Let me start by reminding some basic properties of a support of a module. Let $A$ be a Noetherian local ring and $M$ is a finitely generated $A$=module. Its support is $$ \operatorname{Supp}_ A(M) = \lbrace \mathfrak{p} \in \operatorname{Spec} A \colon M_\mathfrak{p} \neq 0 \rbrace = V(\operatorname{Ann}_ A(M)). $$ In particular, it is closed.
Lemma 1. We are still under these assumptions.
- If $I \subseteq A$ then $\operatorname{Supp}_ {A/I}(M/IM) = \operatorname{Supp}(M) \cap \operatorname{Spec}(A/I)$.
- (Let $L,\mathcal{O},\lambda,\mathbb{F}$ be as before.) Let $A$ be a local $\mathcal{O}$-algebra and assume that the irreducible components $C_1, \dotsc, C_r$ of $\operatorname{Spec} A$ satisfy the condition that $C_i$ are generically in characteristic $0$ and $C_i \times_\mathcal{O} \mathbb{F}$ are irreducible distinct and exhaust the irreducible components $A/\lambda A$. If $M$ is $\mathcal{O}$-torsion free and $\operatorname{Supp}_ {A/\lambda A}(M/\lambda A) = \operatorname{Spec}(A/\lambda A)$ then $\operatorname{Supp}_ A(M) = \operatorname{Spec} A$.
Proof.
The first part is standard by Nakayama. For the second part, let $\mathfrak{P}$ be a minimal prime of $A$ so that $\lambda \notin \mathfrak{P}$. Let $\mathfrak{p} \supseteq (\mathfrak{P}, \lambda)$ be a minimal prime so that $\mathfrak{p} \in \operatorname{Aupp}_ {A/\lambda}(M/\lambda M)$. Then $\mathfrak{P}$ is the unique prime ideal properly contained in $\mathfrak{p}$, by our assumption.
Then $M_\mathfrak{p} \neq 0$, and $\lambda^r \colon M \hookrightarrow M$ implies that $\lambda^r \colon M_\mathfrak{p} \hookrightarrow M_\mathfrak{p}$. This shows that $\lambda \notin \sqrt{\operatorname{Ann}_ {A_\mathfrak{p}}(M_\mathfrak{p})}$ and so there exists a prime ideal $\mathfrak{q} \subseteq A_\mathfrak{p}$ such that $\mathfrak{q} \in \operatorname{Supp}_ A(M)$ and $\lambda \notin \mathfrak{q}$. There are two possibilies $\mathfrak{q} = \mathfrak{p}$ or $\mathfrak{q} = \mathfrak{P}$, but we can rule out the first because $\lambda \notin \mathfrak{q}$. So $\mathfrak{q} = \mathfrak{P}$.
Recall that $\operatorname{depth}(\mathfrak{m}_ A, M)$ is the length of the longest sequence $x_1, \dotsc, x_d \in \mathfrak{m}_ A$ such that $x_i$ is not a zero divisor of $M / (x_1, \dotsc, x_{i-1}) M$ for all $i$. This is then at most the dimension of any irreducible component of $\operatorname{Supp}_ A(M)$.
Corollary 2. If $\operatorname{depth}(\mathfrak{m}_ A, M) \ge \dim A$ then $\operatorname{Supp}_ A(M)$ is a union of irreducible components of $\operatorname{Spec} A$.
Patching
This was over a week ago, but hopefully people remember what was going on. We had $$ r = \dim_\mathbb{F} H_{\mathcal{L}_ \emptyset^\perp}^1(G_F, (\mathrm{ad}^0 \bar{r})(1)). $$ We had auxiliary primes $Q$ and then $$ H_Q = \prod_{v \in Q} \begin{pmatrix} \text{max } \ell\text{-power} \br \text{quotient of } k(v)^\times \end{pmatrix}, \quad \Lambda_Q = \mathcal{O}[[A_{v,i,j} : v \in R, i,j=1,2]] / (A_{v_0,1,1}). $$ Now we are going to look at $$ H_\infty = \mathbb{Z}_ \ell^r, \quad \Lambda_\infty = \Lambda_\emptyset[[H_\infty]], $$ where we note that $\Lambda_\infty$ is a power series ring over $\mathcal{O}$ of (absolute) Krull dimension $4 \lvert R \rvert + r$. We have non-canonical maps $$ H_\infty \twoheadrightarrow H_Q, \quad \Lambda_\infty \twoheadrightarrow \Lambda_Q. $$
We now had for each $N \gt 0$ a set $Q_N$ with size $\lvert Q_N \rvert = r$ and such that for every $v \in Q_N$ we have $q_v \equiv 1 \pmod{\ell^N}$. We now have a diagram $$ R_{\chi,\infty}^\mathrm{loc} = R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{\lvert R \rvert+r-1}]] \twoheadrightarrow R_{Q_N,\chi}^\square \twoheadrightarrow R_{\emptyset,\chi}^\mathrm{univ} $$ with $\Lambda_\infty \to R_{Q_N,\chi}^\square$, where the first map is just calculating the tangent space and lifting the basis elements. The second two rings act on $$ S_{Q_N,\chi}^\square \twoheadrightarrow S_{\emptyset,\chi} \cong S_{Q_N,\chi}^\square / \mathfrak{a}_ {Q_N} $$ where $\mathfrak{a}_ {Q_N} = \langle A_{v,i,j}, h-1 \rangle$ is the augmentation ideal.
Note that this entire picture modulo $\lambda$ is the same as the entire picture for $\chi = 1$ modulo $\lambda$. In fact, we can choose this surjection $R_{\chi,\infty}^\mathrm{loc} \twoheadrightarrow R_{Q_N,\chi}^\square$ so that this map is the same modulo $\lambda$ for both $\chi$ and $\chi = 1$. There we have $$ R_{1,\infty}^\mathrm{loc} \twoheadrightarrow R_{Q_N,1}^\square \twoheadrightarrow R_{\emptyset,1}^\mathrm{univ} $$ acting on $$ S_{Q_N,1}^\square \twoheadrightarrow S_{\emptyset,1}^\mathrm{univ} \cong S_{Q_N,1}^\square / \mathfrak{a}_ \infty $$ where $\mathfrak{c}_ N = \ker(\Lambda_\infty \to \Lambda_{Q_N})$ and $S_{Q_N,1}^\square$ is finite free over $\Lambda_\infty / \mathfrak{c}_ N$.
Now these $Q_N$ have nothing to do with each other. But the idea is that they have finite cardinality modulo some open ideal, and then there are finitely many possibilities. Let $\mathfrak{b}_ N \subseteq \Lambda_\infty$ be an open ideal with $\bigcap \mathfrak{b}_ N = 0$ such that $\mathfrak{b}_ N \supseteq \mathfrak{c}_ M$ for all $M \ge N$.
On the Galois deformation side, let $\mathfrak{d}_ {N,1} \subseteq R_{\emptyset,1}^\mathrm{univ}$ be open ideals such that $$ \mathfrak{d}_ {N,1} \supseteq \mathfrak{d}_ {N+1,1}, \quad \bigcap_N \mathfrak{d}_ {N,1} = 0, \quad \operatorname{Ann}_ {R_{\emptyset,1}^\mathrm{univ}} (S_{\emptyset,1}/\mathfrak{b}_ N) \supseteq \mathfrak{d}_ {N,1} \supseteq \mathfrak{b}_ N R_{\emptyset,1}^\mathrm{univ} $$ and also $\mathfrak{d}_ {N,\chi} \subseteq R_{\emptyset,\chi}^\mathrm{univ}$ with the same properties. Since they might not agree modulo $\lambda$, we may modify them as $$ \mathfrak{e}_ {N,1} = \mathfrak{d}_ {N,1} \cap (\mathfrak{d}_ {N,\chi} + \lambda R_{\emptyset,1}^\mathrm{univ}), \quad \mathfrak{e}_ {N,\chi} = \mathfrak{d}_ {N,\chi} + (\mathfrak{d}_ {N,1} + \lambda R_{\emptyset,\chi}^\mathrm{univ}). $$ This still satisfies everything.
Now we set $$ R_{M,N,\chi} = \im(R_{Q_M,\chi}^\square \to R_{\emptyset,\chi}^\mathrm{univ} / \mathfrak{e}_ {N,\chi} \oplus \operatorname{End}_ {\Lambda_\infty}(S_{Q_M,\chi}^\square / \mathfrak{b}_ N)). $$ for $M \ge N$. Then we have this diagram $$ R_{\chi,\infty}^\mathrm{loc} \twoheadrightarrow R_{M,N,\chi} \to R_{\emptyset,\chi}^\mathrm{univ} / \mathfrak{e}_ {N,\chi} $$ with $\Lambda_\infty \to R_{M,N,\chi}$ acting on $$ S_{Q_m,\chi}^\square / \mathfrak{b}_ N \twoheadrightarrow S_{\emptyset,\chi} / \mathfrak{b}_ N, $$ where the source is finite free over $\Lambda_\infty / \mathfrak{b}_ N$. There is a similar thing for $\chi = 1$.
The point is that if we fix $N$ and $M$ varies over $M \ge N$, there are finitely many possibilities for these diagrams. So we look at diagrams $(1, M)$ and one diagram arises infinitely often. Fix one such $(1, M_1)$ and look at diagrams $(2, M)$ such that $\operatorname{diag}(1, M) \cong \operatorname{diag}(1, M_1)$. One such $(2, M_2)$ arises infinitely often, and we go on. At the end, for all $N \ge 1$ we have a diagram $\operatorname{diag}(N, M_N)$ such that it reduces modulo $\mathfrak{b}_ {N-1}$ and $\mathfrak{e}_ {N-1,\chi}$ to the diagram for $(N-1, M_{N-1})$. Taking the inverse limit of such diagrams, we get $$ \begin{CD} @. \Lambda_\infty \br @. @VVV \br R_{1,\infty}^\mathrm{loc} @>>> R_{\infty,1} @>>> \operatorname{End}(S_{\infty,1}) @. S_{\infty,1} \br @. @VVV @. @VVV \br @. R_{\infty,1}^\mathrm{univ} @>>> \operatorname{End}(S_{\emptyset,1}) @. S_{\emptyset,1} = S_{\infty,1} / \mathfrak{a}_ \infty, \end{CD} $$ where $S_{\infty,1}$ is finite free over $\Lambda_\infty$ and $\mathfrak{a}_ \infty \to (0)$ in $R_{\emptyset,1}^\mathrm{univ}$. There is the same diagram for $\chi$ plus an isomorphism of the two diagrams modulo $\lambda$.
Recall that $\Lambda_\infty$ is a power series ring over $\mathcal{O}$ of dimension $4 \lvert R \rvert + r$. So we can lift it to $$ \Lambda_\infty \to R_{\chi,\infty}^\mathrm{loc}. $$ We see that $$ \operatorname{depty}(\mathfrak{m}_ {\Lambda_\infty}, S_{\infty,\chi}) = 4 \lvert R \rvert + r $$ because it is finite free, and hence $$ \operatorname{depty}(\mathfrak{m}_ {R_{\chi,\infty}^\mathrm{loc}}, S_{\infty,\chi}) \ge 4 \lvert R \rvert + r. $$ But we have seen that $\operatorname{Spec} R_{\chi,\infty}^\mathrm{loc}$ is irreducible. So we have $$ \operatorname{Supp}_ {R_{\chi,\infty}^\mathrm{loc}}(S_{\infty,\chi}) = \operatorname{Spec} R_{\chi,\infty}^\mathrm{loc}. $$ Then modulo $\lambda$, this still has full support. On the other hand, we saw that $R_{1,\infty}^\mathrm{loc}$ has this nice property about components. It follows from the lemma that we have $$ \operatorname{Supp}_ {R_{1,\infty}^\mathrm{loc}}(S_{\infty,1}) = \operatorname{Spec} R_{1,\infty}^\mathrm{loc}. $$ Taking modulo $\mathfrak{a}_ \infty$, we get $$ \operatorname{Supp}_ {R_{1,\infty}^\mathrm{loc} / \mathfrak{a}_ \infty}(S_{\emptyset,1}) = \operatorname{Spec} R_{1,\infty}^\mathrm{loc} / \mathfrak{a}_ \infty. $$ That is, $\operatorname{Supp}_ {R_{\emptyset,1}^\mathrm{univ}}(S_{\emptyset,1}) = \operatorname{Spec} R_{\emptyset,1}^\mathrm{univ}$ and $$ R_{1,\infty}^\mathrm{loc} / \mathfrak{a}_ \infty \twoheadrightarrow R_{\emptyset,1}^\mathrm{univ} \twoheadrightarrow \mathbb{T}_ {\emptyset,1} $$ has nilpotent kernels.