A Simplified Proof of Serre's Conjectures

Home , John Tate, Pierre Deligne

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1. Main results involved in the proof Let us recall the important ingredients used in the proof of Serre’s conjectures. 1.1. Base cases. The base cases needed to propagate modularity are the following. Theorem 1.1. There are no continuous, odd, absolutely irreducible two dimensional Galois representations of the absolute Galois group of Q unramified outside p and with values on a finite field of characteristic p, for p =2 or 3. Proof. The result was proved by Tate ([Tat94]) for p = 2, and later Serre observed that the exact same argument worked for p = 3 ([Ser13, page 710]). Theorem 1.2. There do not exist non-zero semi-stable rational abelian varieties that have good reduction outside ℓ for ℓ =2, 3, 5, 7, 13. Proof. See [Sch05, Theorem 1.1].

Theorem 1.3 (Langlands-Tunnell). Let p be an odd prime and ρ : GQ GL2(Fp) be an odd, continuous, irreducible representation with solvable image.→ Then ρ is modular.

Proof. By Dickson’s classiﬁcation, any solvable subgroup of PGL2(Fp) is isomorphic to a cyclic group, a dihedral group, A4 or S4. In all cases, the representation lifts to an odd representation of GL2(C). The cyclic case gives a reducible representation (and can be omitted). The dihedral case was considered by Hecke, the tetrahedral case was proved by Langlands in [Lan80] and the octahedral by Tunnell in [Tun81]. Although such representations are related to weight one modular forms, a classical result of Deligne-Serre ([DS74]) allows to get a congruence with a cuspidal weight two modular form. Modularity lifting Theorems. ⋆ 1 1.2. If p is an odd prime, let p = −p p, so that the extension Q(√p⋆) is the unique quadratic extension of Q unramiﬁed outside p. ASIMPLIFIEDPROOFOFSERRE’SCONJECTURES 3

1.2.1. Irreducible case.

Theorem 1.4. Let p be an odd prime and ρ : GQ GL2(Qp) be a continuous, odd Galois representation ramiﬁed at ﬁnitely many primes→ and satisfying the following hypothesis:

The residual restriction ρ G is absolutely irreducible, • | Q(√p⋆) The representation ρ G is de Rham with Hodge-Tate weights 0, k 1 , • | Qp { − } with k> 1, The residual representation is modular, i.e. ρ ρf . • ≃ Then ρ is modular.

Proof. The case k = 2 is proven in [Kis09c], while the general case follows from [Kis09a]. There are two extra hypothesis in such result: the second one is removed in [Eme06], while the fourth one is removed in [HT13, Theorem 1.4] (for p 5) and in [Tun18] for p = 3. ≥

We also need a similar (but much weaker) result for p = 2.

Theorem 1.5. Let ρ : GQ GL2(Q2) be a continuous, odd Galois representation ramified at finitely many primes→ and satisfying the following hypothesis: ρ is potentially Barsotti-Tate at 2 and det(ρ) equals the cyclotomic charac- • ter times an even character of finite order, The residual representation ρ is modular and has non-solvable image. • Then ρ is modular.

Proof. See [Kis09b, Theorem 0.1].

1.2.2. Modularity of residually reducible representations.

Theorem 1.6. Let p 5 be a prime number and ρ : GQ GL2(Qp) be a continuous, irreducible odd≥ Galois representation ramiﬁed at ﬁnitely→ many primes and satisfying the following hypothesis:

The representation ρ G is de Rham with Hodge-Tate weights 0, k 1 • | Qp { − } and k> 1, The semisimpliﬁcation of ρ is a sum of two characters χ χ . • 1 ⊕ 2 Then ρ is modular.

Proof. See [SW99] and [Pan19, Theorem 1.0.2].

Theorem 1.7. Let ρ : GQ GL2(Q3) be a continuous, irreducible odd Galois representation ramiﬁed at ﬁnitely→ many primes and satisfying the following hypothesis: ρss 1 χ , • ≃ ⊕ 3 ρ D =1, • | 3 6 ρ I3 ( ∗0 1∗ ), • | ≃ k 1 det(ρ)= ψχ − for some k 2 and is odd. • 3 ≥ Then ρ is modular.

Proof. See [SW99, Theorem]. 4 LUISVICTORDIEULEFAITANDARIELMART´IN PACETTI

1.3. Existence of lifts. Recall that an inertial type at ℓ is a representation τℓ : Iℓ GL (E) (where E/Qp a finite extension), where Iℓ is the inertia subgroup → 2 of Gal(Qℓ/Qℓ). Let ρ : GQ GL2(Fp) be an odd, continuous, irreducible Galois representation with Serre weight→ k(ρ). Let Σ be a finite set of primes containing p and the ones where ρ is ramified, and for each ℓ Σ different from p, let τℓ be an ∈ inertial type compatible with ρ, i.e. such that τℓ = ρ Iℓ . By taking a suitable twist, we can (and will) assume| that the Serre’s weight always satisfies 2 k(¯ρ) p + 1 if p = 2 and 2 k(¯ρ) 4 if p = 2 (twisting is harmless because it≤ preserves≤ modularity).6 ≤ ≤

Theorem 1.8. Let ρ : GQ GL (Fp) be an odd, continuous, representation whose → 2 restriction to GQ(ζp) is absolutely irreducible. Assume furthermore that when p =2 ρ¯ has non-solvable image. Then there exists a lift ρ : GQ GL (E) (for some → 2 finite extension E/Qp) with any of the following prescribed properties: (1) If p = 2 and k(ρ)=2, a minimal crystalline lift with Hodge-Tate weights 0, 1 . (2) If{ p =} 2, a lift with Hodge-Tate weights 0, 1 , minimally ramified outside 2 and the inertial Weil-Deligne parameter{ at} 2 is given by (id,N) when k(ρ)=4. (3) If p> 2, a minimal crystalline lift with Hodge-Tate weights 0, k(ρ) 1 . { − } (4) If p> 2, a lift with Hodge-Tate weights 0, 1 , inertial type condition τℓ at primes ℓ = p in Σ and unramified outside{ Σ.} Furthermore, if k(ρ)=2 the lift can be6 taken to be crystalline at p and if k(ρ) = p +1 the lift can be taken to be Steinberg at p, i.e., its inertial Weil-Deligne parameter at p is given by (id,N) Proof. The first three cases are due to Khare-Wintenberger ([KW09b, Theorem 5.1], its proof given in [KW09c]). Partial results of the last case are also proven in Khare-Wintenberger’s article (same Theorem), the more general case is due to Gee and Snowden ([Gee11]; [Sno09, Theorem 7.2.1]). 1.4. Existence of almost strictly compatible systems. Recall the definition of a strictly compatible system and an almost strictly compatible system of Galois representations (cf. [KW09b]). Definition 1.9. A rank 2 strictly compatible system of Galois representations R of GQ defined over K is a 5-tuple

R = (K, S, Qp(x) , ρλ , k), { } { } where (1) K is a number ﬁeld. (2) S is a ﬁnite set of primes. (3) for each prime p S, Qp(x) is a degree 2 polynomial in K[x]. (4) For each prime λ6∈of K, the representation

ρλ : GQ GL (Kλ), → 2 is a continuous semi-simple representation such that: If p S and p ∤ N(λ), then ρλ is unramiﬁed at p and ρλ(Frobp) has • 6∈ characteristic polynomial Qp(x). If p N(λ), then ρ G is de Rham and furthermore crystalline if p / S. • | | Qp ∈ (5) The Hodge-Tate weights HT(ρλ)= 0, k 1 . { − } ASIMPLIFIEDPROOFOFSERRE’SCONJECTURES 5

(6) For each prime p there exists a Weil-Deligne representation WDp(R) of

WQp over K such that for each place λ of K and every K-linear embedding K-ss . ( ι : K ֒ Kλ, the push forward ι WDp(R) WD(ρλ G → ≃ | Qp An almost strictly compatible system is a 5-tuple satisfying the first five properties, and we also impose condition (6) but with some exceptions: for a prime λ whose residual characteristic is equal to the prime p, if the residual representation ρ¯λ is reducible, then we only impose the compatibility as in condition (6) if this prime p is odd and the representation WDp(R) is unramified.

Theorem 1.10. Let ρ : GQ GL2(Kλ) be an odd, irreducible, continuous Galois representation ramiﬁed at ﬁnitely→ many places and de Rham at p with Hodge-Tate weights 0, k 1 , with k > 1 such that the restriction of ρ¯ to GQ(ζp) is absolutely irreducible.{ Assume− } furthermore that when p =2ρ ¯ has non-solvable image. Then ρ is part of a rank 2 almost strictly compatible system of Galois representations. Proof. See [Die04, Theorem 1.1]. 1.5. Some lemmas on the image of Galois representations. Some of the previous theorems are stated under the hypothesis that the restriction of the residual representation to GQ(ζp) is absolutely irreducible. Let us show that this is in fact equivalent to the condition (introduced by Wiles) of not being bad dihedral.

Lemma 1.11. Let p be an odd prime and ρ¯ : GQ GL2(Fp) be an odd continuous representation. Then the following are equivalent:→

(1) ρ¯ G is irreducible, | Q(√p⋆) (2) ρ¯ G is irreducible. | Q(ζp) Proof. We can assume that p = 3 as otherwise the statement is trivial. Clearly the second condition implies the ﬁrst6 one. For the converse, suppose that the restriction ofρ ¯ to Q(ζp) is reducible. In particular, the image ofρ ¯ is a solvable group. Then either our representation is contained in a Borel group (hence it is reducible), it lies in the normalizer of a Cartan group or its projective image is one of the exceptional groups A4, S4 or A5. Let G denote the image ofρ ¯. Recall that the normalizer of a Cartan group ﬁts into the short exact sequence

φ 1 / T / G / Z/2 / 1 where T is a torus (the map φ is surjective as otherwiseρ ¯ is not irreducible). If

H corresponds to the restriction ofρ ¯ to GalQ(ζp) then the hypothesis that such restriction is reducible implies that H T = H. Let σ GalQ be such that it ∩ ∈ generates the Galois group Gal(Q(ζp)/Q). Then G = H, ρ¯(σ) . The image of the restriction ofρ ¯ to Q(√p⋆) equals H, ρ¯(σ2) which alsoh lies in iT hence it is also a reducible representation. h i If the projective image is A5 (a simple group) the representation cannot become reducible over a cyclotomic extension. In the other two cases, if p = 3 then the projective image being reducible is equivalent to it being decomposable6 , in which case the group of the projective representation has a normal abelian subgroup with abelian quotient. There are no such groups for A4 nor S4.

A residual representationρ ¯ : GQ GL (Fp) which is irreducible but becomes → 2 reducible while restricted to GQ(√p⋆) is what is called in the literature a bad dihedral 6 LUISVICTORDIEULEFAITANDARIELMART´IN PACETTI case. Let us state (and give a detailed proof of) a result that is well-known to experts. Recall that given a residual representation we are assuming that it has been twisted so that its Serre’s weight lies in the range [2,p + 1].

Lemma 1.12. Let ρ¯ : GQ GL2(Fp) be a continuous odd bad dihedral representation. Then either p =2k(¯→ρ) 3 (niveau 2 case) or p =2k(¯ρ) 1 (niveau 1 case). − −

Proof. This is Lemma 6.2 (ii) of [KW09b]. The proof lacks some details, hence the difference between the niveau 1 and the niveau 2 case is hard to see; it follows the lines of [Rib97] (see Proposition 2.2). The fact that the representation has irreducible image together with being dihedral and p 3 imply that all elements are semisimple. The reason is that a dihedral group≥ consists of symmetries and powers of a rotation r. Since symmetries have order two and p > 2, they are semisimple. If r is not semisimple, it has a unique fixed line, hence it is stable under the action of symmetries as well, hence the representation is not irreducible. Then the projective image of inertia is a cyclic abelian group. We claim that if it has projective order n greater than two, then the representation cannot be bad dihedral. Otherwise, the projective image ofρ ¯ is a dihedral group D2m of order 2m where n m. The image of inertia must lie in the unique normal subgroup of rotations (of| index 2), hence if L denotes the field fixed by the projective residual image, it contains a unique quadratic extension which is unramified at p, hence disjoint from Q(√p⋆) and the representation cannot be bad dihedral. Start considering the case of a “niveau 1” character, i.e, the image of inertia is − χk(¯ρ) 1 0 k(¯ρ) 1 of the form p with k(¯ρ) p + 1. But χ − has order at most 2 and 0 1 ≤ k(¯ρ) p+1, so either 2k(¯ρ) 2= p 1or2k(¯ρ) 2=2p 2 or p = 3 and k(¯ρ) = 4. The latter≤ cannot occur as −p = 3 and− Serre weight− 4 implies− that inertia has an element which is not semisimple. Note that if k(¯ρ) = p then the image of inertia has odd order (1 or p) hence the Galois extension corresponding toρ ¯ is disjoint from Q(√p⋆). In particular, the representation is not bad dihedral. k(¯ρ) 1 In the case of a character of niveau 2, the image of inertia is of the form ψ − k(¯ρ) 1 (k(¯ρ) 1)(p 1) ⊕ ψ′ − , hence the projective order equals the order of ψ − − , which is the k(¯ρ) 1-st power of a character of order p +1. It has order at most 2 when p − +1 k(¯ρ) 1 and the condition k(¯ρ) p gives that p =2k(¯ρ) 3. 2 | − ≤ − Remark 1. Over Q, a Galois representation with Serre’s level 1 cannot be bad dihedral in the niveau 2 case due to a Lemma of Wintenberger (see [Kha06, Lemma 6.2] part (i)) hence the bad dihedral case can only occur for p =2k(¯ρ) 1 in level 1. But in this case k(¯ρ) is even, hence a bad dihedral representation of Serre’s− level 1 can only occur when p 3 (mod 4). ≡ 2. The proof of Serre’s modularity conjecture for p =2. 6 We can assume without loss of generality that ρ has non-solvable image, as otherwise modularity follows from Theorem 1.3. Let ρ(0) be a minimal crystalline lift of ρ, which exists by Theorem 1.8 (3) and let ρ(0) be an almost strictly { ℓ } compatible system passing through ρ(0) (Theorem 1.10). Let k = k(ρ) be the weight of the system, i.e., the Hodge-Tate weights are 0, k 1 . { − } Paso 1: change to a weight two system. Let w be a sufficiently large prime, so that it does not belong to the ramification set of the compatible system and w > k. ASIMPLIFIEDPROOFOFSERRE’SCONJECTURES 7

(0) (0) Then ρw is in the Fontaine-Laffaille case, thus the Serre’s weight of ρw coincides with the weight k of the compatible system. By Lemma 1.12 we can take w so that (0) ρw is not residually bad dihedral (since the weight k is fixed). If the residual image (0) of ρw happens to be solvable, then modularity follows from Theorem 1.3 and 1.4 (in the irreducible case) and Theorem 1.6 (in the reducible case). Otherwise, take (1) (0) a weight 2 lift ρw of ρw , which is minimally ramified at all other primes (whose existence is guaranteed by Theorem 1.8 (4)). Note that Theorem 1.4 implies that (0) (1) (1) modularity of ρw is equivalent to that of ρw . Make ρw part of an almost strictly (1) compatible system ρℓ . { } (1) Let p ,...,pr be the primes in the ramification set S of the system ρ . 1 { ℓ } Paso 2: add what is called a “good-dihedral prime”. For this purpose, take a large prime q 1 (mod 4) (greater than 5) which is split in the coefficient field K of the compatible≡ system. As in the previous step, by taking q sufficiently large, we can assume it does not belong to the ramification set S of the compatible system and (1) (1) that ρq is not residually bad dihedral. If the residual image of ρq happens to be solvable, then modularity follows from Theorem 1.3 and 1.4 (in the irreducible case) and Theorem 1.6 (in the reducible case). Then we can assume that the image (1) of the residual representation ρq is non-solvable. Note that the hypothesis q split in the coefficient field implies that the residual image lies in GL2(Fq). By [KW09b, r Lemma 8.2] there exists a prime N which is congruent to 1 modulo 3 8 i=1 pi, such that the restriction of our residual representation to the decomposition· · Q group DN admits a lift that is given by a dihedral representation of order 2q, where the image of the inertia group at N corresponds to a character of order q of the (2) unramified quadratic extension of QN . Take a crystalline lift ρq of weight 2 with such a dihedral image of order 2q at the prime N (whose existence is guaranteed (1) by Theorem 1.8 (4)). Note that Theorem 1.4 implies that modularity of ρq is (2) (2) equivalent to that of ρq . Make ρq part of an almost strictly compatible system ρ(2) . { ℓ } Remark 2. As explained in [KW09b, Lemma 6.3], the existence of the good dihedral prime in our compatible system implies that the residual representation at any ramified prime pi (not at N), at 2 and at 3 is non-solvable. Furthermore, this good dihedral prime survives each step of the “removing the level” process explained in Paso 3 and Paso 4, providing large image all thorough this process.

Paso 3: killing the odd part of the level. Let p ,...,pr,N denote the set of { 1 } primes where the system is ramified, p1 being the smallest (probably equal to 2). (i) For each i =2,...,r, take the pi-th entry of the compatible system ρℓ , consider its reduction (which has large image because of the good-dihedral{ prime),} take (i+1) ρpi a minimal lift unramified at pi (which exists by Theorem 1.8 (3)) and put it (i+1) into an almost strictly compatible system ρℓ . By Theorem 1.4 modularity of (i) { (i+1) } the system ρℓ is equivalent to that of ρℓ . A small{ remark} is needed (for latter purposes):{ } at any point of the reduction procedure, we take a minimal lift whose Hodge-Tate weights are (0,> 0) (to avoid working with modular forms of weight 1). 8 LUISVICTORDIEULEFAITANDARIELMARTÍN PACETTI

If p1 is odd, we do the same procedure at p1, and end with a system unramiﬁed outside N . In such a case, go to Paso 5. { } Paso 4: removing 2 from the level. This step follows the method of [KW09b]. Since Theorem 1.5 is only valid for potentially Barsotti-Tate representations, a congruence modulo 3 is performed (when needed) to reduce to this situation. Take a weight (r+2) (r+1) two minimal lift ρ3 of ρ3 , whose existence is guaranteed by Theorem 1.8 (4) (r+2) and make it part of an almost strictly compatible system ρℓ . Modularity of one system is equivalent to modularity of the other by Theorem{ 1.4.} The following step will prove crucial: (⋆) If the Weil-Deligne representation of the almost strictly compatible system at the prime 2 is (a twist of) Steinberg and has non-solvable residual image at 3 then there exists another almost strictly compatible system which is congruent modulo 3 to the original one, and whose Weil-Deligne type at the prime 2 is given by an order 3 character (in particular has trivial monodromy). Furthermore, modularity of a family is equivalent to modularity of the other one.

Proof: let ρℓ be an almost strictly compatible system such that the Weil-Deligne { } representation of the system at the prime 2 is (a twist of) Steinberg, so ρ3 D2 χ3 | ≃ ( 0 1∗ ) (up to twist). Let χ′ be a character of order 3, and take a weight 2 lift ρ3′ of ρ3 (which exists by Theorem 1.8 (4)) whose restriction to D2 is of the form ′ χ ∗′ (in particular monodromy is trivial) and make it part of an almost strictly 0 χ 2 compatible system ρ′ using Theorem 1.10. Modularity of the system ρ′ is { ℓ} { ℓ} equivalent to that of ρℓ by Theorem 1.4. { } It is important to remark (as explained in the proof of Theorem 9.1 of [KW09b]) that after applying (⋆) to a compatible system, the new family satisfies that k(ρ2′ )= 2. The reason is that the weight 4 case corresponds to a très ramifiée situation; in particular the representation is flat over an extension with even ramification index, but the ramification type at 2 was chosen so that it becomes flat over a cubic one, leading to a contradiction. Tune the family (if needed) via the following steps: (r+2) (r+3) (1) If k ρ2 = 4, take a weight 2 lift ρ2 with Steinberg type at 2 (which exists by Theorem 1.8 (2)) and make it part of an almost strictly compatible (r+2) system ρℓ . Modularity of one family is equivalent to modularity of the { } (r+3) (r+2) other by Theorem 1.5. Otherwise (to unify notation) take ρℓ = ρℓ . (2) If the compatible system ρ(r+3) has Steinberg Weil-Deligne type at two, { ℓ } we apply (⋆) to get an almost strictly compatible system ρ(r+4) whose { ℓ } local type at 2 comes from an order 3 character (otherwise, take ρ(r+4) = { ℓ } ρ(r+3) ). { ℓ } (3) At this step, k ρ(r+4) = 2. Let ρ(r+5) be an almost strictly compatible 2 { ℓ } (r+4) system containing a minimal crystalline weight 2 lift of ρ2 at the prime 2 (whose existence is warranted by Theorem 1.8 (1)). Modularity of one family is equivalent to modularity of the other by Theorem 1.5. Make an extra congruence (if needed) to remove the level at 3 (as in Paso 3) to end with a system ρ′ unramified outside N . { ℓ} { } ASIMPLIFIEDPROOFOFSERRE’SCONJECTURES 9

Paso 5: killing the good-dihedral prime. The reduction modulo N of the system ρ′ lies in one of the following cases: { ℓ} (1) The representation ρN′ is reducible, in which case ρN′ is modular by The- orem 1.6. End of the proof. Incidentally, as the reader may easily check, this is the only step in the whole proof where the fact that the compatible systems that we are considering are only known to be almost strictly compatible is relevant. Since the local parameter at N of this compatible

system is ramified and we are assuming that ρN′ is reducible, the conditions in the definition of ‘almost compatible systems” imply that we do not know the local behavior at N of ρN′ , we only know that it is a de Rham representation. Luckily, this is enough for Theorem 1.6 to hold. (2) The representation is irreducible and not bad-dihedral. Take an almost strictly compatible system ρℓ′′ containing a minimal crystalline lift of it (such a lift exists by Theorem{ 1.8} (3)). Observe that the ramification set S of this system is the empty set. Modularity of the system ρ′′ is equivalent { ℓ } to that of ρ′ by Theorem 1.4. { ℓ} (3) The representation is bad-dihedral. Serre’s level of the representation ρN′ equals 1 (as the system was unramified outside N), then this case cannot occur by Lemma 1.12 and Remark 1, because N 1 (mod 4). ≡ Paso 6: reduction of the weight. Move to the prime p = 5, where the same three possibilities can occur:

(1) The representation ρ5′′ is reducible, in which case ρ5′′ is modular by Theo- rem 1.6. End of the proof. (2) The representation is bad-dihedral. This case cannot happen as 5 1 (mod 4) by Lemma 1.12 and Remark 1. ≡ (3) The representation is irreducible and not bad-dihedral. If Serre’s weight k(ρ ) = 2 or 4, take an almost strictly compatible system ρ˜ℓ , whose 5′′ { } ramification set is empty, containing a minimal crystalline lift of ρ5′′, which exists by Theorem 1.8 (3) and has Hodge-Tate weights (0, k(ρ5′′) 1). If Serre’s weight is 6, take a weight 2 lift, with Steinberg Weil-Deligne− type at p = 5, unramified outside 5, which exists by Theorem 1.8 (4) and make it part of an almost strictly compatible system ρ˜ℓ . In both cases modularity { } of the system ρ′′ follows from that of the system ρ˜ℓ by Theorem 1.4. { ℓ } { } In the cases of weight 2 and 4, we look at the 3-adic representation ρ˜3 and we reduce modulo 3. This residual representation is reducible by Theorem 1.1, andρ ˜3 is ordinary (by [BLZ04, Corollary 4,3], because the weight is either 2 or 4 = 3 + 1). Since Serre’s weight is not 3, the image of ρ˜ D is non-trivial, hence Theorem 1.7 3| 3 implies thatρ ˜3 is modular. The last case to consider is whenρ ˜5 has weight 2, is Steinberg at 5 and unramified at any other prime. In this case, by [Tay02] (Corollary E) we know thatρ ˜5 equals the 5-adic representation of an abelian variety A/Q. Sinceρ ˜5 is semistable at 5 and unramified at any other prime, the variety A has the same properties contradicting Theorem 1.2. We have seen that by a combination of suitable modularity lifting theorems the modularity of the given representation was deduced by propagation from the base cases proved by Serre and Schoof and cases where the residual image is solvable. This concludes the proof of Serre’s conjecture in the odd case. 10 LUISVICTORDIEULEFAITANDARIELMARTÍN PACETTI

3. The case p =2 The projective image is either solvable or non-solvable. Recall that all solvable subgroups of PGL2(F2) in the irreducible case have dihedral projective image (the S4 case does not occur, while the A4 case corresponds to a Borel subgroup over F4 see [KW09b, Lemma 6.1]). The dihedral case was proven in [RT97]. If the image is non-solvable, take an odd weight two lift (guaranteed by Theorem 1.8(1) and (2)), make it part of an almost strictly compatible system and reduce modulo a prime greater than 3 and which is not in the ramiﬁcation set of the compatible system (since the residual Serre weight is 2 under these conditions, p> 2 2 1 = 3 isenough to avoid the bad dihedral situation) to end in a situation covered· − before, namely: the representation is either reducible (so modularity follows from Theorem 1.6) or irreducible in odd characteristic hence it is covered by the cases of Serre’s conjecture we have already solved and Theorem 1.4.

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