We were trying to prove the following.
Theorem 1. If $\bar{r} \colon G_\mathbb{Q} \to \mathrm{GL}_ 2(\bar{\mathbb{F}}_ 3)$ is continuous such that
- $\det \bar{r} = \bar{\epsilon}_ 3$,
- $\bar{r}$ is unramified outside $6$,
- $\bar{r} \vert_{G_{\mathbb{Q}_ 2}} \sim \begin{pmatrix} \bar{\epsilon}_ 3 & \ast \br 0 & 1 \end{pmatrix} \otimes \delta$ with $\delta^2 = 1$ unramified,
- $\bar{r} \vert_{G_{\mathbb{Q}_ 3}} \sim \begin{pmatrix} \bar{\chi}^{-1} \bar{\epsilon}_ 3 & \ast \br 0 & \bar{\chi} \end{pmatrix}$ with $\bar{\chi}$ unramified, $\ast$ integral, and the total thing wildly ramified,
then $\bar{\rho} \sim \begin{pmatrix} \bar{\epsilon}_ 3 & \ast \br 0 & 1 \end{pmatrix}$.
Again, we let $L = \bar{\mathbb{Q}}^{\ker \bar{r}}$. If $u$ is maximal such that $\Gal(L_w^\mathrm{nr} / \mathbb{Q}_ 3)^u$ not the identity, then we have $$ \lvert D_{L/\mathbb{Q}} \rvert \le 2^{2/3} 3^u. $$ Now the claim is that $u = 3/2$. This $L_w$ will be contained in $$ \mathbb{Q}_ 3^\mathrm{un}(\zeta_3, \lbrace \sqrt[3]{\alpha_i} \rbrace), \quad \alpha_i \in 1 + 3 \mathcal{O}_ {\mathbb{Q}_ 3^\mathrm{unr}} \subseteq (\mathbb{Q}_ 3^\mathrm{unr})^\times. $$ We now need to show that $\Gal(\bar{\mathbb{Q}_ 3}/\mathbb{Q}_ 3)^{3/2+\epsilon}$ acts trivially on $L_i = \mathbb{Q}_3^\mathrm{unr}(\zeta_3, \sqrt[3]{\alpha_i})$ for all $i$.
Note that it is not possible that $\alpha_i \equiv 1 \pmod{9}$ because then $\alpha_i$ has a cube root. This $L_i$ has uniformizer $$ \pi = \frac{\zeta_3 - 1}{\sqrt[3]{\alpha_i} - 1} $$ because this has minimal polynomial that is Eisenstein. Now we have $[L_i : \mathbb{Q}_ 3^\mathrm{un}] = 6$ with $\mathcal{O}_ {L_i} \mathcal{O}_ {\mathbb{Q}_ 3^\mathrm{unr}}[\pi]$. Its Galois group is $$ \Gal(L_i / \mathbb{Q}_ 3^\mathrm{unr}) = \langle \sigma, \tau : \sigma^2 = 1, \tau^3 = 1, \sigma \tau \sigma^{-1} = \tau^2 \rangle, $$ where $\sigma \zeta_3 = \zeta_3^{-1}$, $\tau \zeta_3 = \zeta_3$, $\sigma \sqrt[3]{\alpha_i} = \sqrt[3]{\alpha_i}$, and $\tau \sqrt[3]{\alpha_i} = \zeta_3 \sqrt[3]{\alpha_i}$. First note that $i_{L_i/\mathbb{Q}_ 3^\mathrm{unr}}(\sigma) = 1$ and because $\sigma, \sigma \tau, \sigma \tau^2$ haver order $2$, they are not in $\Gal(L_i/\mathbb{Q}_ 3^\mathrm{unr})^{1+\epsilon}$ for any $\epsilon \gt 0$. Then we calculate $$ i_{L_i/\mathbb{Q}_ 3^\mathrm{unr}}(\tau) = v_{L_i}\Bigl( \pi \frac{(\zeta_3-1) \sqrt[3]{\alpha_i}}{\zeta_3 \sqrt[3]{\alpha_i} - 1} \Bigr) = 2, \quad v_{L_i/\mathbb{Q}_ 3^\mathrm{unr}}(\tau^2) = 2. $$ Then we have $$ u_{L/\mathbb{Q}_ 3^\mathrm{unr}}(\tau) = \frac{1}{6} (1 + 1 + 1 + 2 + 2 + 2) = \frac{3}{2} $$ and similarly for $\tau^2$. This shows that $\Gal(L_i/\mathbb{Q}_ 3)^{3/2+\epsilon} = \lbrace 1 \rbrace$ for all $\epsilon \gt 0$ as desired.
This gives us the determinant bound $$ \lvert D_{L/\mathbb{Q}} \rvert^{1/[L:\mathbb{Q}]} \lt 2^{2/3} 3^{3/2} \lt 8.25 $$ and so Poitou gives us $[L : \mathbb{Q}] \le 14$. Because $6$ divides $[L : \mathbb{Q}]$, we have either $[L : \mathbb{Q}] = 6$ or $12$. That is, either $[L : \mathbb{Q}(\zeta_3)] = 3$ or $6$.
This implies that $\Gal(L/\mathbb{Q}(\zeta_3))$ has a unique Sylow $3$-subgroup, say $\langle \sigma \rangle$, which is also normal. If we look at $$ \ker(\bar{r}(\sigma) - 1) = \ker(\bar{r}(\sigma^2) - 1) $$ this is a line in $\bar{\mathbb{F}}_ 3^2$ because we have $$ \tau \ker(\bar{r}(\sigma) - 1) = \ker(\bar{r}(\tau \sigma \tau^{-1}) - 1) = \ker(\bar{r}(\sigma) - 1). $$ So we have $$ \bar{r} \sim \begin{pmatrix} \bar{\chi}_ 1 & \ast \br 0 & \bar{\chi}_ 2 \end{pmatrix}. $$
This $\bar{\chi}_ i$ are unramified away from $3$ and either $\bar{\chi}_ i, \bar{\chi}_ i \bar{\epsilon}_ 3^{-1}$ is unramified at $3$. This implies that $\bar{\chi}_ i = 1$ or $\bar{\epsilon}_ 3$. If we are in the case that $$ \bar{r} \sim \begin{pmatrix} 1 & \ast \br 0 & \bar{\epsilon}_ 3 \end{pmatrix} $$ then if we restrict to $I_{\mathbb{Q}_ 3}$ then it has an invariant line on which $\sigma \in I_{\mathbb{Q}_ 3}$ acts as $\bar{\epsilon}_ 3(\sigma)$. This implies that $\bar{r}_ {I_{\mathbb{Q}_ 3}} = 1 \oplus \bar{\epsilon}_ 3$, which contradicts that $\bar{r}$ is wildly ramified at $3$. This shows that we must have $$ \bar{r} \sim \begin{pmatrix} \bar{\epsilon}_ 3 & \ast \br 0 & 1 \end{pmatrix}. $$
Fontaine–Laffaille theory
We stated that if $E/\mathbb{Q}_ \ell$ has good reduction at $\ell$ then we have $$ \bar{r} \vert_{G_{\mathbb{Q}_ \ell}} = \operatorname{Ind}_ {G_{\mathbb{Q}_ {\ell^2}}}^{G_{\mathbb{Q}_ \ell}} \bar{\theta} \text{ or } \begin{pmatrix} \bar{\chi}^{-1} \bar{\epsilon}_ \ell & \ast \br 0 & \bar{\chi} \end{pmatrix} $$ where the second case is peu ramifiée. There is a better way to think about these things.
Let $F/\mathbb{Q}_ \ell$ be unramified finite, and let $L/\mathbb{Q}_ \ell$ be finite with ring of integer $\mathcal{O}$ and maximal ideal $\lambda$ with $\mathbb{F} = \mathcal{O} / \lambda$.
Definition 2. We define the category $\mathcal{MF}$ of $\mathcal{O}_ F \otimes_{\mathbb{Z}_ \ell} \mathcal{O}$-modules $M$ with a decreasing filtration $\mathrm{Fil}^i M$ and $\mathrm{Frob}_ \ell^{-1} \otimes 1$-linear maps $\Phi^i \colon \mathrm{Fil}^i M \to M$ satisfying
- $\mathrm{Fil}^0 M = M$, $\mathrm{Fil}^{\ell-1} M = 0$,
- $\Phi^i \vert_{\mathrm{Fil}^{i+1} M} = \ell \Phi^{i+1}$,
- $\lbrace \im(\Phi^i) \rbrace_{i=0,\dotsc,\ell-2}$ generates $M$.
Here are some facts.
- $\mathcal{MF}$ is an abelian category.
- There exists an exact covariant functor $\mathbb{G}$ from $\mathcal{MF}$ to the category of finitely generated $\mathcal{O}$-modules with a continuous linear action of $G_F$, compatible with tensor products that are defined in $\mathcal{MF}$.
- This functor $\mathbb{G}$ is fully faithful and commutes with cofiltered inverse limits.
- The essential image is closed under direct sums, subobjects, and quotients.
- The length $\operatorname{lg}_ \mathcal{O}(M)$ is $[F:\mathbb{Q}_ \ell] \operatorname{lg}_ \mathcal{O}(\mathbb{G}(M))$.
- If $\mathrm{Fil}^1(M) = (0)$ then $\mathbb{G}(M)$ is unramified.
- If $0 \le n \lt \ell-1$ then $\mathcal{O}(-n) = \mathbb{G}((\mathcal{O}_ F \otimes \mathcal{O})[-n])$. (This is the object where the filtration is concentrated in degree $n$ and $\Phi^i = \ell^{n-i}(\mathrm{Frob}_ \ell^{-1} \otimes 1)$)
- If $\ell \gt 2$ and $E/F$ is an elliptic curve with good reduction and $\mathcal{O} = \mathbb{Z}_ \ell$, then $$ (T_\ell E)^\vee \cong \mathbb{G}(M_E), $$ where $M_E$ is free of rank $2$ over $\mathcal{O}_ F$ and $\mathrm{gr}^i M_E$ is of rank $1$ if $i = 0, 1$, and $\wedge^2 M_E = (\mathcal{O}_ F \otimes \mathcal{O})[-1]$.
- If $F = \mathbb{Q}_ \ell$ and $M$ is a $\mathcal{O}_ F \otimes \mathbb{F}$-module, with $\mathrm{gr}^i M \cong \mathcal{O}_ F \otimes \mathbb{F}$ only for $i = 0, 1$ and zero otherwise (e.g., the case of $M_E / \ell M_E$) then we have $$ \mathbb{G}(M) \cong \operatorname{Ind}_ {G_{\mathbb{Q}_ {\ell^2}}}^{G_{\mathbb{Q}_ \ell}} \bar{\theta} \text{ or } \begin{pmatrix} \bar{\chi}_ 1 \bar{\epsilon}_ \ell & \ast \br 0 & \bar{\chi}_ 2 \end{pmatrix} $$ where in the second case we are peu ramifée.