On crystabelline deformation rings of $\mathrm {Gal}(\overline{\mathbb {Q}}_p/\mathbb {Q}_p)$ (with an appendix by Jack Shotton)

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.

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 $\dim A+\dim B-1$.

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 $\mathcal {O}=\mathbb {Z}_p$, $A=B=\mathbb {Z}_p[x]/(x^2-p)$ then $A\widehat{\otimes }_{\mathcal {O}} B\cong \mathbb {Z}_p[x, y]/( x^2-p, y^2-p)$ and $(x-y)(x+y)=0$ 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 non-zero, 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 B-module. 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:

$$\begin{aligned} A\otimes _{\mathcal {O}} \mathcal {O}'\hookrightarrow K(A)\otimes _{\mathcal {O}} \mathcal {O}'\cong K(A) \otimes _L L'. \end{aligned}$$

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

$$\begin{aligned} (A\widehat{\otimes }_{\mathcal {O}} B)\otimes _{\mathcal {O}} \mathcal {O}'\cong A'\widehat{\otimes }_{\mathcal {O}'} B' \end{aligned}$$

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 A-algebra, thus

$$\begin{aligned} \dim A\widehat{\otimes }_{\mathcal {O}} B=\dim A + \dim k\otimes _A ( A\widehat{\otimes }_{\mathcal {O}} B) \end{aligned}$$

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 $\dim B-1$, as $\varpi$ is B-regular. $\square$

Appendix B: Local deformation rings for 2-adic 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 pro-l quotient of $I_F$. Let q be the order of the residue field of F. We assume that L contains all $(q^2-1)th$ roots of unity. Choose a pro-generator $\sigma$ 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

$$\begin{aligned} \phi \sigma \phi ^{-1} = \sigma ^q. \end{aligned}$$

(28)

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, p-torsion 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 non-scalar, 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 non-zero. They are determined by the eigenvalues of $\tau (\sigma )$, which must be of 2-power order and either fixed or interchanged by raising to the power q. Writing $a = v_2(q-1)$ and $b = v_2(q^2 - 1)$, if $R^{\square }_{\overline{\rho }}(\tau )$ is non-zero 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 non-zerodivisor — 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

$$\begin{aligned} \{(\mathrm {\Sigma }, \mathrm {\Phi }) \in \mathrm {GL}_2(R) \times \mathrm {GL}_2(R) : \mathrm {\Phi }\mathrm {\Sigma }= \mathrm {\Sigma }^q \mathrm {\Phi }\}. \end{aligned}$$

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 $\mathrm {\Phi }= \begin{pmatrix} P &{} Q \\ R &{} T-P \end{pmatrix}$, so that $\mathcal {A}$ is a quotient of

$$\begin{aligned} \mathcal {S}= \mathcal {O}[A,B,C,D,P,Q,R,T][(\det \mathrm {\Sigma })^{-1}, (\det \mathrm {\Phi })^{-1}]. \end{aligned}$$

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 q-power 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 q-power 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 non-trivial 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

$$\begin{aligned} I_1&= (A,B,C,D) \\ I_N&= (A^2 + BC, CQ + BR, T, A + D). \end{aligned}$$

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

$$\begin{aligned} ( A^2 + BC, CQ + BR, T, A + D) \subset I_N. \end{aligned}$$

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

$$\begin{aligned} \frac{k[A,B,C]}{(A^2 + BC)}[P,Q,R]/(CQ + BR) \end{aligned}$$

which is easily seen to be a 4-dimensional 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

$$\begin{aligned} A + D&= \eta \\ A(A - \eta ) + BC&= \eta . \end{aligned}$$

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

$$\begin{aligned} T = 0. \end{aligned}$$

Lastly, by the Cayley–Hamilton theorem, and the fact that

$$\begin{aligned} X^q \equiv \xi + \xi ^q - X \mod (X - \xi )(X - \xi ^{q}), \end{aligned}$$

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

$$\begin{aligned} (2A - \eta )P + BR + CQ = 0. \end{aligned}$$

Thus, letting

$$\begin{aligned} J = (A+D, T, A(A - \eta ) + BC - \eta , (2A - \eta )P + BR + CQ) \end{aligned}$$

we obtain a surjection $\mathcal {S}/J \twoheadrightarrow \mathcal {A}(\tau )$, and therefore a surjection

$$\begin{aligned} \overline{\mathcal {S}}/J \twoheadrightarrow \overline{\mathcal {A}}(\tau ). \end{aligned}$$

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 4-dimensional domain and $\overline{\mathcal {A}}(\tau )$ is a non-zero 4-dimensional 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 finitely-generated R-module, 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

$$\begin{aligned} e_R(R, \mathfrak {p}_i) \le e_{S^{\wedge }_{\mathfrak {m}_i}}(S^{\wedge }_{\mathfrak {m}_i}, \mathfrak {q}_i) \end{aligned}$$

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,

$$\begin{aligned} e_R(S, \mathfrak {p}_i) = e_{S^\wedge _{\mathfrak {m}_i}}(S^{\wedge }_{\mathfrak {m}_i}, \mathfrak {q}_i). \end{aligned}$$

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

$$\begin{aligned} e_R(S, \mathfrak {p}) = e_S(S, \mathfrak {p}) = e_{S^\wedge }(S^{\wedge }, \mathfrak {q}) \ge e_{R}(R, \mathfrak {p}) \end{aligned}$$

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

$$\begin{aligned} I(\tau ) = I_N \cap I_1 = (A+D, AT, BT, CT, A^2 + BC, BR + CQ). \end{aligned}$$

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

$$\begin{aligned} A + D&= 0 \\ A^2 + BC&= 0 \end{aligned}$$

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

$$\begin{aligned} qBR - CQ + (q - 1)AP&= 0 \\ (q+1)QA + B(qT - (q+1)P)&= 0 \\ (q+1)RA + C(T - (q+1)P))&= 0 \\ qCQ - BR + (q-1)A(P-T)&= 0. \end{aligned}$$

Summing the first and fourth of these gives $(q-1)(BR + CQ + A(2P - T)) = 0$; since $\mathcal {A}(\tau )$ is $(q-1)$-torsion free, we deduce that

$$\begin{aligned} BR + CQ + A(2P - T) = 0 \end{aligned}$$

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

$$\begin{aligned} J= & {} (A\!+D, A^2 \!+\! BC, qBR - CQ + (q - 1)AP, (q+1)QA\! +\! B(qT\! -\! (q+1)P), \\&\quad (q+1)RA + C(T - (q+1)P), CQ + BR + A(2P-T)). \end{aligned}$$

Now, the image of J in $\overline{\mathcal {S}}$ is

$$\begin{aligned} (A+D, A^2+ BC, BR + CQ, BT, CT, BR + CQ + AT) \end{aligned}$$

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

$$\begin{aligned} f:\overline{\mathcal {S}}/(I_N \cap I_1) \twoheadrightarrow \overline{\mathcal {A}}(\tau ). \end{aligned}$$

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

$$\begin{aligned} I(\tau ) = (A+D, BT, CT, CQ+BR, A^2+BC). \end{aligned}$$

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

$$\begin{aligned} A + D&= \mu \\ A(A - \mu ) + BC&= \mu . \end{aligned}$$

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):

$$\begin{aligned} BR - CQ&= 0 \\ Q(2A - \mu )&= B(2P - T) \\ R(2A - \mu )&= C(2P - T). \end{aligned}$$

Let

$$\begin{aligned} J= & {} (A+D - \mu , A(A-\mu ) + BC - \mu , BR-CQ, \nonumber \\&Q(2A - \mu ) - B(2P-T), R(2A - \mu ) - C(2P - T)). \end{aligned}$$

(29)

Let I be the image of J in $\overline{\mathcal {S}}$, so that

$$\begin{aligned} I = \left( A+D, BT, CT, CQ+BR, A^2 + BC\right) . \end{aligned}$$

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

$$\begin{aligned} \frac{\mathcal {O}[[X-1]]}{(X - \zeta _1)(X - \zeta _2)}. \end{aligned}$$

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

$$\begin{aligned} \overline{\mathcal {S}}/I_1 \overset{AT}{\longrightarrow } \overline{\mathcal {S}}/I \longrightarrow \overline{\mathcal {S}}/(I_1 \cap I_N)\rightarrow 0. \end{aligned}$$

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.