On Norm Residue Symbols and Conductors Romyar T. Sharifi April 1998 Abstract Using the norm residue symbol and a reciprocity law of Artin-Hasse, we pn determine the conductors of Kummer extensions of the form K( pa; ζn)=K(ζn) for any unramified extension K of Q , element a K and primitive pnth root p 2 ∗ of unity ζn. We are able to do this without more recent and general reciprocity laws, which were needed in earlier proofs of the case K = Qp. 1 Introduction We are interested in determining the conductors of Kummer extensions of the form pn K( pa; ζn)=K(ζn) for an unramified extension K of Qp, an element a K∗ and a n 2 n primitive p th root of unity ζn. These are exactly the conductors of the p th norm residue symbols (a; )pn;K(ζ ). Using a complete explicit reciprocity law of Coleman's · n [1], the conductors of these symbols have been computed in the case of K = Qp by Coleman and McCallum in [2] for odd primes p and by Prapavessi in [8] for p = 2 (see also [9]). The author has also used Coleman's method to compute the conduc- tors for any unramified K with p odd (unpublished). The same conductors were also computed by Miki in [6] and [7] using Iwasawa's reciprocity law [5] and rami- fication theory. Aside from revealing certain generalizations of classical reciprocity laws, knowledge of the conductors has been used by the aforementioned authors to compute the conductors of Jacobi sum Hecke characters. Our proof of this somewhat more general case depends nevertheless on no more than the most basic of the classical reciprocity laws of Artin and Hasse [3]. Hence, this paper could have been written in the first half of this century. We do not appeal to heavy mathematical machinery. Even the use of ramification theory is avoided for the most part, though it would perhaps shorten some arguments. Let p be a prime number and n a positive integer. In the next two sections, we shall assume that either p is odd or n 2, as the quadratic case is somewhat exceptional and easily handled. In the following,≥ ( ; ) will be understood to mean the pnth · · norm residue symbol for the field K(ζn). We denote by v the normalized additive valuation on K and by vn the normalized additive valuation on K(ζn). Let denote O 1 On Norm Residue Symbols and Conductors 2 the valuation ring of K, n the valuation ring of K(ζn) and p and pn their maximal n k O p − ideals, respectively. We let ζk = ζn and λk = 1 ζk for all 1 k n. For a K∗, let f(a) denote the conductor of (a; ). Let q be the− order of the residue≤ ≤ field of K2, and · let µq 1 denote the set of (q 1)st roots of unity in K. Furthermore, for L=K Galois, − − let TrL=K denote the trace from L to K. In particular, we set Trn = TrK(ζn)=Qp . Let η µq 1 be such that TrK=Qp (η) 1 mod p (by lifting an element with trace 1 from 2 − ≡ the residue field of K). Finally, let ' denote the Frobenius automorphism of K=Qp. 2 Symbols of the Form (α, 1 αpk ) − We first concern ourselves with the evaluation of symbols of the form (α, 1 αpk ) pk − with α pn and 1 k n. Factoring 1 α , we obtain 2 ≤ ≤ − pk pk pk j j j (α, 1 α ) = (α, (1 αζ− )) = (ζ ; 1 αζ− ); Y k Y k k − j=1 − j=1 − and so we have pk pk j j (α, 1 α ) = (ζ ; (1 αζ− ) ): (1) k Y k − j=1 − The following law of Artin-Hasse [3, p. 94] is one of the classical laws of explicit reciprocity. Theorem 1 (Artin-Hasse). For β 1 mod (λn) and 1 k n, we have ≡ ≤ ≤ 1 pn Trn(log β) (ζk; β) = ζk for p odd; n 1 1+2 − 2n Trn(log β) (ζk; β) = ζ for p = 2 and n 2; k ≥ where Trn denotes the trace from K(ζn) to Qp and log denotes the logarithm defined by the usual power series. Let us apply the Artin-Hasse law to (1). First, we have k k p i p j j 1 α ij log (1 αζ− ) = jζ− : (2) Y k X i X k j=1 − − i=1 j=1 In the case pk divides i, we obtain merely pk(pk + 1)=2 as the sum over j. Let us assume now pk does not divide i. We have, in fact, pk pk ij i i j 1 jζ− = ζ j(ζ ) ; (3) X k k− X k− − j=1 j=1 On Norm Residue Symbols and Conductors 3 where the latter sum is recognizable as the derivative of a geometric sum evaluated i at ζk− . We evaluate k p pk+1 k pk pk+1 d 0 j1 d X 1 (p + 1)X (X 1) X + 1 X = − 1! = − − : dX X dX X 1 − (X 1)2 @ j=1 A − − i Letting X = ζk− , we obtain by (3) that k p k k ij i p p jζk− = ζk− i = i : X ζ− 1 1 ζ j=1 k − − k By (2), we conclude that evaluating (1) amounts to determining 1 + δ αi 1 pk + 1 αi Trn + Trn( ) mod Zp; (4) − pn X i 1 ζi 2 X i pk i k pk i 6 j − j n 1 where δ = 0 if p is odd and δ = 2 − if p = 2. Let us concentrate on evaluating the terms of the first sum in (4). Lemma 2. Let β n and set 2 O 1 β Ak = n Trn p λk for 1 k n. ≤ ≤ 1 a: If p is odd or if β 0 mod (2), then we have Ak p− Zp. Otherwise we have 1 ≡ 2 Ak 4− Z2. 2 b: If β 0 mod (λ1λk), then we have Ak Zp. ≡ 2 c: Let 1 l n and ≤ ≤ λ λ β = η 1 k : λl (Recall η µq 1 is such that T rK=Qp (η) 1 mod p.) Then we have Ak 21 − ≡ ≡ 1 mod p− Zp. − Proof. Parts (a) and (b) are immediate from the fact that the different of the extension n 1 K(ζn)=Qp is (p λ1− ). For β as in part (c), we have 1 ηλ1 1 pl 1 1 p 1 − − n Trn = TrK=Qp (η) n TrQp(ζn)=Qp (1 + ζl + + ζl ) − mod Zp: p λl p ··· ≡ p since the cyclotomic trace of any p-power root of unity of order at least p2 is 0. On Norm Residue Symbols and Conductors 4 We will also need the following easy lemma which gives us useful congruences for p-powers of certain elements. Lemma 3. p a: For 2 l n, we have λl λl 1 mod (p). ≤ ≤ ≡ − b: We have p λ1 2 λ1 mod (λ ): p ≡ − 1 Proof. The first statement of part (a) is obvious from the binomial series expansion p i of (1 ζl) . For part (b), note that ζ 1 mod (λ1) for all i. Therefore we have − 1 ≡ p 1 p 1 − i − p 1 ζ1 i 1 p 1 = − = (1 + ζ1 + + ζ1− ) (p 1)! 1 mod (λ1); λ − Y 1 ζ1 Y ··· ≡ − ≡ − 1 i=1 − i=1 where the last congruence is Wilson's theorem. The previous two lemmas lead to the following proposition. Proposition 4. pk a: If α 0 mod (λ1), then (α, 1 α ) is a pth root of unity for all 1 k n. ≡ − ≤ ≤ pk b: Let 1 l n. If α 0 mod (λ1λl), then we have (α, 1 α ) = 1 for all l k ≤n. ≤ ≡ − ≤ ≤ c: Let 1 k n and k l n, and set α = ηλ1λk/λl. Then we have that (α, 1 ≤αpk )≤ = 1 if and≤ only≤ if k 2, or p = 2 and± k = l = 1. − ≥ 2 d: Let p = 2, 1 k n and 2 l n, and set α = 4η/λl . Then we have that (α, 1 α2k ) =≤ 1 if≤ and only if≤k ≤2. − ≥ Proof. The symbols in question are evaluated using (4). We may assume throughout i i 2 that α 0 mod (λ1). Then α =i 0 mod (λ1) for all i 1 and α =i 0 mod (λ1) for all i≡= 1; p. Part (a) now follows≡ from Lemma 2a, and≥ by Lemma 2b≡ equation (4) reduces6 to 1 α αp n Trn + Trn mod Zp (5) −p λk pλk 1 − when k 2, ≥ 1 α n Trn mod Zp (6) −p λ1 when k = 1 and p is odd, and 1 α α2 Trn + Trn mod Z2 (7) 2n 2 4 On Norm Residue Symbols and Conductors 5 when k = 1 and p = 2. For part (b), we have α 0 mod (λ1λk) and ≡ αp 0 mod (λ1λk 1) p ≡ − when k > 1. The result is then immediate from (5), (6), (7) and Lemma 2b. For part (c), we must evaluate (5), (6) and (7). Lemma 2c tells us that the first 1 term of each of these equations is p− . When k 2, there is a second term, and using Lemma 3 it is not hard to see∓ that ≥ p α p λ1λk 1 2 η − mod (λ1): p ≡ ∓ λl 1 − p As TrK=Qp (η ) = TrK=Qp ('(η)) = TrK=Qp (η); we easily see by Lemma 2b,c that this 1 contribution is p− .