IMAGES OF SEMISTABLE GALOIS REPRESENTATIONS

Kenneth A. Ribet

University of California, Berkeley

1. Introduction

Let p be an odd prime number. Suppose first that E is a semistable elliptic curve over Q. The action of Gal(\overline{Q}/Q) on the group of p-division points of E defines a representation

ρ_{E,p}: Gal(\overline{Q}/Q) → GL(2, F_{p}).

Serre has shown that this representation is surjective whenever it is irreducible [20, Prop. 21], [23, §3.1]. Serre’s arguments prove more generally the surjectivity of all continuous irreducible representations

ρ: Gal(\overline{Q}/Q) → GL(2, F_{p})

which are semistable in the sense of [11, §1]. (Recall that ρ is semistable if the determinant of ρ is the mod p cyclotomic character χ, the Serre conductor of ρ is square free, and the Serre weight of ρ is either 2 or p+1. Here, the weight and conductor are the invariants defined in [22].)

In this article, we treat the situation where F_{p} is replaced by a finite field of characteristic p, or more generally by a finite product F = ∏ F_{i} of finite fields of characteristic p. A continuous representation ρ: Gal(\overline{Q}/Q) → GL(2, F) may then be viewed as a product of components ρ_{i}: Gal(\overline{Q}/Q) → GL(2, F_{i}). We shall assume that each ρ_{i} is irreducible and semistable. Since the determinant of each ρ_{i} is then the mod p cyclotomic character, the image of ρ is contained in the group

A := { M ∈ GL(2, F) | det M ∈ F_{p}^{∗} }.

Making the supplementary hypothesis p ≥ 5, we show below that the image of ρ is GL(2, F)-conjugate to

{ M ∈ GL(2, F′) | det M ∈ F_{p}^{∗} },

where F′ is the subalgebra of F generated by the traces tr(ρ(σ)) for σ ∈ Gal(\overline{Q}/Q). (See Theorem 2 and its corollary.) In particular, ρ (Gal(\overline{Q}/Q)) = A if and only if F is generated as an F_{p}-algebra by the traces.

Applying the lifting techniques of [19], we deduce an analogous result for certain p-adic representations of Gal(\overline{Q}/Q).

This article was begun while the author was a research professor at the MSRI, where research was supported in part by NSF grant DMS-9022140. This work was further supported by the investigator’s NSF Grants DMS 88-06815 and DMS 93-06898.

Our results have evident relevance to the study of the Galois representations defined by semistable abelian varieties over Q which are products of abelian varieties with large fields of endomorphisms. (In [17], the author referred to these as abelian varieties of “GL_{2} type.”) While the computation of F′ seems to be difficult to perform in certain cases, it can be carried out for the mod p representations coming from J_{0}(N) if N is a prime and p is a prime number satisfying some mild conditions. In fact, our original goal was to find the image of the p-adic representation attached to J_{0}(N), thus answering questions which were formulated by R. Coleman and B. Kaskel in connection with [2]. It is a pleasure to thank them for their continuing encouragement and interest in this work.

2. Representations over a finite field

Let F be a finite field, and let p be the characteristic of F. We will assume that p is odd; most of our results will require that p be at least 5. Suppose that

ρ: Gal(\overline{Q}/Q) → GL(2, F)

is an irreducible representation whose determinant is the mod p cyclotomic character χ: Gal(\overline{Q}/Q) → F_{p}^{∗} ↪ F^{∗}. Let k(ρ) and N(ρ) be the weight and conductor of ρ in the sense of Serre [22, §§1–2]. As we recalled above, Oesterlé [11, §1] has defined the notion of semistability: ρ is semistable if the conductor N(ρ) of ρ is square free and Serre’s weight k(ρ) is either 2 or p+1.

The definition of the conductor shows that N(ρ) is square free if and only if ρ(σ) is unipotent whenever σ belongs to an inertia subgroup of Gal(\overline{Q}/Q) for a prime ℓ ≠ p. To illuminate the condition on k(ρ), we let I be an inertia subgroup of Gal(\overline{Q}/Q) for the prime p. The semisimplification of ρ_{|I} is described by a pair of characters φ, φ′: I ⇉ \overline{F}^{∗}, cf. [22, §2]. Since det ρ is the cyclotomic character χ, we have in particular φφ′ = χ. If k(ρ) is one of 2, p+1, then {φ, φ′} is either {1, χ} or else the set of fundamental characters ψ, ψ′: I → \overline{F}^{∗} of level 2 (loc. cit.). It follows that the order of φ′/φ^{−1} (a character which is defined only up to inversion) is either p−1 or p+1.

Lemma. Assume that p ≥ 5 and that ρ is a semistable irreducible representation as above. Let ϵ: Gal(\overline{Q}/Q) → F^{∗} be a continuous character. If ϵ ⊗ ρ is semistable, then ϵ is trivial.

Proof. By definition, the representation ϵ ⊗ ρ sends each σ ∈ Gal(\overline{Q}/Q) to the product ϵ(σ)·ρ(σ) ∈ GL(2, F). If ρ and ϵ⊗ρ are both semistable, then ϵ is certainly unramified outside p, since both ρ(σ) and ϵ(σ) · ρ(σ) are unipotent whenever σ belongs to an inertia group for a prime ℓ ≠ p. Moreover, ϵ^{2} is the trivial character, since the determinants of ρ and ϵ⊗ρ coincide. Hence ϵ is either the trivial character, as desired, or the quadratic character χ^{\frac{p−1}{2}} . One checks easily, however, that no pair of characters drawn from the set { 1, χ, ψ, ψ′ } have a quadratic ratio. ■

Remark. The lemma does not extend to cover the case p = 3. To see this, consider the modular forms f = ∑ a_{n}q^{n} and g = ∑ b_{n}q^{n} which correspond respectively to the elliptic curves 89A and 89B in Cremona’s tables [3]. These curves have Weierstrass equations y^{2}+a_{1}xy+a_{3}y = x^{3}+a_{2}x^{2}+a_{4}x+a_{6} with (a_{1}, a_{2}, a_{3}, a_{4}, a_{6}) = (1, 1, 1, −1, 0) and (a_{1}, a_{2}, a_{3}, a_{4}, a_{6}) = (1, 1, 0, 4, 5), respectively. Thus f and g are weight-two newforms on the group Γ_{0}(89). Here are the coefficients a_{p} and b_{p} for primes p ≤ 67:

\begin{matrix}p & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47 & 53 & 59 & 61 & 67 & ⋯ \\ a_{p} & −1 & −1 & −1 & −4 & −2 & 2 & 3 & −5 & 7 & 0 & −9 & −2 & 0 & −7 & −12 & −3 & 4 & 6 & 12 & ⋯ \\ b_{p} & 1 & 2 & −2 & 2 & −4 & 2 & 6 & −2 & 2 & −6 & 6 & 10 & −6 & 2 & 12 & −6 & −10 & −6 & 12 & ⋯ \end{matrix}

Within the range of the table, we have b_{p} ≡ (\frac{p}{3})a_{p} mod 3 for p ≠ 3. A computation using gp [1] shows that the congruence holds more generally for p ≤ 200, p ≠ 3. This suggests that we have the mod 3 congruence

∑_{n=1}^{∞} b_{n}(\frac{n}{3})q^{n} ≡ ∑_{(3,n)=1} a_{n}q^{n}.  (∗)

To confirm this, we may apply the main result of [26]. Indeed, according to [24, Prop. 3.64], the two series appearing in (∗) are weight-two eigenforms on Γ_{0}(3^{2} ·89). On the other hand, the theorem of [26] implies that (∗) holds if and only if it holds “up to order N,” where N = (SL(2, Z) : Γ_{0}(3^{2} · 89)) /6 = 180 < 200.

Now let ρ_{A} and ρ_{B} be the mod 3 representations of Gal(\overline{Q}/Q) associated with 89A and 89B, respectively. The two representations are semistable because the elliptic curves are semistable: their conductor is prime. (We show below that mod p representations coming from semistable abelian varieties GL_{2} type are semistable.) The representations are irreducible, for example because neither a_{2} nor b_{2} is divisible by 3. (If ρ_{A} were reducible, we would have a_{ℓ} ≡ 1 + ℓ mod 3 for all ℓ ≠ 3, 89.) Further, the congruence (∗) translates into the Galois-theoretic statement that ρ_{A} ≈ χ^{\frac{p−1}{2}} ⊗ ρ_{B}. In Serre’s language of “companion forms” (see [7]), the modular forms f and g are mod 3 companions of each other.

It perhaps is worth recalling at this juncture that an irreducible mod p semistable representation ρ is absolutely irreducible. This follows easily from the fact that ρ (Gal(\overline{Q}/Q)) contains a matrix with distinct eigenvalues in F_{p}^{∗}. Such a matrix is obtained by taking ρ(c), where c is a complex conjugation because det ρ is the cyclotomic character, which is odd, the eigenvalues of ρ(c) are +1 and −1. These are distinct because p is odd.

Proposition 1. If ρ is semistable, then the image of ρ has order divisible by p.

Proof. Let G = ρ(Gal(\overline{Q}/Q)), and assume that the order of G is prime to p. Since the semistability hypothesis implies that ramification subgroups of G for primes ℓ ≠ p are unipotent, these ramification groups are forced to to be trivial. In other words, we have N(ρ) = 1 and the representation ρ is unramified outside p. As Serre remarks in a note on page 710 of [21, Vol. III], an analogue of the argument of Tate [27] shows that there are no irreducible representations ρ with this property when p = 3.

Now assume that p ≥ 5 and write \overline{G} for the image of G in PGL(2, F). Group theory shows that \overline{G} is either cyclic or dihedral, or else one of the three exceptional groups S_{4}, A_{4}, A_{5} [20, §2.5]. In fact, \overline{G} cannot be cyclic, since the cyclicity of \overline{G} implies that G is abelian, and hence that ρ is not absolutely irreducible.

To rule out the other cases, one considers an inertia subgroup \overline{I} of \overline{G} for the prime p. We know that \overline{I} is a cyclic group of order either p + 1 or p − 1. Indeed, \overline{I} may be viewed as the image of I ⊂ Gal(\overline{Q}/Q) under the character φ/φ′ which we introduced above. This character has order either p−1 or p+1; its image is cyclic because it is a finite subgroup of \overline{F}^{∗}.

Assume now that \overline{G} is dihedral, and let Z be the center of \overline{G}. It is evident that \overline{I} is contained in Z, since \overline{I} is a cyclic subgroup of a dihedral group and the order of \overline{I} is greater than 2. Accordingly, the quadratic extension of Q corresponding to Z is everywhere unramified this contradiction excludes the dihedral case and shows that \overline{G} must be one of the three exceptional groups.

However, as observed in the proof of [20, Prop. 21], the fact that \overline{I} has an element of order p ± 1 rules out the groups S_{4}, A_{4}, A_{5} in case p ≥ 7. Thus we are left only with the possibility that p = 5, in which case \overline{G} is either S_{4} or A_{4}, since its order is prime to 5 by assumption. The group \overline{I} is then cyclic of order 4, since S_{4} has no element of order 6. Also, we have \overline{G} ≈ S_{4}, since A_{4} has no element of order 4. Consider the quotient S_{3} of S_{4}. This quotient allows us to produce an S_{3}-extension of Q which is ramified only at 5 and such that the inertia groups for 5 in the extension have order 2. However, there certainly is no such extension, since the class number of Q(\sqrt{5}) is 1. ■

Corollary (cf. [23, Prop. 1]). Let ρ be as in Proposition 1, and suppose that p > 2. Then the image of ρ contains a subgroup isomorphic to SL(2, F_{p}). In particular, if F = F_{p}, then ρ is surjective.

Proof. The second statement is a consequence of the first, since the cyclotomic character χ: Gal(\overline{Q}/Q) → F_{p}^{∗} is surjective. To prove the first statement, we proceed as in [20, §2.4]. Namely, let g ∈ G be an element of order p, and let v be a non-zero vector in F ⊕ F which is fixed by g. Since ρ is irreducible, G cannot fix the line generated by v; therefore, there is an r ∈ G such that v and rv form a

basis of F ⊕ F. With with respect to this basis, g has the form ( \begin{matrix}1 & a \\ 0 & 1 \end{matrix} ), while

rgr^{−1} has the form ( \begin{matrix}1 & 0 \\ b & 1 \end{matrix} ). Scaling one of the vectors v, rv, we may assume

that a = 1. Hence, in an appropriate basis, G contains the elements ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) and

( \begin{matrix}1 & 0 \\ b & 1 \end{matrix} ). A well known theorem of L. E. Dickson (cf. [6, Th. 2.8.4]) states that

the group generated by these elements has a subgroup isomorphic to SL(2, F_{p}). In

fact, a more precise statement is true for p ≠ 3: ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) and ( \begin{matrix}1 & 0 \\ b & 1 \end{matrix} ) generate

the group SL(2, F′), where F′ = F_{p}(b). ■

Our aim now is to complement the Corollary by determining the exact image of ρ in a situation generalizing that where F = F_{p}. The situation which we have in mind is that where F is a minimal field of definition for ρ in the sense that it is generated over F_{p} by the numbers tr(ρ(σ)) for σ ∈ Gal(\overline{Q}/Q). We first show in this case that F is also a minimal field of definition for ρ as a projective representation, at least when p ≥ 5.

For the following lemma, let \overline{F} be an algebraic closure of F and consider an arbitrary subfield K of \overline{F}. We view G and \overline{G} inside GL(2, \overline{F}) and PGL(2, \overline{F}), respectively.

Lemma. Suppose that ρ semistable and that p ≥ 5. Then \overline{G} lies in PGL(2, K) if and only if G lies in GL(2, K).

Proof. If G lies in GL(2, K), then it is evident that \overline{G} is contained in PGL(2, K). Conversely, suppose \overline{G} ⊆ PGL(2, K). Then certainly G ⊆ \overline{F}^{∗} · GL(2, K). Let α: Gal(\overline{Q}/Q) → \overline{F}^{∗}/K^{∗} be the composite homomorphism

Gal(\overline{Q}/Q) →^{ρ} G ↪ \overline{F}^{∗}GL(2, K) → (\overline{F}^{∗}GL(2, K))/GL(2, K) = \overline{F}^{∗}/K^{∗}.

Let M be the finite abelian extension of Q cut out by α. The lemma asserts that α is identically 1, i.e., that M = Q. For σ ∈ Gal(\overline{Q}/Q), it is clear that α(σ) = 1 if tr(ρ(σ)) ∈ K^{∗}. In particular, α(σ) = 1 whenever the trace of ρ(σ) is a nonzero element of F_{p}. This remark shows that α is unramified outside p. Indeed, if σ belongs to an inertia subgroup of Gal(\overline{Q}/Q) for a prime ℓ ≠ p, then ρ(σ) is unipotent, so that its trace is 2. Hence α(σ) = 1 because p is odd.

Thus M is a finite abelian extension of Q which is unramified outside p. Moreover, [M: Q] is prime to p, since \overline{F}^{∗} has no elements of order p. Hence one has M ⊆ Q(μ_{p}); equivalently, α factors through the mod p cyclotomic character χ.

Now let I be an inertia group for p in Gal(\overline{Q}/Q). It suffices to show that there is an element σ of I for which α(σ) = 1 and for which χ(σ) is a generator of the cyclic group F_{p}^{∗}. The semisimplification of ρ_{|I} is described by a pair of characters φ, φ′: I ⇉ \overline{F}^{∗}. cf. [22, §2]. As we mentioned above, one has either \{φ, φ′\} = \{1, χ\} or {φ, φ′} = {ψ, ψ^{p}}, where ψ and ψ^{p} are the fundamental characters of level 2. Suppose first that we are in the former case, and let σ ∈ I be such that t = χ(σ) is a generator of F_{p}^{∗}. Then tr (ρ(σ)) = 1 + t is non-zero since p ≠ 3. Thus α(σ) = 1, as required. In the latter case, choose σ ∈ I so that t = ψ(σ) generates ψ(I) ≈ F_{p^{2}}^{∗}. Then χ(σ) = t^{p+1} is a generator of F_{p}^{∗}; note that χ = ψψ′ = ψ^{p+1}. On the other hand, tr (ρ(σ)) = t + t^{p}. The number t^{p−1} cannot be −1, since it has order p + 1. Since tr (ρ(σ)) is non-zero, we may conclude α(σ) = 1. ■

Theorem 1. Assume that ρ is semistable, that p ≥ 5 and that F is generated over F_{p} by the set \{ tr(ρ(σ)) | σ ∈ Gal(\overline{Q}/Q) \}. Then

ρ(Gal(\overline{Q}/Q)) = \{ M ∈ GL(2, F) | det M ∈ F_{p}^{∗} \}.

Proof. It is clear that G = ρ(Gal(\overline{Q}/Q)) is contained in the indicated matrix group because det ρ is F_{p}^{∗}-valued. Since the image of det ρ is precisely F_{p}^{∗}, the theorem amounts to the statement that G contains SL(2, F). An equivalent assertion is that the commutator subgroup of G contains a subgroup isomorphic to SL(2, F).

Let k be a finite extension of F_{p} which contains F and which has even degree over F_{p}. Since det(G) ⊆ F_{p}^{∗}, the subgroup \overline{G} of PGL(2, k) lies in PSL(2, k). Dickson [5, Ch. XII] has enumerated all subgroups of PSL(2, k); his list is summarized in [5, §260].

To situate \overline{G} within Dickson’s list, we recall that the order of \overline{G} is divisible by p by Proposition 1, and that the identity representation G → GL(2, \overline{F}) is irreducible. (The representation ρ is irreducible over F by hypothesis. It is then absolutely irreducible, as was noted above.) It follows that \overline{G} is one of the groups enumerated in [5, §251–§253]. Since we have assumed p ≥ 5, the final conclusion is easy to state: After replacing \overline{G} by a conjugate of \overline{G} inside PGL(2, k), we have either \overline{G} = PSL(2, K) or \overline{G} = PGL(2, K), for some subfield K of k.

Thus, in either case one has \overline{G} ⊆ PGL(2, K). From the Lemma, G ⊆ GL(2, K). In particular, tr(ρ(σ)) ∈ K for all σ ∈ Gal(\overline{Q}/Q). Since these numbers generate F over F_{p}, F ⊆ K. On the other hand, SL(2, K) ⊆ k^{∗} · G because \overline{G} contains PSL(2, K). On taking commutators, we obtain SL(2, K) ⊆ [G, G], and therefore SL(2, F) ⊆ [G, G]. As indicated above, this proves the theorem. ■

Remark. In the context of the Theorem, we may introduce the subfield F′ of F which is generated by the traces tr(ρ(σ)) for σ ∈ Gal(\overline{Q}/Q). Because complex conjugation acts in ρ as a matrix with distinct rational eigenvalues, a well known theorem of I. Schur [18, IX a] (cf. [28, Lemme I.1]) implies that ρ can be conjugated into a representation with values in GL(2, F′). Accordingly, the hypothesis that the traces generate F is always satisfied after replacing F by a minimal field of definition for ρ.

To prove the well known statement that there is a model for ρ over F′, one may proceed alternatively by direct computation, along the lines suggested by Wiles [29, p. 483].

3. Products

We next consider finite products of representations as above. Thus let p be a prime number ≥ 5 and let F_{1}, …, F_{t} (t ≥ 1) be finite fields of characteristic p. Let F be the finite étale F_{p}-algebra F_{1}×⋯×F_{t}. Suppose that ρ: Gal(\overline{Q}/Q) → GL(2, F) is a continuous homomorphism, so that ρ is a product of representations ρ_{i} (i = 1, …, t) as above. We will assume that each ρ_{i} is semistable and irreducible, and also that det ρ_{i} = χ for i = 1, …, t. With the evident convention, the latter hypothesis may be summarized by the formula det ρ = χ.

For each σ ∈ Gal(\overline{Q}/Q), we obtain an element tr(ρ(σ)) of F by considering the trace of the matrix ρ(σ). Motivated by the remark above, we let F′ be the subalgebra of F generated by the tr(ρ(σ)).

Lemma. The representation ρ has a model over F′.

Proof. As we have seen, each component ρ_{i} has a model over the subfield of F_{i} generated by the traces tr(ρ_{i}(σ)). After replacing ρ_{i} by such a model (and F_{i} by the trace field in question), we arrive at a situation in which F′ maps surjectively onto each factor F_{i}.

As one knows, F′ is then isomorphic to a partial product of the factors F_{i}. To see this explicitly, we let π: F → F_{i} be the projections (x_{1}, …, x_{t}) ↦ x_{i} and consider the following relation on the set { 1, …, t }: i ∼ j if and only if the map π_{i} × π_{j} : F′ → F_{i} × F_{j} is not surjective. It is easy to see that this relation is an equivalence relation and that i ∼ j if and only if there is an isomorphism σ: F_{i} →^{∼} F_{j} so that π_{j} = σ^{∘}π_{i} on F′. If there is such an isomorphism, it is unique; we denote it σ_{ji}. One shows that

F′ = { (x_{1}, …, x_{t}) ∈ F | x_{j} = σ_{ji}(x_{i}) for all pairs (i, j) such that i ∼ j }.

In particular, F′ is isomorphic to the product ∏_{i∈I} F_{i}, where I is a set of representatives for the equivalence ∼.

By the Brauer-Nesbitt theorem, ρ_{j} and ^{σ_{ji}}ρ_{i} are isomorphic whenever i and j are equivalent. (The two representations have the same trace and determinant.) Replace ρ_{j} by ^{σ_{ji}}ρ_{i} for all equivalent pairs (i, j) with i ∈ I. Then the representation ρ, a priori with values in GL(2, F), takes values in GL(2, F′). ■

Theorem 2. One has

ρ(Gal(\overline{Q}/Q)) = \{ M ∈ GL(2, F) | det M ∈ F_{p}^{∗} \}

if and only if F′ = F.

Proof. The necessity is clear, since the trace function SL(2, F) → F is surjective. For the sufficiency, as in the proof of Theorem 1, one must show that ρ(Gal(\overline{Q}/Q)) contains SL(2, F). Let H = ρ(Gal(\overline{Q}/Q)) ∩ SL(2, F), so that H is a subgroup of the product SL(2, F_{1})×⋯×SL(2, F_{t}). By Theorem 1, H projects onto each factor SL(2, F_{i}). Because each group SL(2, F_{i}) is its own commutator subgroup, the “two principle” [13, 3.3] implies that H is the full product of the SL(2, F_{i}) if and only if H maps onto each product SL(2, F_{i}) × SL(2, F_{j}) for i ≠ j.

Assume, then, that we have i ≠ j, and suppose that the image of the product ρ_{i} × ρ_{j} does not contain SL(2, F_{i}) × SL(2, F_{j}). By exploiting results of Dieudonné and Hua, one may construct: (i) an isomorphism ω: F_{i} →^{∼} F_{j}, and (ii) a continuous homomorphism ϵ: Gal(\overline{Q}/Q) → F_{j}^{∗} such that ^{ω}ρ_{i} and ρ_{j} ⊗ ϵ are isomorphic representations of Gal(\overline{Q}/Q) over F_{j}. This statement was proved during the course of the proof of [13, Th. 3.8] the theorem itself states merely that ^{ω}ρ_{i} and ρ_{j} ⊗ ϵ have equal traces and determinants.

By the Lemma above, however, we have ϵ = 1; thus ^{ω}ρ_{i} ≈ ρ_{j}. Accordingly, we have ω(tr(ρ_{i}(σ))) = tr(ρ_{j}(σ)) for all σ ∈ Gal(\overline{Q}/Q). This means that the image of tr(ρ(σ)) in F_{i} × F_{j} lies in the subalgebra { (x, ωx) | x ∈ F_{i} } of F_{i} × F_{j}. Thus the quantities tr(ρ(σ)) fail to generate F_{i} ×F_{j} over F_{p}. This contradicts the hypothesis that they generate the full product F over F_{p}. ■

Corollary. Assume that ρ is as in the theorem. Then the image of ρ is conjugate in GL(2, F) to the group

{ M ∈ GL(2, F′) | det M ∈ F_{p}^{∗} }

Proof. By the Lemma, there is a model for ρ over F′: after conjugating by a matrix in GL(2, F), we make ρ take values in GL(2, F′). Applying the Theorem to this model, we arrive at the desired conclusion. ■

4. Lifts

Again suppose that p is a prime ≥ 5 and let O_{1}, …, O_{t} be integer rings of finitedegree unramified extensions of Q_{p}. For each i, let ρ̃_{i}: Gal(\overline{Q}/Q) → GL(2, O_{i}) be a continuous representation whose determinant is the p-adic cyclotomic character χ̃: Gal(\overline{Q}/Q) → Z_{p}^{∗}. Let ρ̃ be the product of the ρ̃_{i}, so that ρ̃ is a p-adic representation Gal(\overline{Q}/Q) → GL(2, O), where O := ∏ O_{i}. The diagonal embedding Z_{p} ↪ O induces an inclusion Z_{p}^{∗} ↪ O^{∗}. We clearly have ρ̃(Gal(\overline{Q}/Q)) ⊆ A, where

A = { M ∈ GL(2, O) | det M ∈ Z_{p}^{∗} }.

Let F = O/pO, and let ρ: Gal(\overline{Q}/Q) → GL(2, F) be the reduction of ρ̃ mod p. We recall the following fact.

Proposition 2. The image of ρ contains SL(2, F) if and only if ρ̃(Gal(\overline{Q}/Q)) coincides with A.

Proof. One has ρ̃(Gal(\overline{Q}/Q)) ⊇ SL(2, O) if and only if ρ(Gal(\overline{Q}/Q)) contains SL(2, F) = SL(2, O/pO). This equivalence, which is proved as Theorem 2.1 of [13], is a mild generalization of a well known lemma of Serre [19, p. IV-23]. The statement to be proved follows from this equivalence, together with the surjectivity of χ̃. ■

As a corollary, we obtain the fact that ρ̃(Gal(\overline{Q}/Q)) = A whenever ρ satisfies the hypotheses of Theorem 2:

Theorem 3. Suppose that the mod p reduction of each ρ̃_{i} is semistable and irreducible. Assume that F is generated as an F_{p}-algebra by the traces tr ρ(g) with g ∈ Gal(\overline{Q}/Q). Then ρ̃(Gal(\overline{Q}/Q)) = A.

Remark. Using Nakayama’s Lemma, one may reformulate the trace hypothesis as the apparently stronger assertion that O is generated as a Z_{p}-algebra by the traces tr ρ̃(g) with g ∈ Gal(\overline{Q}/Q).

5. Semistable abelian varieties of GL_{2}-type over Q

Suppose that A is an abelian variety over Q for which Q⊗End_{Q} A is a number field of degree equal to the dimension of A. Assume further that the ring R = End_{Q} A is the full ring of integers in the field Q ⊗ End_{Q} A. (After replacement of A by an isogenous abelian variety, this second hypothesis is always verified.) Let 𝔪 be a maximal ideal of R and let A[𝔪] be the kernel of 𝔪 on A, i.e., the group of points in A(\overline{Q}) which are annihilated by all elements of 𝔪. It is easy to check that A[𝔪] is free of rank two over F = R/𝔪, cf. [25, Prop. 10, p. 56]. In fact, the representation

ρ: Gal(\overline{Q}/Q) → GL(2, F)

defined by the action of Gal(\overline{Q}/Q) on A[𝔪] is just the mod 𝔪 reduction of the 𝔪-adic representation of Gal(\overline{Q}/Q) which is attached to A. As usual, we denote by p the characteristic of F.

Proposition 3. Assume that p is odd, that ρ is irreducible, and that A is semistable over Q. Then ρ is a semistable representation.

Proof. We first show that the determinant of ρ is the mod p cyclotomic character χ. According to the statement of [17, Lemma 3.1], we have det ρ = ϵχ, where ϵ is a Dirichlet character Gal(\overline{Q}/Q) → R^{∗} which depends only on A (i.e., is independent of 𝔪) and which is ramified only at primes of bad reduction for A. In fact, the proof of this Lemma shows that ϵ is unramified at a prime q whenever the following condition is satisfied: there is a prime λ of R such that the determinant of the λ-adic representation of A is unramified at q. Now fix q and take any prime λ not dividing q. By a well known result of Grothendieck [8, Prop. 3.5], each element of an inertia subgroup for q in Gal(\overline{Q}/Q) acts unipotently in the λ-adic representation for A. In particular, the determinant of this representation is unramified at q. Thus ϵ is unramified at every prime number q, so that ϵ is the trivial character.

The same proposition of Grothendieck, applied to the 𝔪-adic representation for A, shows that the conductor of ρ is square free. We may paraphrase this statement by saying that ρ is semistable outside p. Furthermore, results of Raynaud [12] imply that k(ρ) = 2, so that ρ is semistable at p, whenever ρ is finite at p. (See [22, Prop. 4, p. 189].) It remains only to show that k(ρ) = p+1 if ρ is not finite at p.

For this, we consider A[𝔪] as a subgroup of A[p], and view both as modules for a decomposition group D_{p} in Gal(\overline{Q}/Q) for the prime p. As explained in the proof of [15, Lemma 6.2], one may deduce from Grothendieck’s study in SGA7I that A[p] is an extension of an unramified D_{p}-module by a subgroup A[p]^{f} which is finite, i.e., which extends to a finite flat group scheme over Z_{p}. Were A[𝔪] contained in A[p]^{f}, A[𝔪] would be finite, contrary to assumption. Hence A[𝔪] has an unramified quotient. This implies that the restriction of ρ to D_{p} has the form

( \begin{matrix}θ_{1}χ & ∗ \\ 0 & θ_{2} \end{matrix} ), where the θ_{i} are unramified characters. The recipe for k(ρ) given

in [22, §2] then sets k(ρ) = p+1. (Compare Remarque (1) on page 188 of [22].) ■

6. Application to J_{0}(N) for prime N

Let N be a prime number, and consider the abelian variety J = J_{0}(N) over Q. This abelian variety has been studied extensively by Mazur [10], and by others. Let T be the ring of Hecke operators Z[…, T_{n}, …], considered as a ring of endomorphisms of J_{0}(N). To orient the reader, we recall that T is the full ring of endomorphisms of J_{0}(N) [10, p. 95]. To avoid the situation where N is very small, we will assume that J_{0}(N) ≠ 0. This means that N = 11 or that N ≥ 17.

Let p be a prime number ≥ 5.

Proposition 4. The quotient T/pT is generated by the operators T_{n} with n prime to pN.

Proof. Let S be the space of cusp forms of weight 2 on Γ_{0}(N) over F_{p}. We view S as the mod p reduction of the space of weight-two forms on Γ_{0}(N) with integral q-expansions. (See [10, Ch. II, §4] for a comparison of several possible definitions of S.) We consider the bilinear pairing α: T/pT×S → F_{p} which maps (T, f) to the initial coefficient of q in the Fourier expansion of f|T. It is well known that α is a perfect pairing; this is explained, for example, in [14, §1]. To prove the Proposition, then, it suffices to prove that there is no non-zero element f = ∑ a_{n}q^{n} of S which satisfies a_{n} = 0 for all n prime to pN.

Suppose that f = ∑ a_{n}q^{n} satisfies the condition. We will show first that a_{n} = 0 for all n prime to N. If N = p there is nothing extra to prove, so we will assume for the moment that N and p are distinct. Let g be the form ∑_{(N,n)=1} a_{n}q^{n}, i.e., the sum

∑_{n=1}^{∞} b_{n}q^{n}, b_{n} = \{ \begin{matrix}a_{n} & if (n, N) = 1, \\ 0 & otherwise. \end{matrix}

Then b_{n} = 0 whenever n is not a multiple of p; our aim is to show that g = 0. Now the point is that g may be considered as a weight-two mod p modular form on Γ_{0}(N^{2}); this follows from [24, Prop. 3.64]. The hypothesis about the vanishing of the b_{n} means that g is annihilated by the operator θ = q \frac{d}{dq}. Since p > 2, this forces g = 0 as desired, since θg has “filtration” p + 3 if g is non-zero. (This is Katz’s generalization of the Serre-Swinnerton-Dyer theorem [9, p. 55].)

To complete the proof, we must show that f = 0. This follows from Proposition 4.10, Lemma 5.9 and Lemma 5.10 of [10, Ch. II]. ■

Remark. The argument we have given is essentially that of [10, Ch. II, Prop. 14.13]. The exploitation of q \frac{d}{dq} to deal with the absence of T_{p} is a familiar ploy it was used by the author in [16, Prop. 2] and by Wiles in [29, Lemma, p. 491].

Suppose now that 𝔪 is a maximal ideal of T of residue characteristic p. (We continue to assume p ≥ 5.) Let J[𝔪] denote the kernel of 𝔪 on J(\overline{Q}). By the results of [10], J[𝔪] defines the two-dimensional semisimple representation ρ_{𝔪} of Gal(\overline{Q}/Q) which is associated with 𝔪; this representation is irreducible if and only if 𝔪 is not an Eisenstein prime of T. When 𝔪 is non-Eisenstein, ρ_{𝔪} is a semistable representation: this follows from the results of Deligne-Rapoport [4] to the effect J has multiplicative reduction at N, together with Proposition 3.

Proposition 5. The image of ρ_{𝔪}: Gal(\overline{Q}/Q) → GL(2, T/𝔪) is the group of elements of GL(2, T/𝔪) having determinant in F_{p}^{∗}.

Proof. This mild strengthening of [10, Ch. II, Prop. 14.12] may be derived directly from Theorem 1. Indeed, for each prime ℓ different from p and N, the image under ρ_{𝔪} of a Frobenius element for ℓ in Gal(\overline{Q}/Q) has trace T_{ℓ} mod 𝔪. It follows from the Lemma above that the T_{ℓ} mod 𝔪 generate T/𝔪. ■

In the same vein, suppose that p ≥ 5 is such that none of the 𝔪|p in T is an Eisenstein prime. This means simply that N ≢ 1 mod p. Let F be the F_{p}-algebra ∏_{m|p} T/𝔪, and let

ρ: Gal(\overline{Q}/Q) → GL(2, F)

be the product of the ρ_{𝔪}.

Proposition 6. The image of ρ is the group of matrices in GL(2, F) having determinant in F_{p}^{∗}.

Proof. Indeed, the natural map T/pT → F is surjective by the Chinese remainder theorem. Accordingly, the Lemma implies that F is generated by the images of the T_{ℓ} with ℓ prime to pN. In the language of Theorem 2, this means that F′ = F. Applying that theorem, we find that the image of ρ is as stated. ■

Continuing the discussion, we add the hypothesis that the surjection T/pT → F is an isomorphism, i.e., that p is unramified in T. Then ρ is the representation giving the action of Gal(\overline{Q}/Q) on the group J[p], viewed as a free T/pT-module of rank two. Proposition 6 thus furnishes a description of the Galois group of the field cut out by the p-division points of J. Further, as we have seen in Proposition 2, the result of Proposition 6 is equivalent to an analogous statement about the p-adic Galois representation defined by J. More precisely, because p in unramified in T, T ⊗ Z_{p} is a product of discrete valuation rings. The Tate module Tate_{p}(J) of J is then automatically free of rank two over T⊗Z_{p}. After choosing a basis of Tate_{p}(J), we may summarize the action of Gal(\overline{Q}/Q) on the p-power division points of J by a homomorphism

ρ̃: Gal(\overline{Q}/Q) → GL(2, T ⊗ Z_{p}).

By Propositions 2 and 6, we find that the image of ρ̃ is “as large as possible.” Our conclusion may be recapitulated as follows:

Theorem 4. Suppose that p is unramified in T and that p is prime to 6 · (N−1). Then ρ̃(Gal(\overline{Q}/Q)) = \{ M ∈ GL(2, T ⊗ Z_{p}) | det M ∈ Z_{p}^{∗} \}.

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UC Mathematics Department, Berkeley, CA 94720-3840 USA

E-mail address: ribet@math.berkeley.edu