ACTA ARITHMETICA. Riemann{Hurwitz formula in basic Z S -extensions. Yi Ouyang (Hefei and Minneapolis, Minn.) and Fei Xu (Hefei)

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1 ACTA ARITHMETICA * (199*) Riemann{Hurwitz formula in basic Z S -extensions by Yi Ouyang (Hefei and Minneapolis, Minn.) and Fei u (Hefei) 1. Introduction. Let p be a prime number and F be a CM-eld. Let F 1 be the cyclotomic Z p -extension of F. For every n, wehave a unique subextension F n of degree p n over F in F 1.We denote by F + the maximal be the relative class number of F n =F + n. Then real subeld of F, and let h ; n we have awell known result: ord p (h ; n )=; p n + ; n + ; ; 0, ; 0, and ; are integers, when n is suciently large. Let E be a CM-eld and a p-extension of F. Under the assumption ; F = 0, Y. Kida ([5]) proved a striking analogue of the classical Riemann{ Hurwitz genus formula from the theory of compact Riemann surfaces, by describing the behavior of ; in the p-extension. His result can be stated as follows: Theorem 0 (see [8, Theorem 4.1]). ; F =0if and only if ; E =0,and when this is the case, ; E ; E =[E 1 : F 1 ]( ; F ; F )+!0 (e(!0 = 0 ) ; 1) ;! (e(!=) ; 1) where the summation is taken over all places! 0 on E 1 (resp.! on E + 1 ) which do not lie above p and 0 =! 0 j F 1 (resp. =!j F + 1 ), e(!=) (resp. e(! 0 = 0 )) is the ramication index of! (resp.! 0 ) over (resp. 0 ) and E =1or 0(resp. F =1or 0) according as E (resp. F) contains p (or 4 if p =2)or not. There are several ways to prove this result. K. Iwasawa ([4]) showed us a proof by using Galois cohomology. W. Sinnott ([8]) gave a proof by using the p-adic L-function and J. Satoh ([6]) obtained it by using the theory of ; -transforms of rational functions. In this paper, we generalize the above result to basic Z S -extensions when E and F are abelian. Let S = fp 1 ::: p s g be a nite set of primes, Z S = Q l2s Z l and Q S be the Z S -extension of Q. Then F S = FQ S is called the basic Z S -extension

2 2 Y. Ouyang and F. u of F. Let N = p n 1 1 :::pn s s and F N be the unique subextension of degree N of F S. Let h ; N denote the relative class number of F N =F + N.From a theorem of E. Friedman ([2]), when F is an imaginary abelian number eld, we have ord pi (h ; N )=; (p i S)n i + ; (p i S) where all n i are suciently large and p i 2 S. In this paper, using the relationship between ; (p i S) and the -invariant of the Dirichlet character of F, we obtain the following main result. Theorem 1. For xed p 2 S, let E and F be imaginary abelian number elds and E be ap-extension of F. Wehave ; E (p S) ; E =[E S : F S ]( ; F (p S) ; F ) +!0 (e(!0 = 0 ) ; 1) ;! (e(!=) ; 1) where the summation is taken over all places! 0 on E S (resp.! on E + S ) which do not lie above p and 0 =! 0 j (resp. F S =!j ), F + and e(!=) (resp. S e(! 0 = 0 )) is the ramication index of! (resp.! 0 ) over (resp. 0 ) and E =1or 0(resp. F =1or 0) according as E (resp. F) contains p (or 4 if p =2)or not. 2. Preliminaries. Let p 2 S beaxedprimenumber and put 4 p =2 q = p p 6= 2. Let! p be the Teichmuller character mod q. For every m 2 Zwith (m p)=1 and m 6= 1, we have with m 1 m =! p (m)(1 + m 1 p n m ) 2 Z p (m 1 p)=1andn m being a positive integer. We letq (p) denote the basic Z p -extension on Q and T = S ;fpg. Let O be a ring of integers of a nite extension over Q p and let f()= a 0 + a 1 + :::2O[[]] be a non-zero power series. We dene (f) =minford p a i : i 0g (f) =minfi 0:ord p a i = (f)g: Clearly we have (fg)=(f) +(g) and(fg)=(f)+(g) iff g are non-zero elements of O[[]]. So and can be dened in the quotient eld of O[[]] in a natural way. Let Z S denote the unit group of Z S.So Z S = U S V S where V S is the torsion part of Z S and U S = Q l2s (1+2lZ l). Let hi S and! S denote the projections from Z S to U S and V S respectively. When s =1,! S is the Teichmuller character. Let be an odd primitive Dirichlet character

3 Riemann{Hurwitz formula 3 with values in C p, where C p is a xed completion of the algebraic closure of Q p.any primitive Dirichlet character whose conductor is divisible only by the primes in S can be regarded as a character of Z S. Such acharacter is called of the second kind for S if it is trivial on V S.For a character of the second kind for S, wehave the decomposition = (p) (T ), where (p) (resp. (T ) ) is of the second kind for p (resp. T ) (see [9]). Let be an odd primitive Dirichlet character with values in C p. Fix a generator u of U p. When! p is not of the second kind for p, we dene where () =(g ( ; 1)) g ( ; 1) 2 2O[[ ; 1]] with g (u s ; 1) = L p (s! p ) and L p (s! p )isthep-adic L-function associated with! p. When! p is of the second kind for p, we dene () =;1. The following proposition is Theorem 1 of [6]. Proposition 1. Let be an odd primitive Dirichlet character, be an even primitive Dirichlet character and O be the integer ring of the eld generated overq p by the values of and. Suppose (1) has a p-power order and its conductor l is a prime number, (2) for all a 2 Z (a)=(a)(a). Then (i) if 6=! p ;1, we have ()+p n l =q () = () where } is a prime ideal of O above p, (ii) if =! ;1, we have p () = pn l if (l) 1mod}, if (l) 6 1mod}, q ; 1: Remark 1. This proposition can also be proved by using the p-adic L-function (see [8, x2]). Proposition 2. Let be an odd primitive Dirichlet character of order prime to p, be an even primitive Dirichlet character of p-power order and (a) =(a)(a). Suppose the conductor f() of is prime to p. Write f() = Q l lk l, where k l 1 and l are primes. Then

4 4 Y. Ouyang and F. u (i) k l =1,for all l, (ii) if 6=! p ;1, then if =! p ;1, then () =()+ l (l)=1 () = l p n l q p n l q ; 1: P r o o f. (i) By the Chinese Remainder Theorem, we have = Q l l, where l k l is the conductor of l and l has p-power order. If k l 6= 1, consider the natural map i : Z=(l k l )! Z=(l k l;1 ): For any x 2 ker i, x has l-power order. Thus l (x) isanl-power root of unity. Note l has p-power order and (p l) =1,andsowehave l (x) = 1. This is a contradiction because l k l is the conductor of l. (ii) When 6=! p ;1, the assertion follows from Proposition 1 and (i) since (l) 1mod} if and only if (l) =1.If =! p ;1, then l 1modp since l has p-power order. Therefore (l) 1mod} and we are done by Proposition The number of splitting primes. Let k be a nite abelian extension of Q. In this section, we compute the numberofprimesofk S above a prime number l, which is closely related to the characters of the Galois group. The character group of an abelian pronite group G is the set of continuous homomorphisms from G to the roots of unity inc p with the induced topology. We denote this character group as G^. Now wetake 2 Gal(k S =Q)^. Then ker is a close subgroup with nite index of Gal(k S =Q) (an open subgroup) and is essentially a usual Dirichlet character. Let k be the subeld of k S xed by ker. Then we dene 0 if l is ramied in k (l) =, (Frob l ) if l is unramied in k. Keeping the above notations, we have the following lemma: Lemma 1. For any prime number l, (i) there are nitely many primes in k S above l, (ii) the number of primes above l in k S is equal to #f 2 Gal(k S =Q)^ : (l)=1g:

5 Riemann{Hurwitz formula 5 P r o o f. (i) First consider S = fpg. LetQ be a prime in k above l. If l = p, the assertion is trivial by [10, Lemma 13.3]. If l 6= p, then Q is unramied in k S =k. Write l =! p (l)(1 + p nl l 1 ): Then the number of primes of k above Q is equal to #(Gal(k S =k)=hfrob Q i) #(Gal(Q (p) =Q)=hFrob l i) [k : Q] p n l [k : Q] < 1 and the case s = 1 is proved. If s>1, let D(Q) be the decomposition group of Q. Then D(Q) isa closed subgroup of S and has the form p t 1 1 Z p1 :::p t s s Z ps 0 t i 1 i= 1 ::: s, where p 1 i Z pi =0.Itissucienttoprove that t i < 1, i =1 ::: s. Suppose t i = 1. Let k (p i) L be a basic Z pi -extension of k and D (p i) (Q) be the decomposition group of Q over k (p i).sowehave D (p i) (Q) =D(Q)j Gal(k (pi ) =k) =0: This is a contradiction to the case of s = 1 and (i) is proved. (ii) Let D(l) denote the decomposition group of a prime in k S above l. Then the number of primes in k S above l is equal to #(Gal(k S =Q)=D(l)) = #((Gal(k S =Q)=D(l))^) This is the desired result. =#f 2 Gal(k S =Q)^ : (l) =1g: Remark 2. Lemma 1 is not true for arbitrary Z S -extension (see [10, Ex. 13.2]). From Lemma 1, we immediately have the following lemma: Lemma 2. Suppose k \Q S = Q, p 2 S with p-[k : Q], T = S ;fpg and l is a prime number dierent from p. Then the number of prime ideals above l in kq S is #f 2 Gal(kQ T =Q)^ : (l) =1g#f 2 Gal(Q (p) =Q)^ : (l) =1g P r o o f. By Lemma 1, it is sucient to prove #f 2 Gal(kQ S =Q)^ : (l) =1g Since we have =(p nl =q)#f 2 Gal(kQ T =Q)^ : (l) =1g: =#f 2 Gal(kQ T =Q)^ : (l) =1g#f 2 Gal(Q (p) =Q)^ : (l) =1g: Gal(kQ S =Q) = Gal(kQ T =Q) Gal(Q (p) =Q) Gal(kQ S =Q)^ = Gal(kQ T =Q)^ Gal(Q (p) =Q)^:

6 6 Y. Ouyang and F. u Therefore for any 2 Gal(kQ S =Q)^, wehave = T p, with T 2 Gal(kQ T =Q)^, p 2 Gal(Q (p) =Q)^ and (l) = T (l) p (l). Note p (l) isa p-power root of unity and T (l) isnot,sowehave and Lemma 2 is proved. (l) =1, T (l) = 1 and p (l) =1 4. Proof of Theorem 1. First let k be a nite abelian extension of Q and we use the following notations associated with k: k (resp. k ; ): the set of all (resp. odd) Dirichlet characters associated with k. k (l) (resp. k ; (l)): all the elements of k (resp. k ; ) whose conductors are divisible by a prime number l. J k (l): all the elements of k whose conductors are prime to a prime number l. We write k for an element of k and fk as the conductor of k. Let e, f and g denote the usual meaning as the ramication index, the residue class degree, the number of splitting primes respectively. For a prime number l, by [10, Th. 3.7], we have #J k (l) =f k (l)g k (l) and #( k =J k (l)) = e k (l): Now E, F are the same as in Section 1. Let K be the maximal p-extension of Q in E, andl be the maximal extension of Q in E with p-[l : Q].! (resp.! 0 ) is a prime of E + S (resp. E S ) which does not lie over the prime p, =!j F + S (resp. 0 =! 0 j F S )andu =!j L + S (resp. u0 =!j L S ). Suppose!j Q = l 6= p. Since the residue eld at u or u 0 has no nite p-extensions, it is clear that f(!=u) =f(! 0 =u 0 )=1.Furthermore, We also note the following: e K (l)=e(! 0 =u 0 ) e K + (l) =e(!=u) #J K + = g(!=u) #J K = g(! 0 =u 0 ): 1) It is easy to check that if Theorem 1 holds for two ofe=f K=F and E=K, it holds for the third. This allows us to reduce ourselves to the case where [F : Q] is not divisible by p for p>2. 2) We can also assume E \ F S = F, F \ Q S = Q and the conductor of E is not divisible by qp, since any number eld between E and E S has the same (p S)-invariant as that of E. 3) By the above assumption, wehave [E S : F S ]=[E : F] and E \Q S = Q. With the above notations, we have the following lemma:

7 Riemann{Hurwitz formula 7 Lemma 3.!0 (e(!0 = 0 ) ; 1) ;! = 8P >< >: (e(!=) ; 1) P l pn l;1 # K (l)#f F (T ) : F odd F (T ) (l) =1g if p>2, l 2n l;2 f# ; K (l) ; [E : F]# F\K (l) ; g#f L (T ) : L (T ) (l)=1g, if p =2, where! 0 (resp.!) runs over all the primes in E S (resp. E + S ) which do not lie over p, l runs over all the prime numbers dierent from p and (T ) is taken over the characters of Gal(Q T =Q). () Proof. Wehave!0 (e(!0 =u 0 ) ; 1) ;! = u0 (e(!=u) ; 1) g(! 0 =u 0 )(e(! 0 =u 0 ) ; 1) ; u g(!=u)(e(!=u) ; 1): If p>2, then F = L, = u, 0 = u 0 and K = K +. By Lemma 2, we have () = l6=p = l6=p # K (l) u0 \Q =l 1 ; l6=p # K (l) u\q =l # K (l)#f F (T ) : F (T ) (l) =1gp n l;1 1 ; l6=p # K (l)#f F + (T ) : F + (T ) (l)=1gp n l;1 = l6=p # K (l)p n l;1 #f F (T ) : F odd F (T ) (l)=1g: For p =2,wehave F L and L = L +.So (1) () = # K (l) 1 ; # K + (l) 1 l6=p u0 j Q=l l6=p ujq=l Let E = F. Wehave (2) = l6=p # ; K (l) 2n l;2 #f L (T ) : L (T ) (l)=1g: 0 (e(0 =u 0 ) ; 1) ; (e(=u) ; 1) = 2 n l;2 # ; K \F (l)#f (T L ) : (T ) L (l) =1g: l6=p Since [E S : F S ]=[E : F] andf(! 0 = 0 )=1,wehave

8 8 Y. Ouyang and F. u e(! 0 = 0 )g(! 0 = 0 )=[E : F] e(! 0 =u 0 )=e(! 0 = 0 )e( 0 =u 0 ): Then [E : F] 0 (e(0 =u 0 ) ; 1) = 0 g(! 0 = 0 )(e(! 0 =u 0 ) ; e(! 0 = 0 )) =!0 (e(!0 =u 0 ) ; e(! 0 = 0 )): The same is true for! u. By (1) and (2), we obtain!0 (e(!0 = 0 ) ; 1) ;! (e(!=) ; 1) = 2 nl;2 f# ; (l) ; [E : K F]#; K \F (l)g#f (T L ) : (T ) L (l)=1g: l6=p Now we begin our proof of the main theorem. Proof of Theorem 1. Weknowthatforany imaginary abelian eld k, (p S) satises the following relation (cf. [9]): ; k (p S)= k + (T ) ( (T ) ) where the outer sum is taken over all odd characters of k=q and the inner sum is taken over all (T ) 2 Gal(Q T =Q)^ with ( (T ) ) 6= 0,and k =1if and only if! p is a character of k=q. Therefore () ; E (p S) ; E = ( L (T ) K ) where K L is odd. Eodd ( (T ) E )= (T ) L K When p>2, the conductor of 2 Gal(K=Q)^ is not divisible by p since f E is a not divisible by p 2 and [K : Q]isap-power. Note L = F and K = K + in this case. By Propositions 1 and 2, we have () = ( (T ) F )+ Fodd =[E : F] K Fodd (T ) (T ) ( F (T ) )+ =[E : F]( ; F (p S) ; F )+ l6=p =[E : F]( ; F (p S) ; F ) ljf (K) F (T ) (l)=1 Fodd p n l;1 # K (l) p n l;1 (T ) (T ) l6=p F (T ) (l)=1 # K (l)p n l;1 Fodd F (T ) (l)=1 1

9 Riemann{Hurwitz formula 9 + l6=p p n l;1 # K (l)#f F (T ) : F odd F (T ) (l)=1g =[E : F]( ; F (p S) ; F )+!0 (e(!0 = 0 ) ; 1) ;! (e(!=) ; 1): If p =2,thenL = L +, L F and the conductor of each character of K is not divisible by 8.By[6,Th.1], Kodd ( K )= l6=p 2 n l;2 # ; K (l) ; [K+ : Q]: Since K\F is an imaginary abelian extension of Q,wecanchoose a primitive odd character 0 of Gal((F \ K)=Q) with order 2. Then, for any 2 ; K, we have = 0 e with e 2 K +. By Propositions 1 and 2, we have Kodd Therefore (3) L (T ) 6=1 () = L ( K L (T ) ) = Kodd = l6=p 2 n l;2 # ; K (l)#f L (T ) 6=1: L (T ) (l) =1g +[K + : Q] Kodd =[K + : Q] + l6=p (T ) L (T ) 6=1 L (T ) 6=1 ( K L (T ) ) L (T ) 6=1 ( 0 L (T ) ): ( K L (T ) )+ Kodd ( 0 L (T ) ) ; 1 ( K ) 2 n l;2 # ; K (l)#f L (T ) : L (T ) (l) =1g: If we sete = F in the above equality, then we obtain (4) ; F (2 S) ; F =[K + \ F : Q] + l6=p L (T ) 6=1 ( 0 L (T ) ) ; 1 2 n l;2 # ; K \F (l)#f L (T ) : L (T ) (l) =1g: By (3){(4) we obtain the desired result since [E + : F + ][K + \F : Q]=[K + : Q] and [E : F] =[E + : F + ]=[E S : F S ]. Acknowledgements. Both authors would like to thank the referee for pointing out some mistakes and misprints both in English and in math-

10 10 Y. Ouyang and F. u ematics. The second author was supported by Alexander von Humboldt Foundation and National Natural Science Foundation of China. References [1] N. Childress, -invariants and ; -transforms, Manuscripta Math. 64 (1989), 359{375. [2] E. Friedman, Ideal class groups in basic Z p1 ::: Z ps extensions of abelian number elds, Invent. Math. 65 (1982), 425{440. [3] K. Iwasawa, On ; -extensions of algebraic number elds, Bull. Amer. Math. Soc. 65 (1959), 183{226. [4], Riemann{Hurwitz formula and p-adic Galois representations for number elds, Toh^oku Math. J. (2) 33 (1981), 263{288. [5] Y. Kida, l-extensions of CM-elds and cyclotomic invariants, J. Number Theory 2 (1980), 519{528. [6] J. Satoh, The Iwasawa p-invariants of ; -transforms of the generating functions of the Bernoulli numbers, Japan. J. Math. 17 (1991), 165{174. [7] W. Sinnott, On the -invariant of the ; -transform of a rational function,invent. Math. 75 (1984), 273{282. [8], On the p-adic L-functions and the Riemann{Hurwitz genus formula, Compositio Math. 53 (1984), 3{17. [9], ; -transforms of rational function measures on Z S,Invent. Math. 89 (1987), 139{157. [10] L. C. W a s h i n g t o n, Introduction to Cyclotomic Fields, Grad. Texts in Math. 83, Springer, Department of Mathematics Current address: University of Science and Technology of China School of Mathematics Hefei, Anhui University of Minnesota People's Republic of China Minneapolis, Minnesota U.S.A. Received on and in revised form on (2793)