18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #10 10/8/2013 In this lecture we lay the groundwork needed to rove the Hasse-Minkowski theorem for Q, which states that a quadratic form over Q reresents 0 if and only if it reresents 0 over every comletion of Q (as roved by Minkowski). The statement still holds if Q is relaced by any number field (as roved by Hasse), but we will restrict our attention to Q. Unless otherwise indicated, we use througout to denote any rime of Q, including the archimedean rime =. We begin by defining the Hilbert symbol for. 10.1 The Hilbert symbol Definition 10.1. For a, b Q the Hilbert symbol (a, b) is defined by { 1 ax 2 + by 2 = 1 has a solution in Q, (a, b) = 1 otherwise. It is clear from the definition that the Hilbert symbol is symmetric, and that it only deends on the images of a and b in Q /Q 2 (their square classes). We note that Q /Q 2 Z/2Z if =, (Z/2Z )2 if is odd, (Z/2Z) 3 if = 2. The case = is clear, since R = Q has just two square classes (ositive and negative numbers), and the cases with < were roved in Problem Set 4. Thus the Hilbert symbol can be viewed as a ma (Q /Q 2 ) (Q /Q 2 ) {±1} of finite sets. We say that a solution (x 0,..., x n ) to a homogeneous olynomial equation over Q is rimitive if all of its elements lie in Z and at least one lies in Z. The following lemma gives several equivalent definitions of the Hilbert symbol. Lemma 10.2. For any a, b Q, the following are equivalent: (i) (a, b) = 1. (ii) The quadratic form z 2 ax 2 by 2 reresents 0. (iii) The equation ax 2 + by 2 = z 2 has a rimitive solution. (iv) a Q is the norm of an element in Q ( b). Proof. (i) (ii) is immediate (let z = 1). The reverse imlication is clear if z 2 ax 2 by 2 = 0 reresents 0 with z nonzero (divide by z 2 ), and otherwise the non-degenerate quadratic form ax 2 + by 2 reresents 0, hence it reresents every element of Q including 1, so (ii) (i). To show (ii) (iii), multily through by r, for a suitable integer r, and rearrange terms. The reverse imlication (iii) (ii) is immediate. If b is square then Q ( b) = Q and N(a) = a so (iv) holds, and the form z 2 by 2 reresents 0, hence every element of Q including ax 2 0 for any x 0, so (ii) holds. If b is not square then N(z +y b) = z 2 by ( 2. If a is a norm in Q b) then z 2 ax 2 by 2 reresents 0 (set x = 1), and if z 2 ax 2 by 2 reresents 0 then dividing by x 2 and adding a to both sides shows that a is a norm. So (ii) (iv). 1 Andrew V. Sutherland
Corollary 10.3. For all a, b, c Q, the following hold: (i) (1, c) = 1. (ii) ( c, c) = 1. (iii) (a, c) = 1 = (a, c) (b, c) = (ab, c). (iv) (c, c) = ( 1, c). Proof. Let N denote the norm ma from Q ( c) to Q. For (i) we have N(1) = 1. For (ii), c = N( c) for c Q 2 and c = N( c) otherwise. For (iii), If a and b are both norms in Q( c), then so is ab, by the multilicativity of the norm ma; conversely, if a and ab are both norms, so is 1/a, as is (1/a)ab = b. Thus if (a, c) = 1, then (b, c) = 1 if and only if (ab, c) = 1, which imlies (a, c) (b, c) = (ab, c). For (iv), ( c, c) = 1 by (ii), so by (iii) we have (c, c) = ( c, c) (c, c) = ( c 2, c) = ( 1, c). Theorem 10.4. (a, b) = 1 if and only if a, b < 0 Proof. We can assume a, b {±1}, since {±1} is a comlete set of reresentatives for R /R 2. If either a or b is 1 then ( a, b) = 1, by Corollary 10.3.(i), and ( 1, 1) = 1, since 1 is not a norm in C = Q ( 1). Lemma 10.5. If is odd, then (u, v) = 1 for all u, v Z. Proof. Recall from Lecture 3 (or the Chevalley-Warning theorem on roblem set 2) that every lane rojective conic over F has a rational oint, so we can find a non-trivial solution to z 2 ux 2 vy 2 = 0 modulo. If we then fix two of x, y, z so that the third is nonzero, Hensel s lemma gives a solution over Z. Remark 10.6. Lemma 10.5 does not hold for = 2; for examle, (3, 3) 2 = 1. Theorem 10.7. Let be an odd rime, and write a, b Q α, β Z and u, v Z. Then as a = α u and b = β v, with (a, b) = ( 1) ( β 1 αβ 2 x mod where ( x ) denotes the Legendre symbol ( ). u ) ( v Proof. Since (a, b) deends only on the square classes of a and b, we assume α, β {0, 1}. Case α = 0, β = 0: We have (u, v) = 1, by Lemma 10.5, which agrees with the formula. Case α = 1, β = 0: We need to show that (u, v) = ( v ). Since (u 1, v) = 1, we have (u, v) = (u, v) (u 1, v) = (, v), by Corollary 10.3.(iii). If v is a square then we have (, v) = (, 1) = (1, ) = 1 = ( v ). If v is not a square then z 2 x 2 vy 2 = 0 has no non- v trivial solutions modulo, hence no rimitive solutions. This imlies ( ) ( (, v) = 1 = ( ) ). 1 Case α = 1, β = 1: We must show (u, v) u v = ( 1) 2. Alying Corollary 10.3 we have ) α, (u, v) = (u, v) ( v, v) = ( 2 uv, v) = ( uv, v) = (v, uv) Alying the formula in the case α = 1, β = 0 already roved, we have ( ) ( ) ( ) ( ) ( uv 1 u v u (v, uv) = = = ( 1) 1 2 ) ( v ). 2
Lemma 10.8. Let u, v Z. The equations z 2 ux 2 vy 2 = 0 and z 2 2ux 2 2 2 vy = 0 have rimitive solutions over Z 2 if and only if they have rimitive solutions modulo 8. Proof. Without loss of generality we can assume that u and v are odd integers, since every square class in Z 2 /Z 2 2 is reresented by an odd integer (in fact one can assume u, v {±1, ±5}) The necessity of having a rimitive solution modulo 8 is clear. To rove sufficiency we aly the strong form of Hensel s lemma roved in Problem Set 4. In both cases, if we have a non-trivial solution (x 0, y 0, z 0 ) modulo 8 we can fix two of x 0, y 0, z 0 to obtain a quadratic olynomial f(w) over Z 2 and w 0 Z 2 that satisfies v 2(f(w 0 )) = 3 > 2 = 2v 2 (f (w 0 )). In the case of the second equation, note that a rimitive solution (x 0, y 0, z 0 ) modulo 8 must have y 0 or z 0 odd; if not, then z0 2 and vy2 0, and therefore 2ux2 0, are divisible by 4, but this means x 0 is also divisible by 2, which contradicts the rimitivity of (x 0, y 0, z 0 ). Lifting w 0 to a root of f(w) over Z 2 yields a solution to the original equation. Theorem 10.9. Write a, b Q 2 as a = 2α u and b = 2 β v with α, β Z and u, v Z 2. Then (a, b) 2 = ( 1) ɛ(u)ɛ(v)+αω(v)+βω(u), where ɛ(u) and ω(u) denote the images in Z/2Z of (u 1)/2 and (u 2 1)/8, resectively. Proof. Since (a, b) 2 only deends on the square classes of a and b, It suffices to verify the formula for a, b S, where S = {±1, ±3, ±2, ±6} is a comlete set of reresentatives for Q 2 /Q 2 2. As in the roof of Theorem 10.7, we can use (u, v) 2 = (v, uv) 2 to reduce to the case where one of a, b lies in Z. By Lemma 10.8, to comute (a, b) 2 with one of a, b in Z 2, it suffices to check for rimitive solutions to z 2 ax 2 by 2 = 0 modulo 8, which reduces the roblem to a finite verification which erformed by We now note the following corollary to Theorems 10.4, 10.7, and 10.9. Corollary 10.10. The Hilbert symbol (a, b) is a nondegenerate bilinear ma. This means that for all a, b, c Q we have (a, c) (b, c) = (ab, c) and (a, b) (a, c) = (a, bc), and that for every non-square c we have (b, c) = 1 for some b. Proof. Both statements are clear for = (there are only 2 square classes and 4 combi 1 nations to check). For odd, let c = γ w and fix ε = ( 1) γ 2. Then for a = α u and b = β v, we have ( ) ( ) ( ) ( ) α u w β v w ( a, c) (b, c) = ε ε ( ) γ ( ) uv w α+β = ε α+β = (ab, c). γ α γ β To verify non-degeneracy, we note that if c is not square than either γ = 1 or ( w ) = 1. If γ γ = 1 we can choose b = v with ( v ) = 1, so that (b, c) = ( v ) = 1. If γ = 0, then ε = 1 and ( w ) = 1, so withb = we have (b, c) = ( w ) = 1. 3
For = 2, we have (a, c) (b, c) = ( 1) ɛ(u)ɛ(w)+αω(w)+γω(u) ( 1) ɛ(v)ɛ(w)+βω(w)+γω(v) 2 2 = ( 1) (ɛ(u)+ɛ(v))ɛ(w)+(α+β)ω(w)+γ(ω(u)+ω(v)) = ( 1) ɛ(uv)ɛ(w)+(α+β)ω(w)+γω(uv) = (ab, c) 2, where we have used the fact that ɛ and ω are grou homomorhisms from Z 2 to Z/2Z. To see this, note that the image of ɛ 1 (0) in (Z/4Z) is {1}, a subgrou of index 2, and the image of ω 1 (0) in (Z/8Z) is {±1}, which is again a subgrou of index 2. We now verify non-degeneracy for = 2. If c is not square then either γ = 1, or one of ɛ(w) and ω(w) is nonzero. If γ = 1, then (5, c) 2 = 1. If γ = 0 and ω(w) = 1, then (2, c) 2 = 1. If γ = 0 and ω(w) = 0, then we must have ɛ(w) = 1, so ( 1, c) 2 = 1. We now rove Hilbert s recirocity law, which may be regarded as a generalization of quadratic recirocity. Theorem 10.11. Let a, b Q. Then (a, b) = 1 for all but finitely many rimes and (a, b) = 1. Proof. We can assume without loss of generality that a, b Z, since multilying each of a and b by the square of its denominator will not change (a, b) for any. The theorem holds if either a or b is 1, and by the bilinearity of the Hilbert symbol, we can assume that a, b { 1} {q Z >0 : q is rime}. The first statement of the theorem is clear, since a, b Z for < not equal to a or b, and (u, v) = 1 for all u, v Z when is odd, by Lemma 10.5. To verify the roduct formula, we consider 5 cases. Case 1: a = b = 1. Then ( 1, 1) = ( 1, 1) 2 = 1 and ( 1, 1) = 1 for odd. Case 2: a = 1 and b is rime. If b = 2 then (1, 1) is a solution to x 2 + 2y 2 = 1 over Q for all, thus ( 1, 2) = 1. If b is odd, then ( 1, b) = 1 for {2, b}, while ɛ(b) 1 ( 1, b) (b 1) 2 = ( 1) and ( 1, b) b = ( b ), both of which are equal to ( 1) /2. Case 3: a and b are the same rime. Then by Corollary 10.3, (b, b) = ( 1, b) for all rimes, and we are in case 2. Case 4: a = 2 and b is an odd rime. Then (2, b) = 1 for all {2, b}, while ω(b) 2 (2, b) (b 2 1)/8 2 = ( 1) and (2, b) b = ( ), both of which are equal to ( 1). Case 5: a and b are distinct odd rimes. Then (a, b) = 1 for all {2, a, b}, while ( 1) ɛ(a)ɛ(b) if = 2, ( ) (a, b) a = ( b if = b, b ) if = a. Since ɛ(x) = (x 1)/2 mod 2, we have ( a 1 b 1 a ) ( b ) (a, b) = ( 1) 2 2 = 1, b a by quadratic recirocity. a 4
MIT OenCourseWare htt://ocw.mit.edu 201 For information about citing these materials or our Terms of Use, visit: htt://ocw.mit.edu/terms.