We had $F, D$ as ulsual and $R$ a finite set of places not containing $v \mid \ell$. We were looking at a non-Eisenstein ideal $\mathfrak{m} \subseteq \mathbb{T}(U_\Delta(R), \mathbb{F})_ \chi$, where we extend $L$ so that $k(\mathfrak{m}) = \mathbb{F}$ and $\bar{r}_ \mathfrak{m}(\sigma)$ has eigenvalues in $\mathbb{F}$. We can make sure by solvable base change that
- $\bar{r}_ \mathfrak{m}$ is unramified away from $\ell$,
- $\bar{r}_ \mathfrak{m} \vert_{G_{F_v}} = 1$ for all $v \in R$.
We fix $R$ and $\mathfrak{m}$ for the rest of the argument.
Let $Q$ be a auxiliary set of primes
- disjoint from $R \cup \lbrace v \mid \ell \rbrace$,
- $q_v \equiv 1 \pmod{\ell}$ for all $v \in Q$,
- $\bar{r}_ \mathfrak{m}(\mathrm{Frob}_ v)$ has distinct eigenvalues $\alpha_v, \beta_v$ for $v \in Q$.
For $v \in Q$, we set $\Delta_v$ be the kernel of $k(v)^\times$ to its maximal $\ell$-power quotient and set $\chi_v = 1$. We also write $\Delta_v^0 = k(v)^\times$ and write $$ \Delta_Q = \prod_{v \in R \cup Q} \Delta_v, \quad \Delta_Q^0 = \prod_{v \in R} \Delta_v \times \prod_{v \in Q} \Delta_v^0. $$ We then have $$ \mathbb{T}(U_{\Delta_Q}(R \cup Q), \mathcal{O})_ \chi \twoheadrightarrow \mathbb{T}(U_{\Delta_Q^0}(R \cup Q), \mathcal{O})_ \chi \twoheadrightarrow \mathbb{T}(U_{\Delta_\emptyset}(R), \mathcal{O})_ {\chi,\mathfrak{m}}, $$ because $\mathfrak{m}$ non-Eisenstein implies that the last term is generated by $T_v$ for $v \notin R \cup Q \cup \lbrace v \mid \ell \rbrace$. Then $\mathfrak{m}_ Q \subseteq \mathbb{T}(U_{\Delta_Q}(R \cup Q), \mathcal{O})_ \chi$ and $\mathfrak{m}_ Q^0 \subseteq \mathbb{T}(U_{\Delta_Q^0}(R \cup Q), \mathcal{O})_ \chi$ are maximal ideals. We also have these idempotents $\tilde{e}_ {\alpha_v}$ and $\tilde{e}_ {\beta_v}$ that split $S(U_{\Delta_Q}(R \cup Q), \mathcal{O})_ {\chi,\mathfrak{m}_Q}$ into $2^{\lvert Q \rvert}$ factors.
We now have an action $$ \mathbb{T}(\chi, Q; A) = \mathbb{T}(U_{\Delta_Q}(R \cup Q), A)_ {\chi, \mathfrak{m}_ Q} \curvearrowright S(\chi, Q; A) = ({\textstyle\prod_{v \in Q} \tilde{e}_ {\alpha_v}}) S(U_{\Delta_Q}(R \cup Q), A)_ {\chi, \mathfrak{m}_ Q}. $$ Recall we have a representation $$ r_{Q,\chi}^\mathrm{mod} = r_{\mathfrak{m}_ Q} \colon G_F \to \GL_2(\mathbb{T}(\chi, Q; \mathcal{O})) $$ lifting $\bar{r}_ \mathfrak{m}$, satisfying
- $\det r_{Q,\chi}^\mathrm{mod} = \epsilon_\ell^{-1}$,
- unramified away from $Q \cup R \cup \lbrace v \mid \ell \rbrace$,
- Fontaine–Laffaille with Hodge–Tate numbers $0, 1$ at every $v \mid \ell$,
- if $v \in R$ and $\sigma \in I_{F_v}$ then $r_{Q,\chi}^\mathrm{mod}(\sigma)$ has characteristic polynomial $(X - \chi_v^\mathrm{gal}(\sigma)) (X - \chi_v^\mathrm{gal}(\sigma)^{-1})$,
- if $v \in Q$ then $r_{Q,\chi}^\mathrm{mod} \vert_{G_{F_v}} = \chi_{\alpha_v} \oplus \chi_{\beta_v}$ where $\chi_{\alpha_v}, \chi_{\beta_v} \colon G_{F_v} \to \mathbb{T}(\chi, Q; \mathcal{O})^\times$ satisfy the property that $(\chi_{\alpha_v} \bmod{\mathfrak{m}_ Q})(\mathrm{Frob}_ v) = \alpha_v$ and similarly for $\beta_v$,
- for $v \in Q$ and $a \in F_v^\times \cap \mathcal{O}_ {F,v}$ we have $U_a = \chi_{\alpha_v}(\mathrm{Art}_ a) \in \End(S(\chi, Q; \mathcal{O}))$.
Let us write $H_v = k(v)^\times / \Delta_v$ and $H_Q = \prod_{v \in Q} H_v$, which is isomorphic to $U_{\Delta_Q^0}(R \cup Q) / U_\Delta(R \cup Q)$.
Lemma 1. The space $S(\chi, Q; \mathcal{O})$ is a finite free $\mathcal{O}[H_Q]$-module and if $\mathfrak{a}_ Q \subseteq \mathcal{O}[H_Q]$ is the augmentation ideal (generated by $h-1$) then $$ S(\chi, Q; \mathcal{O}) / \mathfrak{a}_ Q S(\chi, Q; \mathcal{O}) \cong S(\chi, \emptyset; \mathcal{O}). $$ Moreover, the action of $\mathcal{O}[H_Q]$ factors through $$ ({\textstyle \prod \chi_{\alpha_v}}) \circ \mathrm{Art} \colon \mathcal{O}[H_Q] \to \mathbb{T}(\chi, Q; \mathcal{O}). $$
Here, the action is that $h \in H_v$ acts by $U_{\tilde{h}}$ where $\tilde{h} \in \mathcal{O}_ {F,v}^\times$ lifts $h$. We have already seen the finite freeness, because $H_Q$ has $\ell$-power order, and moreover a direct summand of a finite free module over a local ring $\mathcal{O}[H_Q]$ is still finite free. The last assertion is also clear from the construction of the $H_Q$-action. The only nontrivial part is the quotient by the augmentation ideal. The point is that if $M$ is finite free over $\mathcal{O}[H_Q]$ then $$ \tr_{H_Q} = \sum_{h \in H_Q} h \colon M / \mathfrak{a}_ Q M \xrightarrow{\cong} M^{H_Q} $$ is an isomorphism. On the other hand, we have a map $$ S(\chi, \emptyset; \mathcal{O}) \subseteq S(U_{\Delta_Q}(R \cup Q), \mathcal{O})_ {\chi, \mathfrak{m}_ Q}^{H_Q} \xrightarrow{e = \prod_{v \in Q} \tilde{e}_ {\alpha_v}} S(\chi, Q; \mathcal{O})^{H_Q}. $$ We need to show that this is an isomorphism. To see this, we note that there is an map $$ t = \sum_{u \in U_\Delta(R) / U_{\Delta_Q^0}(R \cup Q) = \prod_{v \in Q} \GL_2(\mathcal{O}_ {F,v}) / \mathrm{Iw}_ v} u \colon S(\chi, Q; \mathcal{O})^{H_Q} \to S(\chi, \emptyset; \mathcal{O}). $$
Claim 2. The composition $te \colon S(\chi, \emptyset; \mathcal{O}) \xrightarrow{\cong} S(\chi, \emptyset; \mathcal{O})$ is an isomorphism.
We can compute that this composition is the operator $$ \prod_{v \in Q} \tr_{\GL_2(\mathcal{O}_ {F,v}) / \mathrm{Iw}_ v} \frac{U_{\pi_v} - B_v}{A_v - B_v} = \prod_{v \in Q} \frac{q_v T_v - (q_v + 1) B_v}{A_v - B_v} = \prod_{v \in Q} \frac{q_v A_v - B_v}{A_v - B_v}. $$ This is $1$ modulo $\mathfrak{m}$, and therefore is a unit.
Claim 3. We have $\operatorname{rk} S(\chi, \emptyset; \mathcal{O}) = \operatorname{rk} S(\chi, Q; \mathcal{O})^{H_Q}$.
It now follows that $e$ is an isomorphism.