We want to prove something like $(R^\mathrm{big})^\mathrm{red} = \mathbb{T}$ in the case when $\bar{r} \vert_{G_{F(\zeta_\ell)}}$ is irreducible. Then we will turn to the case when $\bar{r}$ is irreducible but $\bar{r} \vert_{G_{F(\zeta_\ell)}}$ is reducible, the argument that Skinner–Wiles used.
Recall that we have two disjoint sets of primes $R$ and $Q$, where $R$ is the places where $r$ is ramified and $Q$ is some set of auxiliary primes. We are assuming for $v \in R \cup Q$ that $q_v \equiv 1 \pmod{\ell}$, and we have some characters $\chi_v \colon k(v)^\times \to \mathcal{O}^\times$ for $v \in R$ of $\ell$-power order. We had some level group $U_Q^\ell \subseteq \mathrm{GL}_ 2(\mathbb{A}_ F^{\infty,\ell})$ and then we had some space of modular forms $$ M(U_Q^\ell, \mathcal{O})_ \chi = \operatorname{Hom}(S^\mathrm{sm}(U_Q^\ell, L/\mathcal{O})_ \chi, L/\mathcal{O}). $$ This thing had a faithful action of $\mathbb{T}(U_Q^\ell, \mathcal{O})$ and of $\mathrm{PGL}_ 2(F_\ell)$, and in fact finite free over $\mathcal{O}[[U_\ell^1]]$ where $U_\ell^n$ is the kernel of $\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell}) \to \mathrm{PGL}_ 2(\mathcal{O}_ F/\ell^n)$. We can also localize this at some maximal ideal $\mathfrak{m}$.
We also had a pseudo-representation $$ T_\mathfrak{m} \colon G_F \to \mathbb{T}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} $$ with $\det T_\mathfrak{m} = \epsilon_\ell^{-1}$, which is unramified away from $R \cup \lbrace v \mid \ell \rbrace \cup Q$, and with $T_\mathfrak{m}(\mathrm{Frob}_ v) = T_w$. There was a block $\mathcal{B}_ \mathfrak{m}$ associated to $\mathfrak{m}$ which is the product of the blocks $\mathcal{B}_ {T_\mathfrak{m} \vert_{G_{F_v}} \bmod{\mathfrak{m}}}$. This had a corresponding universal ring $$ R_{\mathcal{B}_ \mathfrak{m}}^\mathrm{ps} = \bigotimes^\wedge R_{\bar{T}_ \mathfrak{m} \vert_{G_{F_v}}, \epsilon_\ell^{-1}}^\mathrm{ps} \to \mathbb{T}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}. $$ We saw that this space of modular forms is in the block $$ M(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \in \mathcal{C}_ {\mathrm{GL}_ 2(F_\ell),1}(\mathcal{O})_ {\mathcal{B}_ \mathfrak{m}} \ni P_\mathfrak{m}, $$ where the corresponding module $\operatorname{Hom}(P_\mathfrak{m}, M(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}})$ has two actions of $R_{\mathcal{B}_ \mathfrak{m}}^\mathrm{ps}$ that agree.
- For $v \in R$, it follows from $\chi$-condition on $M(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}$ that $T_\mathfrak{m} \vert_{I_{F_v}} = \chi_v + \chi_v^{-1}$. We can also do base change to ensure that $\bar{r}_ \mathfrak{m} \vert_{G_{F_v}} = 1$. (Given $r$ we can always base change so that it is ramified only at $\lbrace v \mid \ell \rbrace \cup R$ with $\bar{r} \vert_{G_{F_v}} = 1$ and $q_v \equiv 1 \pmod{R}$ if $v \in R$.)
- We choose $Q$ so that $Q \cap (R \cup \lbrace v \mid \ell \rbrack) = \emptyset$ and $q_v \equiv 1 \pmod{\ell}$ for all $v \in Q$ and $(T_\mathfrak{m} \bmod{\mathfrak{m}})(\mathrm{Frob}_ v) = \bar{\alpha}_ v + \bar{\alpha}_ v^{-1}$ with $\bar{\alpha}_ v^2 \neq 1$ for all $v \in Q$.
It follows that $T_\mathfrak{m} = \psi_v + \epsilon_\ell^{-1} \psi_v^{-1}$ for a character $$ \psi_v \colon G_{F_v} \to \mathbb{T}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^\times $$ with $(\psi_v \pmod{\mathfrak{m}})(\mathrm{Frob}_ v) = \bar{\alpha}_ v$. The restriction $\psi_v \vert_{I_{F_v}}$ moreover factors through $$ I_{F_v} \twoheadrightarrow k(v)^\times \twoheadrightarrow H_v $$ where $H_v$ is the maximal quotient of $k(v)^\times$ of $\ell$-power order. Writing $H_Q = \prod_{v \in Q} H-v$, we see that we get $$ \prod_{v \in Q} \psi_v \colon H_Q \to \mathbb{T}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^\times. $$
Given $v \in Q$ and $a \in \mathcal{O}_ {F,v} - \lbrace 0 \rbrace$, we have this operator $$ U_{v,a} \leftrightarrow U_{Q,v}^\ell \begin{pmatrix} a & 0 \br 0 & 1 \end{pmatrix} U_{Q,v}^\ell $$ that commutes with $T_w$ for $w \notin Q \cup R \cup \lbrace w^\prime \mid \ell \rbrace$ and $U_{v^\prime,b}$ for $v^\prime \in Q$ and $b \in \mathcal{O}_ {F,v^\prime} - \lbrace 0 \rbrace$ and $\mathrm{GL}_ 2(F_\ell)$.
If we look at $\phi \in G_{F_v}$ a Frobenius lift corresponding to a uniformizer $\pi_v$, then we see that $$ (U_{\pi_v} - A_v) (U_{\pi_v} - B_v) = 0, \quad A_v = \psi_v(\phi), \quad B_v = \epsilon_\ell^{-1} \psi_v^{-1}(\phi) $$ on $M(U_Q^\ell, \mathcal{O})_ {\chi, \mathfrak{m}}$. This allows us to define integral idempotents $$ e_v = \frac{U_{\pi_v} - B_v}{A_v - B_v}, \quad e_v^2 = e_v, \quad e = \prod_{v \in Q} e_v. $$ Then we can consider $$ \mathbb{T}(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} [\mathrm{GL}_ 2(F_\ell)] \curvearrowright M(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^+ = e M(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}. $$ This is in fact finite free over $\mathcal{O}[[U_\ell^1]][H_Q]$. If we write $\mathfrak{a}_ Q \subseteq \mathcal{O}[H_Q]$ for the augmentation ideal, we get $$ M(U_Q^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^+ / \mathfrak{a}_ Q \cong M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}. $$
Now let us assume that $\bar{r}_ \mathfrak{m} \vert_{G_{F(\zeta_\ell)}}$ is absolutely irreducible. The usual Chebotarev argument shows that for all $N$ there exists $Q_N$ such that
- if $v \in Q_N$ then $q_v \equiv 1 \pmod{\ell^n}$,
- if $v \in Q_N$ then $\bar{r}_ \mathfrak{m}$ is unramified at $v$ with distinct Frobenius eigenvalues,
- the size of $Q_N$ is $r = \dim_ \mathbb{F} H_{\mathcal{L}_ \emptyset^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1))$ (where $\mathcal{L}_ \emptyset^\perp$ had no conditions at $\lbrace v \mid \ell \rbrace \cup R$ and unramified elsewhere),
- we have $H_{\mathcal{L}_ {Q_N}^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1)) = 0$ (where $\mathcal{L}_ {Q_N}^\perp$ is the same condition except that the condition is trivial at $v \in Q_N$).
Now we can look at the deformation ring $R_{Q_N,\chi}^\mathrm{univ}$ for the conditions
- no condition at $\lbrace v \mid \ell \rbrace \cup Q_N$,
- $\tr r \vert_{I_{F_v}} = \chi_v + \chi_v^{-1}$ for $v \in R$,
- unramified elsewhere,
- $\det r = \epsilon_\ell^{-1}$.
We also look at the framed version $$ R_{Q_N,\chi}^\mathrm{univ} \to R_{Q_N,\chi}^\square \cong R_{Q_N,\chi}^\mathrm{univ} \hat{\otimes} \mathcal{O}_ \infty, \quad \mathcal{O}_ \infty = \mathcal{O}[[A_{v,i,j} : v \in R \cup \lbrace v \mid \ell \rbrace, 1 \le i, j \le 2 ]] / (A_{v_0,1,1}). $$
For $v \in Q_N$ we have $r^\mathrm{univ} \vert_{G_{F_v}} = \psi_v \oplus \epsilon_\ell^{-1} \psi_v^{-1}$ with $(\psi_v \bmod{\mathfrak{m}})(\mathrm{Frob}_ v) = \bar{\alpha}_ v$. Then we have $$ \mathcal{O}[H_{Q_N}] \to R_{Q_N,\chi}^\mathrm{univ}, \quad R_{Q_N,\chi}^\mathrm{univ} / \mathfrak{a}_ {Q_N} \cong R_{\emptyset,\chi}^\mathrm{univ}. $$ We can also consider $$ M(U_{Q_N}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square} = M(U_{Q_N}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^+ \otimes_{R_{Q_N,\chi}} R_{Q_N,\chi}^\square, $$ which is finite free over $\mathcal{O}_ \infty[H_{Q_N}][[U_\ell^1]]$. Again if we quotient by the augmentation ideal $\tilde{\mathfrak{a}}_ {Q_N} \subseteq \mathcal{O}_ \infty[H_{Q_N}]$ we get $M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}$.
We can now consider $$ R_\chi^\mathrm{loc} = \bigotimes_{v \in R \cup \lbrace v \mid \ell \rbrace}^\wedge R_{\bar{r}_ \mathfrak{m} \vert_{G_{F_v}}, \epsilon_\ell^{-1}}^\square $$ where we are looking at all liftings for $v \mid \ell$ and liftings with $\tr r \vert_{\sigma_{F_v}} = \chi_v + \chi_v^{-1}$ for $v \in R$.
Then we seek to find a surjection $$ R_\chi^\mathrm{loc}[[x_1, \dotsc, x_t]] \twoheadrightarrow R_{Q_N,\chi}^\square. $$ The dimension of the relative tangent space $t$ can be computed to be $$ \dim H_{\mathcal{L}_ {Q_N}}^1(G_F, \mathrm{ad}^0 \bar{r}) + 3 - \dim H^0(G_F, \mathrm{ad}^0 \bar{r}) + \sum_{v \in R \cup \lbrace v \mid \ell \rbrace}(1 + \dim H^0(G_{F_v}, \mathrm{ad}^0 \bar{r})) - 4, $$ which becomes under duality $$ \dim H_{\mathcal{L}_ Q^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1)) - 1 + \lvert R \rvert + r, $$ because the representation is odd and at $Q_N$ have different Frobenius eigenvalues. Because we have arranged this $H^1$ to vanish, we can set $t = \lvert R \rvert + r - 1$.
Writing $H_\infty = \mathbb{Z}_ \ell^r \twoheadrightarrow H_{Q_N}$ and $\Lambda_ \infty = \mathcal{O}_ \infty[[H_\infty]] \to R_{Q_N,\chi}^\square$, we have the usual diagram $$ \begin{CD} @. \Lambda_\infty \br @. @VVV \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_t]] @>>> R_{Q_N,\chi}^\square @. \curvearrowright @. M(U_{Q_N}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square} \br @. @VVV @. @VVV \br @. R_{\emptyset,\chi}^\mathrm{univ} @. \curvearrowright @. M(U_\emptyset^\ell, \mathcal{O})_ {\chi,\mathfrak{m}} \end{CD} $$ where $M(U_{Q_N}^\ell, \mathcal{O})_ {\chi,\mathfrak{m}}^{+,\square}$ is finite free over $(\Lambda_\infty / f_N)[[U_\ell^1]]$.
The dimension of $R_\chi^\mathrm{loc}[[x_1, \dotsc, x_t]]$ is $6[F:\mathbb{Q}] + 4\lvert R \rvert + r$ and the dimension of $\Lambda_\infty$ is $4 \lvert R \rvert + 4 [F:\mathbb{Q}] + r$. These don’t match, but we expected this because we haven’t taken into account the $U_\ell^1$ group action.
We consider the kernel $$ \mathfrak{c}_ N = \ker(\mathcal{O}[[U_\ell^1]] \to \mathcal{O}[\mathrm{PGL}_ 2(\mathcal{O}_ {F,\ell} / \ell^N)]) $$ and $\mathfrak{b}_ N \subseteq \Lambda_\infty$ open and decreasing with trivial intersection such that $\mathfrak{b}_ N$ contains $A_{v,i,j}$ and $h-1$ for $h \in \ell^N \mathbb{Z}_ \ell^r$. We can now do the same thing as before and choose open ideals $$ \mathfrak{e}_ {N,1} \subseteq R_{\emptyset,1}^\mathrm{univ}, \quad \mathfrak{e}_ {N,\chi_0} \subseteq R_{\emptyset,\chi_0}^\mathrm{univ} $$ that are equal modulo $\lambda$ and contain $\mathfrak{b}_ N R_{\emptyset,\chi}^\mathrm{univ}$.