POWER RECIPROCITY YIHANG ZHU 1. General reciprocity for power residue symbols 1.1. The product formula. Let K be a number eld. Suppose K ⊃ µm. We will consider the m-th norm-residue symbols for the localizations Kv of K. We will omit m from the notation when convenient. Let be a place of . Recall that for ×, we dene v K a; b 2 Kv ρ (a)b1=m (a; b) = (a; b) := v 2 µ ; Kv v b1=m m where ρ : K× ! Gab is the local Artin map. Recall, when v = p is a non- v v Kv archimedean place coprime to m, we have the following formula (tame symbol) bα (q−1)=m (a; b) ≡ (−1)αβ mod p; v aβ where q = N p = jOK =pj ; α = v(a); β = v(b).(v is normalized so that its image is Z.) In particular, if × then ; if is a uniformizer in and a; b 2 Ov ; (a; b)v = 1 a = π Kv ×, b 2 Ov (q−1)=m (π; b)v ≡ b mod p: Suppose v is an archimedean place. If v is complex, then (; )v = 1. If v is real, then by assumption , so . In this case we easily see that R ⊃ µm m = 2 ( −1; a < 0; b < 0 × (a; b)v = ; 8a; b 2 R : 1; otherwise The following statement is an incarnation of Artin reciprocity. Proposition 1.2 (Product formula). Let a; b 2 K×. Then for almost all places v, we have (a; b)v = 1, and we have Y (a; b)v = 1: v Proof. Q 1=m Q 1=m But by the local global compatibility of [ v(a; b)v]b = [ v ρv(a)]b : the Artin map and the fact that the global Artin map factors throught × × , K nAK we conclude that Q ab. v ρv(a) = 1 2 GK 1.3. Calculating the norm residue symbols in the non-tame case. In certain cases, the product formula provides a way of computing the norm-residue symbol when the residue characteristic of divides . (; )Kv ;m Kv m 1 2 YIHANG ZHU Example 1.4. We would like to compute . We have × ∼ K = Q; m = 2 (; )Q2 Q2 =2 = Z=2Z × ∼ Z=2Z ×. To know , it suces to compute the pairing 2 ×Z2 =2 = 2 ×(Z=8Z) (; )Q2;2 between 2; 3; 5. We have Y (2; 2)Q2 = (2; 2)Qp = 1; p6=2 (2; 3)Q2 = (2; 3)Q3 = −1; (2; 5)Q2 = (2; 5)Q5 = −1; (3; 3)Q2 = (3; 3)Q3 = (3; −1)Q3 = −1; (3; 5)Q2 = (3; 5)Q3 (3; 5)Q5 = −1 × −1 = 1; (5; 5)Q2 = (5; 5)Q5 = (5; −1)Q5 = 1: Example 1.5. We would like to compute where is K = Q(ζ3); m = 3 (; )v v the unique place over 3. This is needed for the cubic reciprocity. We partially follow the hints of an exercise in Joe Rabino's notes. Let ζ = ζ3; λ = 1 − ζ, i 2 ηi = 1 − λ . We have 3OK = λ OK , and λ is a uniformizer of Kv. As usual let (i) i (2) ∼ 2 U = 1 + λ Ov. Using the exponential isomorphism, we see that U −! λ Ov, identifying (4) (2) with 4 2 2 . Hence (4) × 3, U ⊂ U λ Ov = 3λ Ov ⊂ λ Ov U ⊂ (Kv ) and × is a -dimensional vector space, with a basis given by . Kv =3 4 F3 λ, η1; η2; η3 We need to compute the pairings (; )v between them. Firstly, −1 is a cubic, so ×. We have (a; a)v = (a; −a)v = 1; 8a 2 Kv j ηi+j = ηj + λ ηi; j 1 = ηj/ηi+j + λ ηi/ηi+j: k k Using (a; a) = 1 and (ηk; λ ) = 1 since ηk + λ = 1, we have j j (1) 1 = (ηj/ηi+j; λ ηi/ηi+j) = (ηj; ηi)(ηi+j; ηj)(ηi; ηi+j)(λ, ηi+j) Therefore if i + j ≥ 4, we have (ηi; ηj) = 1. So the only left case for (ηi; ηj) is . This we can use the product formula to compute. × is a i = 1; j = 2 η1 = ζ 2 OK unit. 2 2 2 η2 = 1 − λ = 1 − (1 − ζ) = 2ζ − ζ = 1 + 3ζ; hence 2 2 NK=Q η2 = (1 + 3ζ)(1 + 3ζ ) = 1 + 3ζ + 3ζ + 9 = 1 − 3 + 9 = 7: Namely, w = (η2) is a split prime over 7. Hence −1 (7−1)=3 (η1; η2)v = (η1; η2)w = (η2; η1)w ≡ η1 2 F7: But (7−1)=3 2. We conclude that 2. Now it remains to compute η1 = ζ (η1; η2)v = ζ i i i (λ, ηi): For i = 1; 2, we have (λ, ηi) = (λ ; ηi) = 1 since λ + ηi = 1. So (λ, ηi) = 1; i = 1; 2: To compute (λ, η3), we set i = 2; j = 1 in formula (1), then 1 = (η1; η2)(η3; η1)(η2; η3)(λ, η3) = (η1; η2)(λ, η3); =) (λ, η3) = ζ: In summary, we have the following table. λ η1 η2 η3 λ 1 1 1 ζ 2 η1 1 1 ζ 1 η2 1 ζ 1 1 2 η3 ζ 1 1 1 POWER RECIPROCITY 3 1.6. The reciprocity law. × Denition 1.7. Let a 2 K ; and p a prime ideal of OK . Suppose a is coprime to p, and suppose p is coprime to m. Then we dene the power residue symbol to be a := (π ; a) p p p for any uniformizer πp of p. That this is well dened follows from the formula for tame symbols. In particular a ≡ a(N p−1)=m mod p: p The power residue symbol is a generalization of the Legendre symbol. Denition 1.8. Let a 2 K×. Let b = Q pep be a fractional ideal of K. Suppose a is coprime to b, and suppose b is coprime to m. We dene the generalized Jacobi symbol to be e a Y a p := : b p p Here when ep = 0, we understand the corresponding term as 1 by conventions. When is principal, we write a a . b = bOK b := b Remark 1.9. The generalized Jacobi symbol is evidently multiplicative in both × variables in K and IK . Theorem 1.10 (Reciprocity law for generalized Jacobi symbols). Let a; b 2 K×. Suppose a and b are coprime to each other. Also suppose both a and b are coprime to m. Then we have −1 a b Y = (a; b) : b a v vjm1 Proof. −v(a) Y av(b) Y b Y Y LHS = = (π ; a)v(b) (π ; b)−v(a) v v v v v v vjb vja vjb vja Y Y −1 Y Y = (b; a)v (a; b)v = (b; a)v = (a; b)v: vjb vja vjab vjm1 The last equality follows from the product formula. Remark 1.11. When m ≥ 3, then the archimedean places make no contribution since they are all complex. Remark 1.12. By the theorem, obtaining a power reciprocity law is equivalent to computing the pairings (; )v for vjm. We have done this in Examples 1.4 and 1.5, which will produce quadratic reciprocity and cubic reciprocity. However, in general, the explicit calculation of (a; b)v for vjm is a highly nontrivial task. Complete answers to this were only obtained no earlier than 1970s. For reference see Neukirch: Algebraic Number Theory or the book Invitation to Higher Local Fields. 4 YIHANG ZHU 2. Cubic reciprocity Using Examples 1.4,1.5, and Theorem 1.10, we can deduce quadratic and cubic reciprocity. We only work out cubic reciprocity. Thus we keep the notation in Example 1.5. Recall that (λ) is the ramied prime over 3. We write v for this × × place. We say a 2 K is primary if a 2 ±1 + 3OK . This implies a 2 K =3 is spanned by η2 and η3. Theorem 2.1 (Cubic reciprocity, Gauss-Eisenstein). Let a; b 2 K× be primary. In particular they are coprime to 3. Suppose a and b are coprime to each other, then a b = : b a Proof. By Example 1.5, we have for the place above . (a; b)v = 1 v 3 2 Remark 2.2. If a 2 OK is coprime to λ, then a 2 a1 +3OK , for a1 2 ±1; ±ζ; ±ζ . 0 0 In other words, aOK = a OK for some a 2 ±1 + 3OK primary. Note that for b coprime to , the symbol b only depends on , so we may always replace a a aOK a by a0 to calculate it. Theorem 2.3 (Complementary laws). Let a 2 K× be primary. Write a = ±(1 + 3(m + nζ)) with m; n 2 Z. Then ζ λ = ζ−m−n; = ζm: a a Proof. Since by denition only depends on , we may assume a aOK a = 1+3(m+ nζ). Since ζ is a unit,by Theorem 1.10 we have ζ = (ζ; a) : a v Applying the logarithm and comparing coecients it is not hard to see a = ηm+nηm 2 2 3 × . Recall , so we obtain the desired value of ζ from the calculation Kv =3 ζ = η1 a in Example 1.5. Next we use the product formula to compute v (a) λ Y λ p Y = = (a; λ) = (a; λ)−1 = (λ, a) = (λ, ηm+nηm) = ζm: a p p v v 2 3 v pja pja Remark 2.4. Using the complementary law and the remark before it, we can reduce the calculation of any b with coprime to and coprime to each other, to a a λ a; b the case a; b are primary.