 2 a 1 1 (mod 8), thus b 2 a 1 is a square in Q 2 (a 2-adic unit ɛ is a square in Q 2 if and only if ɛ 1 (mod 8), see [1]). Note that b and a differ by a square in Q. Thus, for = 2, 1,..., s the quadratic forms reresent zero in Q. ax 2 0 + g and ax 2 0 + h (14) Choosing a > 0, we see that the forms (14) reresent zero in R since a 1 > 0 and a 4 < 0. Suose now that 2, 1,..., s and a, then by Corollary 2 the forms (14) reresent zero in Q. To summarize our current osition, the forms (14) reresent zero in R and all -adic fields excet ossibly when divides 7

a and 2, 1,..., s. Thus, if we can choose a ositive rational integer a that satisfies the congruences (13) and is divisible by only rimes among 2, 1,..., s and ossibly one other rime q, we may aeal to Lemma 3 and conclude that (14) reresents zero in Q for all rimes including = q. We use Dirichlet s theorem on rimes in arithmetic rogressions. Pick a rational integer a > 0 that satisfies the congruences (13). Set d = gcd(a, m), then gcd(a /d, m/d) = 1. By Dirichlet s theorem, there exists a k N such that a d + k m d = q is rime. Take a = a + km = dq. Hence, the forms (14) reresent zero in R and Q for all rimes. By the Hasse Minkowski theorem for three variables, the forms (14) reresent zero in Q. It clearly follows that g and h both reresent a in Q. This roves the Hasse Minkowski theorem for quadratic forms in four variables. Lastly we deal with case n 5. Consider the quadratic form Since the form is indefinite, we may assume a 1 > 0 and a 5 < 0. We set a 1 x 2 1 + a 2 x 2 2 + a 3 x 2 3 + a 4 x 2 4 + a 5 x 2 5. (15) g = a 1 x 2 1 + a 2 x 2 2 and h = a 3 x 2 3 a 4 x 2 4 a 5 x 2 5. Proceeding exactly as in the case n = 4, we find a rational integer a > 0 such that g and h reresent a in R and Q for all rimes with the ossible excetion of one rime q. By Lemma 3, g reresents a in all -adic fields including when = q. To show that h reresents a in all -adic fields we note that h reresents zero in Q q by Corollary 2 (as we saw in the roof of the case n = 4, q a i (3 i 5)). It follows that h reresents all elements of Q q, namely h reresents a. Thus g and h reresent a in R and Q for all rimes. Clearly then, ax 2 0 + g and ax 2 0 + h reresent zero in R and Q for all rimes. By the Hasse Minkowski theorem for forms in three and four variables, ax 2 0 + g and ax 2 0 + h both reresent zero in Q. Hence, g and h both reresent a in Q and the form (15) reresents zero in Q. In order to generalize to n > 5 we first note that any quadratic form over Q with five or more variables always reresents zero in Q (see [1]). So the Hasse Minkowski theorem reads: a quadratic form in five or more variables with rational coefficients reresents zero in Q if and only if it reresents zero in R. An indefinite quadratic form in more than five variable is equivalent to an indefinite diagonal quadratic form f, which may be written as f = f 0 + f 1 where f 0 is an indefinite quadratic form in five variables. Since we are assured that f 0 reresents zero in Q for all (and in R), by what was just roved f 0 reresents zero in Q. It clearly follows that f reresents zero in Q. With that, Theorem 1 has been roved. References: [1] Borevich, Z. I. and Shafarevich, I. R. (translated by Greenleaf, Newcomb), Number Theory, Academic Press Inc., 1966 8