KUMMER S LEMMA KEITH CONRAD

KUMMER S LEMMA KEITH CONRAD Let p be an odd prime and ζ ζ p be a primitive pth root of unity In the ring Z[ζ], the pth power of every element is congruent to a rational integer mod p, since (c 0 + c 1 ζ + + c p 2 ζ p 2 ) p c 0 + c 1 + + c p 2 mod p The number p is not prime in Z[ζ], as (p) (1 ζ), so congruence mod p is much stronger than congruence mod 1 ζ, where all classes have integer representatives Of course not every element of Z[ζ] that is congruent to a rational integer mod p is a pth power, but Kummer discovered a case when this converse statement is true, for certain primes and certain algebraic integers Theorem 1 (Kummer s Lemma) Let p be a regular prime and u Z[ζ] with u a mod p for some rational integer a Then u v p for some v Z[ζ] This was used by Kummer to prove Case II of FLT for regular primes For our purposes, an odd p will be called regular if the Bernoulli numbers B 2, B 4,, B p 3 are all prime to p This is not the most conceptual description of regularity, but it is the form in which we will be using the property in the proof The usual formulation of regularity, in terms of class numbers, allows for other proofs of Kummer s Lemma, by class field theory [3, 6, Chap 13] or by p-adic L-functions [4, Theorem 536] We give a p-adic proof of Kummer s Lemma, modifying the argument of Faddeev from [1, 6, Chap 5] The proof of Kummer s Lemma requires nowing something about the unit group of Z[ζ] Some obvious units in Z[ζ] are ε def ζ ζ mod ζ for 1 p While ε 1 1, the other units are more interesting There is a convenient way to rewrite the ε First, since ζ has odd order, some power of it is a square root of ζ; indeed, ζ (p+1)/2 is a square root of ζ However, η def ζ (p+1)/2 turns out to be the more convenient choice of square root, since when ζ e 2i/p, ζ (p+1)/2 e i/p while η e i/p, the nicer square root of e 2i/p Furthermore, choosing ζ e 2i/p, ε η2 η 2 η η η η η η 1 sin(/p) η 1 sin(/p) Let δ sin(/p)/ sin(/p), so δ is a real (positive) unit The equation ε η 1 δ can be generalized to any unit of Z[ζ] Lemma 2 Every unit u of Z[ζ] has the form u ζ r u, where u is a real unit Proof Suppose u ζ r u Then u ζ r u, so we can divide and get u/u ζ 2r This suggests the idea of considering the ratio u/u and proving it is a root of unity Well, u/u and all of its Q-conjugates have absolute value 1, so it is a root of unity Being in Q(ζ), we 1

2 KEITH CONRAD must have u/u ±ζ a If we can show the plus sign holds, then write a 2b mod p and set u u/ζ b to end the proof Let s wor mod (1 ζ) Since all powers of ζ are congruent to 1, u u, so 1 ±ζ a ±1 Since 1 1 mod (1 ζ), the plus sign holds Let K Q(ζ), K p Q p (ζ), A Z[ζ], A p Z p [ζ], σ j (ζ) ζ j, τ σ 1 is complex conjugation Denote the real elements of A p, ie the elements fixed by τ, as A + p (A similar definition can be made for K + p, but we will only be focusing on real elements of A p ) The Teichmüller lift of an integer to Z p will be written ω() Lemma 3 In Q p (ζ), X + p splits completely and there is a bijection between roots of X + p and nontrivial pth roots of unity ζ, by Proof See [3, Lemma 31, Chap 14] ζ mod 2 Fix a choice of ζ, then fix a root of X + p by ζ mod 2 Write τ() θ, where θ 1 Since τ 2 () θ 2, θ ±1 Since τ(), τ() Thus 2 A + p Note that the minimal polynomial of 2 over Q p is X ()/2 + p and A p Z p [] Lemma 4 A + p Z p [ 2 ] Proof Left to reader So {1, 2,, ( 2 ) } is a Z p -basis for A + p Lemma 5 For x, y 1 mod, log x log y x y Proof Without loss of generality, y 1 Then x 1 mod log x x We want to show that for 1, (x ) x, which is equivalent to x 1/( 1) Since x (1/p) 1/(), we re done Corollary 6 For u Z[ζ] with u a mod p for some rational integer a, log(u ) pa + p Proof Since u 1 mod p, log(u ) u 1/p, so log(u ) pa p Writing u ζ r u for real u by Lemma 2, log(u ) log((u ) ) A + p Given a unit u Z[ζ], write log(u ) i0 b i 2i A trace calculation shows b 0 0: for 1 p, Tr Kp/Qp ( ) σ j ( ) ω(j) 0 j1 So p 2 b 0 Tr Kp/Q p (log(u )) log(n Kp/Q p (u )) log(n K/Q (u )) log(1) 0 j1

KUMMER S LEMMA 3 For 2 m, write log(ε ) c i 2i for some c i Z p The subscripts i and both run over sets of size m The next theorem is the technical heart of our preparations for Kummer s Lemma Theorem 7 The numbers {1, log(ε 2 ),, log(εm )} form a Z p -basis of A + p precisely when p is regular Proof Since {1, 2,, 2() } is a Z p -basis of A + p, we shall chec that the Z p -transition matrix (c i ) between { 2,, 2() } and {log(ε 2 ),, log(εm )} has determinant in Z p, or rather than its determinant mod p is nonzero Write ε ε p ζ ζ Since ε mod, ε p p mod p p mod p mod p Therefore ε (ζ )/(ζ ) mod p, so by Lemma 5, log(ε ) log((ζ )/(ζ )) mod p Let s use the Dwor series E(X) e X+Xp /p to express ζ in terms of [3, Theorem 32, Chap 14]: E() ζ and E(ω()) ζ We will write ω() as So ε ζ ζ mod p E() E( ) ω() E() E( ) mod p E( ) E() Since ζ E( ) 1 + mod 2, we have ( ) ( ) (1) log(ε E() E( ) ) log log mod p Writing E(X) a n X n, (E() )/ n 1 a n n 1 When is a n n 1 0 mod p? In [3, p 322], e n denotes a n n and some lower bound estimates on ord p (e n ) are proved that imply ord p (a n n 1 ) 1 if n p 3 /(p ) 2 But the lower bound is true for n p Rather than wor with the lower bound estimates in [3] to squeeze out this more delicate information, we will use [2, Theorem 35], which gives a sharper lower bound on ord p (a n ) For n p 2, that bound is ord p (a n ) n ( p 2 2 + 1 ) ( p p 2 + 1 ) p p 2,

4 KEITH CONRAD so ord p (a n n 1 ) 2 + (n )/(p ) 2 + (p 2 )/(p ) p + 3 5 If p n p 2 then a n is a p-adic integer since E(X) and the Artin Hasse series have the same coefficients up through degree p 2 1, so ord p (a n n 1 ) (n 1)/() ()/() 1 Therefore E() a n n 1 since a n 1/ for 0 n p Thus ( ) ( E() (2) log log n 1 n 1 ) mod p mod p The sum n 1 / loos lie a truncation of the full series (e )/ But the latter maes no sense, since is not in the disc of convergence of e X Nevertheless, progress will come from looing at the formal power series log((e X )/X) Since we are computing logarithms only modulo p, we can truncate log series To determine how far out we need to go, let s figure out when (x ) n /n p for x Lemma 8 For n p + 1, n /n p Proof We need to determine when n/(p ) ord p (n) + 1 If ord p (n) 0, this holds when n p If ord p (n) 1, this holds as long as n p If ord p (n) 2, this always holds The inequality in the lemma is true for n p, but we won t need this So if x, log x p ( 1)n 1 (x 1) n n mod p The last term in the sum is more subtle than the rest since it has a p in the denominator So let s isolate the last term Let L (1 + T ) def ( 1) n 1 T n n Z p[t ], so x 1 mod log x L (x) + (x ) p /p mod p Writing α n 1 /, by (2) ( ) E() log log(1 + α) L (1 + α) + αp mod p p Let s compute the last term: α p p 1 p ( For 2 n p, n 2 / A p, so ( ) p n 2 n 2 ) p ( ( ) n 2 p mod p ( p ) n 2 mod p n 2 ) p 1 2 mod p since p 0 mod p

KUMMER S LEMMA 5 Thus ( ) E() (3) log L (1 + α) /2 mod p ( ) Since L (1 + T ) Z p [T ], L (1 + α) L n 1 / Z p [] We only care about its expression mod p Let s compute the polynomial L ( T n 1 /) in Z p [T ]/T n 1 Actually, it turns out to be more convenient to wor in Q p [[T ]]/T, where we can use the full exponential and logarithm series, whose coefficients are usually not in Z p Whatever we compute in this larger ring for L ( T n 1 /) must be in Z p [T ]/T For f(t ) 1 + T Q p [[T ]], L (f(t )) log(f(t )) mod T p, so from e T 1 T mod T we get ( L T n 1 ) ( e T ) ( e T ) L log mod T T T T n 1 This last expression is tractable, and is where Bernoulli numbers enter To find the expansion for log((e T )/T ), we differentiate the series: ( d e T ) dt log T T e T T et (e T ) T 2 e T e T T 1 + 1 e T T 1 ( T + T ) T e T 1 T 1 T 1 2 + i 1 Integrating and noting the constant term must vanish, ( e T ) log T T 2 + i 1 T + B n T n n 1 1 2 T + (2i)! T 2i i 1 (2i)! T 2i 1 (2i)!2i T 2i T 2 + (2i)!2i T 2i mod T So L ( T n 1 /) T/2 + (/(2i)!2i)T 2i in Z p [T ]/T, and by (3) ( ) E() log (2i)!2i 2i mod p

6 KEITH CONRAD Similarly, ( ) E( ) log (2i)!2i 2i 2i mod p since mod p Putting these congruences together, we can compute log(ε ) mod p by (1): So c i (2i)!2i (1 2i ) mod p Therefore det(c i ) log(ε ) (2i)!2i (1 2i ) 2i mod p (2i)!(2i) 2 2 3 2 m 2 2 4 3 4 m 4 mod p 2 2() 3 2() m 2() We can rewrite the last determinant in Vandermonde form It equals 1 0 0 1 1 1 1 2 2 m 2 1 2 4 m 4 1 2 2 m 2 1 2 4 m 4 1 2 2() m 2() 1 2 2() m 2() (j 2 i 2 ) 1 i<j m 0 mod p Therefore det(c i ) 0 mod p precisely when none of B 2, B 4,, B 2() B p 3 is divisible by p We now prove Kummer s Lemma Proof First let s show u is real, ie u is fixed by complex conjugation By Lemma 2, u ζ r u for u a real unit So u A + p Z p [ 2 ] Thus u b mod 2 for some rational integer b Since ζ r 1 + r mod 2, we have u b + br mod 2 Since u is congruent to a rational integer mod p, we must have br 0 mod p, so p r, hence u u Without loss of generality, we may tae u to be positive We will wor in the group of real positive units of A Z[ζ], since that group has no torsion Recall δ from before Lemma 2 By Theorem 7, the numbers log(δ ) log(ε ) are linearly independent over Q p for 2 m, so the units δ are multiplicatively independent over Z There are m of them, so by (the easy part of) the Dirichlet Unit Theorem they have finite index in the group of all units In particular, m (4) u n 2 δ c

KUMMER S LEMMA 7 for c Z Since the real positive units have no torsion, we may assume gcd(n, c 2,, c m ) 1 Raising both sides of (4) to the (p )th power and then taing logarithms, we get n log(u ) 2 c log(δ ) By Corollary 6, log(u ) pa + p, so all c lie in pz p Since they are rational integers, all c lie in pz, so u n is a pth power of a unit in Z[ζ] From gcd(n, c 2,, c m ) 1 we see n is prime to p, so u is a pth power of a unit in Z[ζ] This proof of Kummer s Lemma used the units δ rather than ε only because we needed to wor in a group of units where there is no torsion, so the assumption gcd(n, c 2,, c m ) 1 could be used While the regularity assumption implies that the group generated by the ε has no torsion (there is only the trivial linear relation among their p-adic logarithms), it is not clear how to show the group generated by u and the ε has no torsion By woring in the group of positive real units (where we proved u lies) the torsion issue is easily handled In Faddeev s proof of Kummer s Lemma, the Dwor series E(X) e X+Xp /p is not used Instead more systematic use is made of truncations of the series e X (while E(X) is a truncated Artin-Hasse series) For instance, since ζ E(), the congruence (E() )/ n 1 / mod p that we used is the same as ζ e () mod p, where e (T ) n0 T n / This is essentially [1, Lemma 3, p 372] with 1 (and e (T ) denoted there by E(T )) References [1] Z I Borevich and I R Shafarevich, Number Theory, Academic Press, New Yor, 1966 [2] K Conrad, Artin Hasse-type series and roots of unity, http://wwwmathuconnedu/~conrad/blurbs/ gradnumthy/ahrootofunitypdf [3] S Lang, Cyclotomic Fields I and II, Springer-Verlag, New Yor, 1994 [4] L Washington, An Introduction to Cyclotomic Fields, Springer-Verlag, New Yor, 1997