Abstract
We prove that certain crystabelline deformation rings of two dimensional residual representations of \(\mathrm {Gal}(\overline{\mathbb {Q}}_p/\mathbb {Q}_p)\) are Cohen–Macaulay. As a consequence, this allows to improve Kisin’s \(R[1/p]=\mathbb {T}[1/p]\) theorem to an \(R=\mathbb {T}\) theorem.
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Notes
If \(p=2\) then \(\omega \) is trivial and this case is excluded by requiring that \(\rho \) has only scalar endomorphisms. If \(p=3\) then \(\omega ^2=1\) and the case \(\rho \sim \bigl ({\begin{matrix} {\delta }&{} {*}\\ {0}&{}{\delta \omega }\end{matrix}}\bigl )\) is also excluded by our assumption.
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Acknowledgements
YH was partially supported by National Natural Science Foundation of China Grants 11688101; China’s Recruitement Program of Global Experts, National Center for Mathematics and Interdisciplinary Sciences and Hua LooKeng Center for Mathematical Sciences of Chinese Academy of Sciences. VP was partially supported by SFB/TR45 of DFG. The project started when YH visited VP in 2013 supported by SFB/TR45 and he would like to thank the University DuisburgEssen for the invitation and the hospitality. The authors would like to thank Jack Shotton for the appendix to the paper, as well as Toby Gee, James Newton, Shu Sasaki and Jack Thorne for their comments. We also thank the anonymous referee for their careful reading of the paper and pertinent comments.
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Communicated by Toby Gee.
The Appendix section of the article was contributed by Jack Shotton.
Jack Shotton
Department of Mathematics, University of Chicago, 5734 S University Avenue, Chicago, IL 60615, USA
Email: jshotton@uchicago.edu.
Appendices
Appendix A: Equidimensionality of completed tensor products
Let \({\mathcal {C}}(\mathcal {O})\) be the category of local noetherian \(\mathcal {O}\)algebras with residue field k which are \(\mathcal {O}\)torsion free. If \(A, B\in {\mathcal {C}}(\mathcal {O})\) then the completed tensor product \(A\widehat{\otimes }_{\mathcal {O}} B\) also lies in \({\mathcal {C}}(\mathcal {O})\). The purpose of the appendix is to prove the following result, the proof of which we were not able to find in the literature.
Lemma A.1
If A and B in \({\mathcal {C}}(\mathcal {O})\) are both equidimensional then \(A\widehat{\otimes }_{\mathcal {O}} B\) is equidimensional of dimension \(\dim A+\dim B1\).
Recall that a ring of finite Krull dimension is called equidimensional if \(\dim A=\dim A/{\mathfrak {p}}\) for all minimal primes \(\mathfrak {p}\) of A. Recall that \(A\in {\mathcal {C}}(\mathcal {O})\) is called geometrically integral if the algebra \(A\otimes _{\mathcal {O}} \mathcal {O}'\) is an integral domain for all finite extensions \(L'/L\), where \(\mathcal {O}'\) denotes the ring of integers in \(L'\). We note that even if both A and B are integral domains the completed tensor product need not be an integral domain. For example, if \(\mathcal {O}=\mathbb {Z}_p\), \(A=B=\mathbb {Z}_p[x]/(x^2p)\) then \(A\widehat{\otimes }_{\mathcal {O}} B\cong \mathbb {Z}_p[x, y]/( x^2p, y^2p)\) and \((xy)(x+y)=0\) in \(A\widehat{\otimes }_{\mathcal {O}} B\). However, if both A and B are geometrically integral then \(A \widehat{\otimes }_{\mathcal {O}} B\) is also geometrically integral by [2, Lemma 3.3 (3)]. Moreover, Lemma A.1 holds if \(A/\mathfrak {p}\) and \(B/\mathfrak {q}\) are geometrically integral for all the minimal primes \(\mathfrak {p}\) of A and all the minimal primes \(\mathfrak {q}\) of B by [2, Lemma 3.3 (2), (5)]. We will prove the lemma by reducing to this case.
Lemma A.2
If \(A\in {\mathcal {C}}(\mathcal {O})\) then the minimal primes of \(A':=A\otimes _{\mathcal {O}} \mathcal {O}'\) are precisely the prime ideals lying above the minimal primes of A. Moreover, A is equidimensional if and only if \(A'\) is equidimensional.
Proof
The algebra \(A'\) is finite and flat over A. In particular, \(\dim A'=\dim A\) and \(\dim A/\mathfrak {p}=\dim A'/{\mathfrak {P}}\) for all primes \({\mathfrak {P}}\) of \(A'\) with \(\mathfrak {p}= A\cap {\mathfrak {P}}\).
If \(\mathfrak {p}\) is a minimal prime of A then it follows from [31, Theorem 9.3 (i), (ii)] that there is a prime \({\mathfrak {P}}\) of \(A'\) lying above \(\mathfrak {p}\) and any such \({\mathfrak {P}}\) is minimal. If \(A'\) is equidimensional then \(\dim A/\mathfrak {p}=\dim A'/{\mathfrak {P}}= \dim A'=\dim A\), and thus A is equidimensional.
If \({\mathfrak {P}}\) is a minimal prime of \(A'\) then Going down theorem for flat extensions implies that \(\mathfrak {p}:={\mathfrak {P}}\cap A\) is a minimal prime of A. If A is equidimensional then \(\dim A'=\dim A= \dim A/\mathfrak {p}=\dim A'/{\mathfrak {P}}\) and thus \(A'\) is equidimensional. \(\square \)
Lemma A.3
If \(A\in {\mathcal {C}}(\mathcal {O})\) is an integral domain then there is a number n(A) such that for all finite extensions \(L'/L\), \(A\otimes _{\mathcal {O}} \mathcal {O}'\) has at most n(A) minimal primes.
Proof
Since A is \(\mathcal {O}\)torsion free, by Cohen’s structure theorem for complete local rings there is a subring \(B\subset A\) such that A is finite over B and \(B\cong \mathcal {O}[[x_1, \ldots , x_d]]\), see [31, Theorem 29.4 (iii)] and the Remark following it. Let \(A'= A\otimes _\mathcal {O}\mathcal {O}'\) and let \(B'=B\otimes _{\mathcal {O}} \mathcal {O}'\). Then \(B'\subset A'\), \(A'\) is finite over \(B'\) and \(B'\cong \mathcal {O}'[[x_1, \ldots , x_d]]\). In particular, \(B'\) is an integral domain. If \({\mathfrak {P}}\) is a minimal prime of \(A'\) then \({\mathfrak {P}}\cap A\) is a minimal prime of A by Lemma A.2 and hence \({\mathfrak {P}}\cap A=0\), as A is an integral domain. Thus \({\mathfrak {P}}\cap B=0\). Since \(({\mathfrak {P}}\cap B')\cap B=0\) and \(B'\) is an integral domain, Lemma A.2 implies that \({\mathfrak {P}}\cap B'=0\). Hence \(K(B')\otimes _{B'} A'/{\mathfrak {P}}\) is nonzero, where \(K(B')\) denotes the fraction field of \(B'\).
Let \({\mathfrak {P}}_1, \ldots , {\mathfrak {P}}_n\) be the minimal primes of \(A'\). Since every minimal prime ideal is also an associated prime by [31, Theorem 6.5 (iii)] for each i there is an injection of \(A'\)modules \(A'/{\mathfrak {P}}_i\hookrightarrow A'\). The kernel K of the map \(\oplus _{i=1}^n A'/{\mathfrak {P}}_i \rightarrow A'\) is not supported on any \({\mathfrak {P}}_i\). Hence, \(\dim K< \dim B'\) and so \(K(B')\otimes _{B'} K=0\). Hence the map \(\oplus _{i=1}^n K(B')\otimes _{B'} A'/{\mathfrak {P}}_i \rightarrow K(B')\otimes _{B'} A'\) is injective and we conclude that \(A'\) can have at most \(\dim _{K(B')} K(B')\otimes _{B'} A'\) minimal ideals. The dimension of \(K(B')\otimes _{B'} A'\) as \(K(B')\)vector space can be bounded by the number of generators of \(A'\) as a \(B'\)module, which can be bounded by the number of generators of A as a Bmodule. This number does not depend on the extension \(L'/L\). \(\square \)
If \(A\in {\mathcal {C}}(\mathcal {O})\) is a domain with fraction field K(A) then we have injections:
Since \(L'/L\) is a finite separable extension \(K(A)\otimes _L L'\) is a finite product of fields and the map induces a bijection between the minimal primes of \(A\otimes _{\mathcal {O}} \mathcal {O}'\) and the minimal primes of \(K(A)\otimes _L L'\). In particular, we obtain:
Lemma A.4
If \(A\in {\mathcal {C}}(\mathcal {O})\) is an integral domain such that \(A\otimes _{\mathcal {O}} \mathcal {O}'\) has only one minimal prime then \(A\otimes _{\mathcal {O}} \mathcal {O}'\) is an integral domain.
Lemma A.5
Let \(A\in {\mathcal {C}} (\mathcal {O})\) and let \(L'\) be a finite extension of L with the ring of integers \(\mathcal {O}'\) such that the number of minimal primes of \(A':=A\otimes _{\mathcal {O}} \mathcal {O}'\) is maximal. Then \(A'/{\mathfrak {P}}\) is geometrically integral for all minimal primes \({\mathfrak {P}}\) of \(A'\).
Proof
Since A is noetherian, it has only finitely many minimal primes. The minimal primes of \(A\otimes _{\mathcal {O}} \mathcal {O}'\) are precisely the primes lying over the minimal primes of A. It follows from Lemma A.3 that the set of numbers of minimal primes of \(A\otimes _{\mathcal {O}} \mathcal {O}'\) as \(L'\) ranges over all finite extensions of L is bounded from above. Thus we may pick an extension \(L'\) such that this number is maximal. If \({\mathfrak {P}}\) is a minimal prime of \(A'\) and \(L''\) is a finite extension of \(L'\) with the ring of integers \(\mathcal {O}''\) such that \((A'/{\mathfrak {P}})\otimes _{\mathcal {O}'} \mathcal {O}''\) has more than one minimal prime then we would conclude that \(A\otimes _{\mathcal {O}} \mathcal {O}''\) has strictly more minimal primes than \(A\otimes _{\mathcal {O}} \mathcal {O}'\) contradicting the choice of \(L'\). Hence, \((A'/{\mathfrak {P}})\otimes _{\mathcal {O}'} \mathcal {O}''\) has only one minimal prime and Lemma A.4 implies that it is an integral domain. \(\square \)
Proof of Lemma A.1
Let \(L'\) be a finite extension of L with the ring of integers \(\mathcal {O}'\) such that \(A'/{\mathfrak {P}}\) and \(B'/{\mathfrak {Q}}\) are geometrically integral for all the minimal primes \({\mathfrak {P}}\) of \(A'\) and all the minimal primes \({\mathfrak {Q}}\) of \(B'\), where \(A':= A\otimes _{\mathcal {O}} \mathcal {O}'\) and \(B':= B\otimes _{\mathcal {O}} \mathcal {O}'\). The existence of such extension is granted by Lemma A.5. Since \(A'\) and \(B'\) are equidimensional by Lemma A.2, it follows from [2, Lemma 3.3 (2), (5)] that \(A'\widehat{\otimes }_{\mathcal {O}'} B'\) is equidimensional. Since
we use Lemma A.2 again to conclude that \(A\widehat{\otimes }_{\mathcal {O}} B\) is equidimensional.
Since B is \(\mathcal {O}\)flat, \(A\widehat{\otimes }_{\mathcal {O}} B\) is a flat Aalgebra, thus
by [31, Thm.15.1]. The fibre ring \(k\otimes _A ( A\widehat{\otimes }_{\mathcal {O}} B)\) is isomorphic to \(B/\varpi B\), which has dimension equal to \(\dim B1\), as \(\varpi \) is Bregular. \(\square \)
Appendix B: Local deformation rings for 2adic representations of \(G_{\mathbb {Q}_l}\), \(l \ne 2\). By Jack Shotton
Let l and p be distinct primes. Let \(L/\mathbb {Q}_p\) be a finite extension with ring of integers \(\mathcal {O}\), uniformiser \(\varpi \) and residue field k. Let \(F/\mathbb {Q}_l\) be a finite extension with absolute Galois group \(G_F\), inertia group \(I_F\), and wild inertia group \(P_F\). Let \(\tilde{P}_F\) be the kernel of the maximal prol quotient of \(I_F\). Let q be the order of the residue field of F. We assume that L contains all \((q^21)th\) roots of unity. Choose a progenerator \(\sigma \) of \(I_F/\tilde{P}_F\) and \(\phi \in G_F/\tilde{P}_F\) lifting the arithmetic Frobenius element of \(G_F/I_F\). Then we have the relation
If \(\overline{\rho }: G_F \rightarrow \mathrm {GL}_2(k)\) is a continuous homomorphism, let \(R^{\square }_{\overline{\rho }}\) be the universal framed deformation ring for \(\overline{\rho }\) parametrising lifts with coefficients in \(\mathcal {O}\)algebras. By [47] Theorem 2.5, \(R^{\square }_{\overline{\rho }}\) is a reduced, \(\mathcal {O}\)flat complete intersection ring of relative dimension 4 over \(\mathcal {O}\).
If \(\tau : I_F \rightarrow \mathrm {GL}_2(L)\) is a continuous semisimple representation that extends to \(G_F\), let \(R^{\square }_{\overline{\rho }}(\tau )\) be the maximal reduced, ptorsion free quotient of \(R^{\square }_{\overline{\rho }}\) such that, for every \(\mathcal {O}\)algebra homomorphism \(x : R^{\square }_{\overline{\rho }} \rightarrow \overline{L}\), the corresponding representation \(\rho _x : G_F \rightarrow \mathrm {GL}_2(\overline{L})\) satisfies \((\rho _x\mid _{I_F})^{ss} \cong \tau \).
The goal of this appendix is to prove:
Theorem B.1
For any \(\overline{\rho }\) and \(\tau \) as above, the ring \(R^{\square }_{\overline{\rho }}(\tau )\) is either Cohen–Macaulay or zero.
If \(p > 2\), then this is the content of section 5.5 of [46]. If \(p = 2\) and \(\overline{\rho }_{\tilde{P}_F}\) is nonscalar, then the proof of proposition 5.1 of [46] shows that \(R^{\square }_{\overline{\rho }}\) is a completed tensor product of deformation rings of characters, all of whose irreducible components are formally smooth, and that \(R^{\square }_{\overline{\rho }}(\tau )\) is an irreducible component of \(R^{\square }_{\overline{\rho }}\); thus \(R^{\square }_{\overline{\rho }}(\tau )\) is formally smooth in this case. From now on, then, we assume that \(p = 2\) and that \(\overline{\rho }_{\tilde{P}_F}\) is scalar; by twisting, we may and do assume that \(\overline{\rho }_{\tilde{P}_F}\) is trivial. In this case, we may list the semisimple inertial types \(\tau \) for which \(R^{\square }_{\overline{\rho }}(\tau )\) may be nonzero. They are determined by the eigenvalues of \(\tau (\sigma )\), which must be of 2power order and either fixed or interchanged by raising to the power q. Writing \(a = v_2(q1)\) and \(b = v_2(q^2  1)\), if \(R^{\square }_{\overline{\rho }}(\tau )\) is nonzero then either

\(\tau = \tau _\zeta \) is the inertial type in which the eigenvalues of \(\tau (\sigma )\) are both equal to a \(2^a\)th root of unity, \(\zeta \);

\(\tau = \tau _{\zeta _1, \zeta _2}\) is the inertial type in which the eigenvalues of \(\tau (\sigma )\) are equal to distinct \(2^a\)th roots of unity \(\zeta _1\) and \(\zeta _2\);

\(\tau = \tau _{\xi }\) is the inertial type in which the eigenvalues of \(\tau (\sigma )\) are equal to \(\xi \) and \(\xi ^q\) for \(\xi \) a \(2^b\)th root of unity with \(\xi \ne \xi ^q\) (equivalently, with \(\xi \) not a \(2^a\)th root of unity).
We also give a version with fixed determinant:
Corollary B.2
If \(\psi \) is any lift of \(\det \overline{\rho }\) to \(\mathcal {O}^{\times }\) such that \(\psi _{I_F} = \det \tau \), let \(R^{\square , \psi }_{\overline{\rho }}(\tau )\) be the universal framed deformation ring with determinant \(\psi \) and type \(\tau \). Then \(R^{\square , \psi }_{\overline{\rho }}(\tau )\) is Cohen–Macaulay or zero.
Proof
By Theorem B.1, \(R^{\square }_{\overline{\rho }}(\tau )\) is Cohen–Macaulay. If we impose a single additional equation \(\det \rho (\phi ) = \psi (\phi )\), then the ring will still be Cohen–Macaulay provided that \(\det \rho (\phi )  \psi (\phi )\) is a nonzerodivisor — in other words, that it doesn’t vanish on any irreducible components of \({{\mathrm{Spec}}}R^\square _{\overline{\rho }}(\tau )\). This is the case, since the action of \(\mathbb {G}_m^{\wedge }\) on \({{\mathrm{Spec}}}R^\square _{\overline{\rho }}(\tau )\) given by making unramified twists preserves irreducible components but varies the determinant. \(\square \)
Let \(\mathcal {X}\) be the affine \(\mathcal {O}\)scheme whose R points, for an \(\mathcal {O}\)algebra R, are pairs
Then \(\mathcal {X}\) is a reduced, \(\mathcal {O}\)flat complete intersection of relative dimension 4 over \({{\mathrm{Spec}}}\mathcal {O}\) by the proof of Theorem 2.5 of [47]. Let \(\mathcal {A}\) be the coordinate ring of \(\mathcal {X}\). We write \(\mathrm {\Sigma }= \begin{pmatrix} 1+A &{} B \\ C &{} 1+D \end{pmatrix}\) and \(\mathrm {\Phi }= \begin{pmatrix} P &{} Q \\ R &{} TP \end{pmatrix}\), so that \(\mathcal {A}\) is a quotient of
For any continuous \(\overline{\rho }: G_F \rightarrow \mathrm {GL}_2(k)\), the pair of matrices \(\overline{\rho }(\sigma )\) and \(\overline{\rho }(\phi )\) give rise to a closed point of \(\mathcal {X}\), and so a maximal ideal \(\mathfrak {m}\) of \(\mathcal {A}\). Then \(R^{\square }_{\overline{\rho }} = \mathcal {A}^\wedge _{\mathfrak {m}}\). If \(\mathcal {C}\) is a conjugacy class in \(\mathrm {GL}_2(\overline{L})\), then there is a unique irreducible component of \(\mathcal {X}\) such that, for a dense set of geometric points of that component, the corresponding matrix \(\mathrm {\Sigma }\) has conjugacy class \(\mathcal {C}\). This provides a bijection between the irreducible components of \(\mathcal {X}\) and the conjugacy classes of \(\mathrm {GL}_2(\overline{L})\) that are preserved under the qpower map (by [47] Proposition 2.6). If \(\tau \) is one of the above inertial types then we write \(\mathcal {X}(\tau )\) for the union of those irreducible components corresponding to conjugacy classes with the same characteristic polynomial as \(\tau (\sigma )\), with the reduced subscheme structure, and \(\mathcal {A}(\tau )\) for its coordinate ring. Note that, since \(\mathcal {X}\) is \(\mathcal {O}\)flat and \(\mathcal {X}(\tau )\) is an irreducible component of \(\mathcal {X}\), \(\mathcal {X}(\tau )\) is also \(\mathcal {O}\)flat, so that \(\mathcal {A}(\tau )\) is \(\varpi \)torsion free.
Lemma B.3
If \(\tau = \tau _\zeta , \tau _{\zeta _1, \zeta _2},\) or \(\tau _{\xi }\), then \(\mathcal {A}(\tau )_{\mathfrak {m}}^\wedge = R^{\square }_{\overline{\rho }}(\tau )\).
Proof
Since \(\mathcal {A}\) is \(\mathcal {O}\)flat and \(\mathcal {A}(\tau )\) is the quotient of \(\mathcal {A}\) by an intersection of minimal prime ideals, it is also \(\mathcal {O}\)flat. Thus \(\mathcal {A}(\tau )_{\mathfrak {m}}^\wedge \) is also \(\mathcal {O}\)flat, by flatness of localisation and completion. Since \(\mathcal {A}(\tau )\) is of finite type over a DVR it is Nagata by [49, Tag 0335]. Since \(\mathcal {A}(\tau )\) is reduced, the completion \(\mathcal {A}(\tau )_{\mathfrak {m}}^\wedge \) is also reduced by [49, Tag 07NZ]. The composite map \(\mathcal {A}\rightarrow R^{\square }_{\overline{\rho }} \twoheadrightarrow R^{\square }_{\overline{\rho }}(\tau )\) factors through a map \(\mathcal {A}(\tau ) \rightarrow R^{\square }_{\overline{\rho }}(\tau )\), since any function in \(\mathcal {A}\) that vanishes on all \(\overline{L}\)points of type \(\tau \) must vanish in \(R^{\square }_{\overline{\rho }}(\tau )\) by definition. Thus we get a surjection \(\mathcal {A}(\tau )_{\mathfrak {m}}^{\wedge } =\mathcal {A}(\tau ) \otimes _{\mathcal {A}} R^{\square }_{\overline{\rho }} \twoheadrightarrow R^{\square }_{\overline{\rho }}(\tau )\). However, since \(\mathcal {A}(\tau )_{\mathfrak {m}}^{\wedge }\) is reduced and \(\mathcal {O}\)torsion free, and has the property that every \(\overline{L}\)point gives a Galois representation of type \(\tau \), this map is an isomorphism by the definition of \(R^{\square }_{\overline{\rho }}(\tau )\). \(\square \)
Let \(\overline{\mathcal {S}} = \mathcal {S}\otimes _{\mathcal {O}} k\), \(\overline{\mathcal {A}} = \mathcal {A}\otimes _{\mathcal {O}} k\), and \(\overline{\mathcal {X}} = {{\mathrm{Spec}}}\overline{\mathcal {A}}\). Then the irreducible components of \(\overline{\mathcal {X}}\) are in bijection with the conjugacy classes of \(\mathrm {GL}_2(\overline{k})\) that are stable under the qpower map (again by [47] Proposition 2.6). Let \(\overline{\mathcal {X}}_1\) be the irreducible component corresponding to the trivial conjugacy class — this is just the locus where \(\mathrm {\Sigma }= 1\) — and let \(\overline{\mathcal {X}}_N\) be that corresponding to the nontrivial unipotent conjugacy class (we give the irreducible components the reduced subscheme structure). Let \(I_1\) and \(I_N\) be the prime ideals of \(\overline{\mathcal {S}}\) cutting out \(\overline{\mathcal {X}}_1\) and \(\overline{\mathcal {X}}_N\); these correspond to minimal primes of \(\overline{\mathcal {A}}\). If \(\tau \) is one of the above inertial types, then we write \(I(\tau )\) for the ideal of \(\overline{\mathcal {S}}\) cutting out \(\mathcal {A}(\tau ) \otimes _\mathcal {O}k\).
Lemma B.4
The ideals \(I_1\) and \(I_N\) have generators
Proof
The presentation for \(I_1\) is obvious. For \(I_N\), the condition that \(\mathrm {\Sigma }\) is unipotent gives \(A + D \in I_N\) and \(A^2 + BC \in I_N\). If \(N = \mathrm {\Sigma } 1\), then the relation \(\mathrm {\Phi }\mathrm {\Sigma }= \mathrm {\Sigma }^q \mathrm {\Phi }\) becomes \(\mathrm {\Phi }N = q N \mathrm {\Phi }= N\mathrm {\Phi }\) (since we are working mod 2), which implies that \(CQ + BR = 0\). At any closed point of \(\overline{\mathcal {X}}_N\) where \(N \ne 0\), the eigenvalues of \(\mathrm {\Phi }\) must be in the ratio \(1:q = 1:1\), and so \(T = 0\). As such closed points are dense on \(\overline{\mathcal {X}}_N\), we see that \(T \in I_N\). Therefore
The ideal \(I = (A^2 + BC, CQ + BR, T, A+D)\) is prime of dimension 4; indeed, \(\mathcal {S}/I\) is isomorphic to a localisation of
which is easily seen to be a 4dimensional domain. Thus \(I \subset I_N\) are prime ideals of \(\overline{\mathcal {S}}\) of the same dimension, and so must be equal. \(\square \)
Proposition B.5
Let \(\tau = \tau _\xi \). Then \(I(\tau ) = I_N\).
Proof
Write \(\eta = \xi + \xi ^{q}  2\). The condition that \(\mathrm {\Sigma }\) has characteristic polynomial \((X  \xi )(X  \xi ^{q})\) shows that, on \(\mathcal {X}(\tau )\), we have the equations
Using the first of these, we replace D by \(\eta  A\) everywhere. Now, if x is an \(\overline{L}\)point of \(\mathcal {X}(\tau )\) corresponding to a pair of matrices \((\mathrm {\Sigma }_x, \mathrm {\Phi }_x)\), then \(\mathrm {\Phi }_x\) exchanges the \(\xi \) and \(\xi ^q\) eigenspaces of \(\mathrm {\Sigma }_x\) and so must have trace zero. Therefore on \(\mathcal {X}(\tau )\) we have the equation
Lastly, by the Cayley–Hamilton theorem, and the fact that
we see that \(\mathrm {\Sigma }^q = \begin{pmatrix} 1 + \eta  A &{} B \\ C &{} 1 + A \end{pmatrix}\) on \(\mathcal {X}(\tau )\). Equating matrix entries in the relation \(\mathrm {\Phi }\mathrm {\Sigma }= \mathrm {\Sigma }^q \mathrm {\Phi }\), and noting that \(T = 0\), we obtain one new equation
Thus, letting
we obtain a surjection \(\mathcal {S}/J \twoheadrightarrow \mathcal {A}(\tau )\), and therefore a surjection
As \(\eta \) is divisible by \(\varpi \), we see that \(J + (\varpi ) = I_N\), and so we have a surjection \(\overline{\mathcal {S}}/I_N \twoheadrightarrow \overline{\mathcal {A}}(\tau )\). This must be an isomorphism since \(\overline{\mathcal {S}}/I_N\) is a 4dimensional domain and \(\overline{\mathcal {A}}(\tau )\) is a nonzero 4dimensional ring. Therefore \(I_N = I(\tau )\) as required. \(\square \)
For the remaining types the following lemma will be useful. If R is a noetherian ring, \(\mathfrak {p}\) is a minimal prime of R, and M is a finitelygenerated Rmodule, let \(e_R(M, \mathfrak {p}) = l_{R_{\mathfrak {p}}}(M_{\mathfrak {p}})\) (this is a special case of the Hilbert–Samuel multiplicity).
Lemma B.6
Let \(f : R \rightarrow S\) be a surjection of equidimensional rings of the same dimension, and suppose that R is S1 and Nagata. Let \(\mathfrak {p}_1, \ldots , \mathfrak {p}_n\) be the minimal primes of R. Suppose that, for \(i = 1, \ldots , n\), there is a maximal ideal \(\mathfrak {m}_i\) of S such that \(\mathfrak {p}_i \subset \mathfrak {m}_i\) but \(\mathfrak {p}_j \not \subset \mathfrak {m}_i\) for \(i \ne j\). If, for each i, we have
for some minimal prime \(\mathfrak {q}_i\) of \(S^{\wedge }_{\mathfrak {m}_i}\), then f is an isomorphism.
Remark B.7
For those primes \(\mathfrak {p}_i\) such that \(e_R(R, \mathfrak {p}_i) = 1\) — which is all of them if R is reduced — the required inequality is implied simply by the existence of the \(\mathfrak {m}_i\).
Proof
Since R is S1, every associated prime of \(\ker f\) is minimal and so, by [49, Tag 0311], it is enough to show that f induces an isomorphism \(f_{\mathfrak {p}_i} : R_{\mathfrak {p}_i}\rightarrow S_{\mathfrak {p}_i}\) for each i. Since f is surjective and \(R_{\mathfrak {p}_i}\) is artinian, it is enough to show that \(e_R(R,\mathfrak {p}_i) \le e_R(S, \mathfrak {p}_i)\). Let \(i \in \{1, \ldots , n\}\). Choose \(\mathfrak {m}_i\) and \(\mathfrak {q}_i\) as in the hypotheses of the lemma. It is enough to show that for each i,
Since \(\mathfrak {m}_i\) contains a unique minimal prime of R, after localising at \(\mathfrak {m}_i\) we may assume that \(R\rightarrow S\) is a local map of local rings, and that \(\mathfrak {p}_i\) is the unique minimal prime of R, and drop i from the notation. The hypothesis that R and S are equidimensional of the same dimension implies that \(\mathfrak {p}S\) is the unique minimal prime of S, which we also denote by \(\mathfrak {p}\). We have \(e_R(S,\mathfrak {p}) = e_S(S, \mathfrak {p})\) since both are just the length of \(S_{\mathfrak {p}}\). Since \(S \rightarrow S^\wedge \) is flat and \(S^\wedge /\mathfrak {p}= (S/\mathfrak {p})^\wedge \) is reduced because R (and hence S) is Nagata, [49, Tag 02M1] implies that \(e_S(S, \mathfrak {p}) = e_{S^\wedge }(S^{\wedge }, \mathfrak {q})\). So
as required. \(\square \)
The S1 condition holds, in particular, if R is reduced or Cohen–Macaulay, while the Nagata condition holds if R is of finite type over a field or DVR.
Proposition B.8
Let \(\tau = \tau _\zeta \). Then
Proof
For simplicity, we twist so that \(\zeta = 1\). Write \(N = \mathrm {\Sigma } 1 = \begin{pmatrix} A &{} B \\ C &{} D \end{pmatrix}\). On \(\mathcal {A}(\tau )\), \(\mathrm {\Sigma }\) has characteristic polynomial \((X  1)^2\), and so the equations
hold on \(\mathcal {A}(\tau )\). Moreover, since \((\mathrm {\Sigma } 1)^2 = 0\) on \(\mathcal {A}(\tau )\), by the Cayley–Hamilton theorem we have that \(\mathrm {\Sigma }^q = 1 + q(\mathrm {\Sigma } 1) = 1 + qN\) on \(\mathcal {A}(\tau )\). The equation \(\mathrm {\Phi }\mathrm {\Sigma }= \mathrm {\Sigma }^q \mathrm {\Phi }\) becomes \(\mathrm {\Phi }N = q N \mathrm {\Phi }\), and comparing matrix entries we get equations
Summing the first and fourth of these gives \((q1)(BR + CQ + A(2P  T)) = 0\); since \(\mathcal {A}(\tau )\) is \((q1)\)torsion free, we deduce that
in \(\mathcal {A}(\tau )\) and can replace the fourth of the above equations by this.
The ideal cutting out \(\mathcal {A}(\tau )\) therefore contains the ideal
Now, the image of J in \(\overline{\mathcal {S}}\) is
which is equal to \((A + D, A^2 + BC, BR + CQ) + I_1 \cap (T) = I_N \cap I_1\). Therefore there is a surjection
Write \(\tilde{R} = \overline{\mathcal {S}}/(I_N \cap I_1)\). Then \(\tilde{R}\) is reduced with two minimal primes, which we also call \(I_N\) and \(I_1\). Let \(\rho _1 : G_F \rightarrow \mathrm {GL}_2(\mathcal {O})\) be diagonal unramified with distinct eigenvalues of Frobenius, and let \(\rho _N : G_F \rightarrow \mathrm {GL}_2(\mathcal {O})\) send \(\sigma \mapsto \begin{pmatrix} 1 &{} 1 \\ 0 &{} 1 \end{pmatrix}\) and \(\phi \mapsto \begin{pmatrix} q &{} 0 \\ 0 &{} 1 \end{pmatrix}\). Let \(\mathfrak {m}_1\) and \(\mathfrak {m}_N\) be the corresponding maximal ideals of \(\overline{\mathcal {A}}(\tau )\). Then \(I_1 \subset \mathfrak {m}_1\), \(I_N \not \subset \mathfrak {m}_1\), \(I_1 \not \subset \mathfrak {m}_N\) and \(I_N \subset \mathfrak {m}_N\), so f is an isomorphism by the remark following lemma B.6. \(\square \)
Proposition B.9
Let \(\tau = \tau _{\zeta _1, \zeta _2}\). Then
Proof
Write \(\mu = \zeta _1 + \zeta _2  2\). The condition that \(\mathrm {\Sigma }\) has characteristic polynomial \((X  \zeta _1)(X  \zeta _2)\) is equivalent to the equations
As \(X^q \equiv X \mod (X  \zeta _1)(X  \zeta _2)\), we have by the Cayley–Hamilton theorem that \(\mathrm {\Sigma }^q = \mathrm {\Sigma }\) on \(\mathcal {A}(\tau )\). The equation \(\mathrm {\Phi }\mathrm {\Sigma }= \mathrm {\Sigma }^q \mathrm {\Phi }\) therefore becomes \(\mathrm {\Phi }\mathrm {\Sigma }= \mathrm {\Sigma }\mathrm {\Phi }\), and comparing matrix entries we get three equations (the fourth being redundant):
Let
Let I be the image of J in \(\overline{\mathcal {S}}\), so that
We have shown that there is a surjection \(\mathcal {S}/J \twoheadrightarrow \mathcal {A}(\tau )\), and therefore there is a surjection \(f: \overline{\mathcal {S}}/I \rightarrow \overline{\mathcal {A}}(\tau )\). We have to show that f is an isomorphism. Write \(\tilde{R} = \overline{\mathcal {S}}/I\).
Then (see the proof of corollary B.10 below) \(\overline{\mathcal {S}}/I\) is Cohen–Macaulay, with minimal primes \(I_1\) and \(I_N\), and it is easy to see that \(e_{\tilde{R}}(\tilde{R},I_N) = 1\) while \(e_{\tilde{R}}(\tilde{R},I_1) = 2\).
Let \(\rho _1 : G_F \rightarrow \mathrm {GL}_2(\mathcal {O})\) be diagonal such that the eigenvalues of \(\rho _1(\sigma )\) are \(\zeta _1\) and \(\zeta _2\), and the eigenvalues of \(\rho _1(\phi )\) are distinct modulo \(\varpi \). Let \(\rho _N : G_F \rightarrow \mathrm {GL}_2(\mathcal {O})\) send \(\sigma \mapsto \begin{pmatrix} \zeta _1 &{} 1 \\ 0 &{} \zeta _2 \end{pmatrix}\) and \(\phi \mapsto \begin{pmatrix} 1 &{} 0 \\ 0 &{} 1 \end{pmatrix}\). Let \(\mathfrak {m}_1\) and \(\mathfrak {m}_N\) be the corresponding maximal ideals of \(\overline{\mathcal {A}}(\tau )\). Then \(I_1 \subset \mathfrak {m}_1\), \(I_N \not \subset \mathfrak {m}_1\), \(I_1 \not \subset \mathfrak {m}_N\) and \(I_N \subset \mathfrak {m}_N\). By [46] Proposition 5.3, which remains valid when \(p = 2\), \(R^{\square }_{\overline{\rho }_1}(\tau )\) is formally smooth over
Therefore \(R^{\square }_{\overline{\rho }_1}(\tau )\otimes k\) has a unique minimal prime \(\mathfrak {q}\) and its multiplicity is 2. By Lemmas B.3 and B.6, f is an isomorphism. \(\square \)
Corollary B.10
(of Propositions B.5, B.8 and B.9) For \(\tau = \tau _\xi \), \(\tau _\zeta \), or \(\tau _{\zeta _1, \zeta _2}\), \(\mathcal {A}(\tau )\) is Cohen–Macaulay.
Proof
Since \(\varpi \) is a regular element of \(\mathcal {A}(\tau )\), it suffices to prove that \(\overline{\mathcal {A}}(\tau )\) is Cohen–Macaulay. This can easily be checked in magma; we sketch an alternative proof by hand. If \(\tau = \tau _\xi \), then by Proposition B.5, \(I(\tau ) = I_N\). But \(\overline{\mathcal {S}}/I_N\) is a complete intersection ring of dimension 4, and therefore is Cohen–Macaulay. If \(\tau = \tau _\zeta \), then by Proposition B.8, \(I(\tau _\xi ) = I_1 \cap I_N\). Now, \(\overline{\mathcal {S}}/I_1\) and \(\overline{\mathcal {S}}/I_N\) are Cohen–Macaulay of dimension 4 (the latter by the previous case), while \(\overline{\mathcal {S}}/(I_1 + I_N)\) is regular, and so Cohen–Macaulay, of dimension 3. By exercise 18.13 of [15], \(\overline{\mathcal {S}}/(I_1 \cap I_N)\) is also Cohen–Macaulay. Finally, if \(\tau = \tau _{\zeta _1,\zeta _2}\) then by Proposition B.9, \(I(\tau ) = (A + D, A^2 + BC, BR + CQ, BT, CT)\). Let \(I = I(\tau )\). Since \(I + (AT) = I_1 \cap I_N\) and \(AT\cdot I_1 = 0\), there is an exact sequence of \(\overline{\mathcal {S}}/I\)modules
The first map must be injective, since \(I_1\) is prime and \(e_{\overline{\mathcal {S}}/I}(\overline{\mathcal {S}}/I,I_1) = 2 > 1 = e_{\overline{\mathcal {S}}/I}(\overline{\mathcal {S}}/(I_1 \cap I_N), I_1)\). Since we have shown that \(\overline{\mathcal {S}}/I_1\) and \(\overline{\mathcal {S}}/(I_1 \cap I_N)\) are maximal Cohen–Macaulay modules over \(\overline{\mathcal {S}}/I\), so is \(\overline{\mathcal {S}}/I\) (by [53] Proposition 1.3). \(\square \)
Since \(R^{\square }_{\overline{\rho }}(\tau )\) is a completion of \(\mathcal {A}(\tau )\) by Lemma B.3, and a completion of a Cohen–Macaulay ring is Cohen–Macaulay (by [49, Tag 07NX]), we obtain Theorem B.1.
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Hu, Y., Paškūnas, V. On crystabelline deformation rings of \(\mathrm {Gal}(\overline{\mathbb {Q}}_p/\mathbb {Q}_p)\) (with an appendix by Jack Shotton). Math. Ann. 373, 421–487 (2019). https://doi.org/10.1007/s0020801816712
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DOI: https://doi.org/10.1007/s0020801816712
Mathematics Subject Classification
 11F80
 11F85