aa r X i v : . [ m a t h . N T ] S e p On the p -adic Leopoldt transform of a power series Bruno Angl`es Let p be an odd prime number. Let X be the projective limit for thenorm maps of the p -Sylow subgroups of the ideal class groups of Q ( ζ p n +1 ) ,n ≥ . Let ∆ = Gal( Q ( ζ p ) / Q ) and let θ be an even and non-trivial characterof ∆ . Then X is a Z p [[ T ]]-module and the characteristic ideal of the isotypiccomponent X ( ωθ − ) is generated by a power series f ( T, θ ) ∈ Z p [[ T ]] suchthat (see for example [2]): ∀ n ≥ , n ≡ p − , f ((1 + p ) − n − , θ ) = L (1 − n, θ ) , where L ( s, θ ) is the usual Dirichlet L -series. Therefore, it is natural andinteresting to study the properties of the power series f ( T, θ ) . We denote by f ( T, θ ) ∈ F p [[ T ]] the reduction of f ( T, θ ) modulo p. ThenB. Ferrero and L. Washington have proved ([3]): f ( T, θ ) = 0 . Note that, in fact, we have ([1]): f ( T, θ ) F p [[ T p ]] . W. Sinnott has proved the following ([8]): f ( T, θ ) F p ( T ) . But, note that ∀ a ∈ Z ∗ p , F p [[ T ]] = F p [[(1 + T ) a − . Therefore it is naturalto introduce the notion of a pseudo-polynomial which is an element F ( T ) in F p [[ T ]] such that there exist an integer r ≥ , c , · · · c r ∈ F p , a , · · · , a r ∈ Z p , such that F ( T ) = P ri =1 c i (1 + T ) a i . An element of F p [[ T ]] will be called apseudo-rational function if it is the quotient of two pseudo-polynomials. Inthis paper, we prove that f ( T, θ ) is not a pseudo-rational function (part1) of Theorem 4.5). This latter result suggests the following question: is f ( T, θ ) algebraic over F p ( T )? We suspect that this is not the case but we Universit´e de Caen, LMNO CNRS UMR 6139, BP 5186, 14032 Caen Cedex, France.E-mail: [email protected] f ( T, θ ) = T λ ( θ ) U ( T ) , where λ ( θ ) ∈ N and U ( T ) ∈ F p [[ T ]] ∗ . S. Rosenberg has proved that ([6]): λ ( θ ) ≤ (4 p ( p − φ ( p − , where φ is Euler’s totient function. In this paper, we improve Rosenberg’sbound (part 2) of Theorem 4.5): λ ( θ ) < ( p −
12 ) φ ( p − . This implies that the lambda invariant of the field Q ( ζ p ) is less than 2( p − ) φ ( p − (see Corollary 4.6 for the precise statement for an abelian number field).Note that this bound is certainly far from the truth, because accordingto a heuristic argument due to Ferrero and Washington (see [5]) and toGrennberg’s conjecture: λ ( Q ( ζ p )) = X θ ∈ b ∆ , θ =1 and even λ ( θ ) ≤ Log( p )Log(Log( p )) . The author is indebted to Warren Sinnott for communicating some of hisunpublished works (note that Lemma 4.2 is due to Warren Sinnott). Theauthor also thanks Filippo Nuccio for pointing out the work of J. Kraft andL. Washington ([4]).
Let p be an odd prime number and let K be a finite extension of Q p . Let O K be the valuation ring of K and let π be a prime of K. We set F q = O K /πO K , it is a finite field having q elements and its characteristicis p. Let T be an indeterminate over K, we set Λ = O K [[ T ]] . Observe thatΛ /π Λ ≃ F q [[ T ]] . Let F ( T ) ∈ Λ \ { } , then we can write in an unique way([9], Theorem 7.3): F ( T ) = π µ ( F ) P ( T ) U ( T ) , U ( T ) in an unit of Λ , µ ( F ) ∈ N , P ( T ) ∈ O K [ T ] is a monic polynomialsuch that P ( T ) ≡ T λ ( F ) (mod π ) for some integer λ ( F ) ∈ N . If F ( T ) = 0 , we set µ ( F ) = λ ( F ) = ∞ . An element F ( T ) ∈ Λ is called a pseudo-polynomial (see also [6], Definition 2) if there exist some integer r ≥ ,c , · · · , c r ∈ O K , a , · · · , a r ∈ Z p , such that: F ( T ) = r X i =1 c i (1 + T ) a i . We denote the ring of pseudo-polynomials in Λ by A. Let δ ∈ Z / ( p − Z and F ( T ) ∈ Λ , we set: γ δ ( F ( T )) = 1 p − X η ∈ µ p − η δ F ((1 + T ) η − . Then γ δ : Λ → Λ is a O K -linear map and:- for δ, δ ′ ∈ Z / ( p − Z , γ δ γ δ ′ = 0 if δ = δ ′ and γ δ = γ δ , - P δ ∈ Z / ( p − Z γ δ = Id Λ . For F ( T ) ∈ Λ , we set: D ( F ( T )) = (1 + T ) ddT F ( T ) ,U ( F ( T )) = F ( T ) − p X ζ ∈ µ p F ( ζ (1 + T ) − ∈ Λ . Then
D, U : Λ → Λ are O K -linear maps. Observe that:- U = U, - DU = U D, - ∀ δ ∈ Z / ( p − Z , γ δ U = U γ δ , - ∀ δ ∈ Z / ( p − Z , Dγ δ = γ δ +1 D. If F ( T ) ∈ Λ , we denote its reduction modulo π by F ( T ) ∈ F q [[ T ]] . If f : Λ → Λ is a O K -linear map, we denote its reduction modulo π by f : F q [[ T ]] → F q [[ T ]] . For all n ≥ , we set ω n ( T ) = (1 + T ) p n − . Let B be a commutative and unitary ring. We denote the set of invertibleelements of B by B ∗ . We fix κ a topological generator of 1 + p Z p . Let x ∈ Z p and let n ≥ , wedenote the unique integer k ∈ { , · · · , p n − } such that x ≡ k (mod p n ) by[ x ] n . Let ω : Z ∗ p → µ p − be the Teichm¨uller character, i.e. ∀ a ∈ Z ∗ p , ω ( a ) ≡ a (mod p ) . Let x, y ∈ Z p , we write: 3 x ∼ y if there exists η ∈ µ p − such that y = ηx, - x ≡ y (mod Q ∗ ) if there exists z ∈ Q ∗ such that y = zx. The function Log p will denote the usual p -adic logarithm. v p will denote theusual p -adic valuation on C p such that v p ( p ) = 1 . Let ρ be a Dirichlet character of conductor f ρ . Recall that the Bernoullinumbers B n,ρ are defined by the following identity: f ρ X a =1 ρ ( a ) e aZ e fZ − X n ≥ B n,ρ n ! Z n − , where e Z = P n ≥ Z n /n ! . If ρ = 1 , for n ≥ , B n, is the n th Bernoullinumber.Let x ∈ R . We denote the biggest integer less than or equal to x by [ x ] . The function Log will denote the usual logarithm.
Let δ ∈ Z / ( p − Z . In this section, we will recall the construction of the p -adic Leopoldt transform Γ δ (see [5], Theorem 6.2) which is a O K -linearmap from Λ to Λ . First, observe that ( π n , ω n ( T )) = π n Λ + ω n ( T )Λ , n ≥ , is a basis ofneighbourhood of zero in Λ : Lemma 2.1 ∀ n ≥ , ( π, T ) n ⊂ ( π n , T n ) ⊂ ( π, T ) n . ∀ n ≥ , ω n ( T ) ∈ ( p [ n/ , T p [ n/ ) .
3) Let N ≥ , set n = [Log( N ) / Log( p )] . We have: T N ∈ ( p [ n/ , ω [ n/ ( T )) . Proof
Note that assertion 1) is obvious. Assertion 2) comes from the fact: ∀ k ∈ { , · · · , p n } , v p ( p n ! k !( p n − k )! ) = n − v p ( k ) . To prove assertion 3), it is enough to prove the following:4 n ≥ , there exist δ ( n )0 ( T ) , · · · , δ ( n ) n ( T ) ∈ Z [ T ] such that: T p n = X i + j = n ω i ( T ) p j δ ( n ) j ( T ) . Let’s prove this latter fact by recurrence on n. Note that the result is clearif n = 0 . Let’s assume that it is true for n and let’s prove the assertion for n + 1 . Let r ( T ) ∈ Z [ T ] such that: ω n +1 ( T ) ω n ( T ) + pr ( T ) = T p n ( p − . Then: T p n +1 = T p n ω n +1 ( T ) ω n ( T ) + pr ( T ) T p n . Note that there exists q ( T ) ∈ Z [ T ] such that: ω n +1 ( T ) ω n ( T ) = ω n ( T ) p − + pq ( T ) . Thus: T p n +1 = ω n +1 ( T ) δ ( n )0 ( T )+ X i + j = n, j ≥ ( ω n ( T ) p − + pq ( T )) ω i ( T ) p j δ ( n ) j ( T ) + X i + j = n ω i ( T ) p j +1 δ ( n ) j ( T ) r ( T ) . Thus, there exist δ ( n +1)0 ( T ) , · · · , δ ( n +1) n +1 ( T ) ∈ Z [ T ] such that: T p n +1 = X i + j = n +1 ω i ( T ) p j δ ( n +1) j ( T ) . ♦ The following Lemma will be useful in the sequel (for a similar result see[6], Lemma 5):
Lemma 2.2
Let F ( T ) ∈ A. Write F ( T ) = P ri =1 β i (1 + T ) α i , β , · · · , β r ∈ O K , α , · · · , α r ∈ Z p , and α i = α j for i = j. Let N = Max { v p ( α i − α j ) , i = j } . Let n ≥ be an integer. Then: F ( T ) ≡ π n , ω N +1 ( T ))) ⇔ ∀ i = 1 , · · · r, β i ≡ π n ) . roof We have: F ( T ) ≡ r X i =1 β i (1 + T ) [ α i ] N +1 (mod ω N +1 ( T )) . Therefore F ( T ) ≡ π n , ω N +1 ( T ))) if and only if we have: r X i =1 β i (1 + T ) [ α i ] N +1 ≡ π n ) . But for i = j, [ α i ] N +1 = [ α j ] N +1 . Therefore P ri =1 β i (1 + T ) [ α i ] N +1 ≡ π n ) if and only if: ∀ i = 1 , · · · r, β i ≡ π n ) . ♦ Observe that
U, D, γ δ are continuous O K -linear maps by Lemma 2.1 and thefollowing Lemma: Lemma 2.3
Let F ( T ) ∈ Λ and let n ≥ . F ( T ) ≡ ω n ( T )) ⇒ γ δ ( F ( T )) ≡ ω n ( T )) . F ( T ) ≡ ω n ( T )) ⇒ D ( F ( T )) ≡ p n , ω n ( T ))) .
3) If n ≥ , F ( T ) ≡ ω n ( T )) ⇒ U ( F ( T )) ≡ ω n ( T )) . Proof
The assertions 1) and 2) are obvious. It remains to prove 3). Observethat, by [9], Proposition 7.2, we have: ∀ G ( T ) ∈ Λ , G ( T ) ≡ ω n ( T )) ⇔ ∀ ζ ∈ µ p n , G ( ζ −
1) = 0 . Now, let F ( T ) ∈ Λ , F ( T ) ≡ ω n ( T )) . Since the map: µ p n → µ p n ,x ζ x, is a bijection for all ζ ∈ µ p n , we get: ∀ ζ ∈ µ p n , U ( F )( ζ −
1) = 0 . Therefore: U ( F ( T )) ≡ ω n ( T )) . ♦ Let s ∈ Z p . For n ≥ , set: k n ( s, δ ) = [ s ] n +1 + δ n p n +1 ∈ N \ { } , where δ n ∈ { , · · · , p − } is such that [ s ] n +1 + δ n ≡ δ (mod p − . Observethat: 6 ∀ n ≥ , k n ( s, δ ) ≡ δ (mod p −
1) and k n ( s, δ ) ≡ s (mod p n +1 ) , - ∀ n ≥ , k n +1 ( s, δ ) > k n ( s, δ ) , - s = lim n k n ( s, δ ) . In particular: ∀ a ∈ Z p , ∀ n ≥ , a k n +1 ( s,δ ) ≡ a k n ( s,δ ) (mod p n +1 ) . Now, let F ( T ) ∈ A. Write F ( T ) = P ri =1 β i (1 + T ) α i , β , · · · , β r ∈ O K ,α , · · · , α r ∈ Z p . We set:Γ δ ( F ( T )) = X α i ∈ Z ∗ p β i ω δ ( α i )(1 + T ) Log p ( αi )Log p ( κ ) . Thus, we have a surjective O K -linear map: Γ δ : A → A. Note that:
Lemma 2.4
Let F ( T ) ∈ A.
1) Let s ∈ Z p . Then: ∀ n ≥ , Γ δ ( F )( κ s − ≡ D k n ( s,δ ) ( F )(0) (mod p n +1 ) .
2) Let n ≥ . Assume that F ( T ) ≡ ω n ( T )) . Then Γ δ ( F ( T )) ≡ ω n − ( T )) . Proof
For a ∈ Z ∗ p , write a = ω ( a ) < a >, where < a > ∈ p Z p . Let’swrite: F ( T ) = r X i =1 β i (1 + T ) α i ,β , · · · , β r ∈ O K , α , · · · , α r ∈ Z p . We have: D k n ( s,δ ) ( F ( T )) = r X i =1 β i α k n ( s,δ ) i (1 + T ) α i . Thus: D k n ( s,δ ) ( F ( T )) ≡ X α i ∈ Z ∗ p β i ω δ ( α i ) < α i > s (1 + T ) α i (mod p n +1 ) . But recall that: Γ δ ( F )( κ s −
1) = X α i ∈ Z ∗ p β i ω δ ( α i ) < α i > s . F ( T ) ≡ ω n ( T ))for some n ≥ . Then: ∀ a ∈ { , · · · , p n − } , X α i ≡ a (mod p n ) β i = 0 . This implies that: ∀ a ∈ { , · · · , p n − − } , X α i ∈ Z ∗ p , Log p ( α i ) / Log p ( κ ) ≡ a (mod p n − ) ω δ ( α i ) β i = 0 . But recall that: Γ δ ( F ( T )) = X α i ∈ Z ∗ p β i ω δ ( α i )(1 + T ) Log p ( αi )Log p ( κ ) . Thus Γ δ ( F ( T )) ≡ ω n − ( T )) . ♦ Proposition 2.5
Let F ( T ) ∈ Λ . There exists an unique power series Γ δ ( F ( T )) ∈ Λ such that: ∀ s ∈ Z p ∀ n ≥ , Γ δ ( F )( κ s − ≡ D k n ( s,δ ) ( F )(0) (mod p n +1 ) . Proof
Let ( F N ( T )) N ≥ be a sequence of elements in A such that: ∀ N ≥ , F ( T ) ≡ F N ( T ) (mod ω N ( T )) . Fix N ≥ . Then: ∀ m ≥ N, F m ( T ) ≡ F N ( T ) (mod ω N ( T )) . Therefore, by Lemma 2.4, we have: ∀ m ≥ N, Γ δ ( F m ( T )) ≡ Γ δ ( F N ( T )) (mod ω N − ( T )) . This implies that the sequence (Γ δ ( F N ( T ))) N ≥ converges in Λ to somepower series G ( T ) ∈ Λ . Observe that, since Λ is compact, we have: ∀ N ≥ , G ( T ) ≡ Γ δ ( F N ( T )) (mod ω N − ( T )) . In particular: ∀ N ≥ , G ( κ s − ≡ Γ δ ( F N )( κ s −
1) (mod p N ) . ∀ N ≥ , G ( κ s − ≡ D k N − ( s,δ ) ( F N )(0) (mod p N ) . But: ∀ N ≥ , D k N − ( s,δ ) ( F ( T )) ≡ D k N − ( s,δ ) ( F N ( T )) (mod ( p N , ω N ( T ))) . Therfore: ∀ N ≥ , G ( κ s − ≡ D k N − ( s,δ ) ( F )(0) (mod p N ) . Now, set Γ δ ( F ( T )) = G ( T ) . The Proposition follows easily. ♦ p -adic Leopoldt trans-form We need the following fundamental result:
Proposition 3.1
Let δ ∈ Z / ( p − Z and let F ( T ) ∈ Λ . Let m, n ∈ N \ { } . then: Γ δ ( F ( T )) ≡ π n , ω m − ( T ))) ⇔ γ − δ U ( F ( T )) ≡ π n , ω m ( T ))) . Proof
A similar result has been obtained by S. Rosenberg ([6], Lemma 8).We begin by proving that Γ δ is a continuous O K -linear map. By Lemma2.1, this comes from the following fact:Let F ( T ) ∈ Λ . Let n ≥ F ( T ) ≡ ω n ( T )) , thenΓ δ ( F ( T )) ≡ ω n − ( T )) . Indeed, let ( F N ( T )) N ≥ be a sequence of elements in A such that: ∀ N ≥ , F ( T ) ≡ F N ( T ) (mod ω N ( T )) . By the proof of Proposition 2.5: ∀ N ≥ , Γ δ ( F ( T )) ≡ Γ δ ( F N ( T )) (mod ω N − ( T )) . Now, by Lemma 2.4: Γ δ ( F n ( T )) ≡ ω n − ( T )) . δ , γ − δ , U are continuous O K -linear maps, it suffices to provethe Proposition in the case where F ( T ) ∈ A. Write F ( T ) = P ri =1 β i (1 + T ) α i , β , · · · , β r ∈ O K , α , · · · , α r ∈ Z p . Let I ⊂ { α , · · · , α r } be a set ofrepresentatives of the classes of α , · · · , α r for the relation ∼ . For x ∈ I,x p ) , set: β x = X α i ∼ x β i α i x . We get: ( p − γ − δ U ( F ( T )) = X η ∈ µ p − X x ∈ I, x ∈ Z ∗ p η − δ β x (1 + T ) ηx . Now observe that:Γ δ ( F ( T )) = Γ δ γ − δ U ( F ( T )) = X x ∈ I, x ∈ Z ∗ p β x ω δ ( x )(1 + T ) Log p ( x ) / Log p ( κ ) . Therefore Γ δ ( F ( T )) ≡ π n , ω m − ( T ))) if and only if: ∀ a ∈ { , · · · p m − − } , X x ∈ I, x ∈ Z ∗ p , Log p ( x ) / Log p ( κ ) ≡ a (mod p m − ) β x ω δ ( x ) ≡ π n ) . Now, observe that for a ∈ { , · · · , p m − } , there exists at most one η ∈ µ p − such that [ ηx ] m = a, and if such a η exists it is equal to ω ( a ) ω − ( x ) . Therefore Γ δ ( F ( T )) ≡ π n , ω m − ( T ))) if and only if: ∀ a ∈ { , · · · , p m − } , X x ∈ I, x ∈ Z ∗ p , ∃ η x ∈ µ p − , [ η x x ] m = a β x η − δx ≡ π n ) . This latter property is equivalent to γ − δ U ( F ( T )) ≡ π n , ω m ( T ))) . ♦ Now, we can list the basic properties of Γ δ : Proposition 3.2
Let δ ∈ Z / ( p − Z . Γ δ : Λ → Λ is a surjective and continuous O K -linear map.2) ∀ F ( T ) ∈ Λ , Γ δ ( F ( T )) = Γ δ γ − δ U ( F ( T )) . ∀ a ∈ Z ∗ p , Γ δ ( F ((1 + T ) a − ω δ ( a )(1 + T ) Log p ( a ) / Log p ( κ ) Γ δ ( F ( T )) .
4) Let κ ′ be another topological generator of p Z p and let Γ ′ δ be the p -adicLeopoldt transform associated to κ ′ and δ. Then: ∀ F ( T ) ∈ Λ , Γ ′ δ ( F ( T )) = Γ δ ( F )((1 + T ) Log p ( κ ) / Log p ( κ ′ ) − . ) Let F ( T ) ∈ Λ . Then µ (Γ δ ( F ( T ))) = µ ( γ − δ U ( F ( T ))) and: ∀ N ≥ , λ (Γ δ ( F ( T ))) ≥ p N − ⇔ λ ( γ − δ U ( F ( T ))) ≥ p N . Proof
The assertions 1),2),3),4) come from the fact that Γ δ , γ − δ , U are con-tinuous and that these assertions are true for pseudo-polynomials. Theassertion 5) is a direct application of Proposition 3.1 . ♦ Let’s recall the following remarkable result due to W. Sinnott:
Proposition 3.3
Let r ( T ) , · · · , r s ( T ) ∈ F q ( T ) ∩ F q [[ T ]] . Let c , · · · , c s ∈ Z p \ { } and suppose that: s X i =1 r i ((1 + T ) c i −
1) = 0 . Then: ∀ a ∈ Z p , X c i ≡ a (mod Q ∗ ) r i ((1 + T ) c i − ∈ F q . Proof
See [8], Proposition 1. ♦ Let’s give a first application of this latter result:
Proposition 3.4
Let δ ∈ Z / ( p − Z and let F ( T ) ∈ K ( T ) ∩ Λ .
1) If δ is odd or if δ = 0 , then: µ (Γ δ ( F ( T ))) = µ ( U ( F ( T )) + ( − δ U ( F ((1 + T ) − − .
2) If δ is even and δ = 0 , then: µ (Γ δ ( F ( T ))) = µ ( U ( F ( T )) + U ( F ((1 + T ) − − − U ( F )(0)) . Proof
The case δ = 0 has already been obtained by Sinnott ([7], Theorem1). We prove 1), the proof of 2) is quite similar. Now, observe that 1) is aconsequence of Proposition 3.2 and the following fact:Let F ( T ) ∈ K ( T ) ∩ Λ , then µ ( γ − δ ( F ( T ))) = µ ( F ( T )+( − δ F ((1+ T ) − − . Let’s prove this fact. Let r ( T ) ∈ Λ , observe that: γ − δ ( r ( T )) = ( − δ γ − δ ( r ((1 + T ) − − . We can assume that F ( T ) + ( − δ F ((1 + T ) − − = 0 . Write: F ( T ) + ( − δ F ((1 + T ) − −
1) = π m G ( T ) , m ∈ N , and G ( T ) ∈ Λ \ π Λ . Note that G ( T ) ∈ K ( T ) . We mustprove that γ − δ ( G ( T )) π ) . Suppose that it is not the case, i.e. γ − δ ( G ( T )) ≡ π ) . Then: G (0) ≡ π ) . Furthermore, by Proposition 3.3, there exists c ∈ O K such that: G ( T ) + ( − δ G ((1 + T ) − − ≡ c (mod π ) . But, we must have c ≡ π ) . Observe that: G ( T ) = ( − δ G ((1 + T ) − − . Therefore we get G ( T ) ≡ π ) which is a contradiction. ♦ Lemma 3.5
Let F ( T ) ∈ F q ( T ) ∩ F q [[ T ]] . Then F ( T ) is a pseudo-polynomialif and only if there exists some integer n ≥ such that (1+ T ) n F ( T ) ∈ F q [ T ] . Proof
Assume that F ( T ) is a pseudo-polynomial. We can suppose that F ( T ) = 0 . Write: F ( T ) = r X i =1 c i (1 + T ) a i , where c , · · · , c r ∈ F ∗ q , a , · · · a r ∈ Z p and a i = a j for i = j. Since F ( T ) ∈ F q ( T ) there exist m, n ∈ N \ { } , m > Max { v p ( a i − a j ) , i = j } , such that:( T q n − T ) q m F ( T ) ∈ F q [ T ] . Thus: r X i =1 c i (1 + T ) a i + q n + m − r X i =1 c i (1 + T ) a i + q m ∈ F q [ T ] . Observe that:- ∀ i, j ∈ { , · · · , r } , a i + q n + m = a j + q m , - a i + q m = a j + q m ⇔ i = j. Thus, by Lemma 2.2, we get: ∀ i ∈ { , · · · r } , a i + q m ∈ N . Therefore (1 + T ) q m F ( T ) ∈ F q [ T ] . The Lemma follows. ♦ Let’s give a second application of Proposition 3.3:12 roposition 3.6
Let δ ∈ Z / ( p − Z and let F ( T ) ∈ F q ( T ) ∩ F q [[ T ]] . Sup-pose that there exist an integer r ∈ { , · · · , ( p − / } , c , · · · c r ∈ Z p \ { } ,G ( T ) , · · · , G r ( T ) ∈ F q ( T ) ∩ F q [[ T ]] and a pseudo-polynomial R ( T ) ∈ F q [[ T ]] such that: γ δ ( F ( T )) = R ( T ) + r X i =1 G i ((1 + T ) c i − . Then, there exists an integer n ≥ such that: (1 + T ) n ( F ( T ) + ( − δ F ((1 + T ) − − ∈ F q [ T ] . Proof
Note that if η, η ′ ∈ µ p − : η ≡ η ′ (mod Q ∗ ) ⇔ η = η ′ or η = − η ′ . Since r < ( p − / , by Proposition 3.3, there exists η ∈ µ p − such that: η δ F ((1 + T ) η −
1) + − η δ F ((1 + T ) − η −
1) is a pseudo − polynomial . Therefore: F ( T ) + ( − δ F ((1 + T ) − −
1) is a pseudo − polynomial . It remains to apply Lemma 3.5. ♦ Let F ( T ) ∈ Λ . We say that F ( T ) is a pseudo-rational function if F ( T )is the quotient of two pseudo-polynomials. For example, ∀ a ∈ Z p , ∀ b ∈ Z ∗ p , (1+ T ) a − T ) b − is a pseudo-rational function. We finish this section by giving ageneralization of [8], Theorem1: Theorem 3.7
Let δ ∈ Z / ( p − Z and let F ( T ) ∈ F q ( T ) ∩ F q [[ T ]] . Then Γ δ ( F ( T )) is a pseudo-rational function if and only if there exists some in-teger n ≥ such that: (1 + T ) n ( U ( F ( T )) + ( − δ U ( F ((1 + T ) − − ∈ F q [ T ] . Proof
Assume that Γ δ ( F ( T )) is a pseudo-rational function. Then , by 3) ofProposition 3.2 and Proposition 3.1, there exist c , · · · , c r ∈ F ∗ q , a , · · · , a r ∈ Z p , a i = a j for i = j, such that:Γ δ γ − δ U ( r X i =1 c i F ((1 + T ) κ ai − − polynomial . γ − δ U ( r X i =1 c i F ((1 + T ) κ ai − − polynomial . Set: G ( T ) = U ( F ( T )) + ( − δ U ( F ((1 + T ) − − ∈ F q ( T ) ∩ F q [[ T ]] . Now, by Proposition 3.3, there exist d , · · · , d ℓ ∈ F ∗ q , b , · · · b ℓ ∈ Z p , b i = b j for i = j, η , · · · , η ℓ ∈ µ p − , with ∀ i, j ∈ { , · · · , ℓ } , η i κ b i ≡ η j κ b j (mod Q ∗ ) , and η i κ b i = η j κ b j for i = j, such that: ℓ X i =1 d i G ((1 + T ) η i κ bi −
1) is a pseudo − polynomial . For i = 1 , · · · , ℓ, write: η i κ b i = η κ b x i , where x i ∈ Q ∗ ∩ Z ∗ p , and x i = x j for i = j. Since G ( T ) = ( − δ G ((1 + T ) − − , we can assume that x , · · · x ℓ are positives. Now, we get: ℓ X i =1 d i G ((1 + T ) x i −
1) is a pseudo − polynomial . Therefore, there exist N , · · · , N ℓ ∈ N \ { } , N i = N j for i = j, such that: ℓ X i =1 d i G ((1 + T ) N i −
1) is a pseudo − polynomial . Now, by Lemma 3.5, there exists some integer N ≥ T ) N ( ℓ X i =1 d i G ((1 + T ) N i − ∈ F q [ T ] . But, since G ( T ) ∈ F q ( T ) ∩ F q [[ T ]] , d , · · · , d ℓ ∈ F ∗ q , N , · · · N ℓ ∈ N \ { } and N i = N J for i = j, this implies that there exist some integer n ≥ T ) n G ( T ) ∈ F q [ T ] . ♦ Application to Kubota-Leopoldt p -adic L-functions Let θ be a Dirichlet character of the first kind, θ = 1 and θ even. Wedenote by f ( T, θ ) the Iwasawa power series attached to the p -adic L-function L p ( s, θ ) (see [9], Theorem 7.10). Write: θ = χω δ +1 , where χ is of conductor d, d ≥ d p ) , and δ ∈ Z / ( p − Z . Set κ = 1 + pd and K = Q p ( χ ) . We set: F χ ( T ) = P da =1 χ ( a )(1 + T ) a − (1 + T ) d . Let’s give the basic properties of F χ ( T ) : Lemma 4.1
1) If d ≥ , F χ ( T ) ∈ Λ .
2) If d = 1 , ∀ α ∈ Z / ( p − Z , α = 1 , γ α ( F χ ( T )) ∈ Λ . U ( F χ ( T )) = F χ ( T ) − χ ( p ) F χ ((1 + T ) p − .
4) If d ≥ , F χ ((1 + T ) − −
1) = εF χ ( T ) , wher ε = 1 if χ is oddd and ε = − if χ is even.5) If d = 1 , F χ ((1 + T ) − −
1) = − − F χ ( T ) . Proof d = 1 , we have: F χ ( T ) = − P p − a =0 (1 + T ) a − (1 + T ) p . Set: G ( T ) = (1 − (1 + T ) p ) γ α ( F χ ( T )) . Note that: ∀ η ∈ µ p − , − (1 + T ) p − (1 + T ) ηp ≡ η − (mod ω ( T )) . p − G ( T ) ≡ X η ∈ µ p − η α − p − X a =0 (1 + T ) ηa (mod ω ( T )) . Thus: ( p − G ( T ) ≡ X η ∈ µ p − η α − p − X b =0 (1 + T ) b (mod ω ( T )) . Since α = 1 , we get: G ( T ) ≡ ω ( T )) . Therefore γ α ( F χ ( T )) ∈ Λ .
3) For d = 1 , we have: U ( F χ ( T )) = P p − a =1 (1 + T ) a − (1 + T ) p = F χ ( T ) − F χ ((1 + T ) p − . Now, let d ≥ . Set q = κ = 1 + pd. Note that: F χ ( T ) = P q a =1 χ ( a )(1 + T ) a − (1 + T ) q . Therefore: U ( F χ ( T )) = P q a =1 , a p ) χ ( a )(1 + T ) a − (1 + T ) q . But: F χ ( T ) − χ ( p ) F χ ((1+ T ) p −
1) = P q a =1 χ ( a )(1 + T ) a − (1 + T ) q − χ ( p ) P da =1 χ ( a )(1 + T ) pa − (1 + T ) q . The Lemma follows easily. ♦ Lemma 4.2
Assume that d ≥ . The denominator of F χ ( T ) is φ d (1 + T ) where φ d ( X ) is the d th cyclotomic polynomial and the same is true for F χ ( T ) . Proof
Let ζ ∈ µ d . If ζ is not a primite d th root of unity, then, by [9], Lemma4.7, we have: d X a =1 χ ( a ) ζ a = 0 . ζ is a primitive d th root of unity, then by [9], Lemma 4.8, we have: d X a =1 χ ( a ) ζ a e π ) , where e π is any prime of K ( µ d ) . ♦ Lemma 4.3
The derivative of γ − δ ( F χ ( T )) is not a pseudo-polynomial mod-ulo π. Proof
We first treat the case d ≥ . By 3) and 4) of Lemma 4.1, Lemma 4.2and Proposition 3.6, γ − δ U ( F χ ( T )) is not a pseudo-polynomial. But observethat U = D p − . Thus Dγ − δ ( F χ ( T )) is not a pseudo-polynomial.For the case d = 1 . Set ^ F χ ( T ) = F χ ( T ) − F χ ((1 + T ) −
1) = 1 − T . Observe that:- ^ F χ ((1 + T ) − −
1) = 1 − ^ F χ ( T ) , - U ( ^ F χ ( T )) = ^ F χ ( T ) − ^ F χ ((1 + T ) p − . Therefore, as in the case d ≥ , γ − δ U ( ^ F χ ( T )) is not a pseudo-polynomial.Thus γ − δ U ( F χ ( T )) is not a pseudo-polynomial. And one can conclude as inthe case d ≥ . ♦ Lemma 4.4 Γ δ γ − δ ( F χ ( T )) = f ( 11 + T − , θ ) . Proof
We treat the case d = 1 , the case d ≥ T = e Z − . We get: γ − δ ( F χ ( T )) = X n ≥ , n ≡ δ (mod p − B n n ! Z n − . Thus, by [9], Theorem 5.11, we get: ∀ k ∈ N , k ≡ δ (mod p − , D k γ − δ U ( F χ )(0) = L p ( − k, θ ) . But, by Proposition 2.5,we have for s ∈ Z p :Γ δ γ − δ U ( F χ )( κ s −
1) = lim n D k n ( s,δ ) γ − δ U ( F χ )(0) = L p ( − s, θ ) = f ( κ − s − , θ ) . The Lemma follows. ♦ We can now state and prove our main result:17 heorem 4.5 f ( T, θ ) is not a pseudo-rational function.2) λ ( f ( T, θ )) < ( p − φ ( d )) φ ( p − , where φ is Euler’s totient function. Proof
1) Assume the contrary, i.e. f ( T, θ ) is a pseudo-rational function. Then f ( T − , θ ) is also a pseudo-rational function. Thus Γ δ γ − δ U ( F χ ( T )) is apseudo-rational function.We first treat the case d ≥ . By Theorem 3.7, there exists an integer n ≥ T ) n ( U ( F χ ( T )) + ( − δ U ( F χ ((1 + T ) − − ∈ F q [ T ] . This isa contradiction by 3) and 4) of Lemma 4.1 and Lemma 4.2.For the case d = 1 . We work with ^ F χ ( T ) = F χ ( T ) − F χ ((1 + T ) −
1) =1 − T . Then , by Proposition 3.2, Γ δ γ − δ U ( ^ F χ ( T )) is a pseudo-rationalfunction. We get a contradiction as in the case d ≥ .
2) Our proof is inspired by a method introduced by S. Rosenberg ([6]). Wefirst treat the case d = 1 . Note that we can assume that λ ( f ( T, θ )) ≥ . Now, by Lemma 4.3: µ ( γ − δ ( F χ ( T ))) = 0 . Futhermore, we have: γ − δ ( F χ )(0) ≡ π ) . Therefore, by 3) of Lemma 4.1, we get: λ ( γ − δ U ( F χ ( T ))) = λ ( γ − δ ( F χ ( T ))) . Therefore we have to evaluate λ ( γ − δ ( F χ ( T ))) . Set F ( T ) = − T . Since δ isodd, we have: γ − δ ( F χ ( T )) = γ − δ ( F ( T )) . Observe that F ((1 + T ) − −
1) = 1 − F ( T ) . Let S ⊂ µ p − be a set ofrepresentatives of µ p − / { , − } . We have:( p − γ − δ ( F ( T )) = 2 X η ∈ S η − δ F ((1 + T ) η − − X η ∈ S η − δ . Set: G ( T ) = ( Y η ∈ S ((1 + T ) η − γ − δ ( F ( T )) . µ ( G ( T )) = 0 , - λ ( G ( T )) = p − + λ ( γ − δ ( F ( T ))) . For S ′ ⊂ S, write t ( S ′ ) = P x ∈ S ′ x. We can write: G ( T ) = X S ′ ⊂ S a S ′ (1 + T ) t ( S ′ ) , where a S ′ ∈ O K . Set: N = Max { v p ( t ( S ′ ) − t ( S ′′ )) , S ′ , S ′′ ⊂ S, t ( S ′ ) = t ( S ′′ ) } . It is clear that: p N < ( p −
12 ) φ ( p − . But, by Lemma 2.2, we have: λ ( G ( T )) < p N +1 . Thus, by Propositon 3.2, we get: λ ( f ( T, θ )) = λ ( f ( 11 + T − , θ )) < p N < ( p −
12 ) φ ( p − . Now, we treat the general case, i.e. d ≥ . Again we can assume that λ ( f ( T, θ )) ≥ . Thus as in the case d = 1 , we get: λ ( γ − δ U ( F χ ( T ))) = λ ( γ − δ ( F χ ( T ))) . Now, by Lemma 4.2, we can write: F χ ( T ) = P φ ( d ) − a =0 r a (1 + T ) a φ d (1 + T ) , where r a ∈ O K for a ∈ { , · · · , φ ( d ) − } . Let again S ⊂ µ p − be a set ofrepresentatives of µ p − / { , − } . By Lemma 4.1, we have:( p − γ − δ ( F χ ( T )) = 2 X η ∈ S η − δ F χ ((1 + T ) η − . Set: G ( T ) = ( Y η ∈ S φ d ((1 + T ) η ))) γ − δ ( F χ ( T )) .
19e have: G ( T ) = φ ( d ) − X a =0 X η ∈ S X S ′ ⊂ S \{ η } X d =( d η ′ ) η ′∈ S ′ , d η ′ ∈{ , ··· ,φ ( d ) } b S ′ ,d (1 + T ) aη + P η ′∈ S ′ d η ′ η ′ , where b S ′ ,d ∈ O K . Note that again µ ( G ( T )) = 0 and that λ ( G ( T )) = λ ( γ − δ ( F χ ( T ))) . Now, for a, b ∈ { , · · · , φ ( d ) − } , η , η ∈ S, S ∈ S \ { η } ,S ∈ S \ { η } , set: V = aη + X η ∈ S d η η − bη − X η ∈ S d ′ η η, where ∀ η ∈ S , d η ∈ { , · · · , φ ( d ) } , and ∀ η ∈ S , d ′ η ∈ { , · · · , φ ( d ) } . If η = η then we can write: V = ( a − b ) η + X η ∈ S ′ u η η, where | u η |∈ { , · · · , φ ( d ) } and | S ′ |≤ p − . If η = η , we can write: V = a ′ η + b ′ η + X η ∈ S ′ u η η, where | a ′ | , | b ′ | , | u η |∈ { , · · · , φ ( d ) } , and | S ′ |≤ p − . Therefore, if V = 0 , we get: p v p ( V ) < ( p − φ ( d )) φ ( p − . Now, we can conclude as in the case d = 1 . ♦ Let E be a number field and let E ∞ /E be the cyclotomic Z p -extensionof E. For n ≥ , let A n be the p th Sylow subgroup of the ideal class groupof the n th layer in E ∞ /E. Then , by [9], Theorem 13.13, there exist µ p ( E ) ∈ N , λ p ( E ) ∈ N and ν p ( E ) ∈ Z , such that for all sufficiently large n : | A n | = p µ p ( E ) p n + λ p ( E ) n + ν p ( E ) . Recall that it is conjectured that µ p ( E ) = 0 and if E is an abelian numberfield it has been proved by B. Ferrero and L. Washington ([3]).20 orollary 4.6 Let F be an abelian number field of conductor N. Write N = p m d, where m ∈ N and d ≥ , d p ) . Then: λ p ( F ) < p − φ ( d )) φ ( p − . Proof
Set, for all n ≥ , q n = p n +1 d. Then F ⊂ Q ( µ q m ) . It is not difficult tosee that (see the arguments in the proof of Theorem 7.15 in [9]) : λ p ( F ) ≤ λ p ( Q ( µ q m )) . But, note that: λ p ( Q ( µ q m )) = λ p ( Q ( µ q )) . Now, by [9] Proposition 13.32 and Theorem 7.13: λ p ( Q ( µ q )) ≤ X θ even , θ =1 , f θ | q λ ( f ( T, θ )) . It remains to apply Theorem 4.5. ♦ Note that the bound of this latter Corollary is certainly far from thetruth even in the case p = 3 (see [4]). References [1] B. Angl`es, On some p -adic power series attached to the arithmetic of Q ( ζ p ) , J. Number Theory (2007), 221-246.[2] J. Coates and R. Sujatha, Cyclotomic Fields and Zeta Values, Springer-Verlag, 2006.[3] B. Ferrero and L. Washington, The Iwasawa invariant µ p vanishes forabelian number fields, Ann. of Math. (1979), 377-395.[4] J. Kraft and L. Washington, Heuristics for class numbers and lambdainvariants, Math. Comput. (2007), 1005-1023.[5] S. Lang, Cyclotomic fields I and II, Springer-Verlag, 1990.[6] S. Rosenberg, On the Iwasawa invariants of the Γ-transform of a rationalfunction, J. Number Theory (2004), 89-95.217] W. Sinnott, On the µ -invariant of the Γ-transform of a rational function,Invent. Math. (1984), 273-282.[8] W. Sinnott, On the power series attached to p -adic L -functions, J. reineangew. Math.382