The first layers of Z p -extensions and Greenberg s conjecture

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1 The first layers of Z p -extensions and Greenberg s conjecture Hiroki Sumida June 21, 2017 Technical Report No Introduction Let k be a finite extension of Q and p a prime number. We denote by k the cyclotomic Z p -extension of k and by k n its n-th layer. Let A n (k) be the p-sylow subgroup of the ideal class group of k n. We denote respectively by λ p (k) and µ p (k) the Iwasawa λ-invariant and µ-invariant associated to A n (k). Greenberg proposed a conjecture that λ p (k) = µ p (k) = 0 for all p and all totally real number fields k (see [8]). As for µ-invariants, Ferrero and Washington proved that µ p (k) = 0 for all p and all abelian number fields k in [3]. As for λ-invariants, several authors have given sufficient conditions for λ p (k) = 0 and verified it for small primes and real quadratic fields with small discriminants (cf. [2], [4], [5], [6], [8], [9], [10], [14], [15], [17], [19], [20]). In this paper, we study a criterion for λ p (k) = 0 given in [2] and [8]. We briefly explain it in a simple case. Let p be an odd prime number, k a real quadratic field and χ the nontrivial Dirichlet character associated to k. Let µ p be the group of p-th roots of unity and K = k(µ p ). Assume that p does not split in K. By [11], if A 0 (k) is trivial then λ p (k) = 0. So we consider the case that A 0 (k) = p, where A is the cardinality of a finite set A. Fix an ideal class c which generates A 0 (k) and denote by c n the natural image of c in A n (K). Put χ = χ 1 ω, where ω is the Teichmüller character Z/pZ Z p. Let A n = A n (c) be the set of all elements α n of K n satisfying a n c n, a p n = (α n ) and p α n M n (χ ), where M n (χ ) is the χ -part of the maximal abelian p-extension of K n unramified outside p (see 2). In [2] Partly supported by the Grants-in-Aid for Scientific Research, The Ministry of Education, Science and Culture, Japan Mathematics Subject Classification. 11R23 1

2 and [8], if part of the following proposition for p = 3 and n = 0 is proved and used for the verification of the conjecture. We will prove it in 3. Proposition 1. λ p (k) = 0 if and only if there is a Z p -extension of K n which contains p α n for some α n A n and some n 0. The main purpose of this paper is to determine whether the last statement (J n ) of Proposition 1 holds or not for each n. Put Γ = Gal(K /K) and denote by γ 0 the topological generator of Γ such that ζ γ 0 = ζ 1+p for all p n -th roots of unity ζ (n 0). Put A (k) = lim A n (k), where the inverse limit is taken with respect to relative norm maps. As usual, a Z p [[Γ]]-module A (k) can be considered as a Λ = Z p [[T ]]-module by the identification γ 0 = 1 + T. Denote by g χ (T ) Z p [T ] the Iwasawa polynomial associated to the p- adic L-function L p (s, χ) (see details in 2). If the degree of g χ (T ) is zero, A 0 (k) is trivial and the conjecture immediately holds. So, we consider the case that the degree of g χ (T ) is one. This is a simple but interesting case, because A 0 (k) is not always trivial. Then we can uniquely write g χ (T ) = T a χ for a χ pz p. Moreover there uniquely exists a positive integer or infinity m χ such that A (k) is Λ-isomorphic to Λ/(T a χ, p mχ ), where we define p = 0. Put e χ = v p (a χ ), a χ = p a χ and e χ = v p (a χ ), where 1 + a χ v p is the p-adic exponential valuation. In the above setting, we obtain the following theorem. Theorem 1. If m χ =, (J n ) does not hold for all n 0. Otherwise, (J n ) holds if and only if n m χ max{e χ, e χ }. By Iwasawa s formula ( 6 of [12]), we can easily compute e χ = v p (L p (1, χ)) and e χ = v p (L p (0, χ)). Greenberg s conjecture states that m χ is finite for all k. Some authors have given efficient methods not only to verify the finiteness of m χ but also to give an upper bound for m χ (see [10], [15], [17]). However, when m χ e χ is large, it is difficult to compute the real value m χ. 2 Main theorems Let p be an odd prime number and K a finite abelian extension of Q containing µ p. We fix an inclusion map Q Q p, where F is the algebraic closure of a field F. Put = Gal(K/Q), Γ = Gal(K /K) and G = Gal(K /Q). We first assume the following condition. (C1) The exponent of divides p 1. By (C1), the values of any character ψ of are contained in Z p. Put e ψ = 1 ψ(δ)δ 1 Z p [ ] the idempotent of the group ring Q p [ ]. For δ a Z p [ ]-module A, denote by A(ψ) the ψ-component e ψ A of A. 2

3 As in Introduction, we fix the topological generator γ 0 of Γ such that ζ γ 0 = ζ 1+p for all p n -th roots of unity ζ (n 0). We identify, as usual, the complete group ring Z p [[Γ]] with the power series ring Λ = Z p [[T ]] by γ 0 = 1 + T. Thus, for a Z p [[G ]]-module M, M(ψ) is regarded as a Λ-module, where we identify with Gal(K /Q ). Let A = A (K) = lim A n (K), where the inverse limit is taken with respect to relative norm maps. Then A (ψ) is regarded as a Λ-module in a natural way. Let L (n) be the an abelian extension of K n and normal over Q. Then, we can consider Gal(L (n) /K n ) as a G -module in natural manner. We denote by L (n) (ψ) the subfield of L (n) corresponding to Gal(L (n) /K n )(ψ ) ψ ˆ,ψ ψ in Galois theory. We denote by K (n) the composite of all Z p -extensions of K n, which is normal over Q. Let χ be an even character of and we regard χ as a primitive Dirichlet character. Put χ = χ 1 ω, where ω is the Teichmüller character Z/pZ Z p. Assume the following condition. (C2) χ(p) 1 and χ (p) 1. Further we set the third assumption concerning the Iwasawa polynomial. Let L p (s, χ) be the p-adic L-function associated to χ which is constructed by Kubota and Leopoldt in [16]. By [12], if χ is not the trivial character, there uniquely exists G χ (T ) Z p [[T ]] satisfying G χ ((1 + p) 1 s 1) = L p (s, χ) for all s Z p. In [3], it is shown that p does not divide G χ (T ). Therefore, by p-adic Weierstrass preparation theorem, we can uniquely write G χ (T ) = g χ (T )u χ (T ), where g χ (T ) is a distinguished polynomial of Z p [T ] and u χ (T ) is an invertible element of Z p [[T ]]. We call g χ (T ) the Iwasawa polynomial associated to L p (s, χ). Let M be the maximal abelian p-extension of K unramified outside p and L the maximal unramified abelian p-extension of K. We denote by M n (resp. L n ) the maximal abelian extension of K n in M (resp. L). By the Iwasawa main conjecture proved in [18], the characteristic ideal of Gal(M(χ)/K ) (resp. Gal(L(χ )/K )) is (g χ (T )) (resp. (g χ ( T ))), where (1+T )(1+ T ) = 1+p. By Theorem 18 of [13], Gal(M(χ)/K ) Gal(M/K )(χ) has no nontrivial finite Λ-submodule. Hence we have an injective Λ-homomorphism Y = Gal(M(χ)/K ) l Λ/(g i (T )) (1) i=1 with finite cokernel, where g χ (T ) = polynomial of Z p [T ]. l g i (T ) and g i (T ) is a distinguished i=1 3

4 By class field theory, we have X = Gal(L(χ)/K ) A (χ). Hence, if g χ (T ) = 1, then (1) implies that Y and A (χ) are trivial. As a next step, we assume the following condition. (C3) A (χ) 1 and g χ (T ) = T a χ for some a χ pz p. By (1) and (C3), we have Y Λ/(T a χ ). Put I = Gal(M(χ)/L(χ)) Gal(M/L)(χ). Then there exists a positive integer or infinity satisfying I = p mχ Y, i.e. A (χ) Λ/(T a χ, p mχ ). Fix an ideal class c which generates A 0 (χ)[p] the p-torsion subgroup of A 0 (χ). Put A n = {α n K n a n c n, a p n = (α n ) and p α n M n (χ )}, where c n is the natural image of c in A n (χ). We want to determine whether the following statements hold or not for each n 0. (J n ) p α n K (n) (χ ) for some α n A n. (I n ) p α 0 K (n) (χ ) for some α 0 A 0. Put e χ = v p (a χ ), a χ = p a χ 1 + a χ and e χ = v p (a χ ). Our main results are as follow. Theorem 1. Assume (C1), (C2) and (C3). If m χ =, (J n ) does not hold for all n 0. Otherwise, (J n ) holds if and only if n m χ max{e χ, e χ }. Theorem 2. Assume (C1), (C2) and (C3). If m χ =, (I n ) does not hold for all n 0. If m χ max{e χ, e χ }, (I n ) holds for all n 0. Otherwise, (I n ) holds if and only if n max{1, m χ e χ + 1}. In 4, we will prove these theorems. 3 Criteria for Greenberg s conjecture In this section, we deal with an odd prime p and a real quadratic field k. Then (C1) is satisfied for K = k(µ p ). Let χ be the nontrivial even character of = Gal(K/Q) and χ = χ 1 ω. Put A n = A n (K) and A = A (K). It is easy to see that A n (k) A n (χ) since A n (Q) is trivial for all n 0 (see [11]). Assume (C2) and (C3). The prime p does not split in K if and only if (C2) holds. The following statements (H n ) and (H n) hold for some n 0 if and only if λ p (k) = 0. 4

5 (H n ) Gal(M n /L n )(χ) is not trivial. (H n) Ker(i 0,n : A 0 A n )(χ) is not trivial, where i 0,n is induced by the natural inclusion k 0 k n. Using m χ and e χ, we can determine whether (H n ) (resp. (H n)) holds or not for each n 0 (resp. n 1) (see e.g. Proposition 3 of [10]). Proposition 2. Let k be a real quadratic field. Assume (C2) and (C3). If m χ =, (H n ) does not hold for all n 0. If m χ < e χ, (H n ) holds for all n 0. Otherwise, (H n ) holds if and only if n max{1, m χ e χ + 1}. For all n 1, (H n) is equivalent to (H n ). Remark 1. By Theorem 2 and Proposition 2, we see that (H 0 ) implies (I 0 )=(J 0 ). But (I 0 ) does not necessarily imply (H 0 ). For example, for p = 3 and k = Q( 443), (I 0 ) holds but (H 0 ) does not (see 5). In 4, we see that p eχ+n = Gal(M n /K )(χ) and p e χ +n = Gal(M n /K )(χ ) tor, where A tor is the maximal torsion subgroup of a Z p -module A. So e χ (resp. e χ ) is a important invariant of the χ-part (resp. χ -part) of the maximal abelian p- extension of K unramified outside p. These invariants make an interesting contrast in Theorem 2 and Proposition 2 when m χ is sufficiently large. From now, we prove Proposition 1, which give a relation between (J n ) and (H n). Assume (H n) holds, then for α n A n there exists some β n K n such that (α n ) = (β p n). Therefore, p α n K n ( p E n ) = K n ( p ε n ε n E n ), where E n = E n (K) is the group of units of K n. Conversely, if p α n K n ( p E n ), then we have α n = ε n β p n for some ε n E n and β n K n. By (C1), (N Kn/k n α n ) = (N Kn/k n β n ) p implies (H n). Therefore we can write (H n) p α n K n ( p E n )(χ ) for some α n A n. Note that (J n ) p α n K (n) (χ ) for some α n A n. By the argument of [8, p. 273], if K satisfies (C1) and (C2) we have K m ( pm+1 Em )(χ ). (2) n 0 K (n) (χ ) = m 1 Hence (J n ) holds for some n 0 if and only if (H m) holds for some m 1. By Theorem 1 of [8], when A 0 = p, (H n) holds for some n 1 if and only if λ p (k) = 0. Therefore Proposition 1 follows. Remark 2. In the above proof, we can write 5

6 (H n) p α 0 K n ( p E n )(χ ) for some α 0 A 0. Note that (I n ) p α 0 K (n) (χ ) for some α 0 A 0. By (2), we see that (I n ) holds for some n 0 if and only if (H m) holds for some m 1. 4 Proof of theorems Let the notation be the same as in 2. For a Λ = Z p [[Gal(K /Q)]]-module Z, we define a Λ -module Ż by { Ż = Z as a Z p -module, σ x = ϵ(σ)σ 1 x for σ Gal(K /Q) and x Ż, where ϵ is the cyclotomic character. Let E n (K) be the group of units K n and E n(k) the group of p-units of K n. We put E(K) = E n(k), E (K) = n 0 n 0 E n(k), N = K n+1 n( p En ) and N = K n( pn+1 E n 0 n 0 n ). Further put Γ n = Gal(K /K n ). Using the arguments of [13], we obtain the following lemma (cf. [8, p. 274]). For convenience of readers, we will give the outline of the proof. Theorem 3. Assume (C1) and (C2). There is an injective Λ -homomorphism whose cokernel is Λ -isomorphic to Gal(N (χ )/K ) Λ (χ ) Hom(H 1 (Γ n, E (K)), Q p /Z p )(χ ) = Hom(H 1 (Γ n, E (K))(χ), Q p /Z p ) for all sufficiently large n, where σf(x) = f(σx) for σ Gal(K /Q), f Hom(H 1 (Γ n, E (K)), Q p /Z p ), x H 1 (Γ n, E (K)). Proof. (Outline) Put Z = (Gal(N (χ )/K ) ) Gal(N /K )(χ) and D = (E (K) Z Q p /Z p )(χ). Then we have an orthogonal pairing (cf. [13, p. 280]) Z D Q p /Z p. Let Z n denote the annihilator of (E n Z Q p /Z p )(χ) in Z: Z n = (E n Q p /Z p )(χ). We have ω n Z = (D Γn ), where ω n = (1 + T ) pn 1 and D Γn = {d D d γ = d for all γ Γ n }. By (C2), we have (E n Q p /Z p )(χ) (Q p /Z p ) pn and Z/Z n Z pn p. 6

7 By this fact and the argument of [13, p. 281], we have an injective Λ- homomorphism Z Λ (χ) whose cokernel is finite. Put Z = Λ (χ) and identify Z with the image of the above map. By the arguments of [13, p. 283], Z /Z ω n Z /ω n Z = (Z ω n Z )/ω n Z = Z n /ω n Z for all sufficiently large n. On the other hand, by Lemma 7 of [13], we have D Γn /(E n Q p /Z p )(χ) H 1 (Γ n, E (K))(χ). Since Z n /ω n Z Hom(D Γn /(E n Q p /Z p )(χ), Q p /Z p ), the theorem follows. Remark 3. Let I p,n be the group generated by ideals of K n all of whose factors lie over p. By (C2), we see that (I p,n Z Q p /Z p )(χ) is trivial for all n. Hence we have N(χ ) = N (χ ), H 1 (Γ n, E(K))(χ) = H 1 (Γ n, E (K))(χ) and A n (χ) A n(χ), where A n is the p-sylow subgroup of the p-ideal class group of K n. In the following, we assume (C1), (C2) and (C3). Lemma 1. If m χ is finite, the order of the maximal Z p -torsion subgroup of Gal(M n (χ )/K n ) is min{m χ, n + e χ }. Proof. Let c n be an ideal class which generates A n (χ)[p n+1 ]. Then there exists an element α n of k n such that c n c n, c pn+1 n = (α n) and pn+1 α n M n (χ ). By Lemma 9 of [13], we have M(χ ) = n 0 K n( pn+1 En, pn+1 α n)(χ ). Hence the finiteness of m χ implies M(χ ) = N (χ ) (see 3). Therefore by Theorem 3, we have Gal(M(χ )/K ) ( T a χ, p mχ ) (T a χ, p mχ ) as a Λ-module. The maximal Z p -torsion submodule of Λ/ω n (T ȧ χ, p mχ ) is generated by the class of ω n, and its order is p mχ. Since ω n (T a χ, p mχ ) = p max{0,mχ e χ n} ω n, the lemma follows. Let p be a prime ideal of K over p and p n the unique prime ideal of K n over p = p 0. Denote by K pn the completion of K n at p n and by U pn the group of principal units of K pn. We put U n = U pn. Denote by E n the closure of p p the image of E n (K) q 1 under the diagonal embedding En q 1 U n, where q is the cardinality of the residue class field of a prime ideal of K above p. The Leopoldt conjecture proved in [1] implies that E n /En pa E n /En pa for any a 1 (see 5.5 of [21]). Put U = lim U n and E = lim E n, where the inverse limit is taken with respect to relative norm maps. By Theorem 1 of [7], U(χ) is Λ-isomorphic to Λ. Fix an isomorphism φ, then we naturally obtain an isomorphism φ n : U n (χ) Λ/(ω n ) induced by φ. 7

8 Lemma 2. φ n (E n (χ)) = (T a χ, p max{0,n mχ+eχ} )/(ω n ). Proof. By Theorem 2 of [7], φ n (E n ) includes (T a χ, ω n )/(ω n ). By Lemma 3 (ii ) of [9], we have U n (χ)/e n (χ) Gal(M n (χ)/l n (χ)). Here Gal(M n (χ)/k ) Y/ω n Y Z/p n+eχ Z and Gal(L n (χ)/k ) X/ω n X Z/p min{n+eχ,mχ} Z, where Y = Gal(M(χ)/K ) and X = Gal(L(χ)/K ). Hence the lemma follows. For s max{n, t n }, put W s,n = φ s (E s (χ))/p tn+1 φ s (E s (χ)), where t n = min{m χ, n + e χ }. Further, put x p s = [p max{0,s mχ+eχ} ] and x i = [T ps i 1 (T a)] for 1 i p s 1, where [f(t )] is the class of f(t ) in W s,n. We denote by W s,n [ ω n ] the kernel of the map W s,n W s,n (x ω n x), where ω n = ω n ( T ) = (1 + T ) pn 1. Since v p (ω n (a χ )) = n + e χ and v p (ω n (a χ )) = n + e χ, we can uniquely write ω n (a χ ) = u n p n+eχ and ω n (a χ ) = u np n+e χ for u n, u n Z p. Lemma 3. Assume m χ < and s m χ e χ. If n < m χ e χ, p tn (W s,n [ ω n ]) y 1, y 2,..., y p n 1, y p n. If n = m χ e χ, p tn (W s,n [ ω n ]) y 1, y 2,..., y p n 1, y p n u s If n > m χ e χ, p tn (W s,n [ ω n ]) y 1, y 2,..., y p n 1, y p s. u n y p s. Proof. For 1 i p s, let c i be elements of Z p such that [T (T a χ )T ps 2 ] = p s [(T a χ )T ps 1 ω s ] = c i x i. Then we easily see that c i pz p and i=1 c p s = ω s (a χ )/p s mχ+e. Put c c A = c p s c p s p s mχ+eχ c p s a χ the matrix of (p s, p s )-type. Let A l = (E +A) l (1+p) l E = (a (l) i,j ), where E is the matrix unit. Since v p (c p s) = m t n, we have a (l) p s,j lc j c p s mod p tn+1 for 1 j l, a (l) p s,j = 0 for l + 1 j ps 1 and a (l) p s,p s = (1 + a χ) l (1 + p) l. Let b 1 x 1 +b 2 x b p sx p s W s,n [ ω n ] for b j Z p. Here we have ω n = (1+ T ) pn 1 = (1 + T )pn (1 + p) pn. Since p divides p nc j for 1 j p n 1, (1 + T ) pn b p n and b p s satisfy c p sb p n +((1+a χ ) pn (1+p) pn )b p s 0 mod p tn+1. As (1+ ( ) 1 + p p n a χ ) pn (1+p) pn = ((1+a χ ) pn 1), p mχ u s b p n+p n+e χ u 1 + a nb p s χ 8

9 0 mod p tn+1. Since p tn (W s,n [ ω n ]) y 1, y 2,..., y p n 1, y p n, y p s, this equation implies the lemma. We put ν s,n = ω s /ω n. Lemma 4. Assume m χ < and s m χ e χ. If n < m χ e χ, p tn [ν s,n ] = y p n. If n = m χ e χ, p tn [ν s,n ] = y p n + u s u n y p s. If n > m χ e χ, p tn p n mχ+eχ [ν s,n ] = y p s and p tn [(T a χ )ν s,n ] = y p n 1. p s p n 2 Proof. If n m χ e χ, we have p tn [ν s,n ] = p tn [(T ps p n 1 + d j T j )(T j=0 a χ ) + ω s(a χ ) ω n (a χ ) ] = [ptn T ps p n 1 (T a χ ) + p tn p u mχ n eχ s p s mχ+eχ ], where u n d j pz p. Therefore the first and second assertions follow. If n > m χ e χ, we have p tn p n mχ+eχ [ν s,n ] = p tn p n mχ+eχ [(T ps p n 1 + p s p n 2 d j T j )(T a χ ) + ω s(a χ ) ω n (a χ ) ] = [ u s p tn p s mχ+eχ ], where d j pz p. Fur- u n ther we have p tn [(T a χ )ν s,n ] = [p tn (T a χ )T ps p n ]. Therefore the last assertion follows. Let K (n),k be the composite of all k-th layers of Z p -extensions of K n. Proposition 3. Assume (C1), (C2) and (C3). If m χ <, K (n),1 (χ ) = K n ( p E n )(χ ) if and only if n < m χ max{e χ, e χ } or n > m χ min{e χ, e χ }. If m χ =, K (n),1 (χ ) = K n ( p E n )(χ ) for all n. Proof. Assume m χ <. Put D s,n = (E s(k)/e s(k) ptn +1 )(χ) = (E s /Es ptn+1 )(χ) and Z s,n = Gal(K s ( ptn+1 E s)(χ )/K s ) = Gal(Ks ( ptn+1 Es )(χ )/K s ) (see Remark 3). Then we have an orthogonal paring j=0 Z s,n D s,n Z/p tn+1 Z. (3) By Lemma 2, we identify D s,n with W s,n. By (2), we can take an integer s = s n such that K s ( ptn+1 Es )(χ ) includes K (n),tn+1(χ ). Then, by (3), the subgroup of W s,n corresponding to K s K (n),1 (χ ) is p tn (W s,n [ ω n ]). On the other hand, we have φ s (i n,s(u n )) = [ν s,n φ n (u n )] Λ/(ω s ) for u n U n (χ), where i n,s is the natural inclusion U n (χ) U s (χ). Hence by Lemma 3, the subgroup of W s,n corresponding to K s ( p E n )(χ ) is the image of p tn ν s,n (T a χ, p max{0,n mχ+eχ} ) in W s,n. We have v p ((1 + p) pn ) = n + 1 and (1 + p) pn = (1 + a χ ) pn (1 + a χ ) pn = (1 + u n p n+eχ )(1 + u np n+e χ ). Hence u n u n mod p if e χ = e χ (= 1). Using Lemma 3 and Lemma 4, we obtain the first assertion. 9

10 Similarly we can prove the last assertion. Assume m χ =, then E s (χ) Λ/(ω s ) for all s 0. As ω n T pn mod p, we have (1 + T ) pn (E s (χ) ps /E s (χ) ps+1 )[ ω n ] (p s T ps p n, p s+1, ω s )/(p s+1, ω s ). Since p s ν s,n p s T ps p n mod p s+1, (2) implies the last assertion. Let M n,1 be the subfield of M n corresponding to pgal(m n /K n ). Then we have M n,1 (χ ) = K n ( p E n, p α n )(χ ) for any α n A n. By the argument of 3, p α n K n ( p E n )(χ ) if n m χ e χ. Hence for n m χ e χ, (J n ) holds if and only if K s K (n),1 (χ ) K s K n ( p E n )(χ ). By Proposition 3, we obtain Theorem 1. Assume (J 0 )=(I 0 ) does not hold. For n 1, (I n ) holds if and only if K s ( p E 0, p α 0 )(χ ) K s K (n),1. Here the subgroup of W s,n corresponding to K s ( p E 0, p α 0 )(χ ) is the image of y 1, y p s. Hence by Lemma 3, we obtain Theorem 2. 5 Numerical examples Let p = 3 and k = Q( m) with 1 < m < 10 4 satisfying (C2) and (C3). Put n H = n H (k) = min{n (H n ) holds for n}, n I = n I (k) = min{n (I n ) holds for n} and n J = n J (k) = min{n (J n ) holds for n}. In the above range, an upper bound m χ for m χ is given for each k (see [10]). On the other hand, by Fukuda s computation (cf. [5]), it is shown that m χ m χ for m 2 mod 3. Using Proposition 2, Theorem 1 and Theorem 2, we give n H, n I and n J in the following table. For m 0 mod 3, the values are upper bounds. 10

11 m e χ e χ m χ n H n I n J

12 m e χ e χ m χ n H n I n J

13 m e χ e χ m χ n H n I n J

14 m e χ e χ m χ n H n I n J

15 m e χ e χ m χ n H n I n J

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