5. More Examples and Complements 5.1. the Quadratic Reciprocity Law for Number ﬁelds

64 M. FLACH

5. More examples and complements 5.1. The quadratic reciprocity law for number fields. Recall the quadratic reciprocity law first proven by Gauss. For an integer a and prime number p 2a define the Legendre symbol a 1ifa is a square modulo p = p −1 otherwise.

More generally, for an integer b = p1 ···pr relatively prime to 2a one defines the Jacobi symbol a a a = ··· . b p1 pr a Clearly, b only depends on a modulo b. Theorem 5.1. (Gauss) For positive, odd, relatively prime integers a, b one has a b a−1 b−1 =(−1) 2 2 . b a Moreover, for odd b there are the supplementary statements −1 b−1 2 b2−1 =(−1) 2 ; =(−1) 8 . b b We shall reprove the quadratic reciprocity law from the Artin reciprocity law and at the same time generalize it to number fields. Fix a number field K.Forα ∈OK and a prime ideal p 2α one can define an obvious analogue of the Legendre symbol α 1ifα is a square modulo p = p −1 otherwise and an obvious analogue of the Jacobi symbol for β ∈OK relatively prime to 2α α α α = ··· β p1 pr where (β)=p1 ···pr is the prime factorization of β. One clearly has √ α 1ifp splits in K( α)/K = √ p −1ifp is inert in K( α)/K √ with the understanding that any prime splits in K( α)/K if α is a square. This condition has a straightforward reformulation in terms of the Frobenius automor- phism √ Frobp ∈ Gal(K( α)/K) ⊆{±1} which in turn is the image of a local uniformizer πp under the local reciprocity map √ √ × ρp Kp −→ Gal(K( α)P/Kp) ⊆ Gal(K( α)/K) ⊆{±1}. p α∞ So for 2 we have α = ρp(πp). p COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 65

In order to prove a reciprocity√ law one needs to look at ρp for all places p of K. Note that the extension Kp( α)/Kp can be ramified at places p | 2α∞ (or can be trivial if α is locally a square). If p | α but p 2∞ then √ Kp( α)/Kp unramified ⇔ vp(α) ≡ 0mod2 and if vp(α) ≡ 1 mod 2 this extension is tamely ramified. By local class field theory the map ρp then induces an isomorphism √ ρ O× / O× 2 ∼ O /p ×/ O /p × 2 ∼ K α /K ∼ {± } p : Kp ( Kp ) = ( K ) (( K ) ) = Gal( ( )P p) = 1 α ∈O× and so we have for Kp vp(α) α ρp(α )= p for the entirely different reason that the Legendre symbol is the only nontrivial × quadratic character of (OK /p) .Ifp is real then ρp is also easy to analyze. We have √ sgn σp(α)−1 Kp( α) = Kp ⇔ σp(α) < 0 ⇔ (−1) 2 = −1 sgn σp(α )−1 and in that case we have ρp(α )=(−1) 2 . So altogether sgn σp(α )−1 sgn σp(α)−1 ρp(α )=(−1) 2 2 . The hardest primes p to analyze are those dividing 2 since there are several ramified ∼ quadratic extensions of Kp. Let’s do this when Kp = Q2.Wehave × × 2 ∼ Z/2Z Z/2Z Z/2Z Q2 /(Q2 ) = [2] × [−1] × [5] and this group both parametrizes the quadratic extensions by Kummer theory and is the source of the Artin map for all quadratic extension. So we can define a bilinear symbol on this Z/2Z-vector space, the Hilbert symbol α ,α := ρp(α ) ∈{±1}. p √ α ,α Then p = 1 if and only if α is a norm of the quadratic extension Q2( α)/Q2 if and only if x2 − αy2 = α 2 has a solution (x, y) ∈ Q2 which (for nonsquare α) is true if and only if x2 − αy2 − αz2 =0 3 has a nonzero solution (x, y, z) ∈ Z2. But this condition is symmetric in α, α which means that our bilinear form is symmetric. It is given by the following table √ 2 −1 5 Q2(√ 2) 1 1 −1 Q2( √−1) 1 −1 1 Q2( 5) −1 1 1 which is justified by the following computations √ 2 −1 5 2 2 2 2 Q2(√ 2) 2 − 2 · 1 =2 1 − 2 · 1 = −1 −1 2 2 2 2 Q2( √−1) 1 a + b ≡−1mod4 2 +1 =5 2 2 Q2( 5) −1 1 5 − 5 · 2 =5. 66 M. FLACH √ The remaining −1 simply results from the fact that Q2( 5) is the unramified quadratic extension of Q2, so 2 is sent to the Frobenius, i.e. to −1. Of course it also follows from a computation modulo 8. One verifies directly that 2 √ α −1 ρ (α )=(−1) 8 Q2( 2)/Q2 × for α ∈ Z2 and α−1 α −1 ρ √ α − 2 2 Q2( α)/Q2 ( )=( 1) × for α, α ∈ Z2 by checking that both sides are multiplicative and agree on the generators listed in the above table. These formulas are just a neat way to con- dense the table, for otherwise one would have to make many case distinctions. For × × 2 an extension Kp/Q2 of degree d the space Kp /(Kp ) has dimension d +2 over Z/2Z and the Hilbert symbol also gives a symmetric bilinear form which is fairly straightforward to compute in any given example. The quadratic reciprocity law for K would depend on this table as is clear from the following theorem.

Theorem 5.2. Let K be a number ﬁeld. For α, β ∈OK relatively prime to each other and relatively prime to 2 we have α β sgn σp(α)−1 sgn σp(β)−1 = ρp(β) (−1) 2 2 β α p|2 p real √ where ρp is the reciprocity map for the quadratic extension Kp( α)/Kp.If2 splits completely in K/Q this law simpliﬁes to α β σp(α)−1 σp(β)−1 sgn σp(α)−1 sgn σp(β)−1 (38) = (−1) 2 2 (−1) 2 2 β α p|2 p real × and we have the following two supplementary laws. If α ∈OK and β is prime to 2 then α σp(α)−1 σp(β)−1 sgn σp(α)−1 sgn σp(β)−1 (39) = (−1) 2 2 (−1) 2 2 . β p|2 p real If α is 2-primary and β is prime to 2 then 2 −vp(α) α v (α) σp(β) −1 2 σp(α)−1 σp(β)−1 (40) = (−1) p 8 (−1) 2 2 β p|2

sgn σp(α)−1 sgn σp(β)−1 × (−1) 2 2 . p real

Proof. We have α vp(β) = ρp(πp) = ρp(β) β p2α∞ p2α∞ and β β vp(α) = = ρp(β) α p p|α p|α and sgn σp(α)−1 sgn σp(β)−1 (−1) 2 2 = ρp(β). p real p real COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 67

Theorem 5.2 now follows from Theorem 3.1 a)

(41) ρ(β)= ρp(β)=1. p β The supplementary law (39) follows from (38) since α =1if(α)=OK is the unit ideal. The supplementary law (40) is again a direct consequence of (41) but there v (α) are no odd primes dividing α to consider. At each p | 2 we write σp(α)=2 p up × with up ∈ Z2 and note that ρp(β) is multiplicative in σp(α). Remark 5.1.1. Note that (38), (39) and (40) include a generalization of the clas- sical reciprocity and supplementary laws to negative b. For example if K = Q and β = −b<0 we get −1 −b−1 −1·−1 −b+1 b−1 −1 =(−1) 2 (−1) =(−1) 2 =(−1) 2 = −b b a which is in accordance with the fact that b only depends on the ideal generated by b. √ Example 5.1.1. K Q − ω √ In = ( 7) decide if 5+2 is a square modulo 73 where 1+ −7 ω = 2 . Note we have 2 2 NK/Q(a + bω)=a + ab +2b and in particular N(5 + 2ω)=43. The prime 73 ≡ 3mod7is inert in K/Q since 3 is a nonsquare modulo 7.SinceK has no real places and 73 ≡ 1mod4we have 5+2ω 73 30 2 3 5 = = = . 73 5+2ω 5+2ω 5+2ω 5+2ω 5+2ω Since 5 ≡ 1mod4we have with p =(ω) 2 σ ω − 5 5+2ω ω ω¯ σp(¯ω)−1 5−1 52−1 p¯ ( ) 1 5−1 = = =(−1) 2 2 (−1) 8 (−1) 2 2 = −1. 5+2ω 5 5 Since 3 ≡−1mod4and (5 + 2ω − 1)/2=2+ω is even at p =(ω) and odd at ¯p we get 3 2+2ω = − 5+2ω 3

σ ω − 32−1 (1+σp(ω))−1 3−1 3· 32−1 ( p¯ ( )) 3 3−1 = − (−1) 8 (−1) 2 2 · (−1) 8 (−1) 8 2

= − [1 · 1] = −1

(1+σp(ω))−1 since 2 =1/σp(¯ω) is odd and so is

(σp¯(ω)+1)/4 − 1 (σp¯(ω)) − 3 = 2 8 since 32 − 3+2=8≡ 0mod8but not mod 16. Finally (5 + 2ω)2 − 1 = (24 + 20ω +4ω2)/8=(24+20ω +4(ω − 2))/8=2+3ω 8 is even at p =(ω) and odd at p¯ =(¯ω) and hence 2 2 2 (5+2σp(ω)) −1 (5+2σp¯ (ω)) −1 =(−1) 8 (−1) 8 =1· (−1) = −1. 5+2ω 68 M. FLACH

So we ﬁnd 5+2ω =(−1) · (−1) · (−1) = −1. 73 Actually one could have saved quite a bit of work in this example by noting that a a 5+2ω = 43 for a ∈ Z prime to 43. We just use this remark to double check our results. 5 3 = = −1 43 5 3 1 = − = −1 43 3 2 432−1 =(−1) 8 = −1 43 √ −1 Remark 5.1.2. Note that also for K = Q( −7) we can verify that β only depends on the ideal generated by β but this time the compensating sign changes occur at the two primes dividing 2, not at 2 and ∞ as was the case for K = Q. One has −1 −σp(β)−1 −σp¯ (β)−1 σp(β)−1 σp¯ (β)−1 −1 =(−1) 2 (−1) 2 =(−1) 2 (−1) 2 = . −β β

We continue with some remarks on more general reciprocity laws. If a local field Kp contains a primitive n-th root of unity, the combination of Kummer theory and local class field theory leads to the definition of the local Hilbert symbol. For α, α ∈ K p let √ × n ρp : K → Gal(Kp( α)/Kp) → μn p √ n be the reciprocity map for the extension Kp( α)/Kp followed by the canonical homomorphism √ (42) σ → ( n α)σ−1 of Kummer theory, and define the Hilbert symbol α ,α := ρp(α ) ∈ μn. p This symbol is a bilinear map × × n × × n Kp /(Kp ) × Kp /(Kp ) → μn and one can show that it actually coincides with the cup product

1 1 ∪ 2 ⊗2 ∼ 2 inv ⊗μn H (Kp,μn) × H (Kp,μn) −→ H (Kp,μn ) = H (Kp,μn) ⊗ μn −−−−−→ μn.

Apart from bilinearity it has the following properties √ α ,α n a) = 1 if and only if α is a norm from Kp( α)/Kp. p −1 b) α ,α = α,α . p p c) α,1−α = 1 and α,−α =1. p p α,α × × n d) p = 1 for all α ∈ Kp if and only if α ∈ (Kp ) . COURSE NOTES, GLOBAL CLASS FIELD THEORY CALTECH, SPRING 2015/16 69

Theorem 5.3. Let K be a number ﬁeld containing a primitive n-th root of unity. × Then for α, β ∈ K α, β =1. p p

Proof. This is just a rewriting of the Artin reciprocity law

ρ(β)= ρp(β)=1 p √ × n where ρ : AK → Gal(K( α)/K) is the global reciprocity map. ForanumberﬁeldK containing a primitive n-th root of unity one can deﬁne the n-th power residue symbol. Given α ∈OK and p nα let α ∈ μ p n n Np−1 α n be the unique root of unity congruent to α modulo p.Then p =1if n n and only if there is β ∈OK with β ≡ α mod p. From (42) and the fact that ρp(πp)=Frobp where πp is a uniformizer of Kp on deduces that α πp,α = . p n p This is the starting point for deducing a reciprocity law for the n-th power residue symbol from Theorem 5.3. For n = 3 some details are in the exercises.