CYCLOTOMIC UNITS AND DIVISIBILITY OF THE CLASS NUMBER OF FUNCTION FIELDS
Jaehyun Ahn and Hwanyup Jung
Abstract. Let k = F
q(T ) be a rational function field. Let ℓ be a prime number with (ℓ, q −1) = 1. Let K/k be an elementary abelian ℓ-extension which is contained in some cyclotomic function field. In this paper, we study the ℓ-divisibility of ideal class number h
Kof K by using cyclotomic units.
1. Introduction
Let K be a finite abelian number field. When K is a real extension, the class number h K of K is difficult to calculate. If the conductor of K is divisible by many primes q 1 , . . . , q s , where each q i is congruent to 1 modulo a prime ℓ, then h K has a tendency to be divisible by a higher power of ℓ. There are several ways to show results of this type. Cornell and Rosen([3], [4]) showed this result using unramified ℓ-extensions and central extensions of K. There is another way to show similar results using the group of cyclotomic units and its index in the full unit group O K ∗ of K. It is carried out by Kuˇcera ([9]) for p = 2 and compositum K of quadratic fields and by Greither, Hachami and Kuˇcera ([5]) for odd prime p. In function field case, Bae and Jung ([1]) obtained some results of ℓ-rank of ideal class groups of real cyclotomic function fields following Cornell and Rosen. In this paper, we show similar results following the idea in [5] and compare our results with those in ([1]). Now we state our result precisely.
Let A = F q [T ] be the ring of polynomials over a finite field F q with q elements and k = F q (T ). Let q = p f with p = char(k) and f >
0. For each M ∈ A, one uses the Carlitz module to construct a field extension k M , called the M -th cyclotomic function field and its maximal real subfield k + M . For a finite abelian extension F of k which is contained
Received December 7, 2001. Revised March 14, 2002.
2000 Mathematics Subject Classification: 11R58, 11R60.
Key words and phrases: function field, class number, cyclotomic unit.
in some cyclotomic function field, we call M the conductor of F if k M is the smallest cyclotomic function field which contains F . Also we call F real if F is contained in k + M where M is the conductor of F .
Let ℓ be a prime rational number with (ℓ, q −1) = 1. For s ∈ Z, s > 0, we write S = {1, . . . , s}. Let Q 1 , . . . , Q s be distinct monic irreducible polynomials in A. In addition, if ℓ 6= p, we require that each Q i satisfies q deg(Qi) ≡ 1 mod ℓ. We choose e 1 , . . . , e s ∈ Z so that e i = 1 (resp.
e i ≥ 2) for ℓ 6= p (resp. ℓ = p) for each i ∈ S. Let K i be any elementary abelian ℓ-extension of k in k Qei
i
for each i ∈ S and δ i be the ℓ-rank of G i = Gal(K i /k). Since (ℓ, q − 1) = 1, each K i is real, i.e., K i ⊆ k + Qei
i
. It is known ([1, Section 5]) that δ i = 1 if ℓ 6= p and δ i ≤ f × deg(Q i ) × (e i − 1 − [(e i − 1)/p]) if ℓ = p.
Let K be the compositum of all K i , i ∈ S and G = Gal(K/k). Since G ≃ Q s
i=1 G i , G is an elementary abelian ℓ-group of rank δ = P s i=1 δ i . Let O K be the integral closure of A in K. Let h K be the ideal class number of O K . Our main result is the following theorem.
Main Theorem. With notations as above, the ideal class number h K is divisible by ℓ Qsi=1(1+δ
i)−s δ−1 .
Now we assume that K i is the maximal elementary abelian ℓ-extension of k in K Qei
i
. Then δ i equals to f × deg(Q i ) × (e i − 1 − [(e i − 1)/p]) for ℓ = p. In [1], Bae and Jung showed that h K is divisible by ℓ s(s−3)/2 (resp. ℓ Pi<jδ
iδ
j− P
iδ
i) for l 6= p (resp. for ℓ = p). For ℓ 6= p, our Main Theorem says that h K is divisible by ℓ 2s−s
2−1 . Note that 2 s − s 2 − 1 >
−s
2−1 . Note that 2 s − s 2 − 1 >
s(s − 3)/2 for s > 4. For ℓ = p, elementary calculations show that Y s
i=1
(1 + δ i ) ≥ 1 + X
i
δ i + X
i<j
δ i δ j + ( X s
i=3
s i
/s) X
i
δ i , and so Q s
i=1 (1 + δ i ) − sδ − 1 is greater than P
i<j δ i δ j − P
i δ i for s > 4.
Thus our result gives larger ℓ-factor of h K than the result in [1] for s > 4.
In [1], authors also give results for the case that ℓ divides q − 1. These cases corresponds to the case p = 2 in the number field case. Thus it will be interesting to consider the function field analogue of Kuˇcera’s result ([9]).
2. Cyclotomic units
We keep all the notations of the preceding section. In this section,
we give some basic facts of the cyclotomic function fields and cyclotomic
units which are needed in the proof of Main Theorem. Let k ac be an algebraic closure of k. Then k ac becomes an A-module (called Carlitz module) under the following action: For u ∈ k ac and M ∈ A, define
u M = M (ϕ + µ T )(u),
where the map ϕ is defined as ϕ(u) = u q and µ T is defined as µ T (u) = T u. It is known that the set Λ M of roots of u M = 0 generates an abelian extension k(Λ M ) of k, called the M -th cyclotomic function field and we denote it by k M for simplicity. By a primitive M -th torsion point, we mean a generator of Λ M as an A-module. The following facts are basic in the cyclotomic function field theory.
Proposition 2.1 ([8]). (i) Gal(k M /k) ∼ = (A/M ) ∗ .
(ii) k + M is the fixed field of F ∗ q in k M , and it is the maximal subfield of k M , where ∞ = (1/T ) splits completely.
(iii) If (M, N ) = 1, then k M and k N are linearly disjoint over k.
(iv) If M = P n , a power of monic irreducible P , then N kM/k (λ M ) = P . Here λ M is a primitive M -torsion point.
Recall that the group D of cyclotomic numbers of K is defined as the subgroup of K ∗ generated by F ∗ q and all elements N kN/K
N(λ A N ) with N, A ∈ A where K N = k N ∩ K. Let C = D ∩ O K ∗ , called the group of cyclotomic units of K (cf. [7, Section 3]). For ∅ 6= I ⊂ S, we set M I = Q
i∈I Q e i
i, δ I = P
i∈I δ i , K I the compositum of K i , i ∈ I and x I = N k
MI
/K
I(λ I ), where λ I is a fixed primitive M I -th torsion point. It is known that D is generated by F ∗ q ∪ {x σ I : σ ∈ G, ∅ 6= I ⊂ S}, and D (resp. C) has rank ℓ δ + s − 1 (resp. ℓ δ − 1). The index of C in the full unit group O K ∗ is related to the ideal class number h K .
Proposition 2.2 ([2], Corollary 3.11).
[O ∗ K : C] = (q − 1) ℓδ−1 h K . As the classical case, we have the following.
Lemma 2.3. Let N, P, Q ∈ A with P monic irreducible, N = P Q. If P ∤ Q, we let σ = (P, k Q /k) −1 ∈ Gal(k Q /k). Then we have
N kN/k
Q(λ N ) =
( λ P N if P | Q ,
(λ P N ) (1−σ) if P ∤ Q, Q 6∈ F ∗ q .
Note that λ P N is also primitive Q-torsion point and so λ P N = λ τ Q for some
τ ∈ Gal(k Q /k).
Proof. First we note that (k N : k Q ) =
( q deg(P ) if P |Q, q deg(P ) − 1 if P ∤ Q.
Let W = W(QA/N A) be a complete set of representatives of QA/N A consisting of monic polynomials. If P |Q, then
Gal(k N /k Q ) ≃ {(1 + X) + N A ∈ (A/N A) ∗ : X ∈ W}
and so we have
N kN/k
Q(λ N ) = Y
X∈W
λ 1+X N = Y
X∈W
(λ N + λ X N ).
We claim that Λ P = {λ X N : X ∈ W}. Clearly λ X N 6= λ Y N for any distinct X, Y ∈ W and (λ X N ) P = 0. Since |W| = |QA/N A| = |A/P A| = |Λ P |, we get the claim. Since
u P = Y
λ∈Λ
P(u + λ) = Y
X∈W
(u + λ X N ), we have
λ P N = Y
X∈W
(λ N + λ X N ) = N kN/k
Q(λ N ).
Now, suppose that P ∤ Q. There exists X 0 ∈ QA such that X 0 ≡ −1 mod P . Without loss of generality, we may assume that X 0 ∈ W. Then P |(X 0 + 1); so X 0 + 1 is not prime to N . Hence we have
Gal(k N /k Q ) ≃ {(1 + X) + N A ∈ (A/N A) ∗ : X ∈ W, X 6= X 0 } and
N kN/k
Q(λ N ) = Q
X∈W (λ N + λ X N ) λ N + λ X N
0.
Let 1 + X 0 = P B. Since X 0 ∈ QA, (P, k Q /k) −1 = (B, k Q /k) = σ and so λ N + λ X N0 = λ P B N = (λ P N ) B = (λ P N ) (B,kQ/k) = (λ P N ) σ .
/k) = (λ P N ) σ .
Therefore, we get N kN/k
Q(λ N ) = (λ P N ) 1−σ .
Let σ i1 , . . . , σ iδi be the generators of Gal(K S /K S−{i} ). For i ∈ S, 1 ≤ j ≤ δ i , define N i,j = 1 + σ ij + · · · + σ ℓ−1 ij , the norm corresponding to the group hσ ij i. For ∅ 6= I ⊂ S, let e I = {(i, j) : i ∈ I, 1 ≤ j ≤ δ i }. For J ⊂ e I, we define
x I,J = x
Q
(i,j)∈J
N
i,jI .
For any subset J of e I, we say that J satisfies the condition (∗) for I if
|{j : (i, j) ∈ J}| < δ i for each i ∈ I. Note that, if ℓ 6= p, then J = ∅ is the only one which satisfies (∗) and so x I,J is just x I . For ∅ 6= I ⊂ S, J ⊂ e I, define
S I,J = { Y
i∈I,1≤j≤δ
i,(i,j)6∈J
σ a ijij : 0 ≤ a ij ≤ ℓ − 2}.
Let B = {x
Q
1≤j≤δi
σ
ℓ−1ij{i} : 1 ≤ i ≤ s} ∪ {x σ I,J : ∅ 6= I ⊂ S, J ⊂ I, J satisfies(∗), σ ∈ S e I,J }.
Proposition 2.4. B is a Z-basis of D/F ∗ q .
Proof. First we calculate the cardinality of B. We have
|B| = X
∅6=I⊂S
X
J⊂e I,J satisfies (∗)
(ℓ − 1) δI−|J| + s.
Then, X
J⊂ eI Jsatisfies (∗)
(ℓ − 1) δI−|J| = X
J⊂e I
(ℓ − 1) δI−|J| − X
i∈I
X
J⊂e I, f {i}⊂J
(ℓ − 1) δI−|J|
+ X
i,j∈I i<j
X
J⊂e I, ] {i,j}⊂J
(ℓ − 1) δI−|J| − · · ·
= X
I
′⊂I
µ(I ′ , I)g(I ′ ),
where µ(I ′ , I) = (−1) |I|−|I′| , the M¨ obius function on the subsets of S and g(I ′ ) = P
J⊂ e I
′(ℓ − 1) δI′−|J| = ℓ δ
I′. So, by the M¨ obius inversion formula, we have
ℓ δ = g(S) = X
I⊂S
X
J⊂ eI Jsatisfies (∗)
(ℓ − 1) δI−|J| .
Thus, |B| = ℓ δ + s − 1. Therefore, the cardinality of B is equal to the Z-rank of D. It remains to show that B generate D modulo F ∗ q . The remaining part of the proof is so directly analogous to Greither, Hachami and Kuˇcera’s proof in the number field case that the reader should refer to ([5, Proposition 1.2]).
For ℓ 6= p, since J = ∅ is the only subset of e I which satisfies (∗), we have
B = {x σ I : ∅ 6= I ⊂ S, σ ∈ S I } ∪ {x σ {i}i1ℓ−1 : i = 1, . . . , s},
where S I = { Q
i∈I σ i1 a
i: 0 ≤ a i ≤ ℓ − 2}.
3. Proof of the main theorem For ∅ 6= I ⊂ S and J ⊂ e I, we define
y I,J = x e I,JI,J with e I,J = Y
i∈I
Y
1≤j≤δ
i,(i,j)6∈J
(1 − σ ij ) ℓ−2 .
Let L be the subgroup of k ∗ generated by Q 1 , . . . , Q s . For J ⊂ e I, we call J satisfy the condition (∗∗) for I if, for each i ∈ I, |{j ∈ {1, . . . , δ i } : (i, j) ∈ J}| = δ i − 1. Clearly, the condition (∗∗) is a sufficient condition for the condition (∗) and so, if ℓ 6= p, only J = ∅ satisfies the condition (∗∗). We define the subgroup U of C generated by {y σ I,J : ∅ 6= I ⊂ S, J ⊂ e I, J satisfies (∗∗), σ ∈ G}.
Proposition 3.1. For any u ∈ U and σ ∈ G, there are ψ u (σ) ∈ L and f u (σ) ∈ D such that u 1−σ = ψ u (σ)f u (σ) ℓ and ψ u (σ) is uniquely determined modulo L ℓ .
Proof. Since (ℓ, q − 1) = 1, K ℓ ∩ L = L ℓ . From this, the uniqueness of ψ u (σ) modulo L ℓ follows. As in [5], it suffices to prove the proposition for u = y I,J with ∅ 6= I ⊂ S, J ⊂ e I, J satisfies (∗∗) and σ = σ ij with i ∈ I, (i, j) 6∈ J.
We use induction on |I|. Suppose that the proposition is true for all u generated by y I′,J
′ with I ′ I and J ′ ⊆ e I ′ with J ′ satisfying (∗∗). Note that (1 − σ ij ) ℓ−1 (1 − σ ij ) = (1 − σ ij ) ℓ−1 = N i,j + ℓα for some α ∈ Z[σ ij ].
Thus
y I,J 1−σij = x (Ni,j+ℓα)
+ℓα)
Q
i′∈I−{i}
Q
1≤j′≤δi′,(i′,j′)6∈J
(1−σ
i′ j′)
ℓ−2I,J
= (x N I,Ji,j) Qi′Q
j′(1−σ
i′ j′)
ℓ−2w ℓ for some w ∈ D. Note that x N I,Ji,j = x
Q
j′(1−σ
i′ j′)
ℓ−2w ℓ for some w ∈ D. Note that x N I,Ji,j = x
Q
1≤j≤δi
N
i,jI,J− f {i} and that Q
1≤j≤δ
iN i,j is the norm corresponding to Gal(K I /K I−{i} ).
If I − {i} 6= ∅, by Lemma 2.3, we have x N I,Ji,j = (x 1−τ
I−{i},J− f {i} ) τ
′where τ, τ ′ ∈ G with τ = (Q i , K I−{i} /k). Note that J − f {i} ⊂ ^ I − {i} and satisfy (∗∗) for I − {i}. From the induction hypothesis, we have that
y 1−σ I,J ij = (y 1−τ
I−{i},J− f {i} ) τ
′w ℓ
can be written in the desired form and so it remains to prove the case I = {i}. But when I = {i}, we have
x N I,Ji,j = x
Q
1≤j≤δi
N
i,jI = N k
Qei
i
/k (λ Q
eii