We wanted to prove the following automorphy lifting theorems.
Proposition 1. Suppose $\mathfrak{q}$ is a prime of $\mathbb{T}_ {1,\emptyset}/\lambda$ and dimesion $1$ such that $r_\mathfrak{q} \colon G_F \to \mathrm{GL}_ 2(k(\mathfrak{q}))$ is trivial on $G_{F_v}$ for all $v \in R$ and $\mathrm{ad}^0 r_\mathfrak{q}$ is absolutely irreducible. Then $R_{1,\emptyset,\mathfrak{q}}^\mathrm{univ} \twoheadrightarrow \mathbb{T}_ {1,\emptyset,\mathfrak{q}}$ has nilpotent kernel.
Last time we explained how to choose sets of auxiliary primes $Q_N$ with various properties. Now we had
- $U_\ell^1 = \ker(\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell}) \to \mathrm{PGL}_ 2(\mathcal{O}_ F/\ell))$,
- $\Lambda_\infty = (\mathcal{O}[[A_{v,i,j}]] / A_{v_0,1,1})[[H_\infty]] \supseteq \mathfrak{a}_ \infty = \langle A_{v,i,j}, h-1 \rangle$,
- $H_\infty \cong \mathbb{Z}_ \ell^r$,
- $\mathfrak{c}_ N = \ker(\mathcal{O}[[U_\ell^1]] \to \mathcal{O}[\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell} / \ell^N)])$, which is open,
- $\mathfrak{b}_ N \subseteq \Lambda_ \infty$ nested open ideals with trivial intersection, with $\mathfrak{b}_ N \supseteq \langle A_{v,i,j}, h-1 : h \in \ell^N \mathbb{Z}_ \ell^r \rangle$,
- $\mathfrak{e}_ {N,\chi} \subseteq R_{\chi,\emptyset}^\mathrm{univ}$ nested open with trivial intersection such that $\mathfrak{e}_ {N,\chi} \bmod{\lambda}$ is independent of $\chi$ and $\mathfrak{e}_ {n,\chi}$ annihilates $M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N})$.
At this point, we have $$ \begin{CD} @. \Lambda_\infty \br @. @VVV \br @. R_{Q_N,\chi}^\square @. \curvearrowright @. M(U_{Q_N}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \otimes_{\mathcal{O}[[H_\infty]]} \Lambda_\infty \br @. @VVV \br @. R_{\emptyset,\chi}^\mathrm{univ} / e_{N,\chi} @. \curvearrowright @. M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N}) \br @. @AAA @. @AAA \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_g]] @>>> \tilde{R}_ {\chi,N} @. \curvearrowright @. M_{\chi,N} \br @. @AAA \br @. \Lambda_\infty, \end{CD} $$ where
- $M(U_{Q_N}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \otimes_{\mathcal{O}[[H_\infty]]} \Lambda_\infty$ is finite free over $\mathcal{O}[[A_{v,i,j}]] / (A_{v_0,1,1}) [H_{Q_N}][[U_\ell^1]]$,
- the module $M_{\chi,N}$ is finite free over $\Lambda_\infty[[U_\ell^1]] / (\mathfrak{b}_ N + \mathfrak{c}_ {3N})$,
- the map $M_{\chi,N} \to M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N})$ is modding out by $\mathfrak{a}_ \infty$,
- by construction $\tilde{R}_ {\chi,N} \hookrightarrow R_{\emptyset,\chi}^N / \mathfrak{e}_ {N,\chi} \oplus \operatorname{End}(M_{\chi,N})$ is injective,
- the diagram is independent of $\chi$ modulo $\lambda$,
- the map is defined so that $x_i \mapsto \tilde{\mathfrak{q}}_ {N,\chi,i}$ which are some generators,
- $\tilde{\mathfrak{q}}_ {N,\chi} / (\mathfrak{q}_ {N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}, x_1, \dotsc, x_g)$ is killed by some $0 \neq f \in R_{\chi,\emptyset}^\mathrm{univ} / \mathfrak{q}$ independent of $N$ (which was $T^C$ in the previous notation).
If we look at diagram of level $N$, we observe that there are only finitely many possible diagrams, and that from a diagram of level $N$ we can get a diagram of level $N^\prime$. So there is a compatible sequence of diagrams and then we can take the inverse limit.
Now we get a diagram $$ \begin{CD} @. \Lambda_\infty \br @. @VVV \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_g]] @>>> R_{\chi,\infty} @. \curvearrowright @. M_{\chi,\infty} \br @. @VVV @. @VVV \br @. R_{\emptyset,\chi}^\mathrm{univ} @. \curvearrowright @. M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}, \end{CD} $$ where
- $M_{\chi,\infty}$ is finite free over $\Lambda_\infty[[U_\ell^1]]$,
- the map $M_{\chi,\infty} \twoheadrightarrow M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}$ is modding out by $\mathfrak{a}_ \infty$,
- the map $R_{\chi,\infty} \twoheadrightarrow R_{\emptyset,\chi}^\mathrm{univ}$ kills $\mathfrak{a}_ \infty$,
- modulo $\lambda$ this is independent of $\chi$,
- the induced $R_\chi^\mathrm{loc} \to R_{\chi,\emptyset}^\mathrm{univ}$ is the natural one,
- $\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}} \to \mathfrak{q}_ {\chi,\infty}$,
- $\mathfrak{q}_ {\chi,\infty} / (\mathfrak{q}_ {\chi,\infty}^2, \tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}})$ is killed by $0 \neq f \in R_{\emptyset,\chi}/\mathfrak{q}$,
- the representation $M_{\chi,\infty}$ is in the block $\mathcal{C}_ {\mathrm{PGL}_ 2(F_\ell),1}(\mathcal{O})_ {\mathcal{B}_ \mathfrak{m}}$.
Previously, we checked the following bound.
Proposition 2. $\dim \mathbb{T}(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \ge 1 + 2[F:\mathbb{Q}]$.
Using the following commutative algebra fact (together with dimension counting), we get the following:
- every irreducible component of $\operatorname{Supp}_ {R_{\chi,\infty}} M_{\chi,\infty}$ has dimension at least $4 \lvert P \rvert + 4\lvert R \rvert +6[F:\mathbb{Q}] +r$.
Lemma 3. For $A$ a Noetherian local ring and $M$ a finitely generated $A$-module htat is flat, and $a_1, \dotsc, a_r$ and $M$-regular sequence in $\mathfrak{m}$, we have that $M/(a_1, \dotsc, a_r) M$ is flat over $A/(a_1, \dotsc, a_r)$ and the minimal dimension of an irreducible compenent of $\operatorname{Supp}_ A(M)$ is at least $r$ plus the minimal dimension of an irreducible component of $\operatorname{Supp}_ {A/(a_1, \dotsc, a_r)} (M/(a_1, \dotsc, a_r)M)$.
Using this fact, we see that if we pass to the complete localization at $\mathfrak{q}$, we get $$ R_\chi^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge \twoheadrightarrow R_{\chi,\infty,\mathfrak{q}_ {\chi,\infty}}^\wedge \curvearrowright \mathcal{M}_ {\chi,\infty,\mathfrak{q}_ {\chi,\infty}}^\wedge, \quad \mathcal{M}_ {\chi,\infty} = \operatorname{Hom}(P_{\mathcal{B}_ \mathfrak{m}}, M_{\chi,\infty}). $$ Here we are using that $\mathfrak{q}_ {\chi,\infty} / (\mathfrak{q}_ {\chi,\infty}^2, \tilde{\mathfrak{q}}_ {\chi,\infty})$ is torsion. Now using this commutative algebra fact again, we compute that the minimal dimension of an irreducible component of $\operatorname{Supp}_ {R_\chi^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge} (\mathcal{M}_ {\infty,\chi,\mathfrak{q}_ {\chi,\infty}}^\wedge)$ is at least $4 \lvert P \rvert + 4 \lvert R \rvert +6[F:\mathbb{Q}] +r-1$. One other hand, this ring $R_\chi^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge$ is equidimensional of texactly this dimension.
Conclusion 4. This support is a union of the irreducible components of $R_\chi^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge$.
But we are not done yet, because this completion is not necessarily irreducible, and moreover the completion will increase the number of irreducible components. To deal with this, we need to compare $\chi = 1$ with the $\chi \neq 1$ geometry.
Proposition 5. 1. If $\chi_v \neq 1$ for all $v \in R$ then $\operatorname{Spec} R_\chi^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge$ is irreducible.
- If $\mathfrak{p}_ 1$ and $\mathfrak{p}_ 2$ are minimal primes of $R_1^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge$ and if $\mathfrak{P}$ is a minimal prime of $R_1^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge / \lambda$ with $\mathfrak{P}$ containing both $\mathfrak{p}_ 1$ and $\mathfrak{p}_ 2$, then $\mathfrak{p}_ 1 = \mathfrak{p}_ 2$.
Using this, what we do is that once we choose $\chi_v \neq 1$ for all $v \in R$ we can use the first part to conclude that $$ \operatorname{Supp}_ {R_{\chi,\infty,\mathfrak{q}_ {\chi,\infty}}^\wedge}(\mathcal{M}_ {\chi,\infty,\mathfrak{q}_ {\chi,\infty}}^\wedge) = \operatorname{Spec} R_{\chi,\infty,\mathfrak{q}_ {\chi,\infty}}^\wedge. $$ The same is true modulo $\lambda$, and now we use the second part to transfer this information to the $\chi = 1$ case.
Finally, how do we justify this local statement? Note that we have $$ R_\chi^\mathrm{loc}[[\underline{x}]] = S \hat{\otimes} R_\chi, $$ and moreover we have $\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}} = (\mathfrak{q}_ S, \mathfrak{m}_ {R_\chi})$ because we have chosen $\mathfrak{q}$ so that $r_\mathfrak{q}$ is trivial on each $G_{F_v}$ for $v \in R$. This means that we have $$ R_\chi^\mathrm{loc}[[\underline{x}]]_ {\tilde{\mathfrak{q}}_ {\chi,\mathrm{loc}}}^\wedge \cong S_{\mathfrak{q}_ S}^\wedge \hat{\otimes} R_\chi. $$
Lemma 6. If $A/\mathcal{O}$ is a local Noetherian ring and $\operatorname{Spec} A[1/\ell]$ is connected and $A[1/\ell]$ is normal and $A^\mathrm{red}/\mathcal{O}$ is flat, then $A^\mathrm{red}$ is a domain.
Using this, we can use the argument we discussed earlier to show that $\operatorname{Spec} (\hat{S}_ {q_1} \hat{\otimes} R_\chi)[1/\ell]$ is connected. Normality comes from working out some explicit deformation theory that looked like $\mathcal{O}[[x, y, z, w]]/(xy-zw)$. For the second part, we can check that $S_{\mathfrak{q}_ S}^\wedge$ is geometrically irreducible and then there is a bijection between irreducible components of $R_1$ and irreducible ocmponents of $S_{\mathfrak{q}_ S}^\wedge \hat{\otimes} R_1$.