We reduced the problem to the following statement.
Proposition 1. Suppose $\mathfrak{q}$ is a $1$-dimensional prime of $\mathbb{T}_ {1,\emptyset}/\lambda$ such that $r_\mathfrak{q} = r^\mathrm{univ} \bmod{\mathfrak{q}} \colon G_F \to \mathrm{GL}_ 2(k(\mathfrak{q}))$ is trivial on $G_{F_v}$ for all $v \in R$, and $\operatorname{ad}^0 r_\mathfrak{q}$ is absolutely irreducible. Then $$ R_{1,\emptyset,\mathfrak{q}}^\mathrm{univ} \twoheadrightarrow \mathbb{T}_ {1,\emptyset,\mathfrak{q}} $$ has nilpotent kernel.
We were also allowed to assume that the normalization of $\mathbb{T}_ {1,\emptyset} / \mathfrak{q}$ is $A = \mathbb{F}[[T]]$. We had this set of primes $P$ where $r_\mathfrak{q}$ is unramified, where $R_{1,\emptyset,\mathfrak{q}}^\mathrm{univ}$ is topoloigcally generated by the trace of Frobenii at $P$.
Lemma 2. Let $r = \dim_{k(\mathfrak{q})} H_{\mathcal{L}_ \emptyset^\perp}^1(G_F, \mathrm{ad}^0 r_\mathfrak{q}(1))$. Then there exists $C$ such that for all $N$ there exists $Q_N$ a set of primes of $F$ such that
- $\lvert Q_N \rvert = r$,
- for all $v \in Q_N$ we have $N(v) \equiv 1 \pmod{\ell^N}$,
- we have $Q_N \cap (R \cup \lbrace v \mid \ell \rbrace \cup P) = \emptyset$ and for all $v \in Q_N$ we have $r_\mathfrak{q}(\mathrm{Frob}_ v)$ having distinct eigenvalues $\alpha_v, \beta_v$ and $v_T((\alpha_v - \beta_v)^2) \lt C$,
- $\mathfrak{q}_ {\chi,Q_N} / (\mathfrak{q}_ {\chi,Q_N}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}) \otimes A \cong A^g \oplus M_N$ where $g = \lvert R \rvert + \lvert P \rvert + r - 1$ and $\ell_A(M_N) \lt C$,
- there exists generators $R_\chi^\mathrm{loc}[[x_1, \dotsc, x_g]] \twoheadrightarrow R_{\chi,Q_N}^\square$ such that $\ell(\mathfrak{q}_ {\chi,Q_N} / (\mathfrak{q}_ {\chi,Q_N}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}, x_1, \dotsc, x_g)) \lt C$.
We explained that instead of (4) and (5) we can just check the condition
- (4’’) $\ell(H_{\mathcal{L}_ {Q_N}^\perp}^1 (G_F, (\mathrm{ad}^0 r_\mathfrak{q})(1) \otimes A/T^n)) \lt C$ for all $n$.
The main point is that using duality, we can check that $$ \begin{align*} \ell(\mathfrak{q}_ {Q_N,\chi} / (\mathfrak{q}_ {Q_N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}) \otimes (A/T^n)) &= n + \ell(H_{\mathcal{L}_ {Q_N}^\perp}^1(G_F, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A/T^n)) \br &- [F:\mathbb{Q}]n + \lvert S \rvert n + \sum_{v \in Q_N} \ell(H^0(G_{F_v}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A/T^n)), \end{align*} $$ where we write $S = P \cup R \cup \lbrace v \mid \ell \rbrace$. Here, we can use the following lemma to show that this is $ng$ plus something that is bounded by $Cr$.
Lemma 3. For a matrix $\Phi \in M_{n \times n}(A)$ with charactersitic polynomial $\operatorname{char}_ \Phi(x) = x^k P(x)$ with $P(0) \neq 0$, we have $A^d / \Phi A^d \cong A^{\oplus k} \oplus M$ where $\ell(M) \le \operatorname{val}_ T(P(0))$.
In fact, instead of trying to calculate the Selmer condition after modding out by a power of $T$, we can check the following.
- (4’‘’) $\ell(\ker(H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A) \to \bigoplus_{v \in Q_N} (\mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A / (\mathrm{Frob}_ v - 1)))) \lt C$, and similarly for the cokernel.
This is because we can bound $$ \begin{align} \ell(H_{\mathcal{L}_ {Q_N}^\perp}^1(G_F, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A/T^n)) &\le \ell(H^2(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A)^\mathrm{tor}) \br &+ \ell(\ker(H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A) / T^n \to \bigoplus_{v \in Q_N} \frac{\mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A}{\mathrm{Frob} - 1}) / T^n) \end{align} $$ and we have the following fact.
Lemma 4. If $0 \to K \to X \to Y \to Q \to 0$ is an exact sequence of finitely generated $A$-modules and $K, Q$ have finite length then $\ell(\ker(X/T^n \to Y/T^n)) \le \ell(K) + \ell(Q)$.
We now make a reduction to the entire statement.
Claim 5. It suffices to find elements $\sigma_1, \dotsc, \sigma_r \in G_{F(\zeta_{\ell^\infty})}$ such that $r_\mathfrak{q}(\sigma_i)$ have distinct eigenvalues and $$ H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q})) \xrightarrow{\cong} \bigoplus_{i=1}^r \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q}) / (\sigma_i - 1). $$
Proof.
Let $\alpha_i, \beta_i$ be eigenvalues of $\sigma_i$. We choose $C$ such that $C$ is bigger than all of
- $\operatorname{val}_ T((\alpha_i - \beta_i)^2)$,
- $\ell(H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A)^\mathrm{tor})$,
- $\ell(\ker/\coker(H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A) \to \bigoplus_{i=1}^r \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A / (\sigma_i-1)))$.
We also set $\phi_1, \dotsc, \phi_s$ be generators of $H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes A)$ as an $A$-module.
Now given $N$, choose $v_i$ such that
- $\mathrm{Frob}_ {v_i} \in G_{F(\zeta_{\ell^N})}$ (which also implies $q_{v_i} \equiv 1 \pmod{\ell^N}$),
- $\phi_j(\mathrm{Frob}_ {v_i}) \equiv \phi_j(\sigma_i) \pmod{T^{j+1}}$,
- $r_\mathfrak{q}(\mathrm{Frob}_ {v_i}) \equiv r_\mathfrak{q}(\sigma_i) \pmod{T^{C+1}}$.
This can be indeed arranged using Chebotarev, and we can check that it will satisfy (1)–(3) and (4’‘’).
At this point, we have gotton rid of $N$. By the same argument we have seen in the torsion setting, it suffice to show that for every nonzero class $0 \neq [\phi] \in H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q}))$ there exists an element $\sigma \in G_{F(\zeta_{\ell^\infty})}$ such that $r_\mathfrak{q}(\sigma)$ has distinct eigenvalues and $\phi(\sigma) \notin (\sigma - 1) \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q})$. Write $$ F_\infty = \bar{F}^{\ker r_\mathfrak{q}}(\zeta_{\ell^\infty}), \quad \Gamma = \operatorname{Gal}(F_\infty/F) \hookrightarrow \mathrm{PGL}_ 2(A) \times \mathbb{Z}_ \ell^\times. $$
Claim 6. It suffices to show that $H^1(\Gamma, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q})) = 0$.
Proof.
If we know this, then for any nonzero class $[\phi] \in H^1(G_{F,S}, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q}))$ we will have that $\phi \vert_{F_\infty} \neq 0$. Note that $\operatorname{Im}(\phi \vert_{G_{F_\infty}})$ is $G_F$-invariant, because $\phi(\sigma \tau \sigma^{-1}) = \sigma \phi(\tau)$ for $\tau \in G_{F_\infty}$. Now we can chekck that $\langle \operatorname{Im} \phi \vert_{G_{F_\infty}} \rangle_{k(\mathfrak{q})} = \mathrm{ad}^0 r_\mathfrak{q}(1)$. We choose $\tau \in G_{F(\zeta_{\ell^\infty})}$ such that $r_\mathfrak{q}(\tau_0)$ has distinct eigenvalues, because otherwise there exists a nontrivial subspace where $G_{F_{\zeta_{\ell^\infty}}}$ acts by scalar and that line will be preserved by $G_F$.
Proposition 7. We have $H^1(\Gamma, \mathrm{ad}^0 r_\mathfrak{q}(1) \otimes k(\mathfrak{q})) = 0$.
Proof.
It suffices to prove this with $k(\mathfrak{q})$ replaced by $A$, and even $A/T^n$. We then have that this representation is trivial on $\Gamma_n = \ker(\Gamma \to \mathrm{PGL}_ 2(A/T^n) \times (\mathbb{Z}/\ell^n \mathbb{Z})^\times)$ and so it suffices to prove $$ H^1(\Gamma/\Gamma_n, \mathrm{ad}^0 r_\mathfrak{q} \otimes A/T) = 0. $$ Now we can induct on $n$.