Theorem 1. Consider $R_\emptyset^\mathrm{univ} \twoheadrightarrow \mathbb{T}_ \emptyset$ as usual. Suppose that $\mathfrak{p}$ is a pro-modular prime ideal of $R_\emptyset^\mathrm{univ}$ and $\dim R_\emptyset^\mathrm{univ} / \mathfrak{p} \ge [F:\mathbb{Q}] + 3 \lvert R \rvert + 2$. Then any prime $\mathfrak{q} \subseteq \mathfrak{p}$ is also pro-modular.
Lemma 2. Suppose $\mathfrak{p}$ is a prime of $R_\emptyset^\mathrm{univ}$ with $\dim R_\emptyset^\mathrm{univ} / \mathfrak{p} \ge [F:\mathbb{Q}] + 3\lvert R \rvert + 2$. Then there exists a prime $\mathfrak{q} \supseteq \mathfrak{p}$ with $\dim R_\emptyset^\mathrm{univ} / \mathfrak{q} = 1$ such that
- $\ell \in \mathfrak{q}$,
- $r^\mathrm{univ} \bmod{\mathfrak{q}} \vert_{G_{F_v}} = 1$ for all $v \in R$,
- $\mathrm{ad}^0 r^\mathrm{univ} \otimes k(\mathfrak{q})$ is irreducible (i.e., $r^\mathrm{univ} \otimes k(\mathfrak{q})$ it is not dihedral).
The point is that because the dimension is large, we can avoid the dihedral locus and intersect the locus that is trivial on $G_{F_v}$. For each $v \in R$, we choose $a_{v,1}, a_{v,2}, a_{v,3} \in \mathfrak{m}_ {R_v^\square}$ such that $R_v^\square / (a_{v,1}, a_{v,2}, a_{v,3})$ is $0$-dimensional. Then we see that $$ \dim R_\emptyset^\mathrm{univ} / (\mathfrak{p}, \lambda, a_{v,i}) \ge [F:\mathbb{Q}] + 1 \gt R_E^\mathrm{dih}. $$ Then we just need the following little bit of commutative algebra.
Lemma 3. Let $A$ be a Noetherian local ring and $I \subseteq A$ such that $V(I) \subsetneq \operatorname{Spec} A$. Then there exists a $\mathfrak{q} \in \operatorname{Spec} A - V(I)$ with $\dim A/\mathfrak{q} = 1$.
Proof.
If $I$ is nilpotent, then we can choose any $\mathfrak{q}$. If not, choose $a \in I$ such that $a$ is not nilpotent so that $A_a \neq (0)$. Now we can choose $\mathfrak{q}$ corresponding to a maximal ideal of $A_a$.
So it now suffices to show the following.
Theorem 4. If $\mathfrak{q}$ is a pro-modular prime of $R_\emptyset^\mathrm{univ}$ with
- $\ell \in \mathfrak{q}$,
- $\dim R_\emptyset^\mathrm{univ}/\mathfrak{q} = 1$,
- $r^\mathrm{univ} \bmod{\mathfrak{q}} \vert_{G_{F_v}} = 1$ for all $v \in R$,
- $\mathrm{ad}^0 r^\mathrm{univ} \otimes k(\mathfrak{q})$ is absolutely irreducible.
Then the map $R^\mathrm{univ}_ {\emptyset,\mathfrak{q}} \to \mathbb{T}_ {\emptyset,\mathfrak{q}}$ has nilpotent kernel.
The rest of the quarter we are going to talk about this. Write $B = R_\emptyset^\mathrm{univ}/\mathfrak{q}$ and let $A$ be its normalization. Then we have $A \cong \mathbb{F}^\prime[[T]]$ for $\mathbb{F}^\prime/\mathbb{F}$ is finite. (Think of power series ring in $\mathbb{F}^\prime$ whose constant coefficient lies in $\mathbb{F}$.) We now do some observations.
We may assume that $\mathbb{F}^\prime = \mathbb{F}$. This is because we can consider the unramified extension $L^\prime/L$ with residue field $\mathbb{F}^\prime$ and apply the going down theorem to $R_\emptyset^\mathrm{univ} \otimes_{\mathcal{O}} \mathcal{O}^\prime$.
There exists a $v_0 \in R \cup \lbrace v \mid \ell \rbrace$ such that $\tr(r^\mathrm{univ} \bmod{\mathfrak{q}})(\mathrm{Frob}_ {v_0}) \notin \mathbb{F}$ so that $B$ is finite over the closure of $\mathbb{F}[\tr r_\mathfrak{q}(\mathrm{Frob}_ {v_0})$. This means that there exist finitely many $v_1, \dotsc, v_r \notin R \cup \lbrace v \mid \ell \rbrace$ such that $B$ is topologically generated over $\mathbb{F}$ by $\tr r_\mathfrak{q}(\mathrm{Frob}_ {v_i})$. Write $P = \lbrace v_0, \dotsc, v_R \rbrace$.
We can now set $$ R_\chi^\mathrm{loc} = \bigotimes_{v \in R}^\wedge R_{\chi_v}^\square \otimes^\wedge \bigotimes_{v \mid \ell}^\wedge R_v^\square \otimes^\wedge \bigotimes_{v \in P}^\wedge R_v^{\square,\mathrm{unr}} \to R_{Q,\chi}^\square \twoheadrightarrow R_\emptyset^\mathrm{univ}. $$ Here $R_{Q,\chi}^\square$ is not framed at all $v \in R \cup \lbrace v \mid \ell \rbrace \cup P$, and deformations are only ramified at $R \cup \lbrace v \mid \ell \rbrace \cup Q$. Then there is a prime $\mathfrak{q}_ \chi^\mathrm{loc}$ such that $$ R_\chi^\mathrm{loc} / q_{\chi,\mathrm{loc}} \cong B. $$ There is also a corresponding prime ideal $\mathfrak{q}_ {Q,\chi} \subset R_{Q,\chi}^\square$ whose quotient is $B$.
We have $B \supseteq T^c A$ for some $c$. This is because $A$ is finitely generated as a $B$-module.
We can’t patch with $\mathfrak{q}$ because $B$ is not a finite ring. So what one does is patch up to some error.
Chebotarev lemma
We consider $\mathcal{L}_ \emptyset$ which is trivial at all $v \in R \cup P \cup \lbrace v \mid \ell \rbrace$ and unramified elsewhere. Let $$ r = \dim_{k(\mathfrak{q})} H_{\mathcal{L}_ \emptyset^\perp}^1 (G_F, \mathrm{ad}^0 r^\mathrm{univ} \otimes k(\mathfrak{q})(1)) = H^1(G_{F, P \cup R \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 r^\mathrm{univ} \otimes k(\mathfrak{q})). $$
Proposition 5. There exists a $C$ such that for all $N$ there exists a set of primes of $F$ such that
- $v \equiv 1 \pmod{\ell^N}$ for $v \in Q_N$,
- $\lvert Q_N \rvert = r$,
- for $v \in Q_N$ if $r_\mathfrak{q}(\mathrm{Frob}_ v)$ has eigenvalues $\alpha_\mathfrak{q}, \beta_\mathfrak{q}$ then $\ell(A/(\alpha_\mathfrak{q} - \beta_\mathfrak{q})^2) \lt C$,
- the relative tangent space can be controlled as $\mathfrak{q}_ {Q_N,\chi} / (\mathfrak{q}_ {Q_N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}) \otimes_B A \cong A^g \oplus M_N$ where $g = \lvert R \rvert + \lvert P \rvert + r - 1$ and $\ell_A(M_N) \lt C$,
- there exists a map $$ R_\chi^\mathrm{loc}[[x_1, \dotsc, x_g]] \to R_{Q_N,\chi}^\square $$ such that each $x_i$ maps to an element of $\mathfrak{q}_ {Q_N,\chi}$ and $$ \ell(\mathfrak{q}_ {Q_N,\chi} / (\mathfrak{q}_ {Q_N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}, x_1, \dotsc, x_g)) \lt C. $$
Actually, the fifth condition follows from the other four. This is because if $e_i$ is a standard basis of $A^g$ and $B \supseteq T^{c_1} A$ then $$ T^{c_1} e_i \in \mathfrak{a}_ {Q_N,\chi} / (\mathfrak{q}_ {Q_N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}). $$ If we write $T^{c_2} M_N = 0$ and set $x_i$ to map to the lift to $\mathfrak{q}_ {Q_N,\chi}$ of $T^{c_1} e_i$ then for every $m$ in this relative tangent space $$ T^{c_2} m = \sum a_i e_i \in A^g, $$ and so $$ T^{2c_1 + c_2} m = \sum (T^{c_1} a_i) (T^{c_1} e_i) $$ so that $T^{2c_1 + c_2}$ will kill the quotient modulo $x_i$.
Now we replace the fourth condition with this other condition that $$ \lvert \ell_A(\mathfrak{q}_ {Q_N,\chi} / (\mathfrak{q}_ {Q_N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}) \otimes_B A/T^n) - ng \rvert \lt C. $$ It is easy to check that these are equivalent. The next step is to this relative tangent space to something involving Galois cohomology. Consider the dual condition $\mathcal{L}_ Q^\perp$ that now has no condition at $Q$. Then we claim that the condition we want is $$ \ell(H_{\mathcal{L}_ {Q_N}^\perp}^1(G_F, (\mathrm{ad}^0 r_\mathfrak{q})(1) \otimes A/T^n)) \lt C $$ for all $n$. In concrete terms, this Selmer group is just $$ \ker\biggl( H^1(G_{F, P \cup R \cup \lbrace v \mid \ell \rbrace}, (\mathrm{ad}^0 r_\mathfrak{q})(1) \otimes A/T^n) \to \bigoplus_{v \in Q_N} H^1(G_{F_v}, (\mathrm{ad}^0 r_\mathfrak{q})(1) \otimes A/T^n) \biggr). $$ Why is this condition equivalent to the previous one? We are interested in the length of $$ \Hom_ B(\mathfrak{q}_ {Q_N,\chi} / (\mathfrak{q}_ {Q_N,\chi}^2, \mathfrak{q}_ {\chi,\mathrm{loc}}), A/T^n). $$ But this data is the same as a map $$ R_{Q_N,\chi}^\square \to B \oplus (A/T^n) \epsilon $$ lifting $R_{Q_N,\chi}^\square \to B$ and with fixed map $R_\chi^\mathrm{loc} \to B \oplus (A/T^n) \epsilon$. These are parametrized by some $(\phi, \lbrace \alpha_v \rbrace)$ up to elements of $\mathrm{ad} r_\mathfrak{q} \otimes_B A/T^n$, and so we get $$ 0 \to \mathrm{ad} r_\mathfrak{q} \otimes_B A/T^n \to \lbrace (\phi, \lbrace \alpha_v \rbrace) \rbrace \to \Hom_B(\cdots, A/T^n) \to 0. $$ We also have $$ 0 \to \bigoplus_{v \in P \cup R \cup \lbrace v \mid \ell \rbrace} H^0(G_{F_v}, \mathrm{ad}^0 r_\mathfrak{q} \otimes A/T^n) \to \lbrace (\phi, \lbrace \alpha_v \rbrace) \rbrace \to Z^1_{\mathcal{L}_ {Q_N}}(G_F, \mathrm{ad}^0 r_\mathfrak{q} \otimes A/T^n) \to 0, $$ and so we get $$ \begin{align} \ell(\cdots) &= -4n + \sum_{v \in P \cup R \cup \lbrace v \mid \ell \rbrace} \ell(H^0(G_{F_v}, \mathrm{ad}^0 r_\mathfrak{q} \otimes A/T^n)) \br &+ \ell(H_{\mathcal{L}_ {Q_N}}^1(G_F, \mathrm{ad}^0 r_\mathfrak{q} \otimes A/T^n)) +3n - \ell(H^0(G_F, \mathrm{ad}^0 r_\mathfrak{q} \otimes A/T^n)). \end{align} $$