On Kervaire--Murthy conjecture, Bernoulli and Iwasawa numbers, and zeroes of $p$-adic $L$-function

On Kerv aire–Murth y conjecture, Bernoulli and Iw asa w a n um b ers, and zero es of p -adic L - fu nction A. Stolin Department of Mathemat ical Sciences Chalm ers Univ ersi t y of T echnology and Universit y of Gothenburg Gothenburg, Sweden E-mai l : alexand er .stolin@g u. se Abstract The aim of the p resen t pap er is to establish relations b et we en Iw asaw a and Bernoulli num b ers based on some results b y M. Kerv aire and M. P . Mu r th y ab out the structur e of the K 0 groups of the in teger group rings of cyclic groups of prime p o we r order p n . In particular, we will pr o ve that • λ i ≤ p − 1 under assumption t h at the generalized Bernoulli n u m - b er B 1 ,ω − i is not d ivisible b y p 2 . Here ω is the T eic hm ¨ uller c har- acter of Z / ( p − 1) Z . • λ i = 1 if B 1 ,ω − i is divisib le b y p 2 . • W e will pr o ve that S n,i ∼ = Z / ( p n + k i ), where S n is the Sylow p -subgroup of the class group of the field Q ( ζ n ). Here, ζ n is a primitiv e p n +1 -ro ot o f unity , ε i are idemp oten ts in the group ring Z p [Gal( Q ( ζ 0 ) / Q )], S n,i = ε i ( S n ), and k i is the p -adic v aluation of B 1 ,ω − i . • A t the end w e w ill pro ve that k i ≤ 1 and also v p ( L p (0 , ω j )) ≤ 1 for even j under certain conditions on zero es of L p (0 , ω j ) . • Through ou t th e pap er w e assu m e that p satisfies V andiv er’s con- jecture. 1 MSC 2000 Primary 11R23, 11R29. Secondary 19A31. 1 In tro duc tion Let C n denote the cyclic group of o rder p n , where p is an o dd prime. Let Z C n b e the in tegral group ring o f C n . In this pap er we study Pic Z C n and some other groups related to it, in particular, the ideal class group C ( F n ) of the cyclotomic field F n = Q ( ζ n ), where ζ n is a primitiv e p n +1 -st ro ot of unity . Throughout this pap er w e assume that p is semi-regular, that is p do es not divide the order of the ide a l class group of t he maximal real subfield F + 0 = Q ( ζ 0 + ζ − 1 0 ) in F 0 . Let A b e an ab elian group. The follo wing notatio n will b e used in our pap er: • F n has b een already defined, F + n = Q ( ζ n + ζ − 1 n ); • F p = Z /p Z ; • N · A (or sometimes A N , if it is clear fr o m a con text) is the direct sum of N copies of A ; • dA or A d (dep ending on additiv e or multiplic a t iv e o p eration on A ) stands for the subgroup of A whic h consists of t he elemen ts of the form da or a d ; • A ( d ) stands fo r the subgroup o f A which consists of the elemen ts of A suc h that d a = 0 or a d = 1; • A ( p ) denotes the Sylow p - component of A . F or A = C ( F n ) w e use a sp ec ia l notation C ( F n ) ( p ) = S ( F n ) = S n ; • if R is a comm utative ring, then U ( R ) denotes the group o f units of R . • in the sp ecial case R = Z [ ζ n ], we use E n for U ( Z [ ζ n ]); • further, we use notatio n E n,k for the subgroup of E n consisting of units whic h are congruent to 1 mo dulo µ k n = (1 − ζ n ) k . 2 F ollo wing [3] let us consider the fibre pro duct diagram Z C n +1 i 2 / / i 1   Z [ ζ n ] j 2   Z C n j 1 / / F p [ x ] ( x − 1) p n := R n with ob vious maps i 1 , i 2 , j 1 , j 2 . The corresp onding Ma y er-Vietoris exact se- quence can b e written as f ollo ws: U ( Z C n ) × E n j − → U ( R n ) − → Pic( Z C n +1 ) − → Pic( Z C n ) × C ( F n ) − → 0 . One of the main problems in computing Pic( Z C n +1 ) is th us to ev aluate the cok ernel W n of the map j : U ( Z C n ) × E n − → U ( R n ) Instead of W n w e will ev aluate a bigger gro up V n = Cok er { j 2 : E n − → U ( R n ) } . Clearly , W n is a factorgroup of V n . In the calculation of V n a decisiv e role will b e play ed b y the action G n = Gal( F n / Q ) o n the v a rious rings inv o lv ed in the pap er. Let δ : G n − → U ( Z /p n +1 Z ) b e the canonical isomorphism defined by s ( ζ n ) = ζ δ ( s ) n , s ∈ G n . W e will de note by x n the generator in Z [ x ] / ( x p n − 1 ) = Z C n and in F p [ x ] / ( x − 1) p n = R n that corresp onds to x . Since δ ( s ) is an intege r mo dulo p n +1 , prime to p , it is clear that b oth x δ ( s ) n +1 and x δ ( s ) n are w ell-defined. Moreo v er, t he maps in the fibre pro duct ab o v e comm ute with the action of G n . Let c ∈ G n b e the complex conjugation. It is clear that V n = V + n × V − n , where V + n consists of elemen ts suc h that c ( a ) = a and V − n consists of elemen ts suc h that c ( a ) = a − 1 (w e take in to account that V n is a p -group). Similarly , W n = W + n × W − n . F or an y ab elian group A , let us denote by A ∗ the group of characters of A . The ma in results pro v ed b y Kerv aire and Murth y in [3 ] w as Theorem 1.1. If p is a semi-r e gular o dd prime, then ( W + n ) ∗ ⊆ ( V + n ) ∗ ⊆ S − ( F n − 1 ) = S ( F n − 1 ) =: S n − 1 . In other wor ds, ther e is a surje ction S ∗ n − 1 → V + n 3 They a lso conjectured that, in fact, W + n ∼ = V + n ∼ = S ∗ n − 1 . T he first main result of our pap er is a we a k v ersion of the Kerv aire and Murth y conjecture, namely ( S n − 1 ) ( p ) ∼ = ( V + n / ( V + n ) p ) ∗ = (( V + n ) ∗ ) ( p ) Another imp ortan t result prov ed in this pap er (whic h giv es a new link b e- t w een the class groups and the groups V n ) is that there exists a canonical em b edding S ( p ) n − 1 → V − n / ( V − n ) p W orking on the Kerv aire and Murth y conjecture, Ullo m pro ved in [7] that under certain assumptions o n the Iw asaw a num b ers λ i explained later, the group W + n can b e described as follows: W + n ∼ = r 0 · ( Z /p n Z ) ⊕ ( λ − r 0 ) · ( Z /p n − 1 Z ) . Here r 0 = dim F p ( S 0 ) ( p ) = dim F p ( S 0 /S p 0 ) , λ = X λ i . Notice that r 0 also coincides with the n umber of Bernoulli n um b ers among B 2 , B 4 , . . . , B p − 3 whic h are divisible b y p . The Iw asa wa in v arian t λ can b e defined as follows. It is w ell-know n due to Iw asaw a and W ashington (see [8]) that there exist tw o num b ers λ and ν called Iw asa wa in v ariants such that S n has p λn + ν elemen ts for sufficien tly large n . Ullom’s pro of is ba s ed on certain assumptions ab out the Iw asa wa n umber λ . More exactly , G 0 = Gal( F 0 / Q ) ∼ = Z / ( p − 1) Z acts on S n and S n = p − 2 M i =0 S n,i , where S n,i = ε i S n and ε i are idemp oten ts in the group ring Z p [ G 0 ]. Since we w ork with semi-regular p , ε i S 0 ∼ = Z p /B 1 ,ω − i Z p for i = 3 , 5 , . . . , p − 2 . Here B 1 ,ω − i are generalized Bernoulli n um b ers and ω is the T eich m ¨ uller c haracter of Z / ( p − 1 ) Z (see [8]). F urthermore, for each i there exist λ i and ν i suc h that S n,i con tains p λ i n + ν i elemen ts. Ullom’s assumption w as that λ i < p − 1 and he conjectured that 4 it w as true for an y p . In this pap er w e will pro v e that λ i ≤ p − 1 under assumption that B 1 ,ω − i is not divisible by p 2 . T hen w e will pro ve that λ i = 1 if B 1 ,ω − i is divisible b y p 2 that provides almost a complete pro of of Ullom’s inequalit y under the assumption that V andiver’s conjecture is tr ue. Remark 1.2. If G 0 acts on an a belian p -group X , then X = L p − 2 i =0 X i with X i = ε i X . In our pap er we will need the following presen tation of S n,i (see [8] for details). Let ω b e the T eichm¨ uller c haracter and P n ( T ) = ( T + 1) p n − 1. Let f i ( T ) ∈ Z p [[ T ]] b e defined by the relation f i ((1 + p ) s − 1) = L p ( s, ω 1 − i ) , where L p is the p - adic L-function. In this terms the Iw a s aw a n um b er λ i is the first co efficien t of f i ( T ), which is not divisible by p . By the p -adic W eierstrass Preparation Theorem f i ( T ) = U ( T ) p i ( T ), where U ( T ) is an inv ertible elemen t of Z p [[ T ]] suc h that U (0 ) = 1 and p i ( T ) is a unique p olynomial of degree λ i with the leading co efficien t co-prime to p and all o t her co efficien ts divisible b y p . Then S n,i = Z p [[ T ]] / ( f i , P n ) = Z p [ T ] / ( p i ( T ) , P n ( T )) . 2 Second presentation of V n and norm maps The following lemma w as pro ved in [4]. Lemma 2.1. L et A n = Z [ x ] / ( x p n − 1 x − 1 ) . Then Pic Z C n ∼ = Pic A n F rom now on w e will study A n instead of Z C n . Clearly , we ha ve the follo wing fibre pro duct: A n +1 i 2 / / i 1   Z [ ζ n ] j 2   A n j 1 / / F p [ x ] ( x − 1) p n − 1 := R ′ n (1) Lemma 2.2. Cok er { j 2 : Z [ ζ n ] → U ( F p [ x ] / ( x − 1) p n − 1 ) } ∼ = V n . Pr o of. W e ha ve to pro ve that Cok er( U ( Z [ ζ n ]) → U ( F p [ x ] / ( x − 1) p n )) = 5 Cok er( U ( Z [ ζ n ]) → U ( F p [ x ] / ( x − 1) p n − 1 )) . Clearly , it is sufficien t to prov e that the elemen t 1 + ( x − 1 ) p n − 1 ∈ U ( F p [ x ] / ( x − 1) p n ) is the image of some unit of Z [ ζ n ]. It is easy to see that the image of the unit  ζ p n +1 n − 1 ζ n − 1  under the map Z [ ζ n ] → F p [ x ] / ( x − 1) p n , ζ n → x is exactly 1 + ( x − 1 ) p n − 1 , a nd the pro of is complete. Remark 2.3. This lemma justifies an abuse of notatio n j 1 , j 2 , i 1 , i 2 , R ′ n in (1). The map N n : Z [ ζ n ] → A n suc h that N n ( ab ) = N n ( a ) N n ( b ) and the diagram b elo w is comm utative has b een introduced in [5]: A n +1 i 2 / / i 1   Z [ ζ n ] j 2   N n { { ✇ ✇ ✇ ✇ ✇ ✇ ✇ ✇ ✇ A n j 1 / / R n (2) W e would lik e to remin d the reader this construction. T he follo wing fibre pro duct diagra m can b e used for the construction without lost of generality: Z p [ x ] / ( x p n +1 − 1 x − 1 ) i 2 / / i 1   Z p [ ζ n ] j 2   Z p [ x ] / ( x p n − 1 x − 1 ) j 1 / / R n W e construct N n using induction. If n = 1 , then Z p [ x ] / ( x p n − 1 x − 1 ) ∼ = Z [ ζ 0 ] and N 1 is the usual norm map. Comm utativit y of (2) w as prov ed in [4]. The form ula ϕ 1 ( a 1 ) = ( a 1 , N 1 ( a 1 )) ∈ Z p [ x ] / ( x p 2 − 1 x − 1 ) defines a n injectiv e homomorphism ϕ 1 : U ( Z [ ζ 1 ]) → U ( Z p [ x ] / ( x p 2 − 1 x − 1 )). No w w e can define N 2 ( a 2 ) = ϕ 1 (Norm F 2 /F 1 ( a 2 )). 6 Sim ultaneously , N 2 defines ϕ 2 : U ( Z p [ ζ 2 ]) → U ( Z p [ x ] / ( x p 3 − 1 x − 1 )) via ϕ 2 ( a 2 ) = ( a 2 , N 2 ( a 2 )) ∈ Z p [ x ] / ( x p 3 − 1 x − 1 ), and so on. Pro ofs that all of the maps ϕ i , N i are w ell-defined can b e found in [5 ]. They use rings A n,k = Z [ x ] / ( x p n + k − 1 x k − 1 ). Prop osition 2.4. F ormula ϕ n − 1 ( a n − 1 ) = ( a n − 1 , N n − 1 ( a n − 1 )) define s an e m- b e dding E n − 1 → U ( Z [ x ] / ( x p n − 1 x − 1 )) , and Cok er { j 1 : E n − 1 → U ( R n ) } ∼ = V n . Pr o of. Since we deal with semi-regular primes, the fact we need follo ws from that of Norm F n /F n − 1 ( E n ) = E n − 1 and th us, j 2 ( E n ) = j 1 ( E n − 1 ) in U ( R n ). Let us denote b y U n,k the subgroup of U ( Z p [ ζ n ]) := U n , whic h consists of units congruent to 1 mo dulo ( ζ n − 1) k = µ k n . Theorem 2.5. We have V n ∼ = U n / ( U n,p n − 1 · E n ) ∼ = U n − 1 / ( U n − 1 ,p n − 1 · E n − 1 ) Remark 2.6. W e remind the reader that V n ∼ = U n / ( U n,p n · E n ) b y definition. Pr o of. The first isomorphism is clear. Let us prov e that V n ∼ = U n − 1 / ( U n − 1 ,p n − 1 · E n − 1 ). The form ula ϕ n − 1 ( a ) = ( a, N n − 1 ( a )) defines an em b edding ϕ n − 1 : U n − 1 → U ( Z p [ x ] / ( x p n − 1 x − 1 )). It is sufficien t to prov e that the comp osition map ϕ n − 1 · j 1 has the k ernel U n − 1 ,p n − 1 . T o do this, first w e note tha t U ( R n ) and U n − 1 /U n − 1 ,p n − 1 ha v e the same n um b er of elemen ts. Therefore, it is enough to pro ve that U n − 1 ,p n − 1 is con tained in the k ernel. T his w as pro ve d in [5]. W e w ould lik e to demonstrate the case n = 2. F or this, w e should pro ve that ( a, Norm F 1 /F 0 ( a )) ≡ (1 , 1) mo d( p ) in Z p [ x ] / ( x p 2 − 1 x − 1 ) if a ≡ 1 mo d µ p 2 − 1 1 . It is easy to see that ( a, Norm F 1 /F 0 ( a )) ≡ (1 , 1) mo d p is equiv alen t to that of Norm F 1 /F 0 ( a − 1 p ) ≡ Norm F 1 /F 0 ( a ) − 1 p mo d p in Z p [ ζ 0 ]. Since a ≡ 1 mo d µ p 2 − 1 1 , b oth sides are congruen t to 0 mo dulo p . The general case w as pro ve d in [5] using the rings A n,k and induction in n, k . Remark 2.7. In fa ct, it is not difficult to pro ve that ( a, Norm F 1 /F 0 ( a )) ≡ (1 , 1) mo d( p ) in Z p [ x ] / ( x p 2 − 1 x − 1 ) iff a ≡ 1 mo d µ p 2 − 1 1 . 7 In the sequel w e will need the following Corollary 2.8. Supp ose a ∈ U 2 is such that a ≡ 1mo d µ p 2 − 1 2 . The n Norm F 2 /F 1 ( a ) ≡ 1mo d µ p 2 − 1 1 . Pr o of. Consider the dia gram 2 f o r n = 2. Then N 2 ( a ) = (Nor m F 2 /F 1 ( a ) , Norm F 2 /F 0 ( a )) ≡ (1 , 1)mo d( p ) in Z p [ x ] / ( x p 2 − 1 x − 1 ). Consequen tly , Norm F 2 /F 1 ( a ) ≡ 1 mo d µ p 2 − 1 1 . 3 Num b er of elemen ts in V + n Let us in tro duce integers r n as the n umber of elemen ts in E n,p n +1 − 1 /E p n,p n +1 . Similarly , let r n,i b e the num b er of elemen ts in ε i ( E n,p n +1 − 1 /E p n,p n +1 ). In particular, it follows that r n = P r n,i , r 0 ,i = 1 if λ p − i > 0, otherwise r 0 ,i = r k ,i = 0. Lemma 3.1. If ǫ ∈ E n,p n +1 , then ǫ is r e al and ther efor e, E n,p n +1 = E + n,p n +1 . Theorem 3.2. L et α b e an ide al of Z [ ζ n ] s uch that α p = ( q ) . L et q ≡ 1 mo d µ p n +1 − 1 n . Then q ≡ 1 mo d µ p n +1 n . Before w e g iv e a pro of of the theorem, let us form ulate its consequence, whic h w e will need in sequel. Corollary 3.3. E n,p n +1 − 1 = E n,p n +1 +1 . Pr o of of The or em 3.2. Consider the extension F n ( p √ q ) /F n . Only µ n ramifies in this extension. Let ǫ ∈ E n . Then for any v a luation v 6 = µ, ǫ is a norm in the corresponding extens ion of lo cal fields F n,v ( p √ q ) /F n,v . Therefore, the lo cal norm residue sym b ol with v alues in the group of p -th ro ots of unit y ( ǫ, q ) v = 1. By the pro duct form ula, ( ǫ, q ) µ n = 1. Set ǫ = ζ n . If q ≡ 1 mo d µ p n − 1 n but q 6 = 1 mo d µ p n n , then simple lo cal computations (see for instance [1]) show that ( ζ n , q ) µ n 6 = 1. The theorem is prov ed. Theorem 3.4. The numb er of elemen ts in V + n is p r 0 + ... + r n − 1 . Pr o of. If n = 1, then it w as prov ed in [3]. Let us denote the num b er of ele- men ts in group A by | A | . Assume that | V + n | = p r 0 + ... + r n − 1 . Let us prov e that |V + n +1 | = p r 0 + ... + r n − 1 + r n . Indeed, | ( U n / ( U n,p n · E )) + | = p r 0 + ... + r n − 1 . Clearly , 8 ( U n / ( U n,p n · E )) + = U + n / ( U + n,p n · E + ) and U + n,p n = U + n,p n +1 since p is o dd. T aking into accoun t that V + n +1 ∼ = U + n /U + n,p n +1 − 1 · E + n , it remains to prov e t ha t      U + n,p n +1 · E + n U + n,p n +1 − 1 · E + n      = p r n . Let us use the isomorphism U + n,k · E + n E + n ∼ = U + n,k · E + n,k , whic h sho ws that w e hav e to pro ve that      U + n,p n +1 U + n,p n +1 − 1      :      E + n,p n +1 E + n,p n +1 +1      = p r n . It is easy t o see that      U + n,p n +1 U + n,p n +1 +1      = p p n +1 − p n 2 − 1 . The second n umber can b e computed as fo llo ws:      E + n,p n +1 E + n,p n +1 +1      =     E n,p n +1 ( E n,p n +1 ) p     :     E n,p n +1 +1 ( E n,p n +1 ) p     = p p n +1 − p n 2 − 1 : p r n and the theorem is prov ed. Closing this section w e w ould lik e to men tio n the f o llo wing Prop osition 3.5. r 0 ≤ r 1 ≤ . . . ≤ λ = P λ i . Pr o of. Let ǫ ∈ E n,p n +1 +1 / ( E n,p n +1 ) p . Then the extension F n ( p √ ǫ ) /F n is un- ramified, whic h defines an embedding E n,p n +1 +1 / ( E n,p n +1 ) p in to S ∗ n . It is easy to see that the canonical em b edding S ∗ n → S ∗ n +1 defines an em b edding E n,p n +1 +1 / ( E n,p n +1 ) p → E n +1 ,p n +2 +1 / ( E n,p n +1 +1 ) p . Therefore, r n ≤ r n +1 . F urthermore, b ecause of the pro j ection S ∗ n → V + n +1 (see [3]) it is clear tha t p λn + ν ≥ p r 0 + ... + r n , a nd the latter inequalit y implies that r n ≤ λ . Corollary 3.6. If p divide s B 1 ,ω − i , then the numb er of elements in ε i ( V + n ) is p 1+ r 1 ,i + ... + r n − 1 ,i and 1 ≤ r 1 ,i ≤ . . . ≤ r k ,i ≤ λ p − i . 9 4 W eak Kerv aire -Murth y Co njecture and New Link b et w ee n S and V Gro ups In this section let us denote b y ( a, b ) the local norm residue sym b ol with v alues in p -th ro ots of unity . Here ( a, b ) are elemen ts of the completion o f F n with resp ect to µ n . As sume that a ∈ U n,k \ U n,k +1 , b ∈ U n,p n +1 − k \ U n,p n +1 − k +1 , and k is prime to p . Lemma 4.1 (see [1]) . ( a, b ) 6 = 1 . Theorem 4.2. L et α ∈ S ( p ) n and α p = ( q ) . Then the formula f α ( x ) = ( x, q ) , x ∈ V + n +1 defines a non-trivial char acter of V + n +1 (if α is n o t trivial ) . Pr o of. Step 1. If q ≡ 1 mo d µ p n +1 − 1 n , then α = 1 ∈ S n . Indeed, w e a lready kno w that q ≡ 1 mo d µ p n +1 n and hence the extension F n ( p √ q ) /F n is non-ramified. The refo r e, q = ε · a p for some ε ∈ E n , a ∈ F n and consequen tly α = 1 in S n . Step 2. Without lost of generalit y we can a s sume that q ∈ U n,k \ U n,k +1 with k < p n +1 − 1 a nd k b eing prime to p . Indeed, if k = p · s , t hen q = 1 + a 0 µ ps n + tµ ps +1 n , where a 0 is an integer prime to p . Easy computations sho w that q (1 − a 0 µ s n ) p ∈ U n,k +1 . Pro ce eding in this w ay , w e can find q 1 ∈ U n,k 1 suc h that ( q 1 ) = (Γ α ) p , Γ ∈ U ( F n ), and suc h that k 1 is prime to p . Step 3. 1 + µ p n +1 − k n ∈ V n +1 . Indeed, if 1 + µ p n +1 − k n ≡ ε mo d µ p n +1 − k n , ε ∈ E n , then ( ε, q ) = 1. Ho wev er, it is not true by Step 2 and Lemma of this section. Step 4. Since S n = S − n , the c haracter constructed ab o v e is a no n- trivial c haracter of the group V + n +1 . The pro of is complete. Corollary 4.3. (”we ak Kervair e–Murthy c onje ctur e”) S ( p ) n ∼ = ( V + n +1 / ( V + n +1 ) p ) ∗ . Corollary 4.4. S ∗ n / ( S ∗ n ) p ∼ = V + n +1 / ( V + n +1 ) p . Pr o of. This f o llo ws from the existence of the surjection S ∗ n → V + n +1 con- structed in [3]. Theorem 4.5. Ther e exi s ts a c anonic al emb e ddin g i : S ( p ) n − 1 → V − n / ( V − n ) p 10 Pr o of. Let α b e an ideal suc h that α p = ( q ). D efine i ( α ) = q . This map is w ell-defined b ecause the n um b er q is defined up to a t ransformation q → ǫr p q , where ǫ ∈ E n . Cle a rly , the images of q and ǫr p q coincide in V n / ( V n ) p . If α ∈ K e r ( i ), then q ≡ ǫr p mo d (1 − ζ n ) p n − 1 and it f ollo ws fro m the Step 1 of the pro of of the previous theorem that α = 1. Hence, i is an em b edding. Since S = S − , it follows that i maps S n − 1 in to ( V n / ( V n ) p ) − = V − n / ( V − n ) p . 5 Ullom’s inequalit y The aim of t his section is to pro ve the follo wing result (a w eake r v ersion of Ullom’s inequality ): Theorem 5.1. L et the gener alize d Bernoul li n umb er B 1 ,ω − i is div isible b y p but not by p 2 . Then the c orr es p ondin g Iwasawa numb er λ i is les s than p. Pr o of. The pro of will consist of sev eral lemmas. F o r tec hnical reasons it will b e easier to deal with the original definition of V 2 , namely V k = Cok er { j 2 : E k − → U ( R k ) } , k = 1 , 2; see In tro duction. Let us also ma ke an imp ortan t note: R 2 = Z [ ζ 2 ] / ( ζ 2 − 1) p 2 and R 1 = Z [ ζ 1 ] / ( ζ 1 − 1) p Lemma 5.2. The map π : R 2 → R 1 define d as π ( x ) = x p is an surje ctive homomorphism of rings, which in duc e s an epimorphism of the c orr e s p onding gr oups of units π : U ( R 2 ) → U ( R 1 ) and π : V 2 → V 1 . Pr o of. π ( a + b ) = ( a + b ) p = π ( a ) + π ( b ) and π ( m ) = m p = m, m ∈ Z b ecause ( p ) = ( ζ 2 − 1) p 3 − p 2 = ( ζ 1 − 1) p 2 − p = 0 ∈ R 2 . F urthermore, π ( ζ 2 − 1) = ( ζ 2 − 1) p = ζ 1 − 1 and π (1 + ( ζ 2 − 1) k ) = (1 + ( ζ 2 − 1) k ) p = 1 + ( ζ 1 − 1) k what pro ve s that π is surjectiv e homomo r phis m of the rings and the corresp onding groups o f units. T o prov e that π : U ( R 2 ) → U ( R 1 ) induces a surjection π : V 2 → V 1 w e need to prov e that π ( E 2 ) ⊆ E 1 . Since p satisfies V andiv er’s conjecture, w e can use the subgroup of cyclotomic units C ( k ) ⊂ E k insted of E k . This means 11 that we ha ve to sho w t hat π ( C (2)) ⊆ C (1) . Ho w eve r, this is clear b ecause of our previous computations: π  ζ m 2 − 1 ζ 2 − 1  = π (1 + ζ 2 + ... + ζ m − 1 2 ) = 1 + ζ 1 + ... + ζ m − 1 1 = ζ m 1 − 1 ζ 1 − 1 . The latter computation completes the pro of. Corollary 5.3. If x ∈ V 2 is such that x p = 1 , then x ∈ k er ( π ) . Lemma 5.4. I f λ i ≥ p, then S 1 ,i has p gener ators as an ab elian gr oup (we assume that p is semi-r e gular). Pr o of. W e hav e already men tioned in Intro duc t io n that it follow s from results of [8] that S 1 ,i ∼ = Z p [ T ] (( T + 1) p − 1 , p i ( T )) , where p i ( T ) is a p olynomial of degree λ i and suc h that all the co effic ients except of the leading one are divisible b y p. Clearly , S 1 ,i / ( S 1 ,i ) p ∼ = Z p [ T ] ( p, T p , T λ i ) = Z p [ T ] ( p, T p ) , what prov es t he lemma. Lemma 5.5. I f λ i ≥ p and b i := B 1 ,ω − i is divisible by p but not by p 2 , then S 1 ,i ∼ = V 2 ,p − i ∼ = ( F p ) p as ab elian gr oups. Her e V 2 ,p − i = ε p − i V 2 . Pr o of. Consider the f ollo wing fibre pro duct: Z p [ T ] / (( T + 1) p − 1) / / i 1   Z p [ ζ 0 ] = Z p [ T ] / ( ( T +1) p − 1 T ) j 1   Z p [ T ] / ( T ) = Z p / / F p . Here, i 1 ( T ) = 0, j 1 ( ζ 0 ) = 1 and horizon tal maps are defined b y T → T , 1 → 1. Let us write elemen ts of Z p [ T ] / (( T + 1) p − 1) as pairs ( x ∈ Z p , y ∈ Z p [ ζ 0 ]) with clear compatibilit y conditions. In order to prov e that S 1 ,i ∼ = ( F p ) p , it is sufficien t to sho w that ( p i (0) , p i ( ζ 0 − 1) divides ( p, p ) in Z p [ T ] / (( T + 1) p − 1). 12 Indeed, p i (0) = b i and p i ( ζ 0 − 1) = b i + P a k ( ζ 0 − 1) k + ( ζ 0 − 1) λ i . Since p divides a k , ( p ) = ( b i ) = ( ζ 0 − 1) p − 1 , and λ i ≥ p , w e see that p i ( ζ 0 − 1) = b i (1 + ( ζ 0 − 1) X and therefore, ( p i (0) , p i ( ζ 0 − 1)) = ( b i , b i ) × (1 , 1 + ( ζ 0 − 1) X ). It follows that ( p i (0) , p i ( ζ 0 − 1)) divides ( p, p ) and due to the w eak Kerv aire- Murth y conjecture the lemma is completely prov ed. No w, w e can finish pro of of the theorem. Due to 5.2 w e ha ve a surjection π : ε p − i V 2 → ε p − i V 1 . On the other hand, due to 5.3 and 5 .5 π ( ε p − i V 2 ) = 1. This con tradiction completes the pro of of the theorem. 6 F urther relations b et w ee n B ernoulli and Iw a- sa w a n um b ers The aim of this section is to pro ve that if the generalized Bernoulli n umber b i = B 1 ,ω − i is divisible by p 2 , then the Iw asa wa n umber λ i = 1 6.1 Fine structure of V + 2 ,p − i if p 2 divides b i Theorem 6.1. L et b i = p k i t, k i ≥ 2 , wher e t i s c o-prime to p . Then ( V 2 ,p − i ) + ∼ = ( Z / ( p 2 )) ⊕ F k p , wher e k = min ( λ i − 1 , p − 1) . Pr o of. V + 2 ,p − i is a factor of ε p − i ( V 2 ) = ε p − i ( U 1 /U 1 ,p 2 − 1 ) = ε p − i ( U 1 /U 1 ,p 2 +1 ) b ecause p − i is an eve n num b er b et w een 2 and p − 3 . It is easy to prov e that ε p − i ( V 2 ) = ( Z / ( p 2 )) ⊕ F p − 1 . It follows from the w eak Kerv aire–Murth y conjecture and 5.5 that V + 2 ,p − i ∼ = ( Z / ( p 2 )) ⊕ F k p or V + 2 ,p − i ∼ = F k +1 p . So, we ha v e to exclude the second p ossibilit y . Let us denote the lo cal no rm residue symbol with v alues in p -ro ots of unity from the section 4 b y ( a, b ) n, 0 . In particular, we a re in terested in ( a, b ) 1 , 0 and ( c, d ) 0 , 0 . Let us also consider the lo cal nor m residue sym b ol with v alues in p 2 -ro ots of unity , whic h we denote by ( a, b ) n, 1 . Note that it is defined if n > 0 . Let us mak e the followin g easy remarks. T o simplify notatio ns , from now on w e denote V + n,p − i b y V n,p − i • ( a p , b ) 1 , 1 = ( a, b ) 1 , 0 ; 13 • if b ∈ Z p [ ζ 0 ], then ( a, b ) 1 , 0 = (Norm F 1 / F 0 (a) , b) 0 , 0 ; • (1 + ( ζ 0 − 1) i , 1 + ( ζ 0 − 1) j ) 0 , 0 = 1 , if i + j > p ; • (1 + ( ζ 0 − 1) i , 1 + ( ζ 0 − 1) p − i ) 0 , 0 6 = 1 . Let v 2 b e the image o f 1 + ( ζ 1 − 1 ) p − i in V 2 ,p − i . Let v 1 b e the image o f 1 + ( ζ 0 − 1) p − i in V 1 ,p − i ∼ = Z / ( p ). Clearly , Norm F 1 /F 0 ( v 2 ) generates the same elemen t in V 1 ,p − i ∼ = Z / ( p ) as v 1 . W e will write Norm F 1 /F 0 ( v 2 ) = v 1 . Lemma 6.2. L et us assume that p 2 divides b i . Then ther e exists an i d e al α ⊂ Z [ ζ 0 ] , whos e class b elo ngs to S 0 ,i such that α p 2 = ( q ) , q ∈ Z [ ζ 0 ] and α p is not a princip al ide al. Pr o of. The statemen t follows from the fact that S 0 ,i ∼ = Z p / ( b i ) . Lemma 6.3. L et v 2 , q b e as ab o ve. The n ( v 2 , q ) 1 , 1 is a primitive p 2 -r o ot of unity. Pr o of. Since α generates an elemen t o f S 0 ,i , it follow s from results of Section 4 that q can b e chose n suc h tha t q ≡ 1 mo d( ζ 0 − 1 ) i . Let us compute ( v 2 , q ) p 1 , 1 . ( v 2 , q ) p 1 , 1 = ( v p 2 , q ) 1 , 1 = ( v 2 , q ) 1 , 0 = (Norm F 1 /F 0 ( v 2 ) , q ) 0 , 0 = ( v 1 , q ) 0 , 0 6 = 1 Since ( v 2 , q ) p 1 , 1 is a non- trivial p - roo t of unit y , clearly ( v 2 , q ) 1 , 1 is a p 2 -ro ot o f unit y . Lemma 6.4. L et q b e as ab ove. T hen the formula < v , α > = ( v , q ) 1 , 1 defines a char acter of V 2 ,p − i . Pr o of. W e ha ve to pro ve that ( v , q ) 1 , 1 = 1 if v is a unit of Z [ ζ 1 ] o r v ≡ 1 mo d(1 − ζ 1 ) p 2 +1 , the latter b ecause ε p − i ( V 2 ) = ε p − i ( U 1 /U 1 ,p 2 − 1 ) = ε p − i ( U 1 /U 1 ,p 2 +1 ). If v is a unit, then the extension F 1 ( v 1 /p 2 ) /F 1 can ramify at (1 − ζ 1 ) only . F urthermore, for a ny prime θ 6 = (1 − ζ 1 ) w e ha v e q = r p 2 × lo c al unit for some r ∈ ( F 0 ) θ . It follows t ha t ( v , q ) θ = 1, here ( v , q ) θ is the corresp onding lo cal sym b ol with v alues in p 2 -ro ots of unit y . The pro duct form ula implies that ( v , q ) 1 , 1 = 1. It remains to pro ve that ( v , q ) 1 , 1 = 1 if v ≡ 1 mo d(1 − ζ 1 ) p 2 +1 . Indeed, v = t p for some t ∈ Z p [ ζ 1 ] such that t ≡ 1 mo d(1 − ζ 1 ) p+1 and ( v , q ) 1 , 1 = ( t p , q ) 1 , 1 = ( t, q ) 1 , 0 = (Norm F 1 /F 0 ( t ) , q ) 0 , 0 = 1 b ecause Norm F 1 /F 0 ( t ) ≡ 1 mo d(p) and q can b e chos en to satisfy q ≡ 1 mo d (1 − ζ 0 ) 2 . 14 No w w e can finish the pro of of the theorem. W e hav e pro ved that ( v , q ) 1 , 1 is a character of V 2 ,p − i and since ( v 2 , q ) 1 , 1 is a primitive p 2 -ro ot of unit y , w e can exclude the p ossibilit y V 2 ,p − i ∼ = F k p . 6.2 The Main T heorem I Theorem 6.5. Assume p 2 divides b i , q is the same as in the pr evious sub- se ction , and σ is a ge n er ator of G a l(F 1 / F 0 ) such that σ ( ζ 1 ) = ζ p +1 1 . Then σ ( v 2 ) /v 2 = v p 2 , whe r e v 2 is a gener ator o f V 2 ,p − i such that ( v 2 , q ) 1 , 1 = ζ 1 Pr o of. Let us consider ( σ ( v 2 ) , q ) 1 , 1 , where q is the same as in the previous subsection. W e ha v e ( σ ( v 2 ) , q ) 1 , 1 = σ (( v 2 , q ) 1 , 1 ) = ζ p +1 1 = ( v 2 , q ) p +1 1 , 1 . Hence, ( σ ( v 2 ) /v 2 , q ) 1 , 1 = ( v p 2 , q ) 1 , 1 = ζ 0 . Let us consider the annihilator of σ ( v 2 ) /v 2 in the c haracter group ( V 2 ,p − i ) ∗ . W e denote it b y Ann ( σ ( v 2 ) /v 2 ). Lemma 6.6. Ann ( σ ( v 2 ) /v 2 ) ∼ = F k +1 p , whe r e k = min ( λ i − 1 , p − 1) . Pr o of. The pro of consists of three statemen ts below. • σ ( v 2 ) /v 2 has order p . Indeed, ( σ ( v 2 ) /v 2 , q ) 1 , 1 = ζ 0 . Since ( V 2 ,p − i ) ∗ ∼ = ( Z / ( p 2 )) ⊕ F k p and q generates a character of order p 2 , it follows that the v alue of any character on σ ( v 2 ) /v 2 is either a primitiv e p -ro ot of unit y or 1 . Consequen tly , σ ( v 2 ) /v 2 has or der p (it cannot b e 1 b ecaus e ( σ ( v 2 ) /v 2 , q ) 1 , 1 = ζ 0 ). • Therefore, ( V 2 ,p − i ) ∗ / Ann ( σ ( v 2 ) /v 2 ) ∼ = F p . • Since q generates a c haracter of order p 2 , ( σ ( v 2 ) /v 2 , q ) 1 , 1 = ζ 0 , and ( V 2 ,p − i ) ∗ ∼ = ( Z / ( p 2 )) ⊕ F k p , we can deduce that Ann ( σ ( v 2 ) /v 2 ) ∼ = F k +1 p . No w, w e can complete t he pro of of the theorem. By the Kerv aire– Murth y theorem, Ann ( σ ( v 2 ) /v 2 ) is a subgroup of S 1 ,i . By the lemma ab o v e and the w eak Kerv aire–Murth y conjecture, we hav e Ann ( σ ( v 2 ) /v 2 ) ∼ = S ( p ) 1 ,i , the subgroup o f elemen ts of order p . Th us, for an y ideal α m suc h that α p m = ( q m ) ⊂ Z [ ζ 1 ] w e ha ve ( σ ( v 2 ) /v 2 , q m ) 1 , 0 = 1 and ( σ ( v 2 ) , q m ) 1 , 0 = ( v 2 , q m ) 1 , 0 = ( v 2 , q m ) p +1 1 , 0 . It f ollo ws that ( σ ( v 2 ) /v 2 , q m ) 1 , 0 = ( v p 2 , q m ) 1 , 0 . T here- fore, fo r an y character χ ∈ V ∗ 2 ,p − i , w e hav e χ ( σ ( v 2 ) /v 2 ) = χ ( v p 2 ) and conse- quen tly σ ( v 2 ) /v 2 = v p 2 . 15 6.3 Main Theorem I I Lemma 6.7. ε p − i ( V 2 ) and V 2 ,p − i ar e Z p [[ T ]] -mo d ules with one gener ator. Her e the action is define d as fol lows: T · v = σ ( v ) /v and a · v = v a , a ∈ Z p . Pr o of. ε p − i ( V 2 ) = ε p − i ( U 1 /U 1 ,p 2 − 1 ). Since ε p − i ( U 1 ) is an Z p [[ T ]]-mo dules with one generator, it is also true f o r its factors ε p − i ( V 2 ) and V 2 ,p − i b ecause U 1 ,p 2 − 1 and the image of U ( Z [ ζ 1 ]) in V 2 are Z p [[ T ]]-submo dules . Lemma 6.8. ε p − i ( V 2 ) ∼ = Z p [[ T ]] / ( T p , pT , p 2 ) . Pr o of. It is easy to v erify that pT and p 2 annihilate V 2 . F urther, ε p − i ( U 1 ) is annihilated by ( T + 1) p − 1 . Since pT annihilates ε p − i ( V 2 ), w e deduce that T p annihilates it to o. F inally , it is easy to see that b oth ε p − i ( V 2 ) and Z p [[ T ]] / ( T p , pT , p 2 ) con ta in p p +1 elemen ts. The last observ ation complete s the pro of. Theorem 6.9. If the gener alize d Bernoul li numb er b i = B 1 ,ω − i is divis ible by p 2 , then the I w asawa numb er λ i = 1 Pr o of. It f o llo ws from 6.5 that T − p annihilates V 2 ,p − i . Therefore, as a Z p [[ T ]]-mo dule V 2 ,p − i factors throug h Z p [[ T ]] / ( T p , pT , p 2 , T − p ) ∼ = Z p / ( p 2 ). Since w e already kno w that V 2 ,p − i ∼ = Z / ( p 2 ) ⊕ F k , where k = min ( λ i − 1 , p − 1), w e conclude that V 2 ,p − i ∼ = Z / ( p 2 ) and λ i = 1 . Corollary 6.10. V n,p − i ∼ = Z / ( p n ) if p 2 divides b i . Pr o of. It is a n easy consequenc e of Corollary 3 .6. Corollary 6.11. S n,i ∼ = Z / ( p n + k i ) if p 2 divides b i . Her e k i is the p -adic valuation of b i . Pr o of. The statemen t follows from the fact that No rm F n +1 /F n ( i F n +1 /F n )( α ) = α p for a n y ideal α ⊂ Z [ ζ n ] and that of Norm F n +1 /F n : S n +1 → S n is surjectiv e while i F n +1 /F n : S n → S n +1 is injectiv e. 7 Fine str u cture of V n,p − i and S n,i if p 2 do es not divide b i Throughout this section w e assume that the p -adic v alua tion v p ( b i ) = 1 . W e already kno w that if p 2 do es not divide b i , t he n λ i satisfies Ullom’s inequalit y λ i ≤ p − 1 and S 0 ,i ∼ = F p . 16 7.1 Fine structure of V n,p − i Lemma 7.1. L et α ∈ S 0 ,i . Then α = β p , whe r e β ∈ S 1 ,i . Pr o of. W e consider S 1 ,i as a Z p [[ T ]]-mo dule. It follo ws from res ults of [8] that S 1 ,i ∼ = Z p [[ T ]] (( T + 1) p − 1 , f i ( T )) = Z p [ T ] (( T + 1) p − 1 , p i ( T )) , where f i ( T ) , p i ( T ) w ere defined in In tro duction. Clearly , S 1 ,i / ( S 1 ,i ) p ∼ = Z p [ T ] ( p,T p ,T λ i ) = Z p [ T ] ( p,T λ i ) = F p [ T ] / ( T λ i ) , b ecause of Ul- lom’s inequalit y . Let us pro ve that the image of S 0 ,i under the canonical em b edding i F 1 /F 0 : S 0 ,i → S 1 ,i is con tained in S p 1 ,i . Indeed, t his image is gen- erated b y N ( T ) = 1 + ( T + 1) + · · · + ( T + 1) p − 1 = (( T + 1) p − 1 ) /T . Again, b ecause of Ullom’s inequality , t he image of N ( T ) in S 1 ,i / ( S 1 ,i ) p ∼ = Z p [ T ] ( p,T λ i ) is zero. The lemma is pro ve d. The crucial step in computation of V n,p − i is to consider the case n = 2 . F rom the w eak Kerv aire–Murth y conjecture and Ullom’s inequalit y , w e kno w that as an ab elian group V 2 ,p − i has λ i generators. Th us, w e hav e tw o p ossibilities : V 2 ,p − i ∼ = Z / ( p 2 ) ⊕ F λ i − 1 p or V 2 ,p − i ∼ = F λ i p . Theorem 7.2. V 2 ,p − i ∼ = Z / ( p 2 ) ⊕ F λ i − 1 p . Pr o of. It is sufficien t to find an elemen t in V ∗ 2 ,p − i of order p 2 . With some abuse o f notations, let α p = ( q ) , q ∈ Z [ ζ 0 ]. Since β p = α in S 1 ,i , it follows that β p 2 = ( q t p ) , where q ∈ Z [ ζ 1 ] . W e claim that the required c haracter is defined b y ( v , q t p ) 1 , 1 . T o prov e this, w e follow the pro of of Theorem 6.1. Lemma 7.3. ( v , q t p ) 1 , 1 = ζ 1 for som e v . Pr o of. W e ha ve ( v , q t p ) p 1 , 1 = ( v p , q t p ) 1 , 1 = ( v , q ) p 1 , 1 = ( v , q ) 1 , 0 = (Norm F 1 /F 0 ( v ) , q ) 0 , 0 . Clearly , w e can choose v suc h that (Norm F 1 /F 0 ( v ) , q ) 0 , 0 = ζ 0 and t herefore, ( v , q t p ) 1 , 1 = ζ 1 . Lemma 7.4. L et r ∈ Z p [ ζ 1 ] b e such that r ≡ 1mo d(1 − ζ 1 ) p 2 +1 . Then ( r , q t p ) 1 , 1 = 1 . F urther, ( ǫ, q t p ) 1 , 1 = 1 if ǫ ∈ Z [ ζ 1 ] . Pr o of. Since r = r p 1 , where r 1 ∈ Z p [ ζ 1 ], w e can pro ceed exactly as in the pro of of Lemma 6.4. Since ( q t p ) = β p 2 , a gain w e can simply rep eat the argumen ts of the pro of of Lemma 6 .4. 17 Tw o lemmas ab o ve imply that the elemen t q t p induces a character of V 2 ,p − i of order p 2 . The theorem is pro v ed. Corollary 7.5. If b i is not divisible by p 2 , then V n,p − i ∼ = Z / ( p n ) ⊕ ( Z / ( p n − 1 )) λ i − 1 . Pr o of. Corollary 3.6 implies that r m,p − i = λ i for any m ≥ 1 and moreo ve r, the num b er of elemen ts in V n,p − i is p 1+( n − 1) λ i . On the other hand, V n,p − i is a factor of a bigger group ε p − i ( V n ) = ε p − i ( U n − 1 /U n − 1 ,p n − 1 ). It is easy to verify that ε p − i ( V n ) ∼ = Z / ( p n ) ⊕ T , where the ab elian g roup T has exp onen t p n − 1 (an exact form ula can b e deriv ed from [3] but w e do not need it). Comparing the n umber of elemen ts and the num b er of generator s of V n,p − i whic h is λ i ), w e can deduce that V n,p − i ∼ = Z / ( p n ) ⊕ ( Z / ( p n − 1 )) λ i − 1 . 7.2 Fine structure of S n,i Let A b e a finite ab elian group suc h that A ∼ = ⊕ j Z / ( p k j ) . Let us denote the ab elian group ⊕ j Z / ( p k j + m ) by Σ m A. Lemma 7.6. S n +1 ,i ∼ = Σ n S 1 ,i . Pr o of. The fact fo llo ws from the follo wing observ at io ns : • all the gro ups S k ,i , k ≥ 1 , ha v e λ i generators; • Norm F k +1 /F k ( i F k +1 /F k ( α )) = α p , α ∈ S k ,i . Remark: it is w ell-know n that i F k +1 /F k : S k ,i → S k +1 ,i is an em b edding a nd Norm F k +1 /F k : S k +1 ,i → S k ,i is a surjection. It remains to compute S 1 ,i . Theorem 7.7. Assume that 1 ≤ λ i < p − 1 . Then S 1 ,i ∼ = Z / ( p 2 ) ⊕ F λ i − 1 p . Pr o of. S 1 ,i ∼ = Z p [ T ] / (( T + 1) p − 1 , f i ( T )) , see [8 ]. Since p 2 do es not divide b i , the p olynomial f i ( T ) is irreducible. Let a b e its ro ot. Then Z p [ T ] / ( f i ( T )) ∼ = Z p [ a ] , 1 , a, a 2 · · · , a λ i − 1 generate Z p [ a ] as an abelian group, ( p ) = ( a λ i ) in Z p [ a ] , a nd S 1 ,i ∼ = Z p [ a ] / (( a + 1) p − 1) . F urther, (( a + 1 ) p − 1 ) = ( a λ i +1 ) b ecause λ i < p − 1 . It follows tha t the elemen t 1 ∈ Z p [ a ] has exp onen t p 2 and all ot he r generators o f Z p [ a ], a, a 2 , · · · , a λ i − 1 ha v e exp onen t p. The theorem is prov ed. 18 The case λ i = p − 1 is more delicate. T o treat this case w e need the Cartesian square from Lemma 5.5. Let us denote the ring Z p [ T ] / (( T + 1) p − 1)] b y B . W e hav e to study B / ( f i ( T )). W e remind the r eade r that an y elemen t b ∈ B can b e written as a pair ( c, d ) , c ∈ Z p , d ∈ Z p [ ζ 0 ] . In this notations f i ( T ) = ( b i , f i ( ζ 0 − 1)) . A simple analysis of this case sho ws the followin g result: Theorem 7.8. The element 1 ∈ Z p [ T ] / (( T + 1) p − 1 , f i ( T )) ∼ = S 1 ,i has exp o nent p κ with κ = [ k p − 1 ] + 1 . Her e k = v p ( f i ( ζ 0 − 1)) , wh e r e v p is the extension of the p -adic v a l uation on Z p to Z p [ ζ 0 ] . Remark 7.9. v p ( f i ( ζ 0 − 1)) = v p ( L p ( s 0 , ω 1 − i )), where L p is a p -adic L- function, ω is the T eic hm ¨ uller c ha racter of Z / ( p − 1) Z , and s 0 satisfies the follo wing equation: ( p + 1) s 0 = ζ 0 . 7.3 Fine structure of the group ”Lo cal u nits mo dulo closure of the cyc lo tomic units” Our aim is to describe the group ( U n, 1 /C ( n )) + , where U n, 1 is the group of units of Z p [ ζ n ] c o ngruen t 1 modulo ( ζ n − 1) and C ( n ) is the closer of the subgroup of the cyclotomic units. Let G n,k = Gal ( F n /F k ). If A is a G - mo dule, then A G is a submodule of G - in v ariant elemen ts of A . W e begin with the following result: Theorem 7.10. The c anonic al map ( U k , 1 /C ( k )) + → ( U n, 1 /C ( n )) + is an emb e ddin g an d { ( U n, 1 /C ( n )) G n,k } + ∼ = ( U k , 1 /C ( k )) + . Pr o of. Let us consider the follow ing short exact sequenc e: 0 → C ( n ) → U n, 1 → U n, 1 /C ( n ) → 0 . Th us, w e get the corresp onding exact sequence of cohomologies: 0 → C ( n ) G n,k → U G n,k n, 1 → ( U n, 1 /C ( n )) G n,k → H 1 ( G n,k , C ( n )) . Th us, to prov e the theorem w e hav e to sho w that H 1 ( G n,k , C ( n ) + ) = 1. Let us sho w first that the Herbrand index h ( G n,k , C ( n ) + ) = 1 . Indeed, C ( n ) + is a finite index subgroup of the group of g lo bal units E n . Its closure con tains a n op en subgroup of U + n, 1 due to Leop oldt’s conjecture a b out the closure of the group of global units, whic h is true in our case. Th us, C ( n ) + 19 con tains an op en subgroup X of U + n, 1 as w ell. Cons equen tly , h ( G n,k , C ( n ) + ) = h ( X ) · h ( C ( n ) + /X ) = 1 b ecause h ( C ( n ) + /X ) = 1 since the g roup C ( n ) + /X is finite and h ( X ) = 1 b ecause X can b e c hosen as a pro jectiv e G n,k -mo dule, see [1], c hapter 6 . Therefore, it suffices to pro v e that H 2 ( G n,k , C ( n )) = 1 . Ho we ver, it is clear b ecause H 2 ( G n,k , C ( n )) = C ( k ) / N or m F n /F k ( C ( n )) = 1 b ecause the norm is surjectiv e on the group of cyclotomic units. Let us giv e a no ther pro of of the same theorem based on a lemma needed in the sequel. Let us remind the reader that series f i ( T ) and p olynomials p i ( T ) , P n ( T ) w ere defined in Intro duction. L et us define g i ( T ) = f i ( 1+ p 1+ T − 1) and the p olynomial q i ( T ) exactly in the same w ay as p i ( T ) w as defined from f i ( T ) . It is clear that deg ( p i ) = deg ( q i ) = λ i . F urt he mor e, with o ur choice of the p olynomials p i , q i w e hav e p i (0) = f i (0) = L p (0 , ω 1 − i ) = b i and q i (0) = g i (0) = f i ( p ) = L p (1 , ω 1 − i ) := c i . Ho w eve r , generally speaking p i ( p ) 6 = f i ( p ) while we only ha ve v p ( f i ( p )) = v p ( p i ( p )) = v p ( c i ) . As in In tro duction, w e denote ( T + 1) p n − 1 by P n ( T ) and P n ( T ) /P k ( T ) by P n,k ( T ) . Lemma 7.11. We have: • ε p − i U n, 1 ∼ = Z p [[ T ]] / ( P n ( T )) = Z p [ T ] / ( P n ( T )) ; • ε p − i C ( n ) ∼ = ( g i ( T )) / ( g i ( T ) P n ( T )) ; • ε p − i ( U n, 1 /C ( n )) ∼ = Z p [[ T ]] / ( P n ( T ) , g i ( T )) = Z p [ T ] / ( P n ( T ) , q i ( T )) . Pr o of. The first t w o items w ere prov ed in [8], c hapters 13, 15. Let us pro v e the third statemen t. It w as prov ed in [8], c ha pter 15 , that P n and g i ha v e no common ro ots. This implies that ( P n ) ∩ ( g i ) = ( P n · g i ). Th us, ε p − i ( U n, 1 /C ( n )) ∼ = Z p [[ T ]] / ( P n ( T )) ( g i ( T )) / ( g i ( T ) P n ( T )) = Z p [[ T ]] / ( P n ( T )) ( g i ( T )) / ( g i ( T ) ∩ P n ( T )) = Z p [[ T ]] / ( P n ( T ) , g i ( T )) . 20 This lemma enables us to give another pro of of Theorem 7.3. Let R = Z p [[ T ]] / ( g i ( T )) and let M n = R/ ( P n ) b e an R -mo dule . Clearly , the lemma ab o v e sho ws that M n = Z p [[ T ]] / ( P n ( T ) , g i ( T )) ∼ = ε p − i ( U n, 1 /C ( n )) . Let us notice that P n is not a zero divisor in R b ecause P n and g i ha v e distinct ro ots. The n m ultiplication b y P n,k determines a we ll- defin ed R -mo dule map m : M k → M n b ecause m ( P k ) = P n . It is clear that m is an embedding. Indeed, if m ( x ) = 0 , then P n,k · x = P n · y = P n,k P k · y for some y ∈ R . Since P n,k is not a zero divisor in R, w e conclude that x = P k · y = 0 ∈ M k . F urthermore, let us prov e that { y ∈ M n : P k · y = 0 } ∼ = M k . Indeed, P k · y = P n · z = P k · m ( z ) for some z ∈ R and consequen tly y = m ( z ) . Since m is an embedding, the required statemen t f o llo ws. The tw o statemen ts ab o v e are equiv alen t to Theorem 7.3 for mulated in terms of Z p [[ T ]]-mo dules. A t this p oin t w e remind the reader that we assume that p divides b i . Clearly , then p divides c i and vise v ersa. Theorem 7.12. 1. If p 2 divides b i , then λ i = 1 and p 2 do e s not d ivide c i ; 2. If p 2 do e s not divide b i and 2 ≤ λ i ≤ p − 1 , then p 2 do e s not divide c i ; 3. If p 2 divides c i , then p 2 do e s not divide b i and λ i = 1 . Pr o of. 1. W e already know that λ i = 1. Then c i = g i (0) = f i ( p ) = b i + pa 1 + p 2 Z 1 , where p do es not divide a 1 . Then the statemen t is clear. 2. c i = b i + p 2 Z 2 . 3. W e already kno w that if 2 ≤ λ i ≤ p − 1, then p 2 do es not divide b i and consequen tly it do es no t divide c i as w ell. Hence, λ i = 1. Then p 2 do es not divide b i b ecause c i = b i + pa 1 + p 2 Z 1 . Corollary 7.13. Assume that p 2 divides c i . Then ε p − i ( U n, 1 /C ( n )) ∼ = Z / ( p n + l i ) with l i = v p ( c i ) Pr o of. Let us p erform a simple computation: ε p − i ( U 0 , 1 /C (0)) ∼ = Z P [[ T ]] / ( T , g i ( T )) = Z p / ( g i (0)) = Z / ( p l i ) = M 0 . W e already kno w that λ i = 1 and hence, M 1 ∼ = Z / ( p k ) for some k . w e hav e the em b edding m : M 0 → M 1 determined by the formula m (1 M 0 ) = (( T + 1) p /T ) · 21 1 M 1 . Also, w e hav e a canonical pro jection r es : M 1 → M 0 , r es (1 M 1 ) = 1 M 0 . F urther, r es ( m (1 M 0 )) = r es ( { (( T + 1) p − 1 ) / T } · 1 M 1 ) = p · 1 M 0 . This computation show s that k = 1 + l i . Similar computations with m : M 1 → M 2 and r es : M 2 → M 1 yield r es ( m (1 M 1 )) = p · 1 M 1 and M 2 ∼ = Z / ( p 2+ l i ) and so on. Corollary 7.14. Assume that p 2 do e s n ot divide c i and 1 ≤ λ i < p − 1 Then • ε p − i ( U 1 , 1 /C (1)) ∼ = Z / ( p 2 ) ⊕ F λ i − 1 p . • ε p − i ( U n, 1 /C ( n )) ∼ = Z / ( p n +1 ) ⊕ ( Z /p n ) λ i − 1 . Pr o of. Pro ofs are literally the same as the pro ofs of the analogous statemen ts ab out the structure o f the class groups in Subsection 7.2. Remark 7.15. Results of Subsections 7.2 and 7.3 sho w that the class groups S n,i and the groups ε p − i ( U n, 1 /C ( n )) in ma j orit y of cas es are dual to eac h other. Therefore, it is natural to conjecture that these g roups are alw ays dual to each other. Ho wev er, it follo ws f rom Theorem 7 .1 2 that if p 2 divides c i , it do es not divide b i and hence, our conjecture do es not hold in this case. Therefore, if w e an ywa y b eliev e in that conjecture, w e ha v e to exclude the case p 2 divides c i = L p (1 , ω 1 − i ) . L et us do this in the next section. 8 L p (1 , ω 1 − i ) , L p (0 , ω 1 − i ) are n ot divisib l e b y p 2 ? 8.1 V -dualit y In this subsection w e will pro v e t he following result. Theorem 8.1. L et us assume that λ i = 1 and p 2 do e s not div i d e b i . Then the gr oup Gal ( F 1 /F 0 ) acts non-trivial ly on V + 2 . Remark 8.2. In the definition of V 2 (giv en by Kerv aire and Murth y in [3]) has a natural s tr ucture a G al ( F 2 /F 0 )-mo dule. Ho w ev er, they pro v ed that Gal ( F 2 /F 1 ) a cts on V 2 trivially . Consequen tly , V 2 is a Gal ( F 1 /F 0 )-mo dule. Pr o of. Our pro of is based on the main result of [3] (men tioned earlier in the pap er, Theorem 1.1, how ev er we need it in its complete f orm b ecaus e it t races the action of Gal ( F 2 /F 0 ) or equiv alen tly Gal ( F 1 /F 0 )): 22 • Let g b e a generator of G = Gal ( F 1 /F 0 ) ∼ = Z /p Z and let us define the following natural action o f G on V ∗ 2 ( g χ )( v ) = g ( χ ( g − 1 v )) , where χ ∈ V ∗ 2 , v ∈ V 2 . • Then ( V + 2 ) ∗ is isomorphic to a G -submo dule of S ( p ) 1 . • Assuming t he Kerv air e and Murth y conjecture that S ( p ) 1 ∼ = ( V + 2 ) ∗ ∼ = Z /p 2 Z w e get the formula < g s, v > = g ( < s, g − 1 v > ) or equiv alently g ( < g − 1 s, v > ) = < s, g v > . Here s ∈ S ( p ) 1 . No w, let us pro ceed to the pro of o f our theorem. Clearly , without lo osing generalit y w e may assume that S ( p ) 1 = S 1 ,i . Then, since λ i = 1 and p 2 do es not divide b i , o ur previous computatio ns sho w that S ( p ) 1 ∼ = V + 2 ∼ = Z /p 2 Z , i.e. the Kerv aire and Murth y conjecture is true in our situation. Let us choose g ∈ G s uch that g ( ζ 1 ) = ζ 1+ p 1 . It is w ell-kno wn due to Iw asa wa that the subgroup { s ∈ S ( p ) 1 : g ( s ) = s } ∼ = S ( p ) 0 ∼ = Z /p Z . Therefore, G acts o n S ( p ) 1 non-trivially and w e can c ho ose a generator s ∈ S ( p ) 1 suc h that g − 1 ( s ) = s 1+ p . F urther, w e can c ho ose a generator v ∈ V + 2 suc h that < s, v > = ζ 1 . Supp ose G acts on V + 2 trivially . Then ζ 1 = < s, v > = < s, g ( v ) > = g ( < g − 1 ( s ) , v > ) = g ( ζ 1+ p 1 ) = ζ 1+2 p . Clearly , it is imp ossible and hence, G acts on V + 2 non-trivially . 8.2 E -dualit y Let us assume that λ i = 1 and p 2 do es divide c i . Results of this subsection are based on C ha pt er 8 of [8] Let E b e the group o f real units of Z [ ζ 0 ] and in this subsection w e denote U + 0 , 1 b y U. Since p satisfies V andiv er’s conjecture, U / ( C (0) ) + = U / ¯ E , where ¯ E is the closure of E in U. Let us denote E /E p 2 b y E p 2 . It w as pro v ed in [8] t ha t ε p − i E p 2 ∼ = Z / ( p 2 ) and ε p − i U /ε p − i ¯ E ∼ = Z / ( p v p ( c i ) ) = Z p / ( c i ) . according to Coro llary 7.13. Let η ∈ E generate ε p − i E p 2 . Let us consider ε p − i ( η ) ∈ ε p − i ¯ E . Since ε p − i ( E /E p 2 ) = ε p − i ¯ E /ε p − i ¯ E p 2 and E p 2 ⊂ ¯ E p 2 , w e see that η − 1 ε p − i ( η ) ∈ ¯ E p 2 . Th us η = ε p − i ( η ) γ p 2 and w e o bta ined the following 23 Lemma 8.3. η ∈ E is a lo c al p 2 -p ow er. Pr o of. ε p − i ( η ) is a lo cal p 2 -p o w er b ecause ε p − i ( η ) ∈ ε p − i ¯ E = ( ε p − i U ) v p ( c i ) and v p ( c i ) ≥ 2 . Th us, η ∈ E is a lo cal p 2 -p o w er. 8.3 L p (1 , ω 1 − i ) is not divisible b y p 2 ? No w w e can prov e the first main result of this section Theorem 8.4. L p (1 , ω 1 − i ) is not divisible by p 2 . Pr o of. Assuming that L p (1 , ω 1 − i ) is divisible b y p 2 , in sev eral steps we will come to con tradiction. • Clearly , η is not a p -p o we r in the field F 0 b ecause it generates ε p − i ( E /E p 2 ) . • η is not a p -p o w er in the field F 1 b ecause otherwise F 1 = F 0 ( η 1 /p ) and F 1 b ecomes a non-ramified extension of F 0 . • Hence, the extension F 1 ( η 1 /p 2 ) /F 1 is non-ramified and consequen tly η induces a ch a racter of S ( p ) 1 . • Let E 1 b e the subgroup o f the g roup of units of Z [ ζ 1 ] whic h ar e lo cal p 2 -p o w ers. Then E 1 generates a subgroup of the g roup o f characters of S ( p ) 1 . The correspo nding pairing b et w een S ( p ) 1 and E 1 satisfies < g s, ǫ > = g ( < s, g − 1 ǫ > ). • Let X ⊆ ( S ( p ) 1 ) ∗ ∼ = Z / ( p 2 ) b e the subgroup generated b y η . Comparing form ulas < g s, ǫ > = g ( < s, g − 1 ǫ > ) and < g s, v > = g ( < s, g − 1 v > ) w e can conclude that X ∼ = V + 2 as G - modules. • Ho w eve r , t he la tter is imp ossible b ecaus e G acts trivially on X ( η ∈ F 0 ) and no n-trivially on V + 2 . The theorem is pro ve d. 8.4 L p (0 , ω 1 − i ) is not divisible b y p 2 ? No w w e can prov e the second main result of t his section Theorem 8.5. L p (0 , ω 1 − i ) is not divisible by p 2 . 24 Pr o of. Assuming that L p (0 , ω 1 − i ) is divisible b y p 2 , in sev eral steps we will come to con tradiction. W e remind the reader that w e a lr eady kno w that λ i = 1 and c i is divisible b y p but not b y p 2 . L et us assume first that v p ( L p (0 , ω 1 − i )) = v p ( b i ) = 2 . The reader will see that the general case will b e p ossible to treat exactly in the same w ay . • Since S 0 ,i ∼ = Z / ( p 2 ) and S 1 ,i ∼ = Z / ( p 3 ) under our assumptions, we deduce that S G 1 ,i ∼ = Z / ( p 2 ) = S p 1 ,i . • Our previous computations sho w that ε p − i V + 1 ∼ = F p ∼ = ε p − i ( U 0 , 1 /C (0)) and ε p − i V + 2 ∼ = Z / ( p 2 ) ∼ = ε p − i ( U 1 , 1 /C (1)) . • Since there exists a canonical pro jection of G -mo dules U 1 , 1 → V + 2 , w e deduce that ε p − i V + 2 ∼ = ε p − i ( U 1 , 1 /C (1)) as G -mo dules. • ( U 1 , 1 /C (1)) ) G ∼ = U 0 , 1 /C (0)) b y Theorem 7.10 , w e see that G acts on ε p − i V + 2 non-trivially . • By the main Kerv aire–Murth y theorem ([3]), ( ε p − i V + 2 ) ∗ is isomorphic to a subgroup of S 1 ,i as G -mo dules. Comparing orders of the inv olve d groups we get ( ε p − i V + 2 ) ∗ ∼ = ( S G 1 ,i as G -mo dules. • Let us fix g ∈ G suc h that g ( ζ 1 ) = ζ 1+ p 1 . Let us c ho ose v ∈ ε V + 2 so tha t g − 1 ( v ) = v 1+ p and s ∈ S G 1 ,i satisfying < s, v > = ζ 1 . • No w, as b efore let us p erform a simple computation: < s, v > = ζ 1 = < g s, v > = g < s, g − 1 ( v ) > = g ( ζ 1+ p 1 ) = ζ 1+2 p 1 . The latter equalit y is imp ossible and the theorem is prov ed. 8.5 Correction to the p revio u s results of this section The results of this sec tio n mus t b e corrected. What w ent wrong? Let us tak e a lo ok at the pro of of Theorem 8.1. W e c hose a generator of g ∈ G suc h that g ( ζ 1 ) = ζ 1+ p 1 and a nd a generator s ∈ S 1 ,i suc h that g − 1 ( s ) = s 1+ p . Unfortunately , these tw o choice s might b e incompatible. W e can only claim that g ( s ) = s 1+ k p with an in teger k defined mo dulo p. Similarly , g ( v ) = v 1+ lp , where v ∈ V + 2 satisfies < s, v > = ζ 1 . F urther, since < g ( s ) , g ( v ) > = g ( ζ 1 ) , we see t ha t k + l = 1 . Assuming that L p (1 , ω 1 − i ) is divisible b y p 2 , w e can deduce that l = 0 mod ( p ) and hence g ( s ) = s 1+ p . 25 Let us consider S 1 ,i as an Z p [[ T ]] / ( f i ( T ) , (1 + T ) p − 1)-mo dule. It follo ws that T − p acts trivially o n S 1 ,i ∼ = Z / ( p 2 ). Since λ i = 1, w e infer t ha t f i ( T ) has a ro ot of the f o rm T = p + ap 2 for some a ∈ Z p . F urthermore, T = (1 + p ) s − 1 and hence s 0 = log 1+ p (1 + p + ap 2 ) = 1 + bp is a zero o f L p ( s, ω 1 − i ). W e hav e prov ed Theorem 8.6. L et λ i = 1 . If L p ( s, ω 1 − i ) has no zer o es of the form s 0 = l og 1+ p (1 + p + ap 2 ) = 1 + bp, b ∈ Z p , then L p (1 , ω 1 − i ) is not divisible by p 2 . The case k = 0 , l = 1 can b e conditio nally excluded using a similar result. Theorem 8.7. L et λ i = 1 . If L p ( s, ω 1 − i ) has no zer o es of the form s 0 = l og 1+ p (1 + ap 2 ) = bp, b ∈ Z p , then L p (0 , ω 1 − i ) is not divisible by p 2 . Remark 8.8. The previous conditional theorem prov es the K erv aire and Murth y conjecture (conditionally) with only one p ossible exception: λ i = p − 1 . 9 Conclud ing remarks The Kerv aire and Murth y conjecture has another interes ting form. L et us denote b y A ( F n ) the ring of adeles of the field F n . Let w b e a v aluation of F n , differen t from µ n = (1 − ζ n ). Let Q w b e the completion of Z [ ζ n ] at w . Let us consider the f ollo wing subgroup K p n +1 − 1 of GL (1 , A ( F n )), namely K p n +1 − 1 = GL (1 , Q ) × U n,p n +1 − 1 × Y GL (1 , Q w ) . Then the Kerv aire and Murth y conjecture can b e form ulated a s Conjecture 9.1. ( S − n ) ∗ ∼ = { GL (1 , F n ) \ GL (1 , A ( F n )) /K p n +1 − 1 } + ( p ) 10 Afterw o rd This is my last pap er b efore m y retiremen t from the Departmen t o f Mathe- matical Sciences, Chalmers Univ ersit y of T ec hnology/ Univ ersit y of G¨ oteb org, Sw eden after a lmo st 2 5 y ears of professional service. 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