# Modularity lifting: the Taylor–Wiles method in outline *Draft lesson. Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Public domain (CC0).* A deformation ring parametrizes representations; a Hecke algebra parametrizes eigenforms. The map between them can be surjective even when injectivity is a deep arithmetic problem. We first construct the deformation ring and prove the tangent, obstruction and trace arguments. These establish the algebraic part of the comparison. The additional arithmetic that proves a modularity lifting theorem is specified at the end. ## 1. Coefficients, equivalence and first-order deformations Let \(k=\mathbf F_{p^f}\), and let \(\mathcal O\) be a complete discrete valuation ring of characteristic zero with residue field \(k\), uniformizer \(\varpi\), and finite residue quotients. In the unramified case we write \(\mathcal O=W(k)\). Here this denotes the valuation ring of the unramified degree-\(f\) extension of \(\mathbf Q_p\), with its specified residue identification. Its existence, uniqueness and integral model are proved in *Unramified and totally ramified extensions*, Theorem 2.1 and Corollary 3.1. The arguments below also allow a finite ramified coefficient extension. An Artinian coefficient algebra is a commutative Noetherian local Artinian \(\mathcal O\)-algebra \(A\), with residue field identified with \(k\); morphisms induce the identity on \(k\). We use its discrete topology. Such an algebra is finite as a set. Its powers of the maximal ideal stabilize by the descending-chain condition. The stabilized ideal is finitely generated by Noetherianity and equals its product with the maximal ideal. The determinant proof of Nakayama's lemma in Lemma 3.1 therefore makes it zero. The successive ideal quotients are finite-dimensional over the finite field \(k\), since their ideals are finitely generated. This finite filtration proves the assertion. We use complete Noetherian local coefficient algebras with their maximal-ideal topology as well. Fix a profinite group \(G\) and a continuous representation \[ \bar\rho:G\longrightarrow GL_n(k). \tag{1.1} \] For arithmetic applications \(G=G_{F,S}\): the Galois group of the maximal extension of a number field \(F\) unramified outside a finite set \(S\) containing the infinite places and the places above \(p\). We take \(n=2\) when discussing modularity. A framed deformation is a continuous lift \(\rho_A:G\to GL_n(A)\) of the fixed matrix representation (1.1). Two lifts are strictly equivalent if \[ \rho'_A(g)=C\rho_A(g)C^{-1},\qquad C\in1+M_n(\mathfrak m_A). \tag{1.2} \] An unframed deformation is a strict-equivalence class. This remembers the identification of the residual representation, rather than permitting an arbitrary change of residual basis. Write \(D^{\square}(A)\) and \(D(A)\) for these two sets. The adjoint module \(M=\operatorname{ad}\bar\rho=M_n(k)\) has action \(g\cdot X=\bar\rho(g)X\bar\rho(g)^{-1}\). All cochains below are continuous. Define \[ \begin{aligned} Z^1(G,M)&=\{c:G\to M:c(gh)=c(g)+g\cdot c(h)\},\\ B^1(G,M)&=\{g\mapsto g\cdot X-X:X\in M\},\\ H^1(G,M)&=Z^1(G,M)/B^1(G,M). \end{aligned} \tag{1.3} \] These are vector spaces over \(k\). **Theorem 1.1 (the tangent space).** There are natural identifications \[ D(k[\epsilon]/\epsilon^2)=H^1(G,\operatorname{ad}\bar\rho), \qquad D^{\square}(k[\epsilon]/\epsilon^2)=Z^1(G,\operatorname{ad}\bar\rho). \tag{1.4} \] If these spaces are finite-dimensional and \(h^i=\dim_k H^i\), then the framed tangent dimension is \(h^1+n^2-h^0\). **Proof.** Every lift has a unique expression \[ \rho_\epsilon(g)=(1+\epsilon c(g))\bar\rho(g). \tag{1.5} \] Multiplying (1.5) for \(g,h\) gives precisely the cocycle equation in (1.3). Conversely that equation makes (1.5) a continuous representation. Conjugation by \(1+\epsilon X\) replaces \(c(g)\) by \(c(g)+X-g\cdot X\). Thus strict equivalence is exactly quotienting by \(B^1\). The kernel of \(X\mapsto(g\mapsto g\cdot X-X)\) is \(M^G=H^0(G,M)\), so \(\dim B^1=n^2-h^0\). This proves all the assertions. \(\square\) If the determinant is prescribed, its first-order condition is \(\operatorname{tr}c(g)=0\), because \(\det(1+\epsilon X)=1+\epsilon\operatorname{tr}X\). Its tangent space is therefore \[ \ker\bigl(H^1(G,\operatorname{ad}\bar\rho) \xrightarrow{\operatorname{tr}}H^1(G,k)\bigr). \tag{1.6} \] Indeed the scalar coefficient module is trivial, so its one-coboundaries are zero: vanishing of the trace class means vanishing of the trace cocycle itself. All adjoint coboundaries already have zero trace. If \(p\nmid n\), the decomposition into trace-zero matrices and scalars identifies (1.6) with \(H^1(G,\operatorname{ad}^0\bar\rho)\). When \(p\mid n\), scalar matrices belong to the trace-zero submodule, and this identification cannot be inferred by dividing the trace by \(n\). ### Why the arithmetic tangent spaces are finite **Lemma 1.2 (finite cyclic extensions).** For every number field \(L\) and finite set \(T\) of places, there are only finitely many cyclic degree-\(p\) extensions of \(L\) unramified outside \(T\). Consequently \(\operatorname{Hom}_{\rm cont}(G_{L,T},\mathbf F_p)\) is finite. **Proof.** Put \(K=L(\mu_p)\). Its degree over \(L\) divides \(p-1\). Restriction of characters to the absolute Galois group of \(K\) is injective: a character vanishing there factors through \(\operatorname{Gal}(K/L)\), whose order is prime to \(p\). After enlarging a finite set \(T'\) in \(K\), these restricted characters are unramified outside \(T'\). It is enough to bound them. Let \(E/K\) be cyclic of degree \(p\), with generator \(\sigma\), and choose a primitive \(p\)-th root \(\zeta\in K\). There is \(x\ne0\) with \(\sigma x=\zeta x\): the operator \(\sum_{i=0}^{p-1}\zeta^{-i}\sigma^i\) is nonzero, so apply it to a suitable element of \(E\). Here distinct field automorphisms are linearly independent. For clarity, in a shortest nonzero linear relation among them, evaluate at \(ay\) and subtract the value at \(y\) times one selected automorphism's value at \(a\). Distinct automorphisms give an \(a\) making one of the remaining coefficients nonzero; the new relation is shorter, a contradiction. Now \(a=x^p\in K^\times\), \(x\notin K\), and \(E=K(x)\). At a place outside \(T'\), unramifiedness gives \(p\,v_E(x)=v_K(a)\), so \(p\mid v_K(a)\). The possible classes therefore belong to \[ C_{K,T',p}= \{a\in K^\times:p\mid v(a)\text{ outside }T'\}/K^{\times p}. \tag{1.7} \] This group is finite. Work with fractional ideals away from \(T'\). Unique ideal factorization is actually proved in *Discrete valuation rings and Dedekind domains*, Proposition 3.1 and Theorem 3.2. Write \((a)=\mathfrak b^p\) there. Its ideal class lies in the \(p\)-torsion of the class group of the ring of \(T'\)-integers. This class group is a quotient of the ordinary ideal class group: forget the prime factors at \(T'\) in any ideal. Every ideal away from \(T'\) has such a lift, and principal ideals map to principal ideals, proving surjectivity on class groups. The ordinary group is finite by *Idèles and the idèle class group*, Corollary 3.4. The kernel of this class map in (1.7) consists of classes of \(T'\)-units: if \(\mathfrak b=(b)\), then \(a/b^p\) is such a unit. The same earlier proof, equations (23)–(24), proves that the \(T'\)-unit group is finitely generated. Its quotient by \(p\)-th powers is finite. Hence (1.7) is finite. Multiplying \(a\) by a \(p\)-th power does not change \(K(a^{1/p})\). Thus (1.7) bounds the extension fields; each field gives at most \(p-1\) nonzero characters. This proves finiteness and the lemma. \(\square\) **Corollary 1.3.** For \(G=G_{F,S}\), every finite-dimensional continuous \(k[G]\)-module \(M\) has finite-dimensional \(H^1(G,M)\). For every open subgroup \(H\subset G\), its maximal pro-\(p\) quotient is topologically finitely generated. **Proof.** First take an open normal subgroup \(H\) acting trivially on \(M\). Restriction of a cocycle is a homomorphism \(H\to M\). Lemma 1.2, applied to its fixed number field, makes this Hom space finite. A cocycle restricting to zero on \(H\) factors through the finite group \(G/H\), so the kernel of restriction also has finite dimension. Coboundaries vanish on \(H\). This proves the first assertion. Let \(P\) be the maximal pro-\(p\) quotient of any open subgroup \(H\), and let \(\Phi(P)\) be the closed subgroup generated by commutators and \(p\)-th powers. Then \(P/\Phi(P)\) is a profinite \(\mathbf F_p\)-vector space. Its continuous linear functionals are \(\operatorname{Hom}(H,\mathbf F_p)\), which is finite by Lemma 1.2. For an open subgroup, its fixed field is a number field, and its characters are characters of that field unramified outside the finite set above \(S\); passing to the given Galois quotient can only decrease their number. These functionals separate points, since every nonzero point survives in a finite elementary abelian quotient. Their evaluation map embeds \(P/\Phi(P)\) in a finite-dimensional vector space, so this quotient is finite. Lift one of its finite bases to \(P\). Those lifts topologically generate \(P\). Otherwise their closed generated subgroup has proper image in some finite \(p\)-group quotient: separate a point from the closed subgroup by an open normal subgroup. A proper subgroup of a finite \(p\)-group lies in a subgroup of index \(p\). A nontrivial finite \(p\)-group has nontrivial centre: the class equation expresses its order as the size of its centre plus noncentral conjugacy-class sizes divisible by \(p\). Its centre therefore has order divisible by \(p\); powering a nonidentity central element gives a central element of order \(p\). To verify the index assertion, take a maximal subgroup \(Q\); if the centre is not contained in \(Q\), maximality gives the group as \(QZ\), so \(Q\) is normal and the simple abelian quotient has order \(p\). If the centre is contained in \(Q\), apply induction to the quotient by any central subgroup of order \(p\). Composing the resulting index-\(p\) quotient with \(P\) gives a nonzero functional annihilating all the chosen lifts, a contradiction. \(\square\) ## 2. Constructing the universal ring We prove representability, including continuity and the conjugation quotient. The finite-group and trace-ring constructions in [de Smit–Lenstra's freely accessible author edition](https://pub.math.leidenuniv.nl/~smitbde/papers/bu-fermat-1995-desmit-lenstra.pdf), §§2–4, and [Kisin's free Lecture 1](https://people.math.harvard.edu/~kisin/notes/notes.pdf), §§1.2–1.4, provide further reading. **Lemma 2.1 (formal power-series algebra).** The ring \(P=\mathcal O[[X_1,\ldots,X_N]]\) is complete, compact and Noetherian for its maximal ideal \(\mathfrak n=(\varpi,X_1,\ldots,X_N)\). Every ideal is closed, and every quotient is a complete local ring. **Proof.** Its quotients by \(\mathfrak n^r\) are finite, and compatible coefficients identify \(P\) with their inverse limit. This proves completeness and compactness. Its associated graded ring is \[ \operatorname{gr}_{\mathfrak n}P=k[T_0,T_1,\ldots,T_N]. \tag{2.1} \] This polynomial ring is Noetherian. One proof proceeds by induction on the number of variables: in \(B[T]\), the ideals of leading coefficients of elements of an ideal with degree at most \(r\) form an ascending chain in the Noetherian ring \(B\). The chain stabilizes; choose finitely many polynomials giving generators up to that stage. Subtract their multiples to reduce the degree of any polynomial in the ideal, and induct on degree. The finitely many chosen polynomials generate the ideal. Start with the field \(k\). For an ideal \(J\subset P\), its leading forms generate a homogeneous ideal in (2.1). Choose finitely many \(f_1,\ldots,f_a\in J\) whose leading forms generate it. Given \(f\in J\), subtract multiples of the \(f_i\) that remove its leading form, then repeat at increasing orders. For each \(i\) the coefficients added to its multiplier tend to zero in \(\mathfrak n\)-adic order. They converge in \(P\), and separation gives \(f=\sum b_i f_i\). Thus \(J\) is finitely generated. The same leading-form argument applies to an element of the closure of \(J\): at each finite order its leading form is already the leading form of an element of \(J\). It expresses that element of the closure as \(\sum b_i f_i\) too. Hence \(J\) is closed. A quotient by a closed ideal of this compact complete ring is the inverse limit of its finite quotients, and is complete. \(\square\) **Proposition 2.2 (the framed ring).** Suppose the maximal pro-\(p\) quotient of \(H=\ker\bar\rho\) is finitely generated. There is a complete Noetherian local \(\mathcal O\)-algebra \(R^{\square}\) and a continuous universal framed representation, representing \(D^{\square}\) on Artinian and complete Noetherian coefficient algebras. **Proof.** Let \(N\) be the kernel of the maximal pro-\(p\) quotient of \(H\). It is characteristic in \(H\), hence normal in \(G\). Every deformation kills \(N\). Indeed its image of \(H\) lies in \(1+M_n(\mathfrak m_A)\), a finite \(p\)-group: the successive subgroups \(1+M_n(\mathfrak m_A^j)\) have additive quotients in \(M_n(\mathfrak m_A^j/\mathfrak m_A^{j+1})\). For a complete target apply this at every finite quotient. Thus all deformations factor through \(\Pi=G/N\). This profinite group is finitely generated. Its open subgroup \(H/N\) is finitely generated, and adjoining lifts of the finitely many elements of \(G/H\) gives generators \(\gamma_1,\ldots,\gamma_r\) for \(\Pi\). Choose arbitrary matrices \(A_i^0\in GL_n(\mathcal O)\) lifting \(\bar\rho(\gamma_i)\). Put one variable in each entry of each matrix, and write \[ P=\mathcal O[[X_{i,ab}:1\le i\le r,\ 1\le a,b\le n]], \qquad A_i=A_i^0+(X_{i,ab})_{a,b}. \tag{2.2} \] Their determinants are units. Use the free profinite group on these \(r\) generators. Explicitly, start with the abstract free group and take its inverse limit over all its finite quotients. A choice of \(r\) elements in a profinite group gives a continuous homomorphism from this completion, by its maps to every finite quotient. This property also follows by closing the word image in the product of all finite groups with \(r\) marked generators. There is a continuous surjection \(\widehat F_r\to\Pi\), since the word image is dense and the image of a compact group is closed. The matrices (2.2) likewise define a continuous map \(\widehat F_r\to GL_n(P)\), obtained in every finite quotient of \(P\). For each element \(w\) of the kernel of \(\widehat F_r\to\Pi\), impose all the entries of \(w(A_i)-I_n\). Let \(J\) be the ideal they generate. Lemma 2.1 makes it closed. Then \(R^{\square}=P/J\). Its residual representation is (1.1), and its universal matrices factor through \(\Pi\). Any framed lift gives a unique continuous homomorphism from \(P\) by specifying the entries of its generators; it kills \(J\). Conversely every such map from \(P/J\) gives the required continuous lift. Uniqueness follows because the matrices of the \(\gamma_i\) determine a continuous representation on their dense word subgroup, hence on \(\Pi\). This proves the universal property. Lemma 2.1 proves all the asserted ring properties. \(\square\) **Theorem 2.3 (unframed representability).** Under the hypothesis of Proposition 2.2, assume \(\operatorname{End}_{k[G]}(k^n)=k\). Then \(D\) is represented by a complete Noetherian local \(\mathcal O\)-algebra \(R\), with an actual continuous universal representation. In particular this holds for every absolutely irreducible \(\bar\rho\) of \(G_{F,S}\). **Proof.** We give a unique normal form for each strict-equivalence class. The derivative of conjugation is the map \[ M_n(k)/kI_n\longrightarrow M_n(k)^r, \qquad U\longmapsto([U,\bar\rho(\gamma_i)])_{i=1}^r. \tag{2.3} \] It is injective by the endomorphism hypothesis. Represent its source by matrices with \(U_{nn}=0\), a complement to the scalar matrices in every characteristic. Select \(n^2-1\) entry coordinates in the target for which (2.3) is an isomorphism onto \(k^{n^2-1}\); this is possible by choosing a nonsingular square minor of its matrix. Lift those coordinates to \(\mathcal O\). Call a framed lift normal if the selected entries of its \(\rho(\gamma_i)\) equal the corresponding fixed entries of \(A_i^0\). Every lift can be made normal. Suppose these equalities have been achieved modulo \(\mathfrak m_A^j\). Conjugating by \(1+U\), with entries of \(U\) in \(\mathfrak m_A^j\) and \(U_{nn}=0\), changes the selected entries modulo \(\mathfrak m_A^{j+1}\) by the isomorphism (2.3) tensored with \(\mathfrak m_A^j/\mathfrak m_A^{j+1}\). There is a unique correction at that order. Start at \(j=1\) and continue. For Artinian \(A\) this terminates; for complete \(A\) the product of these conjugating matrices converges. Two strictly conjugate normal lifts are equal. Multiply their conjugating matrix by a scalar to make its \(nn\)-entry equal to one. If this matrix differs from the identity, let \(j\) be its first nonzero order. Its leading matrix has \(nn\)-entry zero. The selected entries of both normal lifts agree exactly, so (2.3) kills that leading matrix. Its injectivity is a contradiction. Separation proves the same assertion for a complete target. Thus the normal form is unique and natural under coefficient homomorphisms. In the universal framed ring impose the \(n^2-1\) selected-entry equalities. The resulting quotient \(R\) represents normal framed lifts, and hence \(D\). It has the claimed complete Noetherian properties by Lemma 2.1. The matrices over this quotient provide the actual universal representation. Compatible normal forms show that the same property holds for complete targets, not just Artinian ones. For an absolutely irreducible residual representation the endomorphism condition holds: over an algebraic closure any endomorphism has an eigenvalue and its nonzero eigenspace is invariant, hence the whole irreducible space. It is scalar there, and a scalar matrix defined over \(k\) has its scalar in \(k\). Corollary 1.3 supplies the finite generation hypothesis for \(G_{F,S}\). \(\square\) No division by \(n\), and no hypothesis \(p>n\), occurred in this construction. Notice also that its universal representation is a chosen normal matrix representative; its intrinsic meaning is its deformation class. ## 3. The minimal power-series presentation **Lemma 3.1 (generation from the relative cotangent space).** Let \(B\) be a complete Noetherian local \(\mathcal O\)-algebra with residue field \(k\). If \(u_1,\ldots,u_d\in\mathfrak m_B\) span \[ \mathfrak m_B/(\mathfrak m_B^2+\varpi B), \tag{3.1} \] then the continuous homomorphism \(\mathcal O[[x_1,\ldots,x_d]]\to B\), \(x_i\mapsto u_i\), is surjective. **Proof.** We use the finite-module form of Nakayama's lemma, with proof. If a finitely generated module \(Q\) satisfies \(Q=\mathfrak m Q\), choose generators \(q_i\) and write \(q_i=\sum a_{ij}q_j\) with \(a_{ij}\in\mathfrak m\). The adjugate of \(I-(a_{ij})\) shows that its determinant kills every \(q_i\). That determinant is one modulo \(\mathfrak m\), hence a unit. Thus \(Q=0\). Apply this to \(\mathfrak m_B/(\varpi,u_1,\ldots,u_d)\). Its finite generation follows from Noetherianity; (3.1) says it equals its product with \(\mathfrak m_B\). Consequently \[ \mathfrak m_B=(\varpi,u_1,\ldots,u_d). \tag{3.2} \] Given \(b\in B\), lift its residue by a constant in \(\mathcal O\). At the next stage its error in \(\mathfrak m_B^j\), modulo \(\mathfrak m_B^{j+1}\), is a finite sum of degree-\(j\) monomials in the generators (3.2), with coefficients lifted from \(k\) to \(\mathcal O\). Subtract their corresponding polynomials in \(\varpi,x_i\), then continue at increasing degrees. These corrections converge in the formal power-series ring, and their images converge to \(b\). This proves surjectivity. \(\square\) **Theorem 3.2 (the presentation by tangent variables).** For the ring \(R\) in Theorem 2.3, \[ R\simeq\mathcal O[[x_1,\ldots,x_d]]/I, \qquad d=\dim_k H^1(G,\operatorname{ad}\bar\rho). \tag{3.3} \] For any deformation functor represented by a complete Noetherian local \(\mathcal O\)-algebra, the same conclusion holds with its tangent dimension in place of \(d\). In particular the requested coefficient ring in the unramified case is \(W(k)\). **Proof.** Local \(\mathcal O\)-homomorphisms \(R\to k[\epsilon]/\epsilon^2\) inducing the residual map are exactly the linear functionals on \(\mathfrak m_R/(\mathfrak m_R^2+\varpi R)\). To check this, a homomorphism sends an element of \(\mathfrak m_R\) to \(\epsilon\) times its functional value and kills its square and \(\varpi R\). Conversely such a functional defines a homomorphism on \(R/(\mathfrak m_R^2+\varpi R)\), a \(k\)-algebra whose maximal ideal has square zero; every element is its residue constant plus that ideal element. This proves the identification. Representability and Theorem 1.1 identify this dual with \(H^1(G,M)\). Choose lifts of a basis of the relative cotangent space. Lemma 3.1 gives (3.3), and Lemma 2.1 makes its kernel a closed finitely generated ideal. The same argument uses only representability and the specified tangent space, proving the final assertion. \(\square\) The number \(d\) is the minimum number of relative power-series variables: any surjection with \(e\) variables gives a spanning set of size \(e\) in the relative cotangent space, so \(e\ge d\). The presentation does not assert that \(I\) is generated by a regular sequence; that is an additional complete-intersection assertion. ## 4. The obstruction to a lift A small extension is a surjection of Artinian coefficient algebras \(B\twoheadrightarrow A\) with kernel \(J\) satisfying \(\mathfrak m_BJ=0\). Thus \(J\) is a \(k\)-vector space and \(J^2=0\). Define continuous two-cocycles with coefficients in a \(G\)-module \(N\) by \[ a(g,h)+a(gh,l)=g\cdot a(h,l)+a(g,hl), \tag{4.1} \] and two-coboundaries by \(\delta b(g,h)=b(g)+g\cdot b(h)-b(gh)\). Their quotient is \(H^2(G,N)\). **Theorem 4.1 (the lifting obstruction).** For a specified \(\rho_A\), a small extension has a canonical obstruction \[ o(\rho_A,B)\in H^2(G,\operatorname{ad}\bar\rho\otimes_k J). \tag{4.2} \] It vanishes exactly when \(\rho_A\) lifts. If it vanishes, the lifts up to conjugation by \(1+M_n(J)\), with their specified reduction to \(\rho_A\), form a torsor under \(H^1(G,\operatorname{ad}\bar\rho\otimes_k J)\). **Proof.** Choose a continuous matrix lift \(q(g)\) of \(\rho_A(g)\), with \(q(1)=I_n\). This is possible by selecting one invertible lift for each of the finitely many values of \(\rho_A\). Define \[ q(g)q(h)q(gh)^{-1}=1+a(g,h),\qquad a(g,h)\in M_n(J). \tag{4.3} \] Conjugation on \(M_n(J)\) depends only on \(\bar\rho\), because \(\mathfrak m_BJ=0\). Associativity of three matrices gives (4.1). Replacing \(q(g)\) by \((1+b(g))q(g)\) changes \(a\) by \(\delta b\). Hence its class is independent of the chosen lift. Its vanishing is precisely the possibility of making (4.3) the identity by such a correction. That corrected \(q\) is a continuous representation. Two actual lifts differ by \((1+c(g))\); their representation laws give the one-cocycle equation for \(c\). Conjugation by \(1+X\), \(X\in M_n(J)\), changes \(c\) by the negative of the coboundary of \(X\), just as in Theorem 1.1. Thus \(H^1\) acts freely and transitively on these equivalence classes once one lift has been chosen. This is exactly the torsor assertion. \(\square\) In particular, \(H^2=0\) gives lifting across every small extension. An arbitrary Artinian surjection can be factored into small extensions: successively quotient its kernel along a composition series refined by its products with the maximal ideal. Each successive kernel is killed by that ideal. This proves the corresponding unrestricted lifting property. ## 5. Traces and the map to a Hecke algebra Absolute irreducibility enters a second time here: it lets traces detect an entire deformation, including in residue characteristic two. **Lemma 5.1 (matrix span and inner derivations).** If \(\bar\rho\) is absolutely irreducible, the \(k\)-span of \(\bar\rho(G)\) is \(M_n(k)\). Every \(k\)-linear derivation \(D:M_n(k)\to M_n(k)\otimes_k J\) is inner. **Proof.** Over \(\bar k\), the module \(\bar k^n\) is simple and its endomorphisms are scalars by the eigenvalue argument. Its acting algebra can interpolate any finite independent list of vectors. Here is the induction proof. The image of \(a\mapsto(av_1,\ldots,av_r)\) is a submodule of the direct sum of \(r\) copies. Its projection on the first \(r-1\) coordinates is surjective by induction. Its intersection with the last coordinate is either zero or the whole simple module. In the latter case the image is the whole direct sum. In the former it is the graph of a module map from the first \(r-1\) copies; every coordinate map is scalar, forcing \(v_r\) to be a scalar combination of the previous vectors after evaluation at the identity. This contradicts independence. Taking a basis makes the acting algebra all matrices. Its scalar extension from \(k\) therefore has dimension \(n^2\), so its \(k\)-span already has that dimension and equals \(M_n(k)\). For the derivation let \(E_{ij}\) be matrix units, and set \(U=\sum_i D(E_{i1})E_{1i}\). Differentiate \(E_{ab}E_{i1}=\delta_{bi}E_{a1}\), multiply on the right by \(E_{1i}\), and sum over \(i\). Since \(\sum_i E_{i1}E_{1i}=I_n\), this gives \[ UE_{ab}=D(E_{ab})+E_{ab}U. \tag{5.1} \] Thus \(D(X)=[U,X]\) for every matrix \(X\), as asserted. \(\square\) **Theorem 5.2 (trace detection).** If \(\bar\rho\) is absolutely irreducible and two deformations to an Artinian coefficient algebra have the same trace at every \(g\in G\), they are strictly equivalent. **Proof.** Induct on the length of \(A\), with \(A=k\) the trivial base case. Otherwise choose a nonzero proper ideal \(J\subset A\) killed by \(\mathfrak m_A\), by taking the last nonzero power of the maximal ideal. By induction the representations are strictly equivalent modulo \(J\). Lift their conjugating matrix to \(A\), so they now agree modulo \(J\). Extend both representations to \(A[G]\), and let \(\Delta=\rho'-\rho\), with values in \(M_n(J)\). Equality of traces says \(\operatorname{tr}\Delta(x)=0\) for every \(x\in A[G]\), by linearity. Multiplicativity and \(J^2=0\) give \[ \Delta(xy)=\bar\rho(x)\Delta(y)+\Delta(x)\bar\rho(y). \tag{5.2} \] Here the residual action is used on \(J\), and \(\Delta\) depends only on the coefficients modulo \(\mathfrak m_A\). For \(\bar\rho(x)=0\), apply (5.2) to \(yx\) and take traces: \(\operatorname{tr}(\bar\rho(y)\Delta(x))=0\) for every \(y\). Lemma 5.1 makes these \(\bar\rho(y)\) all matrices. Their trace pairing is nondegenerate, since \(\operatorname{tr}(E_{ij}X)=X_{ji}\). Thus \(\Delta(x)=0\). Consequently \(\Delta\) descends to a derivation of \(M_n(k)\). By Lemma 5.1 it is \([U,-]\) with \(U\in M_n(J)\). Hence \[ \rho'=(1+U)\rho(1-U). \tag{5.3} \] This is a strict equivalence and completes the induction. \(\square\) **Corollary 5.3 (trace generation).** The universal ring \(R\) for an absolutely irreducible residual representation is topologically generated over \(\mathcal O\) by the traces of its universal representation. A finite subset of those traces suffices. **Proof.** For each \(g\), subtract a fixed \(\mathcal O\)-lift of its residual trace to obtain \(t_g\in\mathfrak m_R\). If their classes did not span the relative cotangent space (3.1), a nonzero linear functional would annihilate them all. The corresponding homomorphism \(R\to k[\epsilon]/\epsilon^2\) and the zero tangent homomorphism would give deformations with identical traces. Theorem 5.2 makes them strictly equivalent; representability makes the two ring homomorphisms equal, a contradiction. The cotangent space is finite-dimensional, so finitely many \(t_g\) span it. Lemma 3.1 proves the assertion. \(\square\) **Proposition 5.4 (the surjectivity argument).** Let \(T\) be a complete Noetherian local \(\mathcal O\)-algebra with residue \(k\). Suppose an actual continuous representation \(\rho_T:G\to GL_n(T)\) is a deformation in the problem represented by \(R\). Suppose also that \(T\) is topologically generated over \(\mathcal O\) by a collection of its traces \(\operatorname{tr}\rho_T(g)\). Then the classifying map \[ R\longrightarrow T \tag{5.4} \] is surjective. **Proof.** Representability gives the map, and it sends every universal trace to the specified trace in \(T\). Its image therefore contains the dense algebra generated by the given traces and \(\mathcal O\). The image is closed: \(R\) is compact by its finite-variable presentation and finite residue quotients, the map is continuous, and \(T\) is Hausdorff. A closed dense image is all of \(T\). \(\square\) For a localized Hecke algebra the intended generators include Hecke operators identified with Frobenius traces. Proposition 5.4 proves the surjectivity once the integral representation and those identifications exist. A representation only over \(T[1/p]\) is insufficient for its hypothesis: integral descent and a compatible residual identification must be proved. Nor does surjectivity prove injectivity. The comparison of congruences and deformation relations is needed for \(R=T\). ## 6. An actual ordinary local condition One can impose some local conditions by explicit equations before doing any modularity lifting. Let \(D\subset G\) be a closed subgroup and suppose the residual two-dimensional representation has a \(D\)-stable line, chosen as \(ke_1\). Assume some \(g_0\in D\) has distinct diagonal eigenvalues on that line and its quotient. This is the distinguished-character hypothesis. **Proposition 6.1 (the distinguished line condition).** Deformations possessing a \(D\)-stable direct-summand line lifting \(ke_1\) are represented by a quotient of \(R\). Requiring the quotient character to be trivial on a closed subgroup \(I\subset D\) also gives such a quotient, when the residual quotient character is trivial there. **Proof.** Write the universal matrix of \(g\) as \(\left(\begin{smallmatrix}a_g&b_g\\c_g&d_g\end{smallmatrix}\right)\). Every direct-summand line lifting \(ke_1\) is uniquely generated in normalized coordinates by \((1,z)^t\), \(z\in\mathfrak m\). Its invariance under \(g\) is exactly \[ c_g+(d_g-a_g)z-b_gz^2=0. \tag{6.1} \] For \(g_0\) the derivative at the residual root zero is \(\bar d_{g_0}-\bar a_{g_0}\ne0\). There is a unique solution \(z_0\in\mathfrak m_R\): lift zero successively modulo powers of \(\mathfrak m_R\), correcting each error by division by that unit derivative. Quadratic errors lie in the next power, so the corrections converge; the same first-nonzero-order argument proves uniqueness. This works also in every Artinian target and commutes with coefficient maps. Impose (6.1) for every \(g\in D\) at \(z=z_0\). The closed ideal of these expressions defines the desired quotient. The line is a direct summand, because \((1,z_0)^t,e_2\) is a basis. In that basis the line character is \(a_g+b_gz_0\), and the quotient character is \(d_g-b_gz_0\). To make the latter trivial on \(I\), add the equations \(d_h-b_hz_0=1\), \(h\in I\). They have the specified residual values. These conditions are invariant under strict equivalence: a strict change of basis transports a lifting line to a lifting line, which is unique because it is the unique line for \(g_0\). Thus the quotient represents the unframed subfunctor, not merely a condition on one arbitrary matrix representative. \(\square\) For a decomposition group at \(p\), with \(I\) its inertia group, this is a stable-line deformation condition with unramified quotient. The distinction between the two residual characters is part of the hypothesis. It does not construct a Hida family, impose finite-flat group-scheme conditions, or prove a global modularity theorem. ## 7. Exercises **Exercise 7.1 (easy).** Compute \(\operatorname{ad}(1\oplus\bar\chi)\). What changes in its invariant space when \(\bar\chi=1\)? Explain the role of absolute irreducibility in the two constructions above. **Solution.** On the matrix units the characters are \[ \operatorname{ad}(1\oplus\bar\chi) \simeq 1\oplus1\oplus\bar\chi^{-1}\oplus\bar\chi, \tag{7.1} \] with the first two summands on the diagonal. If \(\bar\chi\ne1\), the invariant space has dimension two; if \(\bar\chi=1\), it has dimension four. In neither case is the full centralizer just the scalars, so the conjugation derivative used for the unique normal form has a larger kernel. Absolute irreducibility also gives the full matrix span needed in trace detection. Reducibility can make trace detection fail: for \(G=C_p\) and residual representation \(1\oplus1\), the lifts of a generator \(I_2\) and \(I_2+\epsilon E_{12}\) are representations over the dual numbers with equal traces and determinants. Their cocycles are zero and nonzero, and all coboundaries vanish for the trivial adjoint action. They are not strictly equivalent. This does not assert that every reducible deformation problem fails representability; it identifies the hypotheses actually used here. \(\square\) **Exercise 7.2 (medium).** Verify (1.4) by direct multiplication, and explain why a change of residual basis is not the equivalence being quotiented. **Solution.** Moving \(\bar\rho(g)\) through \(1+\epsilon c(h)\) in (1.5) gives \(1+\epsilon(c(g)+g\cdot c(h))\). Equality with the lift at \(gh\) is the cocycle equation. Conjugation by \(1+\epsilon X\) adds \(X-g\cdot X\), precisely a coboundary up to sign. The conjugating matrix reduces to the identity, preserving the fixed residual identification. Quotienting by other residual changes would be a different moduli problem. \(\square\) **Exercise 7.3 (medium).** Derive (3.3), and show that no presentation with fewer than \(d\) relative variables is possible. **Solution.** Representability and Theorem 1.1 identify the dual of the relative cotangent space with the \(d\)-dimensional tangent space. Lift a cotangent basis, apply the proved finite Nakayama argument to get (3.2), then approximate any ring element successively by homogeneous polynomials in those generators. The convergent series gives the surjection in (3.3). A presentation with \(e\) variables makes their classes span the same cotangent space, so \(d\le e\). Relations need not form a regular sequence. \(\square\) **Exercise 7.4 (hard).** Produce a nonzero obstruction to lifting a one-dimensional representation of \(C_p\) across a small extension. **Solution.** Set \(A=k[t]/(t^p)\), \(B=k[t]/(t^{p+1})\), and \(J=(t^p)\). Its product with \(\mathfrak m_B\) is zero. Send a generator \(g\) to \(1+t\) over \(A\); its \(p\)-th power is one. Every lift of this element to \(B\) is \(1+t+a t^p\), and its \(p\)-th power is \(1+t^p\ne1\). Thus the representation does not lift. More explicitly choose \(q(g^i)=(1+t)^i\), \(0\le i