Steinitz Class of Mordell Groups of Elliptic Curves

With Complex Multiplication

Tong LIU and Xianke ZHANG

(Tsinghua University, Department of Mathematics, Beijing 100084, P. R. China)

Abstract. Let E be an elliptic curve having Complex Multiplication by the full ring O_{K} of integers of K = Q(\sqrt{−D}), let H = K(j(E)) be the Hilbert class field of K. Then the Mordell-Weil group E(H) is an O_{K}-module, and its Steinitz class St(E) is studied. When D is a prime number, it is proved that St(E) = 1 if D ≡ 3 (mod 4); and St(E) = [P]^{t} if D ≡ 1 (mod 4), where [P] is the ideal class of K represented by prime factor P of 2 in K, t is a fixed integer. General structures are also discussed for St(E) and for modules over Dedekind domain. These results develop the results by D. Dummit and W. Miller for D = 10 and some elliptic curves to more general D and elliptic curves.

1991 Mathematical Subject Classification : 11G05, 11D15, 14H52, 11R37 Keywords: elliptic curve, Mordell group, complex multiplication, Steinitz class,

module

1 Introduction

Let K = Q(\sqrt{−D}) be an imaginary quadratic number field, O_{K} the ring of all integers of K. Let E be an elliptic curve having Complex Multiplication by the ring O_{K}. E is defined over the field F = Q(j(E)), j(E) is the j-invariant of E. So H = K(j(E)) is the Hilbert class field of K^{[1]}, and the Mordell-Weil group E(H) of H-rational points of E is then naturally a module over the Dedekind domain O_{K} (via complex multiplication). By the structure theorem for finitely generated modules over Dedekind domain we have

E(H) ≅ E(H)_{tor} ⊕ O_{K} ⊕ ⋯ ⊕ O_{K} ⊕ A = E(H)_{tor} ⊕ O_{K}^{s−1} ⊕ A, here A is an ideal of O_{K} which is uniquely determined up to a multiplication by a number from K. Thus E(H) uniquely defines an ideal class [A] of K represented by A ; [A] is said to be the Steinitz class of E and denoted by St(E). (Similarly, any module M over a Dedekind domain R defines an ideal class of R, which is said to be the Steinitz class of M and denoted by St(M).) So the structure of the Mordell group E(H), as a module over the Dedekind domain O_{K}, is uniquely determined by its rank s, Steinitz class St(E), and torsion part. Therefore, it is important to determine the Steinitz class St(E). D. Dummit and W. Miller ^{[2]} in 1996 determined the Steinitz class for some specific elliptic curves when D = 10 and also found some properties of them.

Let l = rank_{Z}(E(F)) be the Z-rank of E(F). Let [A] denote the ideal class represented by ideal A of O_{K}. Since St(E) is concerned only with the free part of E(H), we put E(·)_{f} = E(·)/E(·)_{tor}, i.e., the quotient group of the Mordell group E(·) modulo its torsion part. Note that E(·)_{f} is isomorphic to the free part of E(·). This notation will be used also to subgroups of Mordell groups.

We here will analysis the interior structure of E(H), give a general theorem for the structure of modules over Dedekind domain, and then determine Steinitz classes St(E) for some types of elliptic curves. In particular, when D = p is a prime number and p ≡ 3 (mod 4), we will prove that St(E) is the principal class of K; And when the prime number D = p ≡ 1 (mod 4) , we will show that

St(E) = [P]^{t}, t = l + log |H^{1}(G, E(H)_{f})|,

where |H^{1}(G, E(H)_{f})| is the order of the first cohomology group H^{1}(G, E(H)_{f}), and P is any prime factor of 2 in K.

The Weierstrass equation of E could be assumed as ^{[3]}

E : y^{2} = f(x) = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}

with a_{2}, a_{4}, a_{6} ∈ F.

2 The Structure of E(H)

Lemma 1. The degree of the extension H/F is [H : F] = 2.

Proof. Obviously [H : F] ≤ 2. Now if [H : F] = 1, then K ⊂ F. Consider F = Q(j(E)), where E could be any elliptic curve having complex multiplication by O_{K}. By a result in page 12-13 of [4] we know that F = Q(j(E)) has a real embedding into the complex field. Note that K is totally imaginary, so K ⊂ F is impossible. Thus H ⊄ F, [H : F] = [K(j(E)) : Q(j(E))] = 2. This proves the lemma.

Therefore, we throughout assume the Galois group of H/F is G = Gal(H/F) = {1, σ} . For any α ∈ O_{K}, let [α] denote the endomorphism of E corresponding to multiplication by α. The multiplication by \sqrt{−D} will be important to our following proof. Comparing to E, we consider the following elliptic curve

E_{D} : −Dy^{2} = f(x).

Note that E_{D} and E are isomorphic via the map

i : E_{D}(C) → E(C), (x, y) → (x, \sqrt{−D}y).

Then

End(E_{D}) ≅ End(E).

Thus E_{D} also has complex multiplication by O_{K}, and is defined over F. Via the isomorphism i of E and E_{D}, we have obviously that

E_{D}(F) ≅ I = \{(x, \sqrt{−D}y)|(x, \sqrt{−D}y) ∈ E(H), x, y ∈ F \} ⊂ E(H).

The subgroup I of E(H) defined here will be very important in the following analysis.

Lemma 2. The map i ∘ [\sqrt{−D}] is an F -isogeny of E to E_{D}. Thus

rank_{Z}(E_{D}(F)) = rank_{Z}(E(F)) = l

Proof. By [2] we have

[\sqrt{−D}](x, y) = (a(x), y\sqrt{−D}b(x)),

with a(x), b(x) ∈ F (x). So i ∘ [\sqrt{−D}] is an F -isogeny of E to E_{D}.

Lemma 3. (I_{f} : [\sqrt{−D}]E(F)_{f})(E(F)_{f} : [\sqrt{−D}]I_{f}) = D^{l}

Proof.

D^{l} = (E(F)_{f} : [D]E(F)_{f})
= (E(F)_{f} : [\sqrt{−D}]I_{f})([\sqrt{−D}]I_{f} : [D]E(F)_{f})
= (E(F)_{f} : [\sqrt{−D}]I_{f})(I_{f} : [\sqrt{−D}]E(F)_{f}).

Lemma 4. 2E(H)_{f} ⊂ E(F)_{f} ⊕ I_{f} ⊂ E(H)_{f},

rank_{Z}(E(H)) = rank_{Z}(E(F)) + rank_{Z}(E_{D}(F)) = 2 rank_{Z}(E(F)) = 2l.

Proof. If P = (x, y) ∈ E(F)_{f} with P ∈ I_{f}, then y = 0, which means that P is a torsion point. So P = O is the point at infinity, and E(F)_{f} ⊕I_{f} = E(F)_{f} +I_{f} ⊂ E(H)_{f}. For any Q ∈ E(H)_{f}, we have 2Q = (Q + Q^{σ}) + (Q − Q^{σ}). Via the definition of E(F)_{f} and I_{f}, we have

E(F)_{f} = {P |P^{σ} = P, ∀P ∈ E(H)_{f}}, I_{f} = {P |P^{σ} = −P, ∀P ∈ E(H)_{f}}.

So Q + Q^{σ} ∈ E(F)_{f}, Q − Q^{σ} ∈ I_{f} , 2Q ∈ E(F)_{f} ⊕ I_{f}. Thus 2E(H)_{f} ⊂ E(F)_{f} ⊕ I_{f} ⊂ E(H)_{f}. This completes the proof.

As for the index of E(F)_{f} ⊕ I_{f} in E(H)_{f}, we have the following theorem.

Theorem 1.

(E(H)_{f} : E(F)_{f} ⊕ I_{f}) = \frac{2^{l}}{|H^{1}(G, E(H)_{f})|},

where |H^{1}(G, E(H)_{f})| is the order of the cohomology group H^{1}(G, E(H)_{f}).

Proof. Consider H^{1}(G, E(H)_{f}) = Z^{1}(G, E(H)_{f})/B^{1}(G, E(H)_{f}). Let T = {P − P^{σ}|P ∈ E(H)_{f}}. We will prove that Z^{1}(G, E(H)_{f}) ≅ I_{f}, B^{1}(G, E(H)_{f}) ≅ T.

For any cocycle ξ ∈ Z^{1}(G, E(H)_{f}) , let ξ →^{ϕ} ξ_{σ}. By the definition of cocycle we have that 0 = ξ_{e} = ξ_{σ^{2}} = (ξ_{σ})^{σ} + ξ_{σ} , so (ξ_{σ})^{σ} = −ξ_{σ}, thus ξ_{σ} ∈ I_{f}, and ϕ is a map of Z^{1}(G, E(H)_{f}) to I_{f}. Via the map ϕ we could see that Z^{1}(G, E(H)_{f}) ≅ I_{f}, B^{1}(G, E(H)_{f}) ≅ T. Now consider the homomorphism E(H)_{f} ⟶^{ψ=P −P^{σ}} T. Obviously 2I_{f} ⊂ T. Since ψ^{−1}(2I_{f}) = E(F)_{f} ⊕ I_{f}, so

(E(H)_{f} : E(F)_{f} ⊕ I_{f}) = (T : 2I_{f}) = (I_{f} : 2I_{f})/(I_{f} : T),
= 2^{l}/|H^{1}(G, E(H)_{f})|.

3 Main Results and Their Proofs

We first give a general theorem on torsion-free finitely-generated module over Dedekind domain, which establishes a relationship between the Steinitz class and the index of the module in its corresponding free module. This theorem is the key to our final results about Steinitz class.

Theorem 2. Suppose that L is a free O_{K}-module, and M ⊂ L is a submodule, (L : M) < +∞. Then there is an integral O_{K}-ideal A such that [A] is the Steinitz class of M , and N_{Q}^{K}(A) = (L : M) , where N_{Q}^{K}(·) is the norm map of ideals from K to the rationals Q.

Proof. Let L = ⨁_{i=1}^{n} O_{K}e_{i} , so {e_{1}, ⋯, e_{n}} is an O_{K}-basis for L. We will inductively prove that there is O_{K}-ideals B_{i} (i = 1, ⋯, n) such that M ≅ ⨁_{i=1}^{n} B_{i} , and (L : M) =

∏_{i=1}^{n}(O_{K} : B_{i}).

When n = 1, every thing is obvious. Assume then the statement is true for n − 1 and consider the homomorphism of O_{K}-modules: ρ : L → O_{K}, ρ( ∑_{i=1}^{n} r_{i}e_{i}) = r_{n}. Then B = ρ(M) is an ideal of O_{K}, and the sequence

0 → N → M →^{ρ} B → 0

is exact, where N = ker(ρ) ∩ M. Since B is a projective O_{K}-module, there exists O_{K}-module C ⊂ M such that C ≅ B, ρ(C) = B, M = N ⊕ C ≅ N ⊕ B. Thus

(L : M) = (L : N ⊕ C) = (L : ⨁_{i=1}^{n−1} O_{K} + C)(⨁_{i=1}^{n−1} O_{K} + C : N ⊕ C)

where (L : ⨁_{i=1}^{n−1} O_{K} + C) = (ρ^{−1}(O_{K})) : ρ^{−1}(B)) = (O_{K} : B).

Consider C ∩ ⨁_{i=1}^{n−1} O_{K} = C ∩ ker(ρ). When restricted on C, the map ρ is injective, so we have

⨁_{i=1}^{n−1} O_{K} + C = ⨁_{i=1}^{n−1} O_{K} ⊕ C,

(⨁_{i=1}^{n−1} O_{K} + C : N ⊕ C) = (⨁_{i=1}^{n−1} O_{K} ⊕ C : N ⊕ C)
= (⨁_{i=1}^{n−1} O_{K} : N).

Note that N ⊂ ⨁_{i=1}^{n−1} O_{K}. So via the hypothesis of our induction, we know that there are O_{K}-ideals B_{i} (i = 1, ⋯, n − 1) such that N ≅ ⨁_{i=1}^{n−1} B_{i}, and ( ⨁_{i=1}^{n−1} O_{K} : N) =

∏_{i=1}^{n−1} (O_{K} : B_{i}). Thus we have M ≅ ⨁_{i=1}^{n} B_{i} and (L : M) = ∏_{i=1}^{n} (O_{K} : B_{i}) = ∏_{i=1}^{n} N_{Q}^{K}(B_{i}) =

N_{Q}^{K}( ∏_{i=1}^{n} B_{i}), where B_{n} = B. Now the proof is completed by the following lemma.

Lemma 5. Assume A_{1} and A_{2} are two non-zero ideals of Dedekind domain R, then we have isomorphism of R-modules : A_{1} ⊕ A_{2} ≅ R ⊕ A_{1}A_{2}.

Proof. See Lemma 13 in [5] p.168.

We now intend to prove our main results via our Theorem 2. To use Theorem 2, we need first to find the corresponding L and M in the Mordell group E(H) . The corresponding L is given in Lemma 6. While the corresponding M is given in Theorem 4 and 5, i.e., M = [\sqrt{−D}]E(H)_{f} if D ≡ 3 ( mod 4); M = [2\sqrt{−D}]E(H)_{f} if D ≡ 1 ( mod 4).

Lemma 6. L = O_{K} · E(F)_{f} is a free O_{K}-module of rank l.

Proof. Assume P_{1}, ⋯, P_{l} form a Z-basis of E(F)_{f}. We will prove

L = O_{K} · E(F)_{f} = ⨁_{i=1}^{l} O_{K}P_{i}.

Now we suppose that ∑_{i=1}^{l} [α_{i}]P_{i} = 0 for some α_{i} ∈ O_{K} (i = 1, ⋯, l). When D ≡ 3 (mod 4), we have α_{i} = s_{i} + t_{i}(1 + \sqrt{−D})/2 (s_{i}, t_{i} ∈ Z, i = 1, ⋯, l), then via

∑_{i=1}^{l} [α_{i}]P_{i} = 0 we have ∑_{i=1}^{l} [2s_{i} + t_{i}]P_{i} = 0 and ∑_{i=1}^{l} [\sqrt{−D}t_{i}]P_{i} = 0. Thus t_{i} = 0, s_{i} =

0, α_{i} = 0 ( i = 1, ⋯, l). This proves the theorem when D ≡ 3 (mod 4). The case D ≡ 1 (mod 4) goes in the same way.

To determine our corresponding M in the case D ≡ 3 (mod 4), we need the following theorem.

Theorem 3. For D ≡ 3 (mod 4), we have |H^{1}(G, E(H)_{f})| = 1, and E(H)_{f} = O_{K} · E(F)_{f} + I_{f}.

Proof. Let P_{1}, ⋯, P_{l} form a Z-basis of E(F)_{f}, and Q_{1}, ⋯, Q_{l} form a Z-basis of I_{f}. Put α = (1 + \sqrt{−D})/2. We need only to prove that E(H)_{f}/(E(F)_{f} ⊕ I_{f}) = C_{1} ⊕ ⋯ ⊕ C_{l}, where C_{i} = ( \overline{[α]P_{i}} ) is subgroup of order 2 generated by \overline{[α]P_{i}} in the quotient group E(H)_{f}/(E(F)_{f} ⊕ I_{f}). Obviously we have \overline{[α]P_{i}} ≠ \overline{0}; otherwise there would be t_{j}, s_{j} ∈ Z (j = 1, ⋯, l) such that [α]P_{i} = ∑_{j=1}^{l} [t_{j}]P_{j} + ∑_{j=1}^{l} [s_{j}]Q_{j}, then

[1 + \sqrt{−D}]P_{i} = ∑_{j=1}^{l} [2t_{j}]P_{j} + ∑_{j=1}^{l} [2s_{j}]Q_{j}, and P_{i} = ∑_{j=1}^{l} [2t_{j}]P_{j}, giving a contradiction.

Furthermore, if ∑_{i=1}^{l} [u_{i}]\overline{[α]P_{i}} = 0 for some u_{i} ∈ Z (i = 1, ⋯, l), then there are

t_{i}, s_{i} ∈ Z (i = 1, ⋯, l) such that ∑_{i=1}^{l} [u_{i}α]P_{i} = ∑_{i=1}^{l} [t_{i}]P_{i} + ∑_{i=1}^{l} [s_{i}]Q_{i}, so

∑_{i=1}^{l} [u_{i}]P_{i} + ∑_{i=1}^{l} [u_{i}\sqrt{−D}]P_{i} = ∑_{i=1}^{l} [2t_{i}]P_{i} + ∑_{i=1}^{l} [2s_{i}]Q_{i}.

Thus ∑_{i=1}^{l} [u_{i}]P_{i} = ∑_{i=1}^{l} [2t_{i}]P_{i}, which gives u_{i} = 2t_{i} (i = 1, ⋯, l). Hence [u_{i}]\overline{[α]P_{i}} =

\overline{[t_{i}][2α]P_{i}} = \overline{[t_{i}(1 + \sqrt{−D})]P_{i}} = ¯0. This completes the proof.

Now we could prove our main results via Theorem 2.

Theorem 4. Suppose that D = p ≡ 3 (mod 4) is a prime number, and elliptic curve E has complex multiplication by the full ring O_{K} of integers of K = Q(\sqrt{−D}). Then the Steinitz class of E is the principal class, i.e., St(E) = 1.

Proof. Let L = O_{K} · E(F)_{f}, M = [\sqrt{−p}]E(H)_{f}. Since M ≅ E(H)_{f}, so we need only to prove St(M) is the principal class.

By Theorem 3 we have E(H)_{f} = O_{K} · E(F)_{f} + I_{f}. Thus

M = [\sqrt{−p}]E(H)_{f} = E(F)_{f} · (\sqrt{−p}O_{K}) + [\sqrt{−p}]I_{f} ⊂ O_{K} · E(F)_{f} = L;
(L : M) = (O_{K} · E(F)_{f} : [\sqrt{−p}]E(H)_{f})
= \frac{(E(H)_{f} : [\sqrt{−p}]E(H)_{f})}{(E(H)_{f} : O_{K} · E(F)_{f})}
= \frac{p^{l}}{(E(H)_{f} : O_{K} · E(F)_{f})}.

Since p is a prime number, so (L : M) = p^{t} for some t (0 ≤ t ≤ l) . By Theorem 2, the Steinitz class of M is equal to [A] for some O_{K}-ideal A, and p^{t} = (L : M) = N_{Q}^{K}(A). Since p is a prime number, A = (\sqrt{−p}O_{K})^{t} is principal. Thus St(E) = St(M) is the principal class.

Theorem 5. Suppose that D = p ≡ 3 (mod 4) is a prime number, and E is an elliptic curve having complex multiplication by the ring O_{K} of all integers of K = Q(\sqrt{−D}). Then the Steinitz class of E is St(E) = [P^{t}], where [P] is the ideal class of K represented by P the prime factor of 2 in O_{K}, 2^{t} = 2^{l}|H^{1}(G, E(H)_{f})|. In particular, the parity of t determines St(E) , since P is not principal while P^{2} = 2O_{K} is principal.

Proof. Let L = O_{K} · E(F)_{f}, M = [2\sqrt{−p}]E(H)_{f}. Since M ≅ E(H)_{f}, so St(E) = St(M). Note that [2\sqrt{−p}]E(H)_{f} ⊂ [\sqrt{−p}](E(F)_{f} ⊕ I_{f}), [\sqrt{−p}]I_{f} ⊂ E(F)_{f}. Thus we have M ⊂ O_{K} · E(F)_{f} = L, and

(L : M) = (O_{K} · E(F)_{f} : [2\sqrt{−p}]E(H)_{f})
= \frac{(E(H)_{f} : [2\sqrt{−p}]E(H)_{f})}{(E(H)_{f} : O_{K} · E(F)_{f})}
= \frac{(4p)^{l}}{(E(H)_{f} : E(F)_{f} ⊕ I_{f})(E(F)_{f} ⊕ I_{f} : O_{K} · E(F)_{f})}
= \frac{(4p)^{l}}{2^{l}|H^{1}(G, E(H)_{f})|^{−1}(I_{f} : [\sqrt{−p}]E(F)_{f})}
= 2^{l}|H^{1}(G, E(H)_{f})| · p^{l}/(I_{f} : [\sqrt{−p}]E(F)_{f}).

Thus (L : M) = 2^{t}p^{r} for some t, r ≥ 0, since p is a prime number. By Theorem 2 we know that N_{Q}^{K}(A) = 2^{t}p^{r} for some O_{K}-ideal A. Therefore A = P^{t}([\sqrt{−p}]O_{K})^{r}, St(E) = [A] = [P^{t}]. This proves the theorem.

Corollary 1. Suppose as in Theorem 5. If l = rank_{Z}(E(F)) = 1 then H^{1}(G, E(H)_{f}) determines the Steinitz class of E.

Now we analysis the examples of Dummit and Miller in [2] by utilizing our above method . For these examples, we have K = Q(\sqrt{−10}), D = 10, H = K(\sqrt{5}) = Q(\sqrt{−10}, \sqrt{5}). We consider the O_{K}-module L = O_{K}·E(F)_{f} and M = 2[\sqrt{−10}]E(H)_{f}. Then via the same idea in the proof of Theorem 5 we have similar ratiocination for D = 10:

(L : M) = \frac{(E(H)_{f} : 2[\sqrt{−10}]E(H)_{f})}{(E(H)_{f} : O_{K} · E(F)_{f})}
= \frac{(4 · 10)^{l}}{(E(H)_{f} : E(F)_{f} ⊕ I_{f})(E(F)_{f} ⊕ I_{f} : O_{K} · E(F)_{f})}
= \frac{(40)^{l}}{2^{l}|H^{1}(G, E(H)_{f})|^{−1}(I_{f} : [\sqrt{−10}]E(F)_{f})}
= 2^{l}|H^{1}(G, E(H)_{f})|10^{l}/(I_{f} : [\sqrt{−10}]E(F)_{f}).

Thus the Steinitz class of E is determined by the 2-exponent of

2^{l}|H^{1}(G, E(H)_{f})|(I_{f} : [\sqrt{−10}]E(F)_{f}).

(DM1) Consider the following elliptic curve of Dummit and Miller in [2]:

E_{1} : y^{2} = x^{3} + (6 + 6\sqrt{5})x^{2} + (7 − 3\sqrt{5}).

Then l = 1 , |H^{1}(G, E(H)_{f})| = 2 , (I_{f} : [\sqrt{−10}]E(F)_{f}) = 1 . Therefore we know that 2^{l}|H^{1}(G, E(H)_{f})|(I_{f} : [\sqrt{−10}]E(F)_{f}) = 4. Thus the Steinitz class of E_{1} is the principal class, i.e., St(E_{1}) = 1.

(DM2) Consider the following elliptic curve in [2]: E_{1,isog} : y^{2} = x^{3} − (912 + 12\sqrt{5})x^{2} +(188+84\sqrt{5})x. We have l = 1, |H^{1}(G, E(H)_{f})| = 2, (I_{f} : [\sqrt{−10}]E(F)_{f}) = 2, and 2^{l}|H^{1}(G, E(H)_{f})|(I_{f} : [\sqrt{−10}]E(F)_{f}) = 2^{3}. Thus the Steinitz class St(E_{1,isog}) = [P], where P is a prime factor of 2 in O_{K}.

(DM3) For E_{3} : y^{2} = x^{3} + 36x^{2} + (162 − 72\sqrt{5})x, in [2], we have l = 2, |H^{1}(G, E(H)_{f})| = 2, (I_{f} : [\sqrt{−10}]E(F)_{f}) = 1, 2^{l}|H^{1}(G, E(H)_{f})|(I_{f} : [\sqrt{−10}]E(F)_{f}) = 2^{3}. Thus St(E_{3}) = [P], P a prime factor of 2 in O_{K}.

There are still many open problems about the Steinitz classes of elliptic curves. For example, we have the following conjecture.

Conjecture. Both the cases St(E) = 1 and St(E) ≠ 1 exist for some elliptic curves E having complex multiplication by O_{K} with K = Q(\sqrt{−D}) and prime D ≡ 1 (mod 4) .

References

1 Shimura G., Introduction to the Arithmetic Theory of Automorphic Functions, Princeton Univ.

Press. 1971. 2 Dummit D. S. and Miller W.L., The Steinitz Class of the Mordell Weil Group of Some CM

Elliptic Curves, J. Number Theory 56(1996): 52-78. 3 Silverman J. H., The Arithmetic of Elliptic Curves, Spring-Verlag, New York, 1982. 4 Gross B., Arithmetic on Elliptic Curves with Complex Multiplication, SLN, Vol. 776, Springer-

Verlag, Berlin, 1980.

5 Keqin Feng, Introduction to Commutative Algebra, Higher Education Press, Beijing, 1985.

Email: xianke@tsinghua.edu.cn