ON TAMELY RAMIFIED IWASAWA MODULES FOR THE CYCLOTOMIC p -EXTENSION OF ABELIAN FIELDS

Itoh, T. Osaka J. Math. 51 (2014), 513 536 ON TAMELY RAMIFIED IWASAWA MODULES FOR THE CYCLOTOMIC p -EXTENSION OF ABELIAN FIELDS TSUYOSHI ITOH (Received May 18, 2012, revised September 19, 2012) Abstract Let p be an odd prime, and k ½ the cyclotomic p -extension of an abelian field k. For a finite set S of rational primes which does not include p, we will consider the maximal S-ramified abelian pro-p extension M S (k ½ ) over k ½. We shall give a formula of the p -rank of Gal(M S (k ½ )k ½ ). In the proof of this formula, we also show that M {q} (k ½ )L(k ½ ) is a finite extension for every real abelian field k and every rational prime q distinct from p, where L(k ½ ) is the maximal unramified abelian pro-p extension over k ½. 1. Introduction Let k be an algebraic number field, and p a prime number. We denote by k ½ k the cyclotomic p -extension (i.e. the unique p -extension contained in the field generated by all p-power roots of unity over k). Let S be a finite set of rational primes, and MS É (k ½ ) the maximal pro-p extension of k ½ unramified outside S (i.e., the primes of k ½ lying above the primes in S are only allowed to ramify in É MS (k ½ )k ½ ). When p ¾ S, the structure of É XS (k ½ ) Gal( É MS (k ½ )k ½ ) is already studied (see, e.g., Iwasawa [8], Neukirch Schmidt Wingberg [11]). In particular, É XS (k ½ ) is a free pro- p group under certain conditions. Recently, the structure of É XS (k ½ ) for the case that p S is also studied by several authors (Salle [12], Mizusawa Ozaki [10],... ). In this case, it seems that É XS (k ½ ) does not have a simple structure. Then, to study É XS (k ½ ), it is important to study the structure of its abelian quotient. Let M S (k ½ )k ½ be the maximal abelian pro-p extension unramified outside S. In the present paper, we shall consider X S (k ½ ) Gal(M S (k ½ )k ½ ) for the case that p S. Since Gal(k ½ k) acts on X S (k ½ ), we can use Iwasawa theoretic arguments. X S (k ½ ) is called the S-ramified Iwasawa module. (See, e.g., [12], [6]. This is also called a tamely ramified Iwasawa module when p S.) If a p -module M satisfies dim Ép M Å p É p r ½, we say that the p -rank of M is r, and we write rank p M r. Our purpose of the present paper is giving a formula of rank p X S (k ½ ) when k is an abelian extension of É (abelian field) and p is an odd prime. (In this case, we can show that X S (k ½ ) is finitely generated over p.) 2010 Mathematics Subject Classification. Primary 11R23; Secondary 11R18.

514 T. ITOH We shall give a remark about giving a formula of rank p X S (k ½ ). Let L(k ½ ) be the maximal unramified abelian pro-p extension of k ½. We put X(k ½ ) Gal(L(k ½ )k ½ ). Then X(k ½ ) is the (usual) Iwasawa module, and rank p X(k ½ ) is called the Iwasawa -invariant. In general, it is hard to write explicitly. Since X(k ½ ) is a quotient of X S (k ½ ), we consider it is sufficient to obtain a formula including at the present time. (That is, we will only give a formula of rank p Gal(M S (k ½ )L(k ½ )), actually.) However, for abelian fields, the plus part of is conjectured to be 0 (Greenberg s conjecture), and the minus part of can be computed (at least theoretically) from the Kubota Leopoldt p-adic L-functions (Stickelberger elements). We also mention that formulas of rank p X S (k ½ ) are already obtained for several cases. In particular, it can be said that the p -rank of the minus part of X S (k ½ ) for CM-fields is already known (see Section 2). Salle [12] studied X S (k ½ ) for the case that k is an imaginary quadratic field (or É) with p 2. Moreover, when k É, a formula of rank p X S (É ½ ) (including the case that p 2) is shown by Mizusawa, Ozaki, and the author [6] (as a corollary, a general formula for imaginary quadratic fields is also given). In the present paper, we shall extend the method given in [6] for abelian fields. The following theorem is crucial to prove the formula of rank p X S (É ½ ). Theorem A (see [6, Theorem 3.1]). Let q be a rational prime distinct from p. Then M {q} (É ½ )É ½ is a finite extension. At first, we will generalize Theorem A to real abelian fields (under the condition that p is an odd prime). Theorem 1.1. Assume that p is odd. Let k be a real abelian field, and q a rational prime distinct from p. Then M {q} (k ½ )L(k ½ ) is a finite extension. We remark that M {q} (k ½ )L(k ½ ) can be infinite when k is an imaginary abelian filed. (For example, see [12], [9], [6], or Section 6 of the present paper.) Similar to [6], Theorem 1.1 plays an important role to prove our formula of rank p X S (k ½ ) for abelian fields. In Section 2, we shall state some basic facts, and give preparations for proving Theorem 1.1. We will prove Theorem 1.1 in Sections 3 and 4. In Section 5, we shall give a simple remark about a generalization of Theorem 1.1. In Section 6, we shall give a formula of rank p X S (K ½ ) for abelian fields (Theorem 6.4). The formula is given as the -quotient version. We also give examples with applying this formula for some simple cases. 2. Preliminaries Firstly, we shall recall some basic facts from class field theory. Let p be an odd prime number, and k an algebraic number field. (In the following of the present paper, we assume that p is odd.) We denote by k ½ k the cyclotomic p -extension. For a

ON TAMELY RAMIFIED IWASAWA MODULES 515 non-negative integer n, let k n be the nth layer of k ½ k (that is, the unique subfield of k ½ such that k n k is a cyclic extension of degree p n ). Let S be a finite set of rational primes which does not include p. For an algebraic extension (not necessary finite) K of É, let M S (K) be the maximal abelian (pro-)p-extension of K unramified outside S, and L(K) the maximal unramified abelian (pro-) p-extension of K. For an abelian group G, let Ç G be the p-adic completion of G (that is, Ç G lim GG pn ). As noted in Section 1, we shall mainly consider the p -rank of Gal(M S (k ½ )L(k ½ )). We will write several facts which is also stated in [6]. In this paragraph, assume that S is not empty. By class field theory, we have the following exact sequence: ÇE kn n Å q¾s (O kn q) Gal(M S (k n )L(k n )) 0, where E kn is the group of units of k n, O kn is the ring of integers of k n, and n is the natural homomorphism induced from the diagonal embedding. (We will give a remark on the structure of (O kn q). Assume that the prime decomposition of q O kn q e 1 1 q e r r. Then Å (O kn q) r (O kn q i ) i1 because(o kn q e i (O kn q i ).) We put E ½ lim Ç E kn, and R q lim (O kn q), where the projective limits are taken with respect to the natural mappings induced from the norm mapping. Then we obtain the following exact sequence: i ) is E ½ ½ Å q¾s R q Gal(M S (k ½ )L(k ½ )) 0. In the cyclotomic p -extension, all (finite) primes of k are finitely decomposed. Then R q is a finitely generated p -module. From this, we also see Gal(M S (k ½ )L(k ½ )) is finitely generated over p. On the other hand, the theorem of Ferrero Washington [2] implies that Gal(L(k ½ )k ½ ) is a finitely generated p -module, if k is an abelian field. Hence, we see that for every abelian field k, X S (k ½ ) is finitely generated over p. (We can see that the p -rank of X S (k ½ ) is always finite in general.) It seems hard to determine the cokernel of ½ directly. REMARK. When the base field k is a CM-field, the minus part of E ½ is easy to compute (we are also able to compute the minus part of R q ). Hence we can obtain a formula of the p -rank of the minus part of Gal(M S (k ½ )L(k ½ )). This idea is already known (see, e.g., [12], [9], [6]).

516 T. ITOH Secondly, we shall give some preparations to prove Theorem 1.1. Let be a prime number satisfying q p. For simplicity, we will write M p ( ), M q ( ) instead of M {p} ( ), M {q} ( ), respectively. Lemma 2.1. Let k ¼ k be a finite extension of algebraic number fields. If Gal(M q (k ¼ ½ )L(k¼ ½ )) is finite, then Gal(M q(k ½ )L(k ½ )) is also finite. Proof. We may assume that k ¼ k ½ k. Let N n Ï (O k ¼ n q) (O kn q) be the homomorphism induced from the norm mapping. We can see that the order of the cokernel Coker(N n ) is bounded as n ½. (Proof: Since there are only finitely many primes in k ½ lying above q, the p-rank of (O kn q) is bounded. Moreover, the exponent of Coker(N n ), is also bounded.) Since Gal(M q (k ¼ n )L(k¼ n )) (resp. Gal(M q(k n )L(k n ))) is isomorphic to a quotient of (O k ¼ n q) (resp. (O kn q) ), N n induces the homomorphism Gal(M q (kn ¼ )L(k¼ n ))Gal(M q(k n )L(k n )). From the above fact, the order of the cokernel is bounded as n ½. Assume that Gal(M q (k½ ¼ )L(k¼ ½ )) is finite. Then we can show that the order of Gal(M q (kn ¼ )L(k¼ n )) is bounded as n ½. From the above fact, we see that the order of Gal(M q (kn ¼ )L(k¼ n )) is also bounded. Hence Gal(M q(k ½ )L(k ½ )) is finite. From Lemma 2.1 and the theorem of Kronecker Weber, we may replace a real abelian field k to the maximal real subfield of a cyclotomic filed containing k to show Theorem 1.1. For a positive integer d, let d be the set of all dth roots of unity, and É( d ) the dth cyclotomic field. Lemma 2.2. Let f be a positive integer which is prime to p, and m a positive integer. We put K É( f p m ) and k K (the maximal real subfield of K ). If q does not split in K k, then M q (k ½ ) L(k ½ ). Proof. Let q be an arbitrary prime of k lying above q. It is well known that if q does not split in K, then the order of (O k q) is not divisible by p. (Proof: We denote by k q the completion of k at q. Under the assumption, k q does not contain p. By the structure of the group of units in k q, we obtain the assertion.) Since Gal(M q (k)l(k)) is isomorphic to a quotient of (O k q), we see M q (k) L(k). We note that the nth layer k n of k ½ k is the maximal real subfield of É( f p mn). Hence by using the same argument, we also see M q (k n ) L(k n ) for all n 1. This implies that M q (k ½ ) L(k ½ ). From the above arguments, it is sufficient to prove Theorem 1.1 under the following conditions:

ON TAMELY RAMIFIED IWASAWA MODULES 517 (A) k is the maximal real subfield of K É( f p m ), where f and m are positive integers and f is prime to p. Every prime lying above q splits in K k, and is not decomposed in k ½ k (the latter can be satisfied by taking m sufficiently large). 3. Properties of certain Kummer extensions In this section, we shall give some key results to prove Theorem 1.1. Assume that K, k, and q satisfy (A) in Section 2. We will construct certain infinite Kummer extensions over K ½. We shall use some fundamental results given in Khare Wintenberger [9]. We define the terms Case NS and Case S as follows: CASE NS: every prime lying above p does not split in K k. CASE S: every prime lying above p splits in K k. Moreover, we use the following notation (in Sections 3 and 4): J: complex conjugation, q 1,, q r : prime ideals of k lying above q, p 1,, p t : prime ideals of k lying above p, Q i, Q J i (i 1,, r): prime ideals of K lying above q i, P j ( j 1,, t): (unique) prime ideal of K lying above p j (Case NS), P j, P J j ( j 1,, t): prime ideals of K lying above p j (Case S). Following Greenberg [3], we denote by s the number of primes of k which is lying above p and splits in K. Hence we see that s 0 for Case NS, and s t for Case S. Note that every prime lying above p are totally ramified in k ½ k by the assumption on k. Hence s is also the number of primes of k ½ which is lying above p and splits in K ½. By the assumption, K ½ contains all p n th roots of unity. For an element x of K, we define K ½ ( p½ Ô x) n1 K ½ ( pn Ô x). More precisely, K ½ Ô ( p½ x) is the union of all finite Kummer extensions Kn Ô ( pn x) for n 1 (note that K n contains p n). Similarly, for a finitely generated subgroup T of Ô K, we define the extension K ½ ( p½ T )K½ by adjoining all p n th roots of the elements contained in T. As noted in [9], K ½ Ô ( p½ x) K ½ if and only if x is a root of unity. The following result is helpful to prove the results stated in this section. Theorem B (see Khare Wintenberger [9, Lemma 2.5]). Let T be a finitely generated subgroup of K, and S a finite set of ( finite) primes of K. Let I be the subgroup of Gal(K ½ ( p½ T )K½ ) generated by the inertia subgroups for the primes in S. Ô For a prime r ¾ S, let K r be the É completion Ç of K at r. We denote by T the closure of the diagonal image of T in r¾s K Ç r. (Recall that K r is the p-adic completion of K r.) Then rank p I rank p T.

518 T. ITOH We shall construct several Kummer extensions unramified outside {p, q} over K ½ by following the method given in [6]. (See also Greenberg [4].) Let k D be the decomposition field of K É for q. By the assumption, k D is an imaginary abelian field and [k D Ï É] 2r. Let Q 1,, Q r, Q J 1,, Q J r be the primes of k D lying below Q 1,, Q r, Q J 1,, QJ r respectively. We can take a positive integer h such that Q h 1 is a principal ideal generated by «1, and «1 1 ¾ P for every prime ideal P of k D lying above p. We note that Q 1 ( ¾ Gal(k D É)) is the complete set of primes in k D lying above q, and (Q 1 )h («1 ). We write all conjugates of «1 for Gal(k D É) as the following: «1, «2,, «r, «J 1, «J 2,, «J r (these are distinct elements because q splits completely in k D ). Moreover, we put i «i «i J for i 1,, r. For Case S (i.e. p splits in K k), we define 1,, t ¾ K as follows. We can ( j ) for all j 1,, t. We put Û j (O K P j ) take an integer h ¼ such that P h¼ j and j ( j J j )Û j for j 1,, t. DEFINITION. We put T q i i 1,, r, which is a subgroup of (k D ) (and hence also a subgroup of K ). For Case NS, we put T T q. For Case S, we put T i, j i 1,, r, j 1,, t. Moreover, we put N q K ½ ( p½ Ô Tq ) and N K ½ ( p½ Ô T ). (Of course, N N q Case NS.) for Lemma 3.1. (1) N and N q are abelian extensions over k ½. (2) N K ½ and N q K ½ are unramified outside {p, q}. (3) Gal(N K ½ ) rs p and Gal(N q K ½ ) r p. (Recall that s 0 for Case NS, and s t for Case S.) Proof. (1) Since J acts on T as 1, then J acts on Gal(N K ½ ) and the action is trivial. This implies that N k ½ is an abelian extension. The assertion for N q follows similarly. (2) Note that all elements contained in T (resp. T q ) are {p,q}-units. Hence N K ½ (resp. N q K ½ ) is unramified outside {p, q}. (See, e.g., [9, Proposition 2.4].) (3) It is easy to see that T is a free -module of rank r s. Let Ç T be the closure of T in Ç K. Then the p -rank of Ç T is r s. As noted in [9] (see Remarks

ON TAMELY RAMIFIED IWASAWA MODULES 519 after the proof of [9, Lemma 2.2]), this fact implies that rank p Gal(N K ½ ) r s. Since Gal(N K ½ ) is generated by r s elements, we see that Gal(N K ½ ) rs p. The assertion for Gal(N q K ½ ) can be proven quite similarly. Lemma 3.2 (cf. Greenberg [4]). lying above Q i is ramified in K ½ ( p½ Ô i )K ½. (2) N q M p (K ½ ) is a finite extension over K ½. (1) For every i 1,, r, the unique prime Proof. The assertions can be shown easily by using Theorem B. Note that (2) is already mentioned in [4, p. 149]. Proposition 3.3. Let I p be the subgroup of Gal(N K ½ ) generated by all inertia groups for the prime lying above p. Then I p has finite index in Gal(N K ½ ). Proof. First, we consider Case NS. We assume that p does not split in K k. Hence s 0, T T q, N N q, and there are just t primes in K lying above p. By Lemma 3.1, it is sufficient É to show Ç that rank p I p r. Let T q be the closure of the diagonal image of T q in tj1 K P j. By Theorem B, we see that rank p I p rank p T q, hence we shall show rank p T p r. Recall that k D is the decomposition field of K É for q, and T q is also a subgroup of (k D ). We denote by P 1,, P u the primes of k D lying above p (where u t). Let É h1½ be the closure of the diagonal image of T u q in (kp D h ). We claim that rank p Tq ¼ rank p T q. By the definition of T q, every element x of T q satisfies x 1 ¾ P h for all h 1, É, u. Let UP 1 h be the group of principal units of kp D h. We see that Tq ¼ is contained in u h1 U P 1 h. Let be the homomorphism T ¼ q u h1 U 1 P h t induced from the diagonal embedding UP 1 h and (Tq ¼) T q. Then the claim follows. We shall recall the argument given in Brumer s proof of Leopoldt s conjecture for abelian fields (see [1], [15]). Assume that rank p Tq ¼ r. Then there are elements a 1,, a r of p which satisfies j1 U 1 P j É Pp h UP 1. We can see that is injective, a 1 1 a 2 2 a r r 1 in U 1 P h for all h 1,, u, and a i 0 with some i. Since i «i «J i of «1, we also see that ¾Gal(k D É) («1 1 ) x( ) 1 in U 1 P 1 and «i is a conjugate

520 T. ITOH for all ¾ Gal(k D É), where x( ) ¾ p satisfying x( ) 0 with some. Fix an embedding kp D 1 p. By taking the p-adic logarithm of the above equation, we see ¾Gal(k D É) x( ) log p «1 1 0. This implies that the determinant of the matrix (log p «1 1 ), is 0. On the other hand, «1,, «r, «1 J,, «r J are multiplicative independent in (k D ). Then we can see that log p «1,, log p «r, log p «1 J,, log p «r J are linearly independent over É. By Baker Brumer s theorem (see Brumer [1], Washington [15, Theorem 5.29]), they are also linearly independent over É in p. Hence the determinant of the matrix (log p «1 1 ), is not 0. (This follows from the argument given in the proof of [6, Lemma 3.4] which uses [15, Lemma 5.26 (a)].) It is a contradiction. Then we conclude that rank p T ¼ q rank p T q rank p I p r. Next, we shall consider Case S. Assume that p splits in K k. That is, s t and the number of primes of K lying above p is 2t. The outline of the proof is the same as. Case NS. Let T be the closure of the diagonal image of T in É t j1 Ç K P j É t j1 Ç K P J j In this case, we shall show rank p T r t. Assume that rank p T r t. Then there are elements a 1,, a rt of p which satisfies and a 1 1 a 2 2 a r r a r1 1 a r2 2 a rt t 1 in Ç K P j, a 1 1 a 2 2 a r r a r1 1 a r2 2 a rt t 1 in Ç K P J j for all 1 j t, and a i 0 with some i. However, Ú P j ( j ) 0 (for 1 j t) by the definition of j. (Here, Ú P j is the normalized additive valuation of K with respect to P j.) This fact implies that a r j must be 0 for 1 j t. Hence we obtain the equalities and a 1 1 a 2 2 a r r 1 in U 1 P j, a 1 1 a 2 2 a r r 1 in U 1 P J j for all 1 j t. Recall that k D is the decomposition field of K É for q. We denote by P 1,, P u the primes of k D lying above p. As noted before (in the proof for Case NS), we can show UP 1 h É PP h UP 1 is injective. Hence we see a 1 1 a 2 2 a r r 1 in U 1 P h

ON TAMELY RAMIFIED IWASAWA MODULES 521 for all 1 h u, and a i 0 with some i. The rest of the proof is quite same as that of for Case NS. Since N is an abelian extension of k ½, we can take a unique intermediate field N of N k ½ which satisfies Gal(N k ½ ) rs p. Similarly, we are also able to take a unique intermediate field Nq of N q k ½ satisfying Gal(Nq k ½) r p. (Note that N q N, and N q N for Case NS.) Then we obtain the following: Proposition 3.4. N k ½ is unramified outside {p, q}, and a subgroup of Gal(N k ½ ) generated by the inertia groups for the primes lying above p has finite index. Nq M p (k ½ )k ½ is a finite extension. 4. Proof of Theorem 1.1 We will use the same notation and symbols defined in Section 3. Our strategy of the proof of Theorem 1.1 is similar to that of Theorem A. However, our situation has a difficulty which comes from the fact that Gal(M p (k ½ )k ½ ) can be non-trivial. Assume that K, k, and q satisfy (A) stated in Section 2. We shall recall and define the following symbols: M p,q (k ½ ): the maximal abelian pro-p extension of k ½ unramified outside {p, q}, M p (k ½ ): the maximal abelian pro-p extension of k ½ unramified outside p, M q (k ½ ): the maximal abelian pro-p extension of k ½ unramified outside q, L(k ½ ): the maximal unramified pro-p abelian extension of k ½, X p,q (k ½ ) Gal(M p,q (k ½ )k ½ ), X p (k ½ ) Gal(M p (k ½ )k ½ ), X q (k ½ ) Gal(M q (k ½ )k ½ ), X(k ½ ) Gal(L(k ½ )k ½ ). We also define the following notation: ¼ Gal(K ½ K ) (we often identify ¼ with Gal(k ½ k)), : fixed topological generator of ¼, : (p-adic) cyclotomic character of ¼, p [[T ]] p [[¼]]Ï 1 T, È T ( )(1 T ) 1 1 ¾. We note that X p,q (k ½ ), X p (k ½ ), X q (k ½ ), X(k ½ ), and Gal(M p (k ½ )L(k ½ )) are finitely generated torsion -modules. For a finitely generated torsion -module A, we denote by char A the characteristic ideal of A. For finitely generated torsion -modules A and B, we write A B when they are pseudo-isomorphic. We denote by X(K ½ ) Ï X(K ½ ) 1 J the minus part of X(K ½ ). We recall the fact that X p (k ½ ) relates to X(K ½ ) by Kummer duality (see, e.g., [15]). Let f (T ) ¾ be a generator of char X(K ½ ). We note that f (T ) is not divisible by p because K is an abelian field (Ferrero Washington s theorem [2]). It is

522 T. ITOH known that f ( È T ) ¾ generates char X p (k ½ ). By a result of Greenberg [3], we know that the power of T dividing f (T ) is T s, where s 0 for Case NS, and s t for Case S. Hence the power of È T dividing f ( È T ) is just È T s. For Case NS, we see that f ( È T ) is prime to È T. For i 1,,t, let k n,i be the completion of k n at the unique prime lying above p i, and U 1 (k n,i ) the group of principal units in k n,i. Let (E kn ) be the diagonal image of E kn in É i k n,i, and E n the closure of (E kn ) É i U 1 (k n,i ). We put U lim É i U 1 (k n,i ), and E lim E n, where the projective limits are taken with respect to the norm mappings. Recall the exact sequence: 0 Gal(M p (k ½ )L(k ½ )) X p (k ½ ) X(k ½ ) 0, and the fact that Gal(M p (k ½ )L(k ½ )) UE (see [15, Corollary 13.6]). We note that lim U 1 (k n,i ) contains lim È p n T for Case S (see [13], etc.). Hence U contains a submodule which is isomorphic to ( T È ) s. We also note that E n has no non-trivial p -torsion element (see, e.g., [14, Lemma 3.3]). From the above facts, we obtain the following: Lemma 4.1. There is a pseudo-isomorphism of finitely generated torsion -modules: Gal(M p (k ½ )L(k ½ )) ( È T ) s E, where E is an elementary torsion -module (see [7], [11, (5.3.9) Definition], [15, Chapter 15]) whose characteristic ideal is prime to ( È T ). Moreover, the characteristic ideal of X(k ½ ) is prime to ( È T ). (See also [14].) Let N and Nq be extensions over k ½ defined in Section 3 (see the paragraph before Proposition 3.4). Lemma 4.2 (see also Greenberg [4]). M p,q (k ½ ) M p (k ½ )N q. Proof. Although this fact is already shown in [4, pp. 148 149], we will give a detailed proof for a convenient to the reader (and our proof is slightly different). Let ÉM p,q (k ½ ) be the maximal pro-p extension of k ½ unramified outside {p, q}. By Theorem 3 of Iwasawa [8] and Ferrero Washington s theorem [2], we see that Gal( M É p,q (k ½ )k ½ ) is a free pro-p group whose minimal number of generators is r, where rank p X(K ½ ). (Note that every prime lying above q actually ramifies in É M p,q (k ½ )k ½ by Lemma 3.2.) By taking the abelian quotient of Gal( É M p,q (k ½ )k ½ ), we see that X p,q (k ½ ) r p as a p -module.

ON TAMELY RAMIFIED IWASAWA MODULES 523 On the other hand, we see that M p (k ½ ) N q k ½ is a finite extension by Proposition 3.4. Hence rank p Gal(M p (k ½ )N q k ½) rank p X p (k ½ )r r. Since N q k ½ is unramified outside {p, q}, we see M p,q (k ½ ) M p (k ½ )N q. Then we have a surjection of finitely generated p -modules X p,q (k ½ )Gal(M p (k ½ )N q k ½) whose kernel is finite. However X p,q (k ½ ) has no non-trivial p -torsion element, and hence we conclude that M p,q (k ½ ) M p (k ½ )N q. For a finitely generated torsion -module A, we can define the multiplication by ÈT endomorphism of A, and we denote by A[ È T ] (resp. A È T ) its kernel (resp. cokernel): 0 A[ È T ] A È T A A È T 0. Note that ¼ acts on Gal(M p,q (k ½ )L(k ½ )) and then it is also a finitely generated torsion -module. Let M ¼ be the intermediate field of M p,q (k ½ )L(k ½ ) corresponding to È T Gal(Mp,q (k ½ )L(k ½ )). Hence Gal(M ¼ L(k ½ )) is isomorphic to Gal(M p,q (k ½ )L(k ½ )) È T. Lemma 4.3. M q (k ½ ) is contained in M ¼. Proof. By class field theory, Gal(M q (k ½ )L(k ½ )) is isomorphic to a quotient of lim (O kn q). As a -module, lim (O kn q) is isomorphic to ( T È ) r. Hence T È annihilates Gal(M q (k ½ )L(k ½ )), and then Gal(M q (k ½ )L(k ½ )) È T Gal(M q (k ½ )L(k ½ )). From the restriction map we obtain a surjection Gal(M p,q (k ½ )L(k ½ )) Gal(M q (k ½ )L(k ½ )) 0, Gal(M p,q (k ½ )L(k ½ )) È T Gal(M q (k ½ )L(k ½ )) 0. By the definition of M ¼, we see M q (k ½ ) M ¼. Lemma 4.4. L(k ½ )N is contained in M ¼. Proof. Note that Gal(L(k ½ )N L(k ½ )) Gal(N N L(k ½ )), and Gal(N N L(k ½ )) is a subgroup of Gal(N k ½ ). By the construction of N, we see that È T annihilates Gal(N k½ ), and hence it also annihilates Gal(L(k ½ )N L(k ½ )). The rest of the proof is similar to that of Lemma 4.3.

524 T. ITOH Lemma 4.5. M ¼ L(k ½ )N is a finite extension. Proof. We shall show that rank p Gal(M ¼ L(k ½ ))rank p Gal(L(k ½ )N L(k ½ )). By Proposition 3.3, we see that N L(k ½ )k ½ is a finite extension. Hence rank p Gal(L(k ½ )N L(k ½ )) is equal to rank p Gal(N k ½ ) r s. On the other hand, M p,q (k ½ ) M p (k ½ )Nq by Lemma 4.2, and M p (k ½ ) Nq k ½ is a finite extension by Proposition 3.4. By using Lemma 4.1, we can obtain the following pseudo-isomorphisms: X p,q (k ½ ) X p (k ½ ) Gal(N q k ½) ( È T ) rs E ¼, where E ¼ is an elementary torsion -module whose characteristic ideal is prime to ( È T ). Hence, rank p X p,q (k ½ ) È T r s. The following exact sequence: induces the exact sequence: 0 Gal(M p,q (k ½ )L(k ½ )) X p,q (k ½ ) X(k ½ ) 0 X(k ½ )[ È T ] Gal(M p,q (k ½ )L(k ½ )) È T X p,q (k ½ ) È T X(k ½ ) È T 0. Since char X(k ½ ) is prime to ( T È ) by Lemma 4.1, both of X(k½ )[ T È ] and X(k½ ) T È are finite. Hence rank p Gal(L(k ½ )N L(k ½ )) is equal to rank p X p,q (k ½ ) T È rank p Gal(M ¼ L(k ½ )). For a Galois group G appeared below, we denote I(G) by the subgroup of G generated by the inertia groups for all primes lying above p. Lemma 4.6. rank p I(Gal(N L(k ½ )k ½ )) r s. Proof. We shall take a prime P of k ½ lying above p. Let I P be the inertia subgroup of Gal(N k ½ ) for P. Similarly, let I ¼ P be the inertia subgroup of Gal(N L(k ½ )k ½ ) for P. Then the restriction map induces a surjection I ¼ P I P. Hence there is a surjection I(Gal(N L(k ½ )k ½ ))I(Gal(N k ½ )). By Proposition 3.4, rank p I(Gal(N k ½ )) rs. We see that rank p I(Gal(N L(k ½ )k ½ ))rs. Since L(k ½ )k ½ is an unramified extension,i(gal(n L(k ½ )k ½ )) is contained in Gal(N L(k ½ )L(k ½ )). By these results, we see that rank p I(Gal(N L(k ½ )k ½ )) r s. We shall finish to prove Theorem 1.1. By Lemma 4.6, rank p I(Gal(N L(k ½ )k ½ )) r s. Moreover, rank p I(Gal(M ¼ k ½ )) is also r s because M ¼ N L(k ½ ) is a finite extension (Lemma 4.5). Note that rank p Gal(M ¼ L(k ½ )) is r s. Then I(Gal(M ¼ k ½ )) is

ON TAMELY RAMIFIED IWASAWA MODULES 525 a finite index subgroup of Gal(M ¼ L(k ½ )). By Lemma 4.3, M q (k ½ ) is an intermediate field of M ¼ L(k ½ ). Since M q (k ½ )k ½ is unramified at all primes lying above p, we can see that M q (k ½ ) is contained in the fixed field of I(Gal(M ¼ k ½ )). This implies that M q (k ½ )L(k ½ ) is a finite extension. We have shown Theorem 1.1 for k and q satisfying (A). Then, as noted in Section 2, we obtain Theorem 1.1 for general k and q. 5. Slight generalization of Theorem 1.1 In this section, we shall give a simple remark that the converse of Lemma 2.1 holds under a (strict) condition. (In general, the converse of Lemma 2.1 does not hold.) Lemma 5.1. k ¼ k ½ k. Let Let k ¼ k be a finite extension of algebraic number fields satisfying I n Ï (O kn q) (O k ¼ n q) be the homomorphism induced from the natural embedding. If Gal(M q (k ½ )L(k ½ )) is finite and I n is an isomorphism for all n, then Gal(M q (k½ ¼ )L(k¼ ½ )) is also finite. Proof. By the assumption, we obtain the following isomorphism Hence the homomorphism IÏ lim (O kn q) lim (O k ¼ n q). Gal(M q (k ½ )L(k ½ )) Gal(M q (k ¼ ½ )L(k¼ ½ )) induced from I is surjective. The assertion follows. We shall give an example satisfying the assumption that I n is an isomorphism for all n. Let k be an algebraic number field, and k ¼ a quadratic extension of k. Assume that every prime of k lying above q is inert in k½ ¼ (that is, every prime lying above q is inert in k ¼ and k ½ ). Let q 1,, q r be the primes of k lying above q. We also assume that p divides N(q i ) 1 for all i, where N(q i ) is the absolute norm of q i. Then (O kn q i ) is not trivial for all i, n. Under these assumptions, we obtain that N(q i O k ¼ n ) 1 N(q i O kn ) 2 1 (N(q i O kn ) 1)(N(q i O kn )1). We assumed that p is odd, and hence N(q i O kn ) 1 is prime to p. This implies that (O k ¼ n q i ) (O kn q i ) for all i, n. From this, we obtain that (O k ¼ n q) r (O k ¼ n q i ) r (O kn q i ) (O kn q). i1 i1

526 T. ITOH Since I n is injective, we see that I n is an isomorphism for all n. Under the above assumptions, if Gal(M q (k ½ )L(k ½ )) is finite, then Gal(M q (k½ ¼ )L(k¼ ½ )) is also finite by Lemma 5.1. (For the case that k ¼ k is an imaginary quadratic extension of É, see also [12], [6].) The above lemma implies that Theorem 1.1 can be generalized for some non-abelian fields. 6. p -rank of S-ramified Iwasawa modules We shall lead a formula of the p -rank of S-ramified Iwasawa modules (for general S) from Theorem 1.1. As same as Theorem 1.1, the strategy of our proof is quite similar to that of given in [6]. In this section, we will use the following notation (similar to Greither s [5] or Tsuji s [13] but slightly different): p: fixed odd rational prime, S: finite set of rational primes which does not include p, F: finite abelian extension of É unramified at p, K F( p ), K n K ( p n1), K ½ Ë n0 K n: the cyclotomic p -extension of K, G Gal(K ½ É ½ ) Gal(K É), ¼ Gal(K ½ K ), G p : Sylow p-subgroup of G, G 0 : non-p-part of G (the maximal subgroup of G consists of the elements having prime to p order), : fixed topological generator of ¼, : (p-adic) cyclotomic character, : (p-adic) Teichmüller character, J ¾ G: complex conjugation. Let be a p-adic character of G. We denote by É p () the extension of É p by adjoining the values of, and O the valuation ring of É p (). We put d [É p () Ï É p ]. Let O be a free rank one O -module such that ¾ G acts as ( ). For a p [G]-module M, we put M M Å p [G] O, which is called the -quotient in [13] (or the -part in [5, p. 451]). The functor taking the -quotient is right exact. We also put e G 1 tr Ép )É p (( )) ¾G 1 ¾ É p [G]. If p does not divide G, then M e M. In general, we see M Å p É p (M Å p É p ) e (M Å p É p ).

ON TAMELY RAMIFIED IWASAWA MODULES 527 For more informations about the -quotient, see [5], [13] for example. We also give some simple remarks. For a p [G]-module M, we put M M 1 J. Since p is odd, we have a decomposition M M M. For a character of G, we see that M (M M ) (M M )Å p [G] O (M Å p [G] O ) (M Å p [G] O ) M M. We claim that if is odd, then M is trivial. Let (a Å b) ¾ M Å p [G] O M. Note that J acts trivially on M and acts as 1 on O. Hence the equality (a Å b) (Ja Å b) (a Å Jb) (a Å b) implies that (a Å 2b) 2(a Å b) 0. Since p is odd, we obtain the claim. Similarly, we can see that if is even, then M is trivial. For a rational prime q distinct from p, we put R q lim (O Kn q). Let r be the number of primes of K ½ lying above q. Then rank p R q r. Since q is a rational prime, G acts on R q. We shall determine the p -rank of (R q ). First, we assume that q is unramified in K (i.e., the conductor of F is prime to q). Let D be the decomposition subgroup of Gal(K ½ É) for q. Then we can write D D p D 0, where D p p and D 0 is a finite cyclic group whose order is prime to p. We may regard D 0 as a subgroup of G 0. Note that Gal(K ½ É) is isomorphic to ¼ G p G 0. Then we can take a generator of D p of the from pm p with some m 0 and p ¾ G p. We also take a generator 0 ¾ G 0 of D 0. Hence D is a procyclic group generated by pm p 0. In the above choice of the generator of D p, we can see that m and p is uniquely determined. (Since D p p, every generator of D p is written by the from ( pm p ) «with «¾ p.) Lemma 6.1. R q is a cyclic p [G][[¼]]-module. Proof. Fix a sufficiently large integer n 0 such that every prime in K n0 lying above q remains prime in K m for all m n 0. Let n be an integer which satisfies n n 0.

528 T. ITOH We put G (n) Gal(K n É). Let q be a prime in K n lying above q. We remark that p n1 K n and p n2 K n. Under the assumption on n, we can see that the Sylow p-subgroup of (O Kn q) is generated by p n1 (mod q) with a generator p n1 of p n1. Let {q 1, q 2,, q r } be the set of primes of K n lying above q. We assumed that q is unramified in K ½ É, then q O Kn q 1 q 2 q r. We note that the action of G (n) on {q 1, q 2,, q r } is transitive. Take an element «n of O Kn which satisfies «n p n1 (mod q 1 ), «n 1 (mod q 2 ),, «n 1 (mod q r ). Then «n (mod q) is a generator of the Sylow p-subgroup of (O Kn q) as a p [G (n) ]- module. Hence the Sylow p-subgroup of (O Kn q) (which is isomorphic to (O Kn q) ) is a cyclic p [G (n) ]-module. We can choose a suitable set of generators {«n } such that N Km K n («m )«n (mod q) for all m n n 0. Hence we obtain the following commutative diagram with exact raws: p [G (m) ] (O Km q) 0 p [G (n) ] (O Kn q) 0, where the left vertical mapping is induced from the restriction mapping, and the right vertical mapping is induced from the norm mapping. Since lim p[g (n) ] p [G][[¼]], we obtain the assertion. Hence there is a surjection ³Ï p [G][[¼]] R q. We note that pm p 0 acts on R q as ( pm p 0 ). (Recall that is the cyclotomic character.) Then the kernel of ³ contains an ideal generated by pm p 0 ( pm p 0 ). By taking the -quotient, we obtain a surjection ³ Ï O [[¼]] (R q ), and the kernel of ³ contains ( p 0 ) pm ( pm p 0 ). We may regard (R q ) as a O [[T ]]- module via the isomorphism O [[¼]] O [[T ]] with 1 T. We put O [[T ]], and 0 ( ) ¾ 1 p p. Then we see that (R q ) is annihilated by f q, (T ) (1 T ) pm 1 ( p 0 )( p 0 ) pm 0 ¾. Let P be the maximal ideal of O. Since 0 ¾ 1 p p, if 1 ( p 0 )( p 0 ) 1 (mod P), then f q, (T ) is a unit polynomial, and hence (R q ) is trivial. We see 1 ( p 0 )( p 0 ) 1 ( 0 ) 1 ( p ) 1 ( 0 ) 1 ( p ).

ON TAMELY RAMIFIED IWASAWA MODULES 529 Note that 1 ( p ) is a p-power root of unity, and then it is congruent to 1 modulo P. Moreover, 1 ( 0 ) is a root of unity whose order is prime to p. Then 1 ( 0 ) 1 (mod P) if and only if 1 ( 0 ) 1. Consequently, we showed that if 1 ( 0 ) 1, then (R q ) is trivial. We can see that the number of characters satisfying 1 ( 0 ) 1 is just GD 0. (It is equal to the number of characters ¼ of G satisfying ¼ (D 0 )1.) Though if (R q ) is non-trivial, it is annihilated by f q, (T ), and then rank p (R q ) d p m. By considering these facts, we obtain the inequality: r rank p R q rank p (R q ) d p m GD 0 p m, where runs all representatives of the conjugacy classes satisfying 1 ( 0 ) 1 in the above sums. (We give some remarks. The second equation follows from the fact that R q Å p É p Ä (R q) Å p É p. Moreover, È d GD 0, and m is independent of.) We claim that r GD 0 p m. Let K D be the decomposition field of K ½ É for q. Then the p-part of [K D Ï É] is equal to the p-part of [K m Ï É]. Hence this is equal to G p p m. On the other hand, the non-p-part of [K D Ï É] is equal to G 0 D 0. We see [K D Ï É] G p p m G 0 D 0 GD 0 p m. Since r [K D Ï É], the claim follows. From this claim, we see that the above inequality is just an equality. Hence for all character satisfying 1 ( 0 ) 1, the p -rank of (R q ) is d p m. We also note that ( p ) 1 because p fixes all elements of p n for all n. Hence, when 1 ( 0 ) 1, we can write f q, (T ) (1 T ) pm 1 ( p ) pm 0. Next, we consider the case that q is ramified in K. Let I be the inertia subgroup of Gal(K É) for q, and K I the inertia field of K É for q. We remark that all primes lying above q are totally ramified in K n Kn I. Hence (O Kn q) (O K I n q) for all n. We put Rq I lim (O K I n q). Since q is unramified in É( p ½) (where p ½ p ½. Then the proof of Lemma 6.1 also works for K I ½. Ë n1 p n), we see that K I ½ contains Let be a character of G. If (I ) 1, then is also a character of Gal(K I É), and hence (R q ) (Rq I ). From this, if (I ) 1, we see that (R q ) is finite because rank p R q rank p Rq I rank p (Rq I ) rank p (R q ). (I )1 (I )1

530 T. ITOH We also determine the structure of (R q ) for a general case. Assume that (I )1. Then ( ) for ¾ Gal(K I É) GI is well defined. Repeating the argument given in the unramified case for K I, we can take 0 and p for q. (They are determined modulo I, and p (mod I ) is uniquely determined. Hence ( p ) is dependent only on q.) We also assume that 1 ( 0 ) 1. Since p [GI ] O, we can take f q, (T ) (1 T ) pm 1 ( p ) pm 0 as an element of O [[T ]]. We see that (R q ) is annihilated by f q, (T ) because (R q ) (Rq I ). As a consequence, we obtained the following result (see also [6, Lemma 2.1]). Proposition 6.2. Let be a character of G. Then (R q ) Å p É p is non-trivial if and only if satisfies (I ) 1 and 1 ( 0 ) 1. Moreover, if (R q ) Å p É p is non-trivial, then and rank p (R q ) d p m. (R q ) Å p É p f q, (T )Å p É p, By class field theory, we have the following exact sequence: E ½ R q Gal(M q (K ½ )L(K ½ )) 0, where E ½ lim Ç E Kn. Assume that is a non-trivial even character of G satisfying (I ) 1 and 1 ( 0 ) 1. By taking the -quotient (it is right exact), we see (E ½ ) (R q ) Gal(M q (K ½ )L(K ½ )) 0 is exact. Since (R q ) is annihilated by f q, (T ), we obtain the exact sequence: (E ½ ) f q, (T ) (R q ) Gal(M q (K ½ )L(K ½ )) 0. By tensoring with É p, we also obtain the exact sequence: (E ½ ) f q, (T )Å p É p (R q ) Å p É p Gal(M q (K ½ )L(K ½ )) Å p É p 0. We shall show the following result (see also [6]). Proposition 6.3. For every non-trivial even character of G satisfying (I )1 and 1 ( 0 ) 1, the mapping appeared above is an isomorphism. (E ½ ) f q, (T )Å p É p (R q ) Å p É p

ON TAMELY RAMIFIED IWASAWA MODULES 531 Proof. As we noted before, we have a decomposition Gal(M q (K ½ )L(K ½ )) Gal(M q (K ½ )L(K ½ )) Gal(M q (K ½ )L(K ½ )). Moreover, we already know that Gal(M q (K ½ )L(K ½ )) is trivial for every even character. Let K be the maximal real subfield of K. Since p is odd, we see that Gal(M q (K ½ )L(K ½ )) is isomorphic to Gal(M q (K½ )L(K ½ )). Hence we obtain Gal(M q (K ½ )L(K ½ )) Gal(M q (K ½ )L(K ½ )) for every satisfying the assumption. (We note that can be viewed as a character of Gal(K½ É ½).) By Theorem 1.1, we see that Gal(M q (K½ )L(K ½ )) is finite, and hence we see that Gal(M q (K ½ )L(K ½ )) Å p É p is trivial. From this, we have a surjection (E ½ ) f q, (T )Å p É p (R q ) Å p É p. By Proposition 6.2, dim Ép (R q ) Å p É p d p m. We shall calculate the dimension of (E ½ ) f q, (T )Å p É p. Let U be the projective limit of semi local units in K ½ K for the primes lying above p, and E the closure of the diagonal image of global units. (For the precise definition, see Section 4.) Since Leopoldt s conjecture is valid for all K n (see [1], [15]), we see that E is isomorphic to E ½. It is known that E Å p É p is a free cyclic Å p É p -module. (See [14, Lemma 3.5], [5].) Then we see (E f q, (T ))Å p É p (E Å p É p )( f q, (T )Å1) ( Å p É p )( f q, (T )Å1) ( f q, (T ))Å p É p. Hence we showed that dim Ép (E ½ ) f q, (T )Å p É p d p m. This implies the assertion. Here we shall state our main result. Let be an arbitrary character of G. Let S be a finite set of rational primes which does not include p. We put X S (K ½ ) Gal(M S (K ½ )K ½ ), where M S (K ½ )K ½ is the maximal abelian pro-p extension unramified outside S. In this case, G acts on X S (K ½ ) and hence its -quotient can be considered. We shall give a formula of rank p X S (K ½ ). For a prime q ¾ S, let I q be the inertia subgroup of G for q. We also write p,q, 0,q, m q as p, 0, m for q (defined before), respectively. (Recall that p,q and 0,q are determined modulo I q.) We also recall that 0 ( ).

532 T. ITOH Theorem 6.4. We put and If S is not empty, then S {q ¾ S (I q ) 1, 1 ( 0,q ) 1} f q, (T ) (1T ) pmq 1 ( p,q ) pmq 0 ¾ O [[T ]]. rank p X S (K ½ ) rank p X(K ½ ) q¾s d p m q P, where P 1 0 ( ), (: odd, ), d deg F(T ) (: even). and F(T ) lcm q¾s f q, (T ). If S is empty, then rank p X S (K ½ ) rank p X(K ½ ). Proof. We may assume that S is not empty. Recall the following exact sequence: E ½ Å q¾s R q Gal(M S (K ½ )L(K ½ )) 0 which is stated in Section 2. By Proposition 6.2, we see that (R q ) Å p É p is nontrivial if and only if q ¾ S. Hence, if S is empty, then Gal(M S (K ½ )L(K ½ )) Å p É p is trivial, and rank p X S (K ½ ) rank p X(K ½ ). In the following, we assume that S is not empty. By taking the -quotient and tensoring with É p, we have the exact sequence: (E ½ ) Å p É p Å q¾s (R q ) Å p É p Gal(M S (K ½ )L(K ½ )) Å p É p 0. It is sufficient to determine the cokernel of. Since p is odd, we have a decomposition E ½ E½ E ½, and we can see E ½ lim p n. We also note that (E ½) (E½ ) (E ½ ) for a character of G. It was already shown that if is an odd character, then (E½ ) is trivial, and hence (E ½ ) (lim p n). Assume that. Then (E ½ ) lim pn, and the natural mapping lim p n R q is injective. We also note that d 1. From these facts, we see that Ä q¾s rank p X S (K ½ ) rank p X(K ½ ) q¾s rank p (R q ) 1 rank p X(K ½ ) q¾s p m q 1.

ON TAMELY RAMIFIED IWASAWA MODULES 533 Assume that is odd and. Then (E ½ ) Å p É p is trivial. Hence we see rank p X S (K ½ ) rank p X(K ½ ) q¾s d p m q. Let be the trivial character. Then we can see that X S (K ½ ) Å p É p X S (É ½ )Å p É p, X(K ½ ) Å p É p X(É ½ )Å p É p. Hence rank p X(K ½ ) 0. We also note that S {q ¾ S ( 0,q ) 1} {q ¾ S q 1 (mod p)}. By the results for É ½ (see [6]), we see that X S (É ½ ) X S (É ½ ), and rank p X S (K ½ ) q¾s p m q max{p m q q ¾ S }. Since d 1 and f q, (T ) (1 T ) pmq pmq 0, we see that From these facts, the formula max{p m q q ¾ S } d deg F(T ). rank p X S (K ½ ) rank p X(K ½ ) q¾s d p m q d deg F(T ) is certainly satisfied. Finally, assume that is non-trivial and even. By Proposition 6.3, we see that (E ½ ) f q, (T )Å p É p (R q ) Å p É p is an isomorphism. This isomorphism implies that (E ½ ) F(T )Å p É p Å q¾s (R q ) Å p É p is injective (see also [6]). By using the same argument stated in the proof of Proposition 6.3, we obtain that dim Ép ((E ½ ) F(T ))Å p É p d deg F(T ). Hence we see that rank p X S (K ½ ) rank p X(K ½ ) q¾s d p m q d deg F(T ). We have shown the formula for all cases.

534 T. ITOH REMARK. Assume that p does not divide G. Then f q, (T )(1 T ) pmq pmq 0, and hence deg F(T ) max{p m q q ¾ S }. Moreover, we can see that p m q is equal to the number of primes of É ½ lying above q. EXAMPLE 6.5. We put K É( p ), the pth cyclotomic field (recall that p is an odd prime). In this case, every character of G Gal(K É) is written by the form i with 0 i p 2. Note also that q ¾ S is unramified in K. We may identify G with (p), and 0,q with q (mod p). Then we can see S i {q ¾ S 1 i ( 0,q ) 1} {q ¾ S i 1 (mod f q )}, where f q is the order of q in (p). Assume that S i P (i) 1 (i 1), 0 (i: odd, i 1), max{p m q q ¾ S i} (i: even). is not empty. We put By Theorem 6.4, we see rank p X S (K ½ ) i i q¾s i p m q P (i), where i rank p X(K ½ ) i is the i -part of the (unramified) Iwasawa -invariant of K ½ K. EXAMPLE 6.6. Let k be a real quadratic field with conductor d (the case that p divides d is allowed). We put K k( p ). Let be the quadratic character of G Gal(K É) corresponding to k. We may regard (resp. ) as a Dirichlet character modulo d (resp. modulo p). In this case, we see Hence S S {q ¾ S (q) 0, (q) (q)}. consists of the primes in S which satisfy: q 1 (mod p) and q splits in k, or q 1 (mod p) and q is inert in k. Assume that S. We put P max{p m q q ¾ S }, then we obtain the formula rank p X S (K ½ ) rank p X(K ½ ) q¾s p m q P. Note that rank p X(K ½ ) is equal to the (unramified) Iwasawa -invariant of k ½ k. (If Greenberg s conjecture is true for k and p, then rank p X(K ½ ) 0.) Since rank p X S (k ½ ) rank p X S (K ½ ) rank p X S (É ½ )

ON TAMELY RAMIFIED IWASAWA MODULES 535 (where is the trivial character), we can obtain a formula of the p -rank of X S (k ½ ) (including rank p X(K ½ ) ). EXAMPLE 6.7. Let FÉ be a cyclic extension of degree p. Assume that p is unramified in F. We put K F( p ), and fix a character of G Gal(K É) satisfying K ker() F. Let be a fixed generator of Gal(F ½ É ½ ) Gal(FÉ), and F (i) the fixed field of F ½ by i for 0 i p 1 (hence F (0) F). For simplicity, we assume that every prime of S is not decomposed in É ½. Hence if q ¾ S is unramified in F, then the splitting field of F ½ É for q must be one of F (0),, or F (p 1). By the definition of, we see that 1 ( ) is defined, and we put 1 ( ) (note that is a primitive pth root of unity). In this case, we obtain that S {q ¾ S q 1 (mod p), q is not ramified in F}. Under the assumption for S, we see m q 0 for all q ¾ S. When q ¾ S splits in F (i), we see that 1 ( p,q ) 1 ( i ) i, and hence f q, (T ) (1T ) i 0. We note that if i j, then (1 T ) i 0 and (1 T ) j 0 are relatively prime. We put S,i {q ¾ S q splits in F (i) }, for 0 i p 1. Assume that S. From the above facts, we see that deg F(T ) is equal to the number of non-empty S,i s. That is, deg F(T ), where {i 0 i p 1, S,i }. Since d p 1, we see rank p X S (K ½ ) rank p X(K ½ ) (p 1) i¾(s,i 1). ACKNOWLEDGEMENTS. The basic ideas of our proof of the main theorem are due to the previous paper [6]. The author express his thanks to Professors Yasushi Mizusawa and Manabu Ozaki for giving many suggestions during (and after) the collaborative work. The author also express his thanks to Dr. Takae Tsuji for answering a question, and to the referee for his/her useful comments. The author was supported by research grant from the Research Institute, Chiba Institute of Technology. References [1] A. Brumer: On the units of algebraic number fields, Mathematika 14 (1967), 121 124.

536 T. ITOH [2] B. Ferrero and L.C. Washington: The Iwasawa invariant p vanishes for abelian number fields, Ann. of Math. (2) 109 (1979), 377 395. [3] R. Greenberg: On a certain l-adic representation, Invent. Math. 21 (1973), 117 124. [4] R. Greenberg: On p-adic L-functions and cyclotomic fields. II, Nagoya Math. J. 67 (1977), 139 158. [5] C. Greither: Class groups of abelian fields, and the main conjecture, Ann. Inst. Fourier (Grenoble) 42 (1992), 449 499. [6] T. Itoh, Y. Mizusawa and M. Ozaki: On the p -ranks of tamely ramified Iwasawa modules, Int. J. Number Theory 9 (2013), 1491 1503. [7] K. Iwasawa: On Z l -extensions of algebraic number fields, Ann. of Math. (2) 98 (1973), 246 326. [8] K. Iwasawa: Riemann Hurwitz formula and p-adic Galois representations for number fields, Tôhoku Math. J. (2) 33 (1981), 263 288. [9] C. Khare and J.-P. Wintenberger: Ramification in Iwasawa modules, arxiv:1011.6393, (2010). [10] Y. Mizusawa and M. Ozaki: On tame pro-p Galois groups over basic p -extensions, Math. Z. 273 (2013), 1161 1173. [11] J. Neukirch, A. Schmidt and K. Wingberg: Cohomology of Number Fields, second edition, Grundlehren der Mathematischen Wissenschaften 323, Springer, Berlin, 2008. [12] L. Salle: On maximal tamely ramified pro-2-extensions over the cyclotomic 2 -extension of an imaginary quadratic field, Osaka J. Math. 47 (2010), 921 942. [13] T. Tsuji: Semi-local units modulo cyclotomic units, J. Number Theory 78 (1999), 1 26. [14] T. Tsuji: On the Iwasawa -invariants of real abelian fields, Trans. Amer. Math. Soc. 355 (2003), 3699 3714. [15] L.C. Washington: Introduction to Cyclotomic Fields, second edition, Graduate Texts in Mathematics 83, Springer, New York, 1997. Division of Mathematics Education Center Faculty of Social Systems Science Chiba Institute of Technology 2-1-1 Shibazono, Narashino, Chiba 275-0023 Japan e-mail: tsuyoshi.itoh@it-chiba.ac.jp