Reciprocity laws for Legendre symbols of the type (a+b m − − √ /p) - Long Version
aa r X i v : . [ m a t h . N T ] A ug RECIPROCITY LAWS FOR LEGENDRE SYMBOLS OFTHE TYPE (cid:16) a + b √ mp (cid:17) - LONG VERSION CONSTANTIN N. BELI
Note
This paper is an announcement. We do not prove the mainresult, Theorem 2.5 (as well as Theorem 2.10). We only show howTheorem 2.5 can be used to produce reciprocity laws for Legendresymbols of the type (cid:16) a + b √ mp (cid:17) . Note that in the applications we won’tuse Theorem 2.10, which is stronger than Theorem 2.5. Thereforereaders who are not interested in details may skip Theorem 2.10 andeverything related to it. (Lemma 1.1 and the part of § Introduction and notations
Given p > a, b, m ∈ Z /p Z such that (cid:16) mp (cid:17) = (cid:16) a − mb p (cid:17) =1, the Legendre’s symbol (cid:16) a + b √ mp (cid:17) is defined as (cid:16) a + bαp (cid:17) , where α ∈ Z /p Z satisfies α = m . The definition is independent of the choiceof the square root α of m because ( a + bα )( a − bα ) = a − b m so (cid:16) a + bαp (cid:17) (cid:16) a − bαp (cid:17) = (cid:16) a − mb p (cid:17) = 1 so (cid:16) a + bαp (cid:17) = (cid:16) a − bαp (cid:17) . One may aslodefine (cid:16) a + b √ mp (cid:17) if a − mb = 0 and a = 0. In this case the two radicalsof m are α = ± ab . If we take α = − ab then (cid:16) a + bαp (cid:17) = (cid:16) p (cid:17) , whichis not convenient. Hence we will take α = ab and we get (cid:16) a + b √ mp (cid:17) = (cid:16) a + bαp (cid:17) = (cid:16) ap (cid:17) .In the particular case when a = 0, b = 1 we get (cid:16) √ mp (cid:17) , defined when (cid:16) mp (cid:17) = (cid:16) − m · p (cid:17) = 1, i.e. when p ≡ (cid:16) mp (cid:17) = 1. Thiscoincides with the rational quartic residue symbol (cid:16) mp (cid:17) . THIS PAPER CONTAINS SOME RATHER LENGHTY DISCUTIONSINVOLVING 2-ADIC HILBERT SYMBOLS. READERS WHO WANTTO SKIP THIS TECHNICAL PART MAY CONSIDER THE SHORTVERSION.
The Legendre symbol (cid:16) a + b √ mp (cid:17) can also be defined when a, b, m are p -adic integers satisfying (cid:16) mp (cid:17) = (cid:16) a − mb p (cid:17) = 1 or p | a − mb , p ∤ a .For any prime p of Q , including the archimedian prime p = ∞ , wedenote by (cid:16) · , · p (cid:17) : Q × p / ( Q × p ) × Q × p / ( Q × p ) → {± } the Hilbert symbol.(At p = ∞ we have Q ∞ = R .)Similarly as for (cid:16) a + b √ mp (cid:17) we can introduce the Hilbert symbol (cid:16) a + b √ m,cp (cid:17) .It is defined for a, b, c, m ∈ Q p satisfying m ∈ ( Q × p ) and (cid:16) a − mb ,cp (cid:17) =1. If ± α ∈ Q × p are the two square roots of m then (cid:16) ( a + bα )( a − bα ) ,cp (cid:17) = (cid:16) a − mb ,cp (cid:17) = 1 so (cid:16) a + bα,cp (cid:17) = (cid:16) a − bα,cp (cid:17) so there is no ambiguity in thisdefinition. If a − mb = 0 and a = 0, same as for (cid:16) a + b √ mp (cid:17) , we define (cid:16) a + b √ m,cp (cid:17) = (cid:16) a,cp (cid:17) . Throughout this paper we will consider (cid:16) a + b √ m,cp (cid:17) only for p = 2.Various mathematicians have obtained results involving Legendresymbols of the type (cid:16) a + b √ mp (cid:17) . Most of this results involve the bi-quadratic symbols (cid:16) mp (cid:17) and they are called biquadratic or quarticreciprocity laws. In this paper we will give some very general reci-procity laws, which generalize all the existing results. In § Definition 1.
For any A ∈ Q × / ( Q × ) we denote by A the only square-free integer such that A = A ( Q × ) , The algebras T ( V ) , S ( V ) and S ′ ( V ) . For convenience we denote by V := Q × / ( Q × ) and for any prime p we denote V p := Q × p / ( Q × p ) . In particular, V ∞ = R × / ( R × ) .Now V and V p are Z / Z -vector spaces. Therefore we may considerthe corresponding tensor algebras T ( V ) and T ( V p ) and the symmetricalgebras S ( V ) and S ( V p ). Note that, while V has a multiplicativenotation, T ( V ) and S ( V ) have an additive one. In order to preventconfusion for any A , . . . , A n ∈ V we will denote by A • A • · · · • A n their product in S n ( V ) (instead of simply A · · · A n ). For example, thedistributivity law on S ( V ) will be written as AB • C = A • C + B • C .We also define the algebra S ′ ( V ) = T ( V ) / h A ⊗ · · · ⊗ A n − A π (1) ⊗ · · · ⊗ A π ( n ) | π ∈ A n i . ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) (Same definition as for S ( V ) but this time the products are invariantonly under even permutations of the factors.)We denote by A ⊙· · ·⊙ A n the image in S ′ ( V ) of A ⊗· · ·⊗ A n ∈ T ( V ).We will only be concerned with the homogenous component of degree3, S ′ ( V ) = T ( V ) / h A ⊗ B ⊗ C − B ⊗ C ⊗ A i . We have A ⊙ B ⊙ C = B ⊙ C ⊙ A = C ⊙ A ⊙ B and B ⊙ A ⊙ C = A ⊙ C ⊙ B = C ⊙ B ⊙ A .However A ⊙ B ⊙ C = B ⊙ A ⊙ C (unless A, B, C are linearly dependent).Similarly we define S ′ ( V p ).For any ξ ∈ V , T ( V ), S ( V ) or S ′ ( V ) and any prime p we denoteby ξ p the image of ξ in V p , T ( V p ), S ( V p ) or S ′ ( V p ). When there is nodanger of confusion we simply write ξ instead of ξ p . Lemma 1.1.
We have an exact sequence S ( V ) τ −→ S ′ ( V ) ρ −→ S ( V ) → , where τ is given by A • B • C A ⊙ B ⊙ C + B ⊙ A ⊙ C and ρ by A ⊙ B ⊙ C A • B • C . P roof.
The mapping ρ is well defined and surjective because S ( V ) = T ( V ) /I and S ′ ( V ) = T ( V ) /I ′ where I, I ′ are bubgroups of T ( V )with I ′ ⊂ I . For τ we note that the mapping : V → S ′ ( V ) given by( A, B, C ) A ⊙ B ⊙ C + B ⊙ A ⊙ C is trilinear and also symmetric.(Recall that A ⊙ B ⊙ C = B ⊙ C ⊙ A = C ⊙ A ⊙ B and B ⊙ A ⊙ C = A ⊙ C ⊙ B = C ⊙ B ⊙ A .) Hence τ is well defined.We have ρ ( τ ( A • B • C ))) = A • B • C + B • A • C = 2( A • B • C ) = 0so ρ ◦ τ = 0 so Im τ ⊆ ker ρ . For the reverse inclusion take ξ = P i A i ⊙ B i ⊙ C i ∈ ker ρ . Then 0 = ρ ( ξ ) = P i A i • B i • C i so thereare A j, , A j, , A j, ∈ V and π j ∈ S such that P i A i ⊗ B i ⊗ C i = P j ( A j, ⊗ A j, ⊗ A j, − A j,π j (1) ⊗ A j,π j (2) ⊗ A j,π j (3) ). This implies that ξ = P i A i ⊙ B i ⊙ C i = P j ( A j, ⊙ A j, ⊙ A j, − A j,π j (1) ⊙ A j,π j (2) ⊙ A j,π j (3) ).But A j,π j (1) ⊙ A j,π j (2) ⊙ A j,π j (3) equals A j, ⊙ A j, ⊙ A j, or A j, ⊙ A j, ⊙ A j, if π j is even or odd, respectively. So A j, ⊙ A j, ⊙ A j, − A j,π j (1) ⊙ A j,π j (2) ⊙ A j,π j (3) is 0 if π j ∈ A and it is A j, ⊙ A j, ⊙ A j, − A j, ⊙ A j, ⊙ A j, = τ ( A j, • A j, • A j, ) otherwise. It follows that ξ = τ ( η ), where η = P j,π j / ∈ A A j, • A j, • A j, . (cid:3) Main result
Definition 2.
We denote by D the set of all ( B, A, C ) ∈ V such that1) (cid:16) A,Bp (cid:17) = (cid:16) A,Cp (cid:17) = (cid:16) B,Cp (cid:17) = 1 for every prime p , including thearchimedian prime p = ∞ .2) ( A, B, C ) = 1 . CONSTANTIN N. BELI
3) At least one of
A, B, C is ≡ . Definition 3.
We define f : D → {± } as follows. If ( B, A, C ) ∈D we take x, y, z ∈ Z with ( x, y, z ) = 1 such that x − Ay = Bz .(The existence of such x, y, z is ensured by Minkovski’s theorem since (cid:16) A,Bp (cid:17) = 1 for all p .) Then define f ( B, A, C ) = α ∞ Q p | C α p , where α ∞ = ( if C > sgn ( x ) if C < ,α p = (cid:18) x + y √ Ap (cid:19) if p ∤ A (cid:18) x + z √ B ) p (cid:19) if p ∤ B if p | C , p > and α = if C ≡ (cid:18) x + y √ A,C (cid:19) if A ≡ (cid:18) x + z √ B ) ,C (cid:19) if B ≡ (cid:0) z, (cid:1) if A ≡ C ≡ − (cid:0) y, (cid:1) if B ≡ C ≡ − (cid:18) x + z √ B, − (cid:19) if A ≡ B ≡ , C ≡ − − (cid:18) x + z √ B, (cid:19) if A ≡ B ≡ , C ≡ . Definition 4.
We define f : D → {± } as follows. If ( B, A, C ) ∈ D we take x, y, z ∈ Z with ( x, y, z ) = 1 such that x − By = − ABCz .(The existence of such x, y, z is ensured by Minkovski’s theorem since (cid:16) A,Bp (cid:17) = (cid:16) C,Bp (cid:17) = 1 so (cid:16) − ABC,Bp (cid:17) = 1 for all p .) Then define f ( B, A, C ) = β ∞ Q p | C β p , where β ∞ = ( if C > sgn ( x ) if C < ,β p = (cid:16) xp (cid:17) if p ∤ A (cid:18) x + y √ B ) p (cid:19) if p ∤ B ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) if p | C , p > and β = if C ≡ (cid:0) x,C (cid:1) if A ≡ (cid:18) x + y √ B ) ,C (cid:19) if B ≡ (cid:18) y + z q ABCB , (cid:19) if A ≡ C ≡ − (cid:0) z, (cid:1) if B ≡ C ≡ (cid:18) − x + y √ B ) ,C (cid:19) if A ≡ B ≡ . If x = 0 and A ≡ β we don’t use the option β = (cid:0) x,C (cid:1) = (cid:0) ,C (cid:1) ,which is not defined. More precisely, if x = 0 then 0 − By = − ABCz ,which implies that B = ABC so A = C . Hence if A ≡ C ≡ β = 1. (See thespecial case 2.) In the definition of f and f we may replace the con-dition that x, y, z are relatively prime integers by the condition that x, y, z are relatively prime elements of Z [ ]. Indeed, if we replace x, y, z by 2 k x, k y, k z for some k ∈ Z then α ∞ is not changed, α is re-placed by (cid:16) k ,C (cid:17) α and for any p > p | C α p is replced by (cid:16) k p (cid:17) α p .Since (cid:16) k ,C (cid:17) Q p | C,p> (cid:16) k p (cid:17) = Q p (cid:16) k ,Cp (cid:17) = 1 the product α ∞ Q p | C α p is not changed. Similarly for the product β ∞ Q p | C β p , which defines f ( B, A, C ). In the definition of α , β for C (cid:16) a + b √ m,C (cid:17) ,where a, b ∈ Q and m ∈ Z . Here we have the liberty of choosingeither of the two quadratic roots of m . Let m = 1 + 8 x . We make theconvention that √ m is the quare root of m that is ≡ √ m +(1 − x ) ≡ √ m ) − (1 − x ) = − x ( x − √ m − (1 − x )...16 so √ m ≡ − x = − m (mod 16).Consider the ± sign such thatord ( a ± b · − m { ord ( a + b · − m , ord ( a − b · − m } . It follows that ord ( a ± b · − m ) ≤ ord (( a + b · − m ) + ( a − b · − m )) =ord b · − m = ord b . (Recall that − m = 1 − x is an odd 2-adic CONSTANTIN N. BELI integer.) Hence a ± b · − m | b . Since also b ( √ m − − m )...16 b we get a ± b √ ma ± b · − m − ± b ( √ m − − m ) a ± b · − m ... 16 b b = 8 . It follows that a ± b √ ma ± b · − m is a square so (cid:16) a ± b √ m,C (cid:17) = (cid:16) a ± b · − m ,C (cid:17) .If C ≡ (cid:16) a ± b √ m, (cid:17) . By thesame reasoning as above, since √ m ≡ ± signsuch that ord ( a ± b ) = min { ord ( a + b ) , ord ( a − b ) } then a ± b √ ma ± b ≡ (cid:18) a ± b √ ma ± b , (cid:19) = 1, which implies that (cid:16) a ± b √ m, (cid:17) = (cid:0) a ± b, (cid:1) .Although calculating the Hasse symbols from the definition of α , β can be done very easily in numerical cases by using 2.3, proving generalresults may be quite labourios. Readers who want to skip this part maycheck the short version of this paper. Special cases1. B = − A and we want to calculate f ( − A, A, C ). Condition 1)from the definition of D means that (cid:16) A,Cp (cid:17) = (cid:16) − ,Cp (cid:17) = 1 ∀ p andcondition 2) means that ( A, C ) = 1. We have B = − A and we cantake x = 0, y = z = 1. Since (cid:0) − ,C ∞ (cid:1) = 1 we have C > α ∞ = 1. If p | C , p > α p = (cid:18) √ Ap (cid:19) = (cid:16) Ap (cid:17) . Also α = C ≡ (cid:18) √ A,C (cid:19) if A ≡ (cid:18) √ − A,C (cid:19) if A ≡ − A ≡ C ≡ − A ≡ , C ≡ . (The proof follows straight-forward from the definition of α . Since B = − A the cases with A ≡ B ≡ B ≡ A ≡ − B ≡ C ≡ A ≡ C ≡ A = C and we want to calculate f ( B, C, C ). Condition 1) inthe definition of D means (cid:16) B,Cp (cid:17) = (cid:16) − ,Cp (cid:17) = 1 ∀ p and condition2) means that ( B, C ) = 1. We have
ABC = BC = B so we arelooking for x, y, z relatively prime, such that x − By = − Bz . One ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) obvious choice is ( x, y, z ) = (0 , ,
1) but it is more convenable to choose( x, y, z ) = (0 , , ). (See Remark 2.2.) Since (cid:0) − ,C ∞ (cid:1) = 1 we have C > β ∞ = 1. For p | C , p > p ∤ B so we take β p = (cid:18) x + y √ B ) p (cid:19) = (cid:18) √ B ) p (cid:19) = (cid:18) √ Bp (cid:19) = (cid:16) Bp (cid:17) . If A = C is odd then (cid:0) C, − (cid:1) = 1 implies C ≡ C ≡ β = 1. If A = C ≡ β = (cid:18) y + z q ABCB , (cid:19) = (cid:18) + q BB , (cid:19) = (cid:0) , (cid:1) = 1. If A = C is even then by the definition of D we have B ≡ (cid:0) B,C (cid:1) = 1, implies B ≡ k C .) It follows that β = (cid:18) x + y √ B ) ,C (cid:19) = (cid:18) √ B ) ,C (cid:19) = (cid:18) √ B,C (cid:19) . Note that √ B is an odd 2-adic integer soby multiplying with ± √ B ≡ β = (cid:18) √ B,C (cid:19) is 1 or − √ B ≡ √ B ≡ − B (mod 8) so β = (cid:18) √ B,C (cid:19) is 1 if B ≡ − B ≡ C is even then β = (cid:16) B (cid:17) , where (cid:0) · (cid:1) : 1 + 8 Z → {± } is given by (cid:0) a (cid:1) = 1 if a ≡ (cid:0) a (cid:1) = − a ≡ C is odd then β = 1. In conclusion: f ( B, C, C ) = Y p | C (cid:18) Bp (cid:19) =: (cid:18) BC (cid:19) . ABC = −
1, i.e. A = − BC , and we want to calculate f ( B, − BC, C ).In this case condition 1) in the definition of D is equivalent to (cid:16) B,Cp (cid:17) =1 ∀ p , while condition 2) is vacuous. We have ABC = 1 so we need x, y, z such that x − By = z . An obvious choice is x = z = 1, y = 0.Since x > β ∞ = 1. If p | C , p > p ∤ A = − BC is equivalent to p | B . In this case β p = (cid:16) xp (cid:17) = 1. If p ∤ B then CONSTANTIN N. BELI β = (cid:18) x + y √ B ) p (cid:19) = (cid:16) p (cid:17) . Also β = (cid:0) ,C (cid:1) if B ≡ (cid:0) − ,C (cid:1) if B ≡ . (Since x = z = 1, y = 0 and ABC = − β = 1 if C ≡ A ≡ β = (cid:0) ,C (cid:1) if B ≡ β = (cid:18) q − B , (cid:19) = 1 if A ≡ C ≡ B ≡ − − B is a unit in Z and so is q − B ); β = − B ≡ C ≡ β = (cid:0) − , − (cid:1) = − A ≡ B ≡ C ≡ − B ≡ (cid:0) B,C (cid:1) = 1 we get that C is odd. If B ≡ , C ≡ β = 1; if B ≡ , C ≡ β = −
1; if B ≡ , C ≡ − B ≡ A ≡ β = −
1; if B ≡ , C ≡ B ≡ A ≡ β = 1. Hence if B ≡ β = (cid:0) − ,C (cid:1) . If B ≡ β = (cid:0) ,C (cid:1) .In all the other cases β = 1.)Note that Q p | C,p> β p = Q p | C,p ∤ B (cid:16) p (cid:17) = ( − n − , where n = C ( C, B ) .Since also β ∞ = 1 we get f ( B, − BC, C ) = ( − n − β . If p | C , p > p ∤ A, B then in the definition of α p and β p we can choose the formula from both the p ∤ A case and the p ∤ B case. However the outcome is the same.Indeed, for α p we have x − Ay = Bz so ( x + y √ A ) · x + z √ B ) =( x + y √ A + z √ B ) . It follows that (cid:18) ( x + y √ A ) · x + z √ B ) p (cid:19) = 1 so (cid:18) x + y √ Ap (cid:19) = (cid:18) x + z √ B ) p (cid:19) .For β p we have x − By = ABCz so x · x + y √ B ) = ( x + y √ B ) − ABCz ≡ ( x + y √ B ) (mod p ). It follows that (cid:18) x · x + y √ B ) p (cid:19) = 1 so (cid:16) xp (cid:17) = (cid:18) x + y √ B ) p (cid:19) .We have similar redundancies in the definition of α , β but again allthe cases of the definition which apply produce the same outcome.We now state our main result. ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) Theorem 2.5. (i) The functions f and f are well defined, i.e. theyare independent of the choice of x, y, z , and they are equal. We denote f = f = f .(ii) f is symmetric in all three variables.(iii) If ( B i , A i , C i ) ∈ D for ≤ i ≤ n and P i B i ⊙ A i ⊙ C i = 0 then Q i f ( B i , A i , C i ) = 1 . Theorem 2.5 justifies the following definition.
Definition 5.
We denote by W the subspace of S ′ ( V ) generated by B ⊙ A ⊙ C with ( B, A, C ) ∈ D . We define the group morphism γ : W → {± } by B ⊙ A ⊙ C f ( B, A, C ) . Part of Theorem 2.5(ii) follows from Theorem 2.5(iii).Namely, since B ⊙ A ⊙ C = A ⊙ C ⊙ B = C ⊙ B ⊙ A , we have f ( B, A, C ) = f ( A, C, B ) = f ( C, B, A ), i.e. f has a circular symmetry.However from Theorem 2.5(ii) we know that f is symmetric, not merelycircular symmetric. Hence one may assume that we have a more preciseresult, namely Q i f ( B i , A i , C i ) = 1 whenever P i B i • A i • C i = 0 in S ( V ). The reason that this doesn’t happen is the following. Assumethat ( B i , A i , C i ) ∈ D and P i B i • A i • C i = 0. Then ξ := P i B i ⊙ A i ⊙ C i can be written as ξ = P j ( B ′ j ⊙ A ′ j ⊙ C ′ j + A ′ j ⊙ B ′ j ⊙ C ′ j ) for some A ′ j , B ′ j , C ′ j ∈ V . We have Q i f ( B i , A i , C i ) = γ ( ξ ). If ( B ′ j , A ′ j , C ′ j ) ∈ D for all j then γ ( ξ ) = Q j f ( B ′ j , A ′ j , C ′ j ) f ( A ′ j , B ′ j , C ′ j ). But from Theorem2.5(ii) we have f ( B ′ j , A ′ j , C ′ j ) = f ( A ′ j , B ′ j , C ′ j ). Hence we get γ ( ξ ) = 1,as expected. The reason why this “proof” is wrong is that not always B ′ j , A ′ j , C ′ j can be chosen such that ( B ′ j , A ′ j , C ′ j ) ∈ D . Assume that p, q, r, s, a, b, c, d are odd primes with p ≡ q ≡ r ≡ s ≡ (cid:16) ap (cid:17) = (cid:16) bp (cid:17) = (cid:16) cp (cid:17) = (cid:16) dp (cid:17) = (cid:16) aq (cid:17) = (cid:16) dq (cid:17) = (cid:0) br (cid:1) = (cid:0) dr (cid:1) = (cid:0) cs (cid:1) = (cid:0) ds (cid:1) = − (cid:16) bq (cid:17) = (cid:16) cq (cid:17) = (cid:0) ar (cid:1) = (cid:0) cr (cid:1) = (cid:0) as (cid:1) = (cid:0) bs (cid:1) = 1. Then ( ad, pq, pq ) , ( bd, pr, pr ) , ( cd, ps, ps ) , ( a, rs, rs ) , ( b, qs, qs ) , ( c, qr, qr ) , ( abcd, pqrs, pqrs ) ∈ D . Indeed, in all cases the condition 2)from the definition of D is trivial and the conditon 3) follows from p ≡ q ≡ r ≡ s ≡ (cid:0) ad,pqt (cid:1) = 1 and (cid:0) pq, − t (cid:1) = 1 for any prime t . The second conditionfollows from the fact that pq is a sum of two squares. The first conditionat t = ∞ follows from pq > t = 2 from the fact that pq ≡ ad is odd. At t = p, q, a and d it means (cid:16) adp (cid:17) = (cid:16) adq (cid:17) = (cid:0) pqa (cid:1) = (cid:0) pqd (cid:1) = 1 and it follows from (cid:16) ap (cid:17) = (cid:0) pa (cid:1) = (cid:16) dp (cid:17) = (cid:0) pd (cid:1) = (cid:16) aq (cid:17) = (cid:0) qa (cid:1) = (cid:16) dq (cid:17) = (cid:0) qd (cid:1) = −
1. For t = ∞ , , p, q, a, d our statement is trivial.Similarly for the other triplets.We also have ad • pq • pq + bd • pr • pr + cd • ps • ps + a • rs • rs + b • qs • qs + c • qr • qr + abcd • pqrs • pqrs = 0. However the product f ( ad, pq, pq ) f ( bd, pr, pr ) f ( cd, ps, ps ) f ( a, rs, rs ) f ( b, qs, qs ) f ( c, qr, qr ) f ( abcd, pqrs, pqrs ) is −
1. Indeed, by the special case 2 our product isequal to (cid:16) adpq (cid:17) (cid:16) bdpr (cid:17) (cid:16) cdps (cid:17) (cid:0) ars (cid:1) (cid:16) bqs (cid:17) (cid:16) cqr (cid:17) (cid:16) abcdpqrs (cid:17) = (cid:16) a b c d p (cid:17) (cid:16) a b c d q (cid:17) (cid:16) a b c d r (cid:17) (cid:16) a b c d s (cid:17) = (cid:16) abcp (cid:17) (cid:16) abcdq (cid:17) (cid:0) abcdr (cid:1) (cid:0) abcds (cid:1) = ( − · · · − Lemma 2.8. If U = τ − ( W ) then for every p the image U p of U ⊆ S ( V ) in S ( V p ) is generated by elements of the form A • A • B with A, B ∈ V p . With the exception of the case p = 2 we have dim Z / Z V p ≤ U p = S ( V p ). If p > V p = h ∆ p , p i , where ∆ p is a nonsquareunit in Z p so S ( V p ) = h ∆ p • ∆ p • ∆ p , ∆ p • ∆ p • p, p • p • ∆ p , p • p • p i = U p .If p = ∞ then V ∞ = R × / ( R × ) = h− i so S ( V ∞ ) = h ( − • ( − • ( − i = U ∞ .If p = 2 then V = h− , , i so a basis for S ( V ) is { A • B • C | A, B, C ∈ {− , , }} . A basis of U is made of all the elements of thebasis for S ( V ) except ( − • • U ⊂ S ( V ). (The dimensionsof S ( V ) and U are 10 and 9, respectively.) Definition 6.
We define δ : U → {± } by δ ( ξ ) = Q p δ p ( ξ p ) , where theproduct is taken over all primes, including p = ∞ , and δ p : U p → {± } is given by A • A • B (cid:16) A,Bp (cid:17) . Theorem 2.10. γ ◦ τ = δ . Theorem 2.10 provides a generalization of Theorem 2.5(iii). Namely,if ( B i , A i , C i ) ∈ D and P i B i • A i • C i = 0 then we write ξ = P i B i ⊙ A i ⊙ C i ∈ W and we have ρ ( ξ ) = 0. It follows that ξ ∈ ker ρ = Im τ .(See Lemma 1.1.) Then ξ = τ ( η ) for some η ∈ τ − ( W ) = U and wehave Q i f ( B i , A i , C i ) = γ ( ξ ) = γ ( τ ( η )) = δ ( η ). Note that, in principle,calculating δ ( η ) is easier than γ ( ξ ) as it only involves usual Legendresymbols. In order to calculate γ ( ξ ) one has to compute f ( B i , A i , C i ),which involves finding nontrivial zeros for some ternary quadratic formand calculating Legendre symbols of the type (cid:16) a + b √ mp (cid:17) .Some more explicit formulas for δ p , so for δ are given bellow: ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) If p = ∞ then δ ∞ : U ∞ → {± } is given by δ ∞ ( A • B • C ) = ( − A, B, C <
01 otherwise . If p > δ p : U p → {± } is obtained as follows. For any A, B, C ∈ V if we write A = p r a , B = p s b and C = p t c , where r, s, t ∈ { , } and p ∤ abc then δ p ( A • B • C ) = (cid:16) − p (cid:17) rst (cid:16) ap (cid:17) st (cid:16) bp (cid:17) rt (cid:16) cp (cid:17) rs ·· (cid:16)(cid:16) bp (cid:17) ∗ (cid:16) cp (cid:17)(cid:17) r (cid:16)(cid:16) cp (cid:17) ∗ (cid:16) ap (cid:17)(cid:17) s (cid:16)(cid:16) ap (cid:17) ∗ (cid:16) bp (cid:17)(cid:17) t , where ∗ : {± } × {± } → {± } is given by ε ∗ η = ( − ε = η = −
11 otherwise . If p = 2 and η ∈ U then η = P i A i • B i • C i with A i , B i , C i ∈ V . We write A i , B i , C i in terms of the basis − , , V . We have A i = ( − r i, r i, r i, , B i = ( − s i, s i, s i, and C i = ( − t i, t i, t i, with r i,j , s i,j , t i,j ∈ { , } . Then δ ( η ) = Y i Y j =1 ( − r i,j s i,j t i,j . Note that we may extend δ : U → {± } to the whole S ( V ) bysetting arbitrarily δ (( − • •
6) = 1. This way δ : U → {± } extendsto the whole S ( V ). The formula above for δ will hold for for all η ∈ S ( V ). If A = ( − r r r , B = ( − s s s and C = ( − t t t then δ ( A • B • C ) = Q j =1 ( − r j s j t j .In most applications we won’t need Theorem 2.10. In fact in allproofs from the next section we don’t even need the full strength ofTheorem 2.5(iii). It is enough to use a weaker version of Theorem2.5(iii) where the condition P i B i ⊙ A i ⊙ C i = 0 is replaced by P i B i ⊗ A i ⊗ C i = 0. (Note that P i B i • A i • C i = 0 = ⇒ P i B i ⊙ A i ⊙ C i =0 = ⇒ P i B i ⊗ A i ⊗ C i = 0.) The only symmetry properties we needare the ones following from Theorem 2.5(ii).3. Applications
We now recover some results regarding quartic reciprocity that canbe found in [L, §
5] or on wikipedia athttp://en.wikipedia.org/wiki/Quartic reciprocity
Formulas for (cid:16) mp (cid:17) . By the special case 2 if p ≡ m = ± q · · · q k with (cid:16) q i p (cid:17) = 1 then ( m, p, p ) ∈ D and f ( m, p, p ) = (cid:18) mp (cid:19) . By Theorem 2.5(i) and (ii) we have (cid:16) mp (cid:17) = f ( m, p, p ) = f ( p, p, m ).We have p = a + b with 2 | b . Then p − b p = a p so in the definitionof f ( p, p, m ) we may take ( x, y, z ) = ( p, b, a ). We have f ( p, p, m ) = α ∞ Q q | m α q . Now x = p > α ∞ = 1 and if q | m , q > q ∤ p = A so α q = (cid:18) x + y √ Aq (cid:19) = (cid:16) p + b √ pq (cid:17) .Assume first that m = 2. Then (cid:16) p (cid:17) = 1 so p ≡ | b .We have (cid:16) p (cid:17) = f ( p, p,
2) = α ∞ α = α . Since A = p ≡ α = (cid:18) x + y √ A (cid:19) = (cid:16) p + b √ p, (cid:17) . Now √ p is an odd integer in Z and so 4 | b so b √ p ≡ b (mod 8) so p + b √ p ≡ p + b ≡ b (mod 8). If b ≡ p + b √ p is ≡ Z × so (cid:16) p + b √ p, (cid:17) = 1 so (cid:16) p (cid:17) = 1. If b ≡ p + b √ p ≡ (cid:16) p + b √ p, (cid:17) = (cid:0) , (cid:1) = − (cid:16) p (cid:17) = −
1. In conclusion (cid:16) p (cid:17) = 1 iff 8 | b , which is one of Euler’s conjectures, proved by Gaussin 1828.Take now m = q ∗ := ( − q − q , where q > (cid:16) qp (cid:17) = 1.Then (cid:16) q ∗ p (cid:17) = f ( p, p, q ∗ ) = α ∞ α q α = α q α . We have α q = (cid:16) p + a √ pq (cid:17) and since A = p and C = q ∗ and p ≡ q ∗ ≡ α . If q ∗ ≡ α = 1. If p ≡ α = (cid:18) x + y √ A,C (cid:19) = (cid:16) p + b √ p,q ∗ (cid:17) . But p is odd and b is even so p + b √ p is an odd 2-adic integer. Since also q ∗ ≡ α = 1. If p ≡ q ∗ ≡ α = (cid:0) x, (cid:1) = (cid:0) p, (cid:1) = 1. So α = 1and (cid:16) q ∗ p (cid:17) = α q = (cid:16) p + b √ pq (cid:17) . This result belongs to Lehmer. (See [L, § ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) Burde.
Assume that p ≡ q ≡ (cid:16) pq (cid:17) = 1 and write p = a + b , q = c + d with b and d even. Then pq = e + f , where e = ac − bd , f = ad + bc and e is odd and f even.Now p ⊙ p ⊙ q + p ⊙ q ⊙ q = p ⊙ pq ⊙ q so by Theorem 2.5(iii) (cid:18) pq (cid:19) (cid:18) qp (cid:19) = f ( p, p, q ) f ( q, q, p ) = f ( p, pq, q ) . Now f ( p, pq, q ) = f ( pq, p, q ). In the definition of f we have A = p , B = pq and C = q so ABC = 1. Since e − pq = − f we may take( x, y, z ) = ( e, , f ). We have C = q > β ∞ = 1 and the only primesdividing 2 C = 2 q are 2 , q so f ( pq, p, q ) = β q β . Now q ∤ p = A so β q = (cid:16) xq (cid:17) = (cid:16) eq (cid:17) . We have A = pq , C = q so A ≡ C ≡ C ≡ β = 1. If C ≡ β = (cid:0) x,C (cid:1) = 1. ( x = e is odd and C ≡ A ≡ C ≡ β = (cid:18) y + z q ABCB , (cid:19) = (cid:0) y ± z, (cid:1) . (See 2.3.) Here the ± sign is taken such that ord ( y ± z ) is minimum. But for both choicesof the ± sign y ± z = 1 ± f is an odd integer so again β = 1. Hence (cid:16) pq (cid:17) (cid:16) qp (cid:17) = β q β = (cid:16) eq (cid:17) · (cid:16) ac − bdq (cid:17) , which is Burde’s law. (See [L, § (cid:16) pq (cid:17) (cid:16) qp (cid:17) from [L, Theorem5.7].We write (cid:16) pq (cid:17) (cid:16) qp (cid:17) = f ( p, pq, q ) = f ( p, pq, q ). We have A = pq , B = p , and C = q so ABC = 1. Since a − p = − b we can take( x, y, z ) = ( a, , b ). Again f ( p, pq, q ) = β q β . Since q ∤ B = p wehave β q = (cid:18) x + y √ B ) q (cid:19) = (cid:16) ( a + √ p ) q (cid:17) and, by the same proof as inthe previous case, β = 1. Hence (cid:16) pq (cid:17) (cid:16) qp (cid:17) = (cid:16) ( a + √ p ) q (cid:17) . To obtain (cid:16) pq (cid:17) (cid:16) qp (cid:17) = f ( p, pq, q ) = (cid:16) b + √ pq (cid:17) we write b − p = − a and we have( x, y, z ) = ( b, , a ). Same as before, β q = (cid:16) ( b + √ p ) q (cid:17) , but this time β = (cid:16) q (cid:17) , i.e. β = 1 if C = q ≡ β = − C ≡ Z q a + √ p )( b + √ p ) = ( a + b + √ p ) so (cid:16) ( a + √ p ) q (cid:17) = (cid:16) b + √ pq (cid:17) . Similarly (cid:16) pq (cid:17) (cid:16) qp (cid:17) = (cid:16) ( c + √ q ) p (cid:17) = (cid:16) d + √ pq (cid:17) . In order to obtain (cid:16) pq (cid:17) (cid:16) qp (cid:17) = f ( p, pq, q ) = (cid:16) a + b √− q (cid:17) = (cid:16) c + d √− p (cid:17) we note that p ⊙ pq ⊙ q = p ⊙ ( − pq ) ⊙ q + p ⊙ ( − ⊙ q so (cid:18) pq (cid:19) (cid:18) qp (cid:19) = f ( p, pq, q ) = f ( p, − pq, q ) f ( p, − , q ) . Now f ( p, − pq, q ) = f ( − pq, p, q ) and since A = p , B = − pq and C = q we have ABC = − n = C ( C, B ) = 1 and, since B = − pq β = 1. It follows that f ( − pq, p, q ) = ( − n − β = 1. So f ( p, pq, q ) = f ( p, − , q ) = f ( p, − , q ) . We have A = − B = p and C = q . Since a + b = p we may take( x, y, z ) = ( a, b, p ). We have C = q > α ∞ = 1 so f ( p, − , q ) = α q α . Since q ∤ − A we have α q = (cid:16) a + b √− q (cid:17) . We have B = p an C = q ao B ≡ C ≡ C ≡ α = 1. If B ≡ α = (cid:18) x + z √ B ) ,C (cid:19) = (cid:16) a + √ p ) ,C (cid:17) . But C ≡ α = (cid:16) a ± ,C (cid:17) ,where the ± sign is chosen such that ord ( a ±
1) is minimum. But a is odd so we must have a ± ≡ ( a ±
1) = 1).Then a ± is odd and since C ≡ α = (cid:16) a ± ,C (cid:17) = (cid:16) a ± ,C (cid:17) = 1. If B ≡ C ≡ α = − (cid:0) y, (cid:1) = − (cid:0) b, (cid:1) .But a + b = p = C ≡ b is even so b ≡ (cid:0) b, (cid:1) = − α = 1. Since α = 1 in all cases f ( p, − , q ) = α q = (cid:16) a + b √− q (cid:17) . Note that since q ∤ p = B we may alsotake α q = (cid:16) a + √ p ) q (cid:17) and thus we recover the equality (cid:16) pq (cid:17) (cid:16) qp (cid:17) = (cid:16) ( a + √ p ) q (cid:17) from above. (Alernatively we may use the relation 2( a + b √− a + √ p ) = ( a + b √− √ p ) from Remark 2.4, which impliesthat (cid:16) a + b √− q (cid:17) = (cid:16) a + √ p ) q (cid:17) so the two statements are equivalent.)To prove [L, Ex. 5.5, p. 176] we note that we also may write (cid:18) pq (cid:19) (cid:18) qp (cid:19) = f ( p, − , q ) = f ( p, q, − . We have A = q , B = p and C = − f ( p, q, −
1) = α ∞ α . If e = pf + qg then e − qg = pf so we may take ( x, y, z ) = ( e, g, f ). ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) We want to prove that f ( p, q, −
1) = ( − fg (cid:0) − e (cid:1) . By permuting, ifnecessary p and q we may assume that if p ≡ q ≡ f is even and g is odd and if p q (mod 8) then p ≡ , q ≡ A = q , B = p so A ≡ B ≡ C = −
1. Since
C < α ∞ = sgn ( e ). There are two cases: q ≡ q ≡ p ≡ g is even. Inthe first case A ≡ α = (cid:18) x + y √ A,C (cid:19) = (cid:16) e + g √ q, − (cid:17) . If g iseven then α = (cid:0) e, − (cid:1) (cid:16) ge √ q, − (cid:17) . Now e √ q is an odd 2-adic integerand g is even. If 2 k g then 1 + ge √ q ≡ (cid:16) ge √ q, − (cid:17) = −
1. If 4 | g then 1 + ge √ q ≡ (cid:16) ge √ q, − (cid:17) = 1. Hence (cid:16) ge √ q, − (cid:17) = ( − g = ( − fg and so α = ( − fg (cid:0) e, − (cid:1) . If g isodd so f is even note that pf = e − qf ≡ − | f . Now e and g √ q are odd 2-adic integers so by replacing,if necessary √ q by −√ q we have e ≡ g √ q (mod 4). It follows that2 k e + g √ q . Since also ( e − g √ q )( e + g √ q ) = e − qg = pf ...16 we get e − g √ q ...8 so e + g √ q ≡ e (mod 8). It follows that e + g √ q e ≡ (cid:18) e + g √ q e , − (cid:19) = 1, which implies that α = (cid:16) e + g √ q, − (cid:17) = (cid:0) e, − (cid:1) = (cid:0) e, − (cid:1) = ( − fg (cid:0) e, − (cid:1) . (Recall, 4 | f .) In the second case A ≡ B ≡ C = − α = − (cid:18) x + z √ B, − (cid:19) = − (cid:16) e + f √ p, − (cid:17) = − (cid:0) e, − (cid:1) (cid:16) f e √ q, − (cid:17) . Now pf = e − qg ≡ − ≡ k f . Since also e √ p is an odd 2-adic integer we have 1 + f e √ p ≡ (cid:16) f e √ q, − (cid:17) = −
1. Since also − (cid:0) e, − (cid:1) = (cid:0) e, − (cid:1) we get α = − (cid:0) e, − (cid:1) = ( − fg (cid:0) e, − (cid:1) . In conclusion (cid:16) pq (cid:17) (cid:16) qp (cid:17) = α ∞ α = sgn ( e ) · ( − fg (cid:0) e, − (cid:1) = ( − fg (cid:0) − e (cid:1) .Next we prove that if p = r + qs then (cid:16) pq (cid:17) (cid:16) qp (cid:17) = (cid:16) q (cid:17) s (see [L,Ex. 5.6, p. 176]). This time we use (cid:18) pq (cid:19) (cid:18) qp (cid:19) = f ( p, − , q ) = f ( p, q, − . We have A = q , B = p and C = − ABC = − pq . Since p − pr = pqs we may take ( x, y, z ) = ( p, r, s ). We have f ( p, q, −
1) = β ∞ β . Since x = p > β ∞ = 1 so we have to prove that β = (cid:16) q (cid:17) s .Since C = − β . If q = A ≡ β = (cid:0) x,C (cid:1) = (cid:0) p, − (cid:1) = 1 = (cid:16) q (cid:17) s . If q ≡ p = B ≡ (cid:16) q (cid:17) s = ( − s and β = (cid:18) x + y √ B ) ,C (cid:19) = (cid:16) p + r √ p ) , − (cid:17) = (cid:16) p + r √ p, − (cid:17) . If s is odd and r is even then r = qs − p ≡ − ≡ k r . We have β = (cid:0) p, − (cid:1) (cid:18) r q p , − (cid:19) . But 2 k r and q p is an odd integer so 1 + r q p ≡ (cid:18) r q p , − (cid:19) = −
1. Itfollows that β = (cid:0) p, − (cid:1) · ( −
1) = − (cid:16) q (cid:17) s . (Recall that s is oddand q ≡ s is even and r odd then qs = p − r ≡ − | s . Since p , r √ p are 2-adic odd integers, bymultiplying √ p with ± p + r √ p ≡ k p + r √ p . Together with p − pr = pqs ...16 we get that 8 | p − r √ p so p + r √ p ≡ p (mod 8) so p ( p + r √ p ) ≡ (cid:18) p ( p + r √ p ) , − (cid:19) = 1, which implies that β = (cid:0) p, − (cid:1) = 1 = (cid:0) s (cid:1) s ( s iseven). Now assume that p ≡ q ≡ A ≡ B ≡ β = (cid:18) − x + y √ B ) ,C (cid:19) = (cid:16) − p + r √ p ) , − (cid:17) = − (cid:16) p + r √ p, − (cid:17) . If s isodd and r even then r = qs − p ≡ − | r so5 p + r √ p ≡ p ≡ β = − (cid:16) p + r √ p, − (cid:17) = − (cid:16) q (cid:17) s .(Recall that s is odd and q ≡ s is even and r is oddthen qs = p − r ≡ − k s . By multiplying √ p with ± k p + r √ p . We also have 25 p − pr =20 p + 5 pqs ≡ ≡ p ≡ p ≡ ≡ pq ( s ) ≡ pqs ≡ p − r √ p ≡ p + r √ p ≡ p + 4 ≡
50 + 4 ≡ (cid:16) p + r √ p, − (cid:17) = − β = 1 = (cid:16) q (cid:17) s ( s is even). Scholz.
Assume that p ≡ q ≡ (cid:16) pq (cid:17) = 1and assume that ε is a unit of Q ( √ p ) of norm −
1. Then Scholz’s lawstates that (cid:16) εq (cid:17) = (cid:16) pq (cid:17) (cid:16) qp (cid:17) . (See [L, p. 167].) We could proveScholz’s law directly if ε ∈ Z [ (cid:16) √ p (cid:17) ] \ Z [ √ p ]. However in order to ECIPROCITY LAWS FOR THE LEGENDRE SYMBOLS (cid:16) a + b √ mp (cid:17) overcome the 2-adic complications we replace ε by ε and we mayassume that ε = t + u √ p ∈ Z [ √ p ]. We have (cid:18) pq (cid:19) (cid:18) qp (cid:19) = f ( p, − , q ) = f ( − , p, q )so A = p , B = − C = q . Since t − pu = − x, y, z ) = ( t, u, f ( − , p, q ) = α ∞ α q α . But C > α ∞ = 1 and q ∤ p = A so α p = (cid:18) x + y √ Aq (cid:19) = (cid:16) t + u √ pp (cid:17) = (cid:16) εp (cid:17) . Sowe have to prove that α = 1. If C = q ≡ α = 1by definition. If C = q ≡ A = p ≡ α = (cid:18) x + y √ A,C (cid:19) = (cid:16) t + u √ p, (cid:17) . But t and u have opposite parities so t + u √ p is an odd 2-adic integer, which implies that α = 1. If p ≡ q ≡ A ≡ C ≡ α = (cid:0) z, (cid:1) = (cid:0) , (cid:1) = 1.A similar result holds if q = 2. Namely, if p ≡ (cid:16) √ p (cid:17) = (cid:16) p (cid:17) (cid:0) p (cid:1) . (See [L, p. 169].) We repeat the reasoning usedto prove that (cid:16) pq (cid:17) (cid:16) qp (cid:17) = f ( p, − , q ). By the special case 2 we have f (2 , p, p ) = (cid:16) p (cid:17) and f ( p, ,
2) = (cid:0) p (cid:1) so (cid:16) p (cid:17) (cid:0) p (cid:1) = f (2 , p, p ). Wehave f (2 , p, p ) = f (2 , − p, p ) f (2 , − , p ). But f (2 , − p, p ) = f (2 , − p, p )can be calculated using the special case 3 with B = 2, C = p . Wehave n = (cid:16) CC, B (cid:17) = p and B = 2 β = 1. Thus f (2 , − p, p ) = ( − p − · (cid:16) p (cid:17) (cid:0) p (cid:1) = f (2 , p, p ) = f (2 , − , p ) = f ( − , , p ). We have A = 2, B = − C = p so f ( − , , p ) = α ∞ α p α .We have 1 − − x, y, z ) = (1 , , C = p > α ∞ = 1 and since p ∤ A we may take α p = (cid:18) x + y √ Ap (cid:19) = (cid:16) √ p (cid:17) so we have to prove that α = 1. But thisfollows from C = p ≡ References [L] Lemmermeyer, Franz,