Math. J. Okayama Univ. 58 (2016), 125–132

ON A DUALITY OF GRAS BETWEEN TOTALLYPOSITIVE AND PRIMARY CYCLOTOMIC UNITS
Humio Ichimura
Abstract. Let K be a real abelian field of odd degree over Q, and C
the group of cyclotomic units of K. We denote by C+ and C0 the totally
positive and primary elements of C, respectively. G. Gras found a duality between the Galois modules C+ /C 2 and C0 /C 2 by some ingenious
calculation on cyclotomic units. We give an alternative proof using a
consequence (=“Gras conjecture”) of the Iwasawa main conjecture andthe standard reflection argument. We also give some related topics.

1. Introduction
Let K be a real abelian field with [K : Q] odd, and let ∆ = Gal(K/Q). Let
E be the group of units of K, E+ the subgroup consisting of totally positive
units, and E0 the subgroup consisting of units ǫ satisfying ǫ ≡ u2 mod 4 forsome u ∈ K × . (A unit satisfying this congruence is often called a “primary”
unit.) Denote by C the group of cyclotomic units in the sense of Sinnott [11,page 209]. We put C+ = C ∩ E+ and C0 = C ∩ E0 . Let χ be a nontrivial
Q̄2 -valued character of ∆, and O = Oχ the subring of Q̄2 generated overZ2 by the values of χ. Here, Q̄2 is a fixed algebraic closure of the 2-adic
rationals Q2 , and Z2 is the ring of 2-adic integers. For a Z2 [∆]-module M ,we denote by M (χ) = M ⊗ O the χ-part of M , where the tensor product istaken over Z2 [∆] regarding O as a Z2 [∆]-module via χ. We naturally regard
M (χ) as an O-module. We see that (E/E 2 )(χ) ∼= O/2O by a theorem of
Minkowski on units of a Galois extension over Q (cf. Narkiewicz [9, Theorem
3.26]). Since the index [E : C] is finite ([11, Theorem 4.1]), it follows
that (C/C 2 )(χ) ∼= O/2O. Therefore, each of the O-modules (E+ /E 2 )(χ),(E0 /E 2 )(χ), (C+ /C 2 )(χ) and (C0 /C 2 )(χ) is either trivial or isomorphic to
O/2O. In [4, Théorème III.2], Georges Gras found the following beautiful
relation between the Galois modules C+ /C 2 and C0 /C 2 .
Theorem 1 (Gras). Under the above setting, we have (C+ /C 2 )(χ) ∼= O/2O
2−1∼
if and only if (C0 /C )(χ ) = O/2O.
Gras proved this duality by some ingenious calculation on the “Fermat quotient” of certain cyclotomic units. The main purpose of this paper is to givea modern alternative proof using a consequence (=“Gras conjecture”) of theMathematics Subject Classification. Primary 11R18.
Key words and phrases. cyclotomic units, reflection argument, ideal class group.
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H. ICHIMURA

Iwasawa main conjecture and the standard reflection argument.
As for the groups E+ and E0 , Taylor [12] proved the following
Theorem 2 (Taylor). Under the above setting, the following conditions areequivalent:(I) At least one of (E+ /E 2 )(χ) and (E+ /E 2 )(χ−1 ) is nontrivial.
(II) At least one of (E0 /E 2 )(χ) and (E0 /E 2 )(χ−1 ) is nontrivial.
Taylor proved this theorem by effectively using some properties of norm
residue symbols including the product formula. We give an alternative proof
using the reflection argument.
Remark 1. Assertion (∗) of [12, page 157] reads that if both of (E+ /E 2 )(χ)
and (E+ /E 2 )(χ−1 ) are nontrivial, then the condition (II) in Theorem 2
holds. However, what Taylor actually proved in his paper is the above
stronger version.
In §2, we recall some fundamental facts on the reflection argument, which
is a basic tool of this paper. In §3, we prove Theorem 1. In §4, we prove
Theorem 2 and give an example which shows that a duality corresponding
to Theorem 1 does not hold in general for E+ and E0 . In §5, we give a slight
generalization of a theorem of Cornacchia [2, Theorem 1] on the trivialityof the minus and plus class groups of a cyclotomic field of prime conductor
by using a reflection theorem and “genus theory”,
2. Reflection
In this section, we recall some fundamental facts on the reflection argument for the 2-parts of ideal class groups. Typical literatures on this topic
are [8, 10, 12]. We use the same notation as in §1. In particular, K denotes
a real abelian field of odd degree over Q. For a number field N , let AN (resp.
ÃN ) be the 2-part of the ideal class group of N in the ordinary (resp. narrow) sense. We often abbreviate as A = AK and Ã = ÃK in the following.
For an ideal A of K, we denote by [A]0 (resp. [A]∞ ) the ideal class in theordinary (resp. narrow) sense containing A. Let H/K and H̃/K be the classfields corresponding to the quotients A/A2 and Ã/Ã2 , respectively. Thenthe Galois group Gal(H/K) (resp. Gal(H̃/K)) is canonically isomorphic to
A/A2 (resp. Ã/Ã2 ) via the reciprocity law map which is compatible withthe action of ∆. For a multipicative abelian group X and an element x ∈ X,
we denote by [x] the class in the quotient X/X 2 represented by x. Let V
and Ṽ be the subgroups of K × /(K × )2 such that
H = K(v 1/2 [v] ∈ V )and H̃ = K(v 1/2 [v] ∈ Ṽ ),DUALITY OF GRAS ON CYCLOTOMIC UNITS

127

respectively. Clearly, we haveV = {[v] ∈ Ṽv ≫ 0}.

Here, we write x ≫ 0 when an element x ∈ K × is totally positive. We
can naturally regard the groups V and Ṽ as modules over Z2 [∆]. It is wellknown that
(E(K × )2 /(K × )2 ) ∩ Ṽ = E0 (K × )2 /(K × )2 (= E0 /E 2 )and that(E(K × )2 /(K × )2 ) ∩ V = (E+ ∩ E0 )(K × )2 /(K × )2 (= (E+ ∩ E0 )/E 2 )(cf. Washington [13, Excercise 9.3]). We have a nondegenerate pairing
Ã/Ã2 × Ṽ → {±1};([c], [v]) → hc, vi = (v 1/2 )ρc −1 ,where ρc denotes the automorphism of H̃/K corresponding to the ideal classc. It is well known and easy to show that the pairing satisfies hcδ , v δ i = hc, vi
for any c ∈ Ã, [v] ∈ Ṽ and δ ∈ ∆. Because of this relation, the above pairing
induces a nondegenerate subpairing(1)(Ã/Ã2 )(χ) × Ṽ (χ−1 ) → {±1}for each Q̄2 -valued character χ of ∆. Similarly, we have a nondegeneratepairing(2)(A/A2 )(χ) × V (χ−1 ) → {±1}.

For each element [v] of V or Ṽ , we have vOK = A2 for some ideal A of K
where OK is the ring of integers of K. By mapping [v] to the ideal class[A]0 ∈ A, we obtain exact sequences(3){0} → E0 /E 2 → Ṽ → 2 A

and
(4)

{0} → (E+ ∩ E0 )/E 2 → V → 2 A

compatible with the action of ∆. Here, 2 A denotes the elements c of A
with c2 = 1. Let K>0 be the subgroup of K × consisting of totally positiveelements. We also need the following natural exact sequence
(5)

f

g

{0} → K × /EK>0 → Ã → A → {0}which is compatible with the action of ∆. For an element α ∈ K × , let [α]∞be the class in K × /EK>0 containing α. The maps f and g are defined byf ([α]∞ ) = [αOK ]∞ and g([A]∞ ) = [A]0 , respectively. It is known that the
O-module (K × /EK>0 )(χ) is either trivial or isomorphic to O/2O and thatthe following equivalence holds.
(6)(K × /EK>0 )(χ) = {0} ⇐⇒ (E+ /E 2 )(χ) = {0}.

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This is because the Galois module K × /K>0 is isomorphic to F2 [∆] via thesign map where F2 is the finite field of 2 elements.
Lemma 1. Under the above setting, if A(χ) is trivial and A(χ−1 ) is nontrivial, then we have (E+ /E 2 )(χ) ∼= O/2O.
Proof. As A(χ−1 ) is nontrivial, V (χ) is nontrivial by (2). Then, since A(χ)
is trivial, it follows from (4) that((E+ ∩ E0 )/E 2 )(χ) = V (χ) ∼= O/2O.
Hence, we obtain (E+ /E 2 )(χ) ∼= O/2O.

Theorem 3. Under the above setting, the following conditions are equivalent.
(I) At least one of Ã(χ) and Ã(χ−1 ) is trivial.
(II) The groups A(χ) and A(χ−1 ) are both trivial.
Proof. The implication (II) ⇒ (I) follows immediately from [10, Théorème2]. To show (I) ⇒ (II), assume that Ã(χ) is trivial. We see that A(χ) is also
trivial, and that (E+ /E 2 )(χ) = {0} from (5) and (6). Then it follows from
Lemma 1 that A(χ−1 ) is trivial.
Remark 2. Theorem 3 is a refinement of [10, Corollary 2c].
3. Proof of Theorem 1
We use the same notation as in the previous sections. In particular, χdenotes a nontrivial Q̄2 -valued character of ∆. The following consequenceof the Iwasawa main conjecture was proved in Greither [5, Theorem 4.14]:(7)|(E/C)(χ)| = |A(χ)|,

which is called a conjecture of Gras. Here, we abbreviate the χ-part of theZ2 [∆]-module (E/C)⊗ Z2 as (E/C)(χ), the tensor product being taken overZ.
Lemma 2. If |(E/C)(χ)| > 1, then ((C+ ∩ C0 )/C 2 )(χ) ∼= O/2O.
Proof. Assume that |(E/C)(χ)| > 1. Then, as (E/E 2 )(χ) ∼= O/2O, it
2follows that there exists a nontrivial class [c] ∈ (C/C )(χ) for which c is asquare in E. Hence, c ∈ C+ ∩ C0 , and the assertion follows.
Proof of Theorem 1. First we assume that (C+ /C 2 )(χ) ∼= O/2O and show
the “only if” part. Namely, we show that (C0 /C 2 )(χ−1 ) ∼= O/2O. If A(χ−1 )is nontrivial, then it follows from (7) and Lemma 2 that (C0 /C 2 )(χ−1 ) ∼=−1
O/2O. Therefore, we may further assume that A(χ ) is trivial. Let us
show that Ã(χ) is nontrivial. For this, we choose a cyclotomic unit η ∈ C+
for which the class [η] generates (C+ /C 2 )(χ) (∼= O/2O). If η is a squareDUALITY OF GRAS ON CYCLOTOMIC UNITS

129

in K, then |(E/C)(χ)| > 1. Therefore, it follows from (7) that A(χ) is
nontrivial, and hence so is Ã(χ). If η is not a square in K, then, as(E/E 2 )(χ) ∼= O/2O, (E/C)(χ) is trivial. As η is totally positive, this implies that (E/E 2 )(χ) = (E+ /E 2 )(χ) ∼= O/2O. Then it follows from (5) and(6) that Ã(χ) is nontrivial also when η is nonsquare. Now, we see from (1)that Ṽ (χ−1 ) is nontrivial. However, since A(χ−1 ) is trivial, it follows from
(3) that(E0 /E 2 )(χ−1 ) = Ṽ (χ−1 ) ∼= O/2O.
−1
Again as A(χ ) is trivial, we obtain from this and (7) that (C0 /C 2 )(χ−1 ) ∼=O/2O.
We assume that (C0 /C 2 )(χ) ∼== O/2O and show that (C+ /C 2 )(χ−1 ) ∼O/2O. Similarly as in the proof of the “only if” part, we may as wellassume that A(χ−1 ) is trivial because of (7) and Lemma 2. We choose a
cyclotomic unit η ∈ C0 for which the class [η] generates (C0 /C 2 )(χ). First,
we deal with the case where η is a square in K. In this case, we have(E/C)(χ) 6= {0} and hence A(χ) is nontrivial by (7). Then, by Lemma
1, we see that (E+ /E 2 )(χ−1 ) ∼= O/2O. Since A(χ−1 ) is trivial, it follows
from this and (7) that (C+ /C 2 )(χ−1 ) ∼= O/2O. Next, we deal with thecase where η is not a square. Then, [η] is a nontrivial element of Ṽ (χ). Ifη ≫ 0, then as [η] ∈ V (χ), it follows from (2) that A(χ−1 ) is nontrivial, acontradiction. If η is not totally positive, we have [η] ∈ Ṽ (χ) \ V (χ). Hence,|(Ã/Ã2 )(χ−1 )| > |(A/A2 )(χ−1 )| by (1) and (2). Then, by (5) and (6), we
see that (E/E 2 )(χ−1 ) = (E+ /E 2 )(χ−1 ) ∼= O/2O. Since A(χ−1 ) is trivial, itfollows from this and (7) that (C+ /C 2 )(χ−1 ) ∼= O/2O.
4. Proof of Theorem 2
In this section, we show Theorem 2 by using the reflection argument.
Proof of (II) ⇒ (I). Assume that (E0 /E 2 )(χ) ∼= O/2O but (E+ /E 2 )(χ−1 )is trivial. Then we see that Ã(χ−1 ) = A(χ−1 ) from (5) and (6), and henceṼ (χ) = V (χ). Since (E0 /E 2 )(χ) is contained in Ṽ (χ), this implies that
(E0 /E 2 )(χ) = ((E+ ∩ E0 )/E 2 )(χ) ⊆ (E+ /E 2 )(χ).
Thus we obtain (E+ /E 2 )(χ) ∼= O/2O.



Proof of (I) ⇒ (II). Assume that (E+ /E 2 )(χ) ∼= O/2O but that both of22−1(E0 /E )(χ) and (E0 /E )(χ ) are trivial. Then, from (3), we obtain inclusions(8)

V (χ) ⊆ Ṽ (χ) ֒→ 2 A(χ) andV (χ−1 ) ⊆ Ṽ (χ−1 ) ֒→ 2 A(χ−1 ).

Combining these inclusions with (1) and (2), we see thatrk(2 A(χ)) = rk(V (χ−1 )) ≤ rk(2 A(χ−1 )) = rk(V (χ)) ≤ rk(2 A(χ)).

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H. ICHIMURA

Here, for an O-module M , rk(M ) denotes the dimension of M/M 2 over thefinite field O/2O. Thus, the inclusions (8) yield the isomorphisms(9)

V (χ) = Ṽ (χ) ∼= 2 A(χ)

and V (χ−1 ) = Ṽ (χ−1 ) ∼= 2 A(χ−1 )and the equalities(10)

rk(2 A(χ)) = rk(2 Ã(χ))

andrk(2 A(χ−1 )) = rk(2 Ã(χ−1 )).

Since we are assuming that (E+ /E 2 )(χ) ∼= O/2O, we see from (6) that×∼
(K /EK>0 )(χ) = O/2O. Hence, there exists a nontrivial class [α]∞ in(K × /EK>0 )(χ) with α ∈ K × . By the exact sequence (5) and the firstequality in (10), we see that f ([α]∞ ) = [αOK ]∞ = [A]2∞ for some ideal A of
K with [A]0 ∈ A(χ). Hence, αxOK = A2 for some totally positive elementx ∈ K × . By [A]0 ∈ 2 A(χ), it follows from the first isomorphism in (9) that
there exists a unit ǫ for which [αxǫ] ∈ V (χ). This implies that αxǫ is totally
positive, and hence the class [α]∞ ∈ (K × /EK>0 )(χ) is trivial. This is acontradiction.

Example. Let K be the real abelian field of degree 7 and conductor 491.
It is known that A(χ−1 ) ∼= O/2O but A(χ) is trivial for some nontrivial Q̄2 -valued character χ of ∆ = Gal(K/Q). (Note that χ and χ−1 arenot conjugate over Q2 .) For this, see Cornacchia [3, §5], or Koyama andYoshino [7, §8] combined with [13, page 421]. It follows from (2) that V (χ)
is nontrivial and V (χ−1 ) is trivial. From the latter, it follows that ((E+ ∩E0 )/E 2 )(χ−1 ) is trivial. Since V (χ) is nontrivial and A(χ) is trivial, we see
from (4) that ((E+ ∩ E0 )/E 2 )(χ) ∼= O/2O. Therefore, both of (E+ /E 2 )(χ)
and (E0 /E 2 )(χ) are nontrivial, while at least one of (E+ /E 2 )(χ−1 ) and(E0 /E 2 )(χ−1 ) is trivial. Thus, a duality corresponding to Theorem 1 does
not hold in general for E+ and E0 .
5. Minus class group
We use the same notation as in the previous sections. Let k/Q be an
imaginary abelian extension of 2-power degree, and put L = Kk. Let k +and L+ be the maximal real subfields of k and L, respectively. Let A−
L be
the kernel of the norm map AL → AL+ . We naturally identify the Galois
groups Gal(L/k) and Gal(L+ /k + ) with ∆ = Gal(K/Q), and regard A−
L and
AL+ as modules over ∆. When L coincides with a cyclotomic field of prime
conductor, Cornacchia [2, Theorem 1] used Theorem 1 and (7) to obtaina relation between the triviality of A−
L and AK . We slightly generalize his
result as follows. Let S be the set of prime numbers ℓ such that a prime
ideal of k + over ℓ ramifies in k. Let χ be a nontrivial Q̄2 -valued characterof ∆, which we also regard as a primitive Dirichlet character.

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Theorem 4. Under the above setting, assume further that at most one
prime ideal of K ramifies in L+ . Then the following conditions are equivalent.
− −1(I) At least one of A−
L (χ) and AL (χ ) is trivial.
(II) The groups AK (χ) and AK (χ−1 ) are both trivial, and χ(ℓ) 6= 1 forall ℓ ∈ S.
The case L = Q(ζp ) is due to Cornacchia, where p is an odd prime number.
The assumption in Theorem 4 on ramification in L+ /K is rather strong.
Actually, when the assumption is satisfied, we see that the conductor of
the real abelian field k + is an odd prime number or a power of 2, and inparticular k + /Q is cyclic. To show the general case, we use a reflection
theorem (Theorem 3) and “genus theory”, and do not use Theorem 1 nor(7).
Proof. In [6, Corollary 2], we showed that A−
L (χ) is trivial if and only if thenarrow class group ÃL+ (χ) is trivial and χ(ℓ) 6= 1 for all ℓ ∈ S. We obtained
this by using the exact hexagon of Conner and Hurrelbrink [1, Theorem 2.3],
which is a kind of genus theory. By the assumption on ramification, L+ /K
is a cyclic 2-extension in which exactly one prime ideal of K is ramified.
Therefore, we see that ÃL+ (χ) is trivial if and only if so is ÃK (χ). Now we
obtain the assertion from Theorem 3.
References[1] P. E. Conner and J. Hurrelbrink, Class Number Parity, World Sci. Publishing, Singapore, 1988.
[2] P. Cornacchia, The parity of the class number of the cyclotomic fields of prime conductor, Proc. Amer. Math. Soc., 125 (1997), 3163-3168.
[3] P. Cornacchia, The 2-ideal class groups of Q(ζℓ ), Nagoya Math. J., 162 (2001), 1-18.
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H. ICHIMURA[11] W. Sinnott, On the Stickelberger ideal and the circular units of an abelian field,
Invent. Math., 62 (1980), 181-234.
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1997.
Humio Ichimura
Faculty of Science,
Ibaraki University
Bunkyo 2-1-1, Mito, 310-8512, Japane-mail address: hichimur@mx.ibaraki.ac.jp(Received September 2, 2013 )(Accepted November 8, 2013 )