Proof of De Smit's Conjecture: a Freeness Criterion

Proof of de Smit’s conjecture: a freeness criterion Sylvain Brochard

To cite this version:

Sylvain Brochard. Proof of de Smit’s conjecture: a freeness criterion. Compositio Mathematica, Foundation Compositio Mathematica, 2017, 153 (11), pp.2310 - 2317. 10.1112/S0010437X17007370. hal-01819651

HAL Id: hal-01819651 https://hal.archives-ouvertes.fr/hal-01819651 Submitted on 20 Jun 2018

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Theorem 1.1. Let ϕ : A → B be a local morphism of Noetherian local rings. Assume the following: (i) edim(B) 6 edim(A) (ii) There is a nonzero A-flat B-module M. (iii) A and B are Artin or M is of finite type as a B-module. Then: (1) The morphism A → B is flat and edim(B) = edim(A).

(2) The ring B/mAB is a complete intersection of dimension zero. (3) M is ﬂat over B.

Corollary 1.2 (de Smit’s conjecture). Let A → B be a morphism of Artin local rings with the same embedding dimension. Then any A-ﬂat B-module is B-ﬂat.

Corollary 1.3 (see Theorem 2.5). Let f : X → S be a morphism of locally Noetherian schemes such that for any x ∈ X, the tangent spaces satisfy dim Tx 6 dim Tf(x). Then any S-ﬂat coherent OX -module is also ﬂat over X.

Since its publication, Diamond’s freeness criterion and the related patching method have been used quite a lot. For instance, here is a list of papers (nonexhaustive, in chronological order) that directly use Diamond’s Theorem 2.1: [6], [9], [12], [1], [20], [5], [10], [18], [13], [3]. A variant of this freeness criterion, due to Kisin (see [17, (3.3.1)]), was also used in several major results, for instance the proof of Serre’s modularity conjecture (see [17], [14] and [16]) or the Fontaine–Mazur conjecture for GL2 (see [15]). A more functorial approach of the patching method is proposed in [11]. The paper is organized as follows: in section 2 we prove Theorem 1.1. We also give a global version for schemes that are not necessarily local (see Theorem 2.5). In section 3 we brieﬂy explain how this new freeness criterion allows one to simplify Wiles’s proof of Fermat’s Last Theorem and avoid the use of the patching method.

2. Proof of Theorem 1.1

The following lemma is the heart of the proof.

Lemma 2.1. Let B be a ring. Let (x1,...,xn) and (u1, . . . , un) be sequences of elements of B and let us denote by Jx and Ju the ideals they generate. Let M be a B-module. Assume that:

(i) We have the inclusion of ideals Jx ⊂ Ju, in other words, if x and u denote the column T T matrices (x1,...,xn) and (u1, . . . , un) , there is a square n × n matrix W = (cij) with entries in B such that x = W u. Let ∆ = det W . n (ii) For any m1,...,mn ∈ M, if we have the relation i=1 ximi = 0, then mi ∈ JxM for all i. Then, for any m ∈ M, if ∆m ∈ JxM, then m ∈ JuM. P

Proof. We will need some notations for the minors of W . For any integer ℓ, let Eℓ be the I set {1,...,ℓ} and let Pℓ be the set of subsets I ⊂ En such that |I| = ℓ. For I, J ∈ Pℓ, let ∆J denote the minor of W obtained by deleting the rows (respectively the columns) whose index is I ∅ En in I (respectively J). In other words, ∆J = det(cij) i∈EnrI . In particular, ∆∅ =∆ and ∆En = 1. j∈EnrJ

2 Proof of De Smit’s conjecture: a freeness criterion

i+j i The comatrix is Com (W ) = ((−1) ∆j). If I ⊂ En and i ∈ En, we write I − i for I r {i}. For i 6= j ∈ En, let ε(i, j) = 1if i < j and ε(i, j) = −1 if i > j. For I ⊂ En and i ∈ I, p let ε(i, I) = (−1) , where p is the position of i in the ordered set En r (I − i). Note that i ε(i, I) = (−1) j∈I−i ε(i, j). For any i, j ∈ I with i 6= j, we have the relation Q ε(i, I)ε(i, I − j)= ε(i, j) (1)

Eℓ Now, for any 0 6 ℓ 6 n − 1, if I ∈ Pℓ+1 and i ∈ I, the expansion of the minor ∆I−i along the column i is n Eℓ ℓ k Eℓ∪{k} ∆I−i = (−1) ε(i, I) (−1) cki∆I (2) k=Xℓ+1 Eℓ If i∈ / I, pick an element j ∈ I and consider the expansion along the column j of the minor ∆I−j in which we have replaced the column j by the column i. Since the latter minor has two identical columns, it is zero and we see that for any i∈ / I,

n k Eℓ∪{k} (−1) cki∆I = 0 (3) k=Xℓ+1

Now let us come back to our module M. Let m ∈ M be such that ∆m ∈ JxM. For 0 6 ℓ 6 n, let (Aℓ) denote the following statement.

ℓ (Aℓ): There is a family (aI )I∈Pℓ−1 of elements of M such that for any I ∈ Pℓ, the element

Eℓ ℓ gI =∆I m + ε(i, I)uiaI−i Xi∈I belongs to JxM.

We will prove that (Aℓ) holds for any 0 6 ℓ 6 n by induction on ℓ. The statement (A0) means that ∆m ∈ JxM and holds by assumption. Assume that (Aℓ) holds for some 0 6 ℓ 6 n−1 and let k us prove the statement (Aℓ+1). For any I ∈ Pℓ, since gI ∈ JxM, we can write gI = k∈En xkaI for some elements ak ∈ M. Let I ∈ P . We will compute m′ = ε(i, I)u g . For this, we I ℓ+1 i∈I i I−i P ﬁrst compute P

Eℓ ℓ ε(i, I)ui(gI−i − ∆I−im)= ε(i, I)ui ε(j, I − i)ujaIr{i,j} Xi∈I Xi∈I jX∈I−i ℓ = uiujaIr{i,j}S(i, j) i,j∈I Xi6=j where S(i, j) = ε(i, I)ε(j, I − i). Now the sum vanishes because S(i, j) = −S(j, i). Indeed, S(i, j)S(j, i) = ε(i, j)ε(j, i) by equation (1) above. Hence,

′ Eℓ m = ε(i, I)ui∆I−im Xi∈I

3 Sylvain Brochard

Eℓ We expand ∆I−i along the column i: n ′ ℓ k Eℓ∪{k} m = ε(i, I)ui(−1) ε(i, I) (−1) cki∆I m Xi∈I k=Xℓ+1 n ℓ k Eℓ∪{k} = (−1) (−1) ∆I ckiui m ! k=Xℓ+1 Xi∈I n ℓ k Eℓ∪{k} = (−1) (−1) ∆I xk − ckiui m (because x = W u) k=ℓ+1 i/∈I ! Xn X ℓ k Eℓ∪{k} = (−1) (−1) ∆I xkm (using (3) for each i∈ / I) k=Xℓ+1 On the other hand, by definition ′ k m = ε(i, I)uigI−i = ε(i, I)ui xkaI−i Xi∈I Xi∈I kX∈En Hence, n k ℓ k Eℓ∪{k} ε(i, I)ui xkaI−i − (−1) (−1) ∆I xkm = 0 . Xi∈I kX∈En k=Xℓ+1 By our assumption (ii), the coefficient of xℓ+1 in the above equation belongs to JxM. But this coefficient is ℓ+1 Eℓ+1 ε(i, I)uiaI−i +∆I m . i∈I X n This proves the statement (Aℓ+1). In particular, (An) holds and hence there are elements aI ∈ M for I ∈ Pn−1 such that the element

En n gEn =∆En m + ε(i, En)uiaEn−i iX∈En En belongs to JxM. Since ∆En = 1, this proves that m ∈ JuM. Before proceeding to the proof of Theorem 1.1, let us recall a result from [4]. It will provide a useful characterization for the flatness of M over B. Definition 2.2 [4, 3.1]. Let R be a local ring and M be an R-module. We say that M is weakly torsion-free if, for every λ ∈ R and every m ∈ M, the relation λm = 0 implies that λ = 0 or m ∈ mRM. Proposition 2.3 [4, 3.7]. Let R be an Artin local ring. Assume that R is Gorenstein and contains a field. Then an R-module M is flat over R if and only if it is weakly torsion-free. Proof of Theorem 1.1. Let M be a nonzero A-flat B-module. We will apply Lemma 2.1 with (x1,...,xn) the image in B of a minimal system of generators of mA, and (u1, . . . , un) a system of generators of mB, so that Jx = mAB and Ju = mB. The assumption (i) holds because the morphism A → B is local, and the assumption (ii) holds by [4, 4.2 and 4.3] because M is flat over A. Then ∆ ∈/ mAB, otherwise we would have ∆M ⊂ mAM, which would imply that M = mBM by Lemma 2.1, hence M = 0, which is a contradiction. In particular, for any choice of the matrix W and the generators u1, . . . , un of mB such that x = W u, we have det W 6= 0.

4 Proof of De Smit’s conjecture: a freeness criterion

This proves that edim(B) = edim(A), because if edim(B) < edim(A) we can choose W with a zero column. Moreover, in the ring B/mAB, the matrix W satisﬁes W u = 0 and det W 6= 0. In the language of [19, Deﬁnition 2.6], this means that W is (the transpose of) a u-Wiebe matrix for the ring B/mAB. By [19, 2.7], this implies that the ring B/mAB is a complete intersection of dimension zero, and proves (2).

Let us prove that M is B-flat. By [4, 2.3], it suffices to prove that M/mAM is flat over B/mAB. We have seen that B/mAB is a complete intersection of dimension zero and hence a Gorenstein Artin local ring. Moreover, it contains the field κ(A). Hence, by Proposition 2.3, it suffices to prove that M/mAM is weakly torsion-free over B/mAB, i.e. that for any λ ∈ B and any m ∈ M, the relation λm ∈ mAM implies that λ ∈ mAB or m ∈ mBM. So, let λ ∈ B and m ∈ M be such that λm ∈ mAM and λ∈ / mAB. Let us prove that m ∈ mBM. By [19, 2.7], we know that the image ∆ of ∆ is a generator of the socle of the Gorenstein ring B/mAB. Hence, (∆) ⊂ (λ) in B/mAB. Since λm ∈ mAM, this implies that ∆m ∈ mAM. By Lemma 2.1, we get that m ∈ mBM, as required. Lastly let us prove that A → B is flat. Let E → F be an injection of A-modules. Let K be the kernel of E ⊗A B → F ⊗A B. Since M is B-flat, K ⊗B M is the kernel of E ⊗A M → F ⊗A M, which is zero because M is A-flat. But M is faithfully flat over B (because it is free and nonzero) and hence K = 0.

Remark 2.4. If we replace the assumption (iii) of Theorem 1.1 with “A and B are Artin or M is of ﬁnite type over A”, then we do not need to assume that the morphism ϕ : A → B is local: it is then a consequence of the other hypotheses. Indeed, if there is an element a ∈ mA such that × ϕ(a) ∈ B , then M = ϕ(a)M and hence M = mAM. Then M = 0 by Nakayama’s lemma (or because mA is nilpotent if A is Artin), which is a contradiction. On the other hand, if A is Artin, I am not sure that we really need the assumption that B is Artin too.

The following is a global version of Theorem 1.1.

Theorem 2.5. Let f : X → S be a morphism of locally Noetherian schemes. Assume that for any x ∈ X, the dimensions of the tangent spaces Tx and Tf(x) satisfy

dim Tx 6 dim Tf(x)

Let M be a coherent OX -module, flat over S. Then: (1) M is flat over X. (2) For each point x ∈ Supp(M), f is flat at x, the fiber f −1(f(x)) is a complete intersection of dimension zero at x (i.e. its local ring at x is a complete intersection of dimension zero) and dim Tx = dim Tf(x).

In particular, if there exists a coherent OX -module M flat over S whose support is X, then f is a complete intersection morphism of relative dimension zero (i.e. f is flat and its fibers are zero-dimensional complete intersections).

Proof. Note that the local ring of X at a point x is unaltered through the base change along Spec OS,f(x) → S. By [2, Ch. II, §3, no 4, Prop. 14], the module Mx is flat over OS,f(x). Since dim Tx = edim(OX,x) and dim(Tf(x)) = edim(OS,f(x)), we can apply Theorem 1.1 and Mx is flat over OX,x. This proves (1). If x ∈ Supp(M) then Mx 6= 0 and by Theorem 1.1 we deduce that OX,x is flat over OS,f(x), that these rings have the same embedding dimension and that OX,x/mOX,x is a zero-dimensional complete intersection (where m is the maximal ideal

5 Sylvain Brochard of OS,f(x)). Since the latter ring OX,x/mOX,x identiﬁes with the local ring at x of the ﬁber Xf(x), this proves (2).

3. Application to the proof of Fermat’s Last Theorem

To conclude his proof, Wiles had to prove that a certain morphism of O-algebras ΦΣ : RΣ → TΣ is an isomorphism and that TΣ is a complete intersection over O. Here O is the ring of integers of a finite extension of Qℓ, RΣ is the universal deformation ring for a Galois representation ρ : Gal (Q/Q) → GL 2(k) (where k is the residue field of O) and TΣ is a certain Hecke algebra. The proof proceeds in two steps: first it is proved for Σ = ∅ (the “minimal case”), then the general case is deduced from this one using some numerical criterion for complete intersections. To handle the minimal case, Wiles constructs a system of sets Qn, n > 1, where Qn is a set of primes congruent to 1 mod ℓn with some other ad hoc properties. Then, by considering a kind of “patching” of the morphisms ΦQn : RQn → TQn , Wiles manages to prove the desired result for Σ = ∅. Among other things, the proof relied on the fact (due to Mazur and Ribet) that the homology of the modular curve is a free module of rank 2 over T∅ (known as a “multiplicity one” result). Diamond improved the method in [8]: by patching the modules as well as the algebras, he managed to remove multiplicity one as an ingredient of the proof of Fermat’s Last Theorem. His proof relies on his freeness criterion [8, 2.1]. Since we only want to illustrate how our Theorem 1.1 can be used to avoid the patching method and the use of Taylor-Wiles systems, we will work in the rather restrictive setting of [7] and [8]. We use their notations and statements in the sequel.

Let us prove that Φ∅ is an isomorphism. Let λ be a uniformizer of O. By [7, 2.49] (note that we only use it for n = 1, so here also the proof can be simpliﬁed), there exists a ﬁnite set of prime numbers Q such that:

– edim(RQ) 6 #Q =: r, where RQ = RQ ⊗O k = RQ/λRQ

– For any q ∈ Q, q ≡ 1 (ℓ), ρ is unramiﬁed at q and ρ(Frobq) has distinct eigenvalues. × Let G be the ℓ-Sylow subgroup of q∈Q(Z/qZ) . We endow RQ with the structure of an O[G]- algebra as in [7, §2.8]. Let N be the integer N of [7, (4.2.1)] and M = p2 q, where p is a Q ∅ q∈Q well-chosen auxiliary prime (see [7, §4.3]). Consider the group Q Γ=Γ0(N) ∩ Γ1(M) ′ and let XΓ be the associated modular curve. By [7, 4.10], there is an isomorphism TQ → Tm, ′ 1 where T is the algebra of Hecke operators acting on H (XΓ, O) and m is a certain maximal ideal ′ ′ of T . Consider the Tm-module 1 − H := H (XΓ, O)m of elements on which complex conjugation acts by -1. We can view H as a module over O[G], RQ, or TQ via the O-algebra morphisms ′ O[G] → RQ → TQ → Tm . Theorem 3.1. With the above notation:

(i) H is free over TQ,

(ii) TQ is free over O[G],

(iii) TQ is a relative complete intersection over O[G] (hence also over O),

(iv) ΦQ : RQ → TQ is an isomorphism.

6 Proof of De Smit’s conjecture: a freeness criterion

Proof. By [8, Lemma 3.2], H = H/λH is free over k[G]. Note that there is an isomorphism O[[S1,...,Sr]] O[G] ≃ α αr , where the α are the cardinalities of the l-Sylow subgroups ((1+S1) 1 −1,...,(1+Sr) −1) i × of the (Z/qZ) for q ∈ Q. In particular, αi > ℓ > 2 and it follows that edim(k[G]) = r. We can apply Theorem 1.1 to the morphism k[G] → RQ and the module H. We get that H is free over RQ, and that RQ is free and is a relative complete intersection over k[G]. In particular, RQ → TQ must be injective (otherwise H would have torsion over RQ). Since it is already known to be surjective, it is an isomorphism. We have proved all the statements after ⊗Ok. Since TQ is free over O, the theorem now follows from Lemma 3.2 below.

Lemma 3.2. Let O be a Noetherian local ring with maximal ideal m and residue field k. (i) Let f : M → N be a morphism of finite-type O-modules. Assume that N is free over O and that f ⊗O k is an isomorphism. Then f is an isomorphism. (ii) Let A be a finite O-algebra and M be an A-module which is free of finite rank over O. If M ⊗O k is free over A ⊗O k, then M is free over A. (iii) Assume that O is complete. Let A → B be a morphism of finite local O-algebras, with B free over O. If B ⊗O k is a relative complete intersection over A ⊗O k, then B is a relative complete intersection over A (i.e. the ring B/mAB is a complete intersection). Proof. These are standard consequences of Nakayama’s Lemma.

Remark 3.3. By [7, Cor. 2.45 and 3.32], the canonical morphisms RQ → R∅ and TQ → T∅ induce isomorphisms RQ/aQRQ ≃ R∅ and TQ/aQTQ ≃ T∅, where aQ is the augmentation ideal of O[G] (i.e. the ideal generated by S1,...,SR). Hence, from Theorem 3.1, it follows immediately that Φ∅ : R∅ → T∅ is an isomorphism and T∅ is a complete intersection over O.

Acknowledgments. I warmly thank Bart de Smit for raising his conjecture and for pointing out to me the freeness criterion [8, Thm 2.1]. I also thank Michel Raynaud and Stefano Morra for interesting conversations, and the anonymous referee for valuable comments that helped to clarify the exposition.

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Sylvain Brochard [email protected] Institut Montpelli´erain Alexander Grothendieck, CNRS, Univ. Montpellier