We had a complicated situation. For each $M \ge N$ we had this diagram $$ \begin{CD} @. \Lambda_\infty = \mathcal{O}[[…]] \br @. @VVV \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_t]] @>>> R_{M,N,\chi} @. \curvearrowright @. M(U_{Q_M}^\ell, \mathcal{O})^{+,\square}_ {\chi,\mathfrak{m}} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N}) \br @. @VVV @. @VVV \br @. R_{\emptyset,\chi}^\mathrm{univ} / \mathfrak{e}_ {N,\chi} @. \curvearrowright @. M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N}), \end{CD} $$ where
- $r = \dim H_{\mathscr{L}_ \emptyset^\perp}^1(G_F, \mathrm{ad}^0 \bar{r}(1))$,
- $\Lambda_\infty$ is a power series ring in $4\lvert R \rvert + 4[F:\mathbb{Q}] + r - 1$ variables,
- $t = \lvert R \rvert + r - 1$ and $R_\chi^\mathrm{loc}[[x_1, \dotsc, x_t]]$ has dimension $3\lvert R \rvert + 6[F:\mathbb{Q}] + 1$,
- $\mathfrak{b}_ N \subseteq \Lambda_\infty$ is an open ideal,
- $\mathfrak{c}_ N \subseteq \mathcal{O}[[U_\ell^1]]$ is an open two-sided ideal given as the kernel of $\mathcal{O}[[U_\ell^1]] \to \mathcal{O}[U_\ell^1 / U_\ell^N]$,
- $U_\ell^N$ is the kernel of $\mathrm{GL}_ 2(\mathcal{O}_ {F,\ell}) \to \mathrm{GL}_ 2(\mathcal{O}_ F / \ell^N)$,
- $\mathfrak{e}_ {N,\chi} \subseteq R_{\emptyset,\chi}^\mathrm{univ}$ are open ideals,
- $R_{M,N,\chi}$ is the image of $R_{Q_M,\chi}^\square \to \operatorname{End}(…) \oplus R_{\emptyset,\chi}^\mathrm{univ} / \mathfrak{e}_ {N,\chi}$.
This space $$ M(U_{Q_M}^\ell, \mathcal{O})^{+,\square}_ {\chi,\mathfrak{m}} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N}) $$ is actually finite free over the ring $\Lambda_\infty / \mathfrak{b}_ N \otimes_ {\mathcal{O}} \mathcal{O}[[U_\ell^1]] / \mathfrak{c}_ {3N}$. The observation to make is that
- the bottom left square consists of finite objects with cardinality bounded only in terms $N$,
- modulo $\lambda$ this is independent of $\chi$.
But in this picture we are throwing away the action of $\mathrm{GL}_ 2(F_\ell)$ and only looking at the action of $U_\ell^1$. Consider the increasing exhaustive filtration $$ \mathrm{GL}_ 2(F_\ell) \supseteq G_N = \bigcup_{a,d \in F_\ell^\times, \lvert v(a/d) \rvert \le N \text{ for all } v \mid \ell} \mathrm{GL}_ 2(\mathcal{O}_ {F,\ell}) \begin{pmatrix} a & 0 \br 0 & d \end{pmatrix} \mathrm{GL}_ 2(\mathcal{O}_ {F,\ell}). $$ Then for $g \in G_N$ we have a diagram $$ \begin{CD} M(U_{Q_M}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square} / (\mathfrak{b}_ N + \mathfrak{c}_ {3N}) @>>> M(U_{Q_M}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square} / (\mathfrak{b}_ N + \mathfrak{c}_ i) \br @V{g}VV @V{g}VV \br M(U_{Q_M}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square} / (\mathfrak{b}_ N + \mathfrak{c}_ {2N}) @>>> M(U_{Q_M}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square} / (\mathfrak{b}_ N + \mathfrak{c}_ {i-N}) \end{CD} $$ for $N \le i \le 3N$. This is some shadow of the $\mathrm{GL}_ 2(F_\ell)$-action, and there are actually a finite number of choices because $U_\ell^1 \backslash G_N / U_\ell^1 F_\ell^\times$ is finite.
Now we need to patch these things together. Because one diagram for occurs infinitely many times for a fixed $N$, we can find an infinite sequence $$ (M_1, N_1), (M_2, N_2), \dotsc $$ where the diagram for $(M_{i+1}, N_i)$ is isomorphic to the diagram for $(M_i, N_i)$ (and the diagram for $(M_{i+1}, N_{i+1})$ restricts to the diagram for $(M_{i+1}, N_i)$). Taking the limit, we get $$ \begin{CD} @. \Lambda_\infty \br @. @VVV \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_t]] @>>> R_{\chi,\infty} @. \curvearrowright @. M_{\chi,\infty} \br @. @VVV @. @VVV \br @. R_{\emptyset,\chi}^\mathrm{univ} @. \curvearrowright @. M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \end{CD} $$ where the $M_{\chi,\infty}$ has a $\mathrm{GL}_ 2(F_\ell)$-action and finite free over $\Lambda_\infty \otimes^\wedge_{\mathcal{O}} \mathcal{O}[[U_\ell^1]]$ and we have $M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \cong M_{\chi,\infty} / \mathfrak{a}_ \infty$. Moreover, modulo $\lambda$ this is independent of $\chi$.
We now apply $\operatorname{Hom}(P_{\mathcal{B}_ \mathfrak{m}}, -)$ to the space of modular forms. Then we get $$ \begin{CD} R_{\chi,\infty} @. \curvearrowright @. \mathcal{M}_ {\chi,\infty} @. \curvearrowleft @. \Lambda_\infty \otimes_{\mathcal{O}}^\wedge \bigotimes_{v \mid \ell}^\wedge R_{\bar{r}_ \mathfrak{m} \vert_{G_{F_v}}, \epsilon_\ell^{-1}}^\mathrm{ps} \br @VVV @. @VVV @. @VVV \br R_{\emptyset,\chi}^\mathrm{univ} @. \curvearrowright @. \mathcal{M}(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} @. \curvearrowleft @. \bigotimes_{v \mid \ell}^\wedge R_{\bar{r}_ \mathfrak{m} \vert_{G_{F_v}}, \epsilon_\ell^{-1}}^\mathrm{ps}. \end{CD} $$ The top right action comes from the $\mathrm{GL}_ 2(F_\ell)$-action, and the bottom right action comes from both the $\mathrm{GL}_ 2(F_\ell)$-action and the action from Galois theory, by local-global compatibility.
Now this $\mathcal{M}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}$ is finite over $R_{\chi,\infty}$, and so $\mathcal{M}_ {\chi,\infty}$ is finite over $R_{\chi,\infty}$. On the other hand, it follows from some formalism that $M_{\chi,\infty}$ being flat over $\Lambda_\infty$ implies $\mathcal{M}_ {\chi,\infty}$ being flat over $\Lambda_\infty$. Because $\Lambda_\infty$ is just a power series ring, we see that $$ \dim_ {R_\chi^\mathrm{loc}[[\underline{x}]]}(\mathcal{M}_ {\chi,\infty}) = r + 4\lvert R \rvert + 4[F:\mathbb{Q}] - 1 + \dim_{R_\chi^\mathrm{loc}[[\underline{x}]]} \mathcal{M}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \ge r + 4\lvert R \rvert + 6[F:\mathbb{Q}]. $$ Because this is an equality, we see that the support of $M_{\chi,\infty}$ is a union of irreducible components of $\operatorname{Spec} R_\chi^\mathrm{loc}[[\underline{x}]]$.
Now the argument is the same as before. When $\chi = \chi_0$ then this is irreducible and so $\mathcal{M}_ {\chi_0,\infty}$ needs to have full support. This means that $$ \operatorname{Supp}_ {R_{\chi_0}^\mathrm{loc}/\lambda[[\underline{x}]]} (\mathcal{M}_ {\chi_0,\infty}/\lambda) = \operatorname{Supp}_ {R_1^\mathrm{loc}/\lambda[[\underline{x}]]} (\mathcal{M}_ {1,\infty}/\lambda) $$ and we can compute that reduction modulo $\lambda$ gives a bijection between the set of irreducible components of $R_1^\mathrm{loc}$. Using this, we can see that $M_{1,\infty}$ also has full support. This means that $$ \ker(R_{\emptyset,1}^\mathrm{univ} \twoheadrightarrow \mathbb{T}(U_Q^\ell, \mathcal{O})_ {1,\mathfrak{m}}) $$ is nilpotent.
Now we want to consider the case when we can’t use the Chebotarev argument.
Connectedness dimension
Let $R$ be a complete local Noetherian ring, so that $\operatorname{Spec} R$ is connected. Then the connectedness dimension $R$ is $$ C(R) = \min(\dim W : W \subseteq \operatorname{Spec} R \text{ is closed}, \operatorname{Spec} R - W \text{ disconnected}). $$ Here, we say that $\emptyset$ is disconnected, so that when $R$ is a domain then $C(R) = \dim R$.
Proposition 1. If $R$ is a complete local Noetherian domain and $f_1, \dotsc, f_r \in \mathfrak{m}$ then we have $c(R / (f_1, \dotsc, f_r)) \ge (\dim R)-r-1$.
The proof uses local cohomology. If $I \subseteq R$ is an ideal and $M$ is an $R$-module, then we define $$ \Gamma_I(M) = \lbrace m \in M : I^n m = (0) \rbrace = \varinjlim_n \Hom(R/I^n, M). $$ This has right derived functors $$ H_I^i(M) = \varinjlim_n \operatorname{Ext}_ R^i(R/I^n, M). $$ It is then pretty clear that this only depends the closed subset $\operatorname{Spec} R/I$.
Facts 2. 1. This can be computed using some Čech complex. If $I = (f_1, \dotsc, f_r)$ then $H_I^i(M)$ is the homology of $$ M \to \bigoplus_{i_0} M_{f_{i_0}} \to \bigoplus_{i_0 \lt i_1} M_{f_{i_0} f_{i_1}} \to \dotsb. $$ In particular, if $V(I) = V(f_1, \dotsc, f_r)$ then $H_I^i(M) = 0$ for $i \gt r$.
- (Hartshorne–Lichtenbaum) If $R$ is a domain and $I \subseteq R$ is an ideal with $\sqrt{I} \neq \mathfrak{m}$, then $H_I^i(M) = 0$ for $i \ge \dim R$.
- If $M \neq 0$ is finitely generated over $R$ then $H_\mathfrak{m}^{\dim R}(M) \neq 0$.
- If $J_1, J_2 \subseteq R$ then there is a long exact sequence $$ \dotsb \to H_{J_1+J_2}^i(M) \to H_{J_1}^i(M) \oplus H_{J_2}^i(M) \to H_{J_1 \cap J_2}^i(M) \to \dotsb. $$