We were working with $F$ and $D$ and $L / \mathbb{Q}_ \ell$ finite with $\mathcal{O}, \lambda, \mathbb{F}$. Let $\ell$ be unramified in $F$, and let $R$ be a finite set of places of $F$ not dividing $\ell$. We were interested in representations $$ r \colon G_F \to \GL_2(\mathcal{O}) $$ be a representation
- unramified outside $R$ and $\lbrace v \mid \ell \rbrace$,
- if $v \mid \ell$ then $r \vert_{G_{F_v}}$ is Fontaine–Laffaille with Hodge–Tate numbers $0, 1$,
- if $v \in R$ then $r \vert_{I_{F_v}}$ is unipotent,
- $\det r = \epsilon_\ell^{-1}$.
Here, $\bar{r} = r \bmod{\lambda}$ is absolutely irreducible, satisfies $\bar{r} \vert_{G_{F_v}} = 1$ for $v \in R$, and if $v \in R$ then $q_v \equiv 1 \pmod{\ell}$. We also increased $L$ so that for all $\sigma \in G_F$ the eigenvalues of $\bar{r}(\sigma)$ are in $\mathbb{F}$.
To deal with this situation, we had to fix some auxiliary data. We consider $\chi = \prod_v \chi_v$ and $\chi_v \colon k(v)^\times \to \mathcal{O}^\times$, and $Q$ a finite set of places of $F$ disjoint from $R \cup \lbrace v \mid \ell \rbrace$ such that for $v \in Q$ we have $q_v \equiv 1 \pmod{\ell}$ and $\bar{r}(\mathrm{Frob}_ v)$ has distinct eigenvalues $\alpha_v \neq \beta_v$.
With this auxiliary data, we defined a universal deformation ring $R_{\chi,Q}^\mathrm{univ}$ parametrizing deformations of $\bar{r}$ such that
- $\det r = \epsilon_\ell^{-1}$,
- $r$ is unramified away from $Q \cup R \cup \lbrace v \mid \ell \rbrace$,
- over $v \mid \ell$ it is Fontaine–Laffaille with weights $0, 1$,
- for $v \in R$ and $\sigma \in I_{F_v}$ the matrix $r(\sigma)$ has characteristic polynomial $(X - \chi_v^\mathrm{gal}(\sigma)) (X - \chi_v^\mathrm{gal}(\sigma)^{-1})$.
Then this $r$ corresponds a quotient $R_{1,\emptyset}^\mathrm{univ} \to \mathcal{O}$.
For each $v \in Q$, let $H_v$ be the maximal $\ell$-power quotient of $k(v)^\times$ and let $H_Q = \prod_{v \in Q} H_v$. For each $v \in Q$, we had $$ r_{\chi,Q}^\mathrm{univ} \vert_{G_{F_v}} = \chi_{\alpha_v} \oplus \chi_{\beta_v}, $$ where $\chi_{\alpha_v} \pmod{\mathfrak{m}} = \alpha_v$, and similarly for $\beta$. So $\chi_{\alpha_v} \vert_{I_{F_v}}$ factors through $H_v$, and therefore we considered the $H_Q$-action $$ \prod_{v \in Q} \chi_{\alpha_v} \colon \mathcal{O}[H_Q] \to R_{\chi,Q}^\mathrm{univ}. $$ We also saw that $R_{\chi,\emptyset}^\mathrm{univ} = R_{\chi,Q}^\mathrm{univ} / \mathfrak{a}_ Q$.
On the automorphic side, we have $$ R_{\chi,Q}^\mathrm{univ} \to \End(S(\chi,Q)), $$ where $S(\chi,Q)$ is a finite free $\mathcal{O}[H_Q]$-module and $S(\chi,Q) / \mathfrak{a}_ Q = S(\chi, \emptyset) \neq (0)$ because $\bar{r}$ is automorphic.
Want To Prove 1. If $\mathfrak{p} = \ker(R_{1,\emptyset}^\mathrm{univ} \twoheadrightarrow \mathcal{O})$ corresponds to $r$, then $\mathfrak{p}$ is in the support of the $R_{1,\emptyset}^\mathrm{univ}$-module $S(1, \emptyset)$.
Here, we have isomorphisms $$ S(\chi, Q) \otimes_\mathcal{O} \mathbb{F} = S(1, Q) \otimes_\mathcal{O} \mathbb{F}, \quad R_{\chi,Q}^\mathrm{univ} \otimes_\mathcal{O} \mathbb{F} = R_{1,Q}^\mathrm{univ} \otimes_\mathcal{O} \mathbb{F} $$ compatible with the isomorphism $S(\chi, Q) / \mathfrak{a}_ Q = S(\chi, \emptyset)$.
Claim 2. Under these properties, together with the fact that $\bar{r} \vert_{G(F_{\zeta_\ell})}$ is absolutely irreducible, we can prove that the support of $S(1, \emptyset)$ is all of $\Spec R_{1,\emptyset}^\mathrm{univ}$.
We treat the case when $R = \emptyset$ first, even if it is not logically necessary. We suppress $\chi$ from the notation in this case. We have seen that $R_Q^\mathrm{univ}$ is topologically generated over $\mathcal{O}$ by $$ \dim_\mathbb{F} H_{\mathscr{L}_ Q}^1(G_{F, Q \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 \bar{r}) $$ many elements, where we used the local condition
- for $v \mid \ell$ the subspace $L_v = H_f^1(G_{F_v}, \mathrm{ad}^0 \bar{r})$, corresponding to Fontaine–Laffaille deformations,
- for $v \in Q$ no condition $L_v = H^1(G_{F_v}, \mathrm{ad}^0 \bar{r})$.
This dimension we calculated using global duality as $$ \begin{align} -\sum_{v \mid \infty} &\dim (\mathrm{ad}^0 \bar{r})^{c=1} + \dim H^0(G_F, \mathrm{ad}^0 \bar{r}) + \dim H_{\mathscr{L}_ Q}^2(G_{F, Q \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 \bar{r}) \br &\qquad -\dim_{\mathscr{L}_ Q}^3(G_{F, Q \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 \bar{r}) + \sum_{v \mid \ell}(\dim L_v - \dim H^0(G_{F_v}, \mathrm{ad}^0 \bar{r})) \br &\qquad + \sum_{v \in Q} (\dim L_v - \dim H^0(G_{F_v}, \mathrm{ad}^0 \bar{r})) \br &= -[F:\mathbb{Q}] + 0 + \dim H_{\mathscr{L}_ Q}^2(G_{F,Q \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 \bar{r}) - 0 + \sum_{v \mid \ell} [F_v : \mathbb{Q}_ \ell] + \lvert Q \rvert \br &= \dim H_{\mathscr{L}_ Q^\perp}^1(G_{F, Q \cup \lbrace v \mid \ell \rbrace}, (\mathrm{ad}^0 \bar{r})(1)) + \lvert Q \rvert. \end{align} $$
Key Lemma 3. Assume that $\bar{r} \vert_{G_{F(\zeta_\ell)}}$ is absolutely irreducible. Let $s = \dim H_{\mathscr{L}_ \emptyset}^\perp(G_{F, \lbrace v \mid \ell \rbrace}, (\mathrm{ad}^0 \bar{r})(1))$. Then for all $N \gt 0$ a positive integer, there exists a finite set of primes $Q_N$ disjoint from $R \cup \lbrace v \mid \ell \rbrace$ such that
- $\lvert Q_N \rvert = s$,
- for $v \in Q_N$ we have $q_v \equiv 1 \pmod{\ell^N}$,
- $H_{\mathscr{L}_ {Q_N}^\perp}^1(G_{F, Q_N \cup \lbrace v \mid \ell \rbrace}, (\mathrm{ad}^0 \bar{r})(1)) = 0$,
- for $v \in Q_N$ the matrix $\bar{r}(\mathrm{Frob}_ v)$ has distinct eigenvalues.
When $\lvert H_v \rvert = \ell^r$, we can identify $$ \mathcal{O}[H_v] \cong \mathcal{O}[[s]] / ((1 + s)^{\ell^r} - 1). $$ This means that we can write $$ \begin{CD} \mathcal{O}[[s_1, \dotsc, s_s]] @>>> \mathcal{O}[H_{Q_N}] \br @. @VVV \br \mathcal{O}[[x_1, \dotsc, x_s]] @>>> R_{Q_N}^\mathrm{univ}, \end{CD} $$ where we have a $R_{Q_N}^\mathrm{univ}$-module $S(Q_N)$, where if we quotient both sides by $\mathfrak{a}_ Q$ we get the $R_\emptyset^\mathrm{univ}$-module $S(\emptyset)$.
Now there is some way we can take a limit over $N$ and get
- a ring homomorphism $\mathcal{O}[[s_1, \dotsc, s_s]] \to \mathcal{O}[[x_1, \dotsc, x_s]]$,
- a $R_\infty = \mathcal{O}[[x_1, \dotsc, x_s]]$-module $S(\infty)$, which is a finite free $\mathcal{O}[[s_1, \dotsc, s_s]]$-module,
- $S(\infty) \twoheadrightarrow S(\emptyset)$ that is modding out by $\mathfrak{a}_ \infty = (s_1, \dotsc, s_s)$,
- a ring homomorphism $R_\infty \twoheadrightarrow R_\emptyset^\mathrm{univ}$ that factors through $\mathcal{O}[[x_1, \dotsc, x_s]] / \mathfrak{a}_ \infty$.
Let $I \subseteq R_\infty$ be the annihilator of $S(\infty)$, i.e., $I = \ker(R_\infty \to \End_{\mathcal{O}[[s_1, \dotsc, s_s]]}(S(\infty)))$. Then we have $$ \dim R_\infty \ge \dim R_\infty / I \ge \operatorname{depth}_ {\mathfrak{m}_ {R_\infty}}(S(\infty)) \ge s+1 $$ as $s_1, \dotsc, s_r, \pi$ form a regular sequence for $S(\infty)$. This implies that we have equality everywhere, so that $\dim R_\infty = \dim R_\infty / I$ implies $I = 0$. This shows that $$ \operatorname{supp}_ {R_\infty}(S(\infty)) = \Spec R_\infty, $$ and so $$ \operatorname{supp}_ {R_\infty/\mathfrak{a}_ Q}(S(\emptyset)) = \Spec (R_\infty / \mathfrak{a}_ \infty) \supseteq \Spec R_\emptyset^\mathrm{univ}. $$ This shows that $$ \operatorname{supp}_ {R_\emptyset^\mathrm{univ}}(S(\emptyset)) = \Spec R_\emptyset^\mathrm{univ}. $$
What do we do when $R$ is nonempty? If we calculate the number of generators of $R_{1,Q}^\mathrm{univ}$ over $\mathcal{O}$, then this will be $$ \dim_{\mathscr{L}_ Q^\perp}^1 + \lvert Q \rvert + \sum_{v \in R} \dim H^0(G_{F_v}, (\mathrm{ad}^0 \bar{r})(1)) = \dim_{\mathscr{L}_ Q^\perp}^1 + \lvert Q \rvert + 3 \lvert R \rvert. $$ Again, we can choose that $Q$ so that the first term vanishes, but we are still left with $\lvert Q \rvert + 3 \lvert R \rvert$ generators. Mark Kisin observe that the origin of these singularities are local, and so one should work over something like $$ \bigotimes_{v \in R} R_{\bar{r} \vert_{G_{F_v}}, \chi_v}^\mathrm{univ}. $$ These deformation rings don’t actually exist, so one works with framed deformation rings $R_{\bar{r} \vert_{G_{F_v}}, \chi_v}^\square$, which doesn’t quotient out by conjugation. This parametrizes lifts $r$ of $\bar{r} \vert_{G_{F_v}}$ with condition $$ \operatorname{char}_ {r(\sigma)}(X) = (X - \chi_v^\mathrm{gal}(\sigma)) (X - \chi_v^\mathrm{gal}(\sigma)^{-1}) $$ for all $\sigma \in I_{F_v}$. Now we can take $$ R_\chi^\mathrm{loc} = \bigotimes_{v \in R}^\wedge R_{\bar{r} \vert_{G_{F_v}}, \chi_v}^\square. $$
Then we have $$ R_\chi^\mathrm{loc} \to R_{\chi,Q}^\square, $$ where both of the singularities can be bad, but the map itself isn’t too singular. Here, we can non-canonically identify $$ R_{\chi,Q}^\square = R_{\chi,Q}^\mathrm{univ} [[A_{v,i,j}]]_ {v \in R, 1 \le i, j \le 2} / (A_{v_0,1,1}). $$
Claim 4. The ring $R_{\chi,Q}^\square$ is topologically generated over $R_\chi^\mathrm{loc}$ by $$ \lvert R \rvert - 1 + \lvert Q \rvert + \dim H_{\mathscr{L}_ Q^\perp}^1(G_{F, Q \cup R \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 \bar{r}(1)) $$ elements, where we can choose $Q$ suitably so that this $H_{\mathscr{L} Q^\perp}^1$ term vanishes.
Now we choose $Q_N$ such that $\lvert Q_N \rvert = s = \dim H_{\mathscr{L}_ \emptyset^\perp}^1(G_{F, R \cup \lbrace v \mid \ell \rbrace}, \mathrm{ad}^0 \bar{r}(1))$ consisting of primes $q_v \equiv 1 \pmod{\ell^N}$. Setting $$ S(\chi, Q)^\square = S(\chi, Q) \otimes_{R_{\chi,Q}^\mathrm{univ}} R_{\chi,Q}^\square, $$ we now have $$ \begin{CD} \mathcal{O}[[s_1, \dotsc, s_{s + 4\lvert R \rvert - 1}]] @>>> \mathcal{O}[[A_{v,i,j}]] / (A_{v_0,1,1}) [\Delta_{H_{Q_N}}] \br @. @VVV \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s + \lvert R \rvert-1}]] @>>> R_{\chi,Q_N}^\square. \end{CD} $$ Writing $\mathfrak{a}_ {Q_N} = \langle A_{v,i,j}, h-1 : h \in H_{Q_N} \rangle$ now, we see that $R_{\chi,Q_N}^\square$ acting on $S(\chi, Q_N)^\square$ becomes $R_{\chi,\emptyset}^\mathrm{univ}$ acting on $S(\chi, \emptyset)$. Here, each $R_{\bar{r} \vert_{G_{F_v}}, \chi_v}^\square$ has Krull dimension $4$, so miraculously both $\mathcal{O}[[s_1, \dotsc, s_{s + 4\lvert R \rvert - 1}]]$ and $R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s + \lvert R \rvert - 1}]]$ have the same Krull dimension. By the same depth argument, we see that $$ \dim \operatorname{supp}_ {R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s+\lvert R \rvert-1}]]} S(\chi,\infty)^\square = \dim R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s+\lvert R \rvert-1}]]. $$
So if $\Spec R_\chi^\mathrm{loc}$ is irreducible, then we would have $$ \operatorname{supp}_ {R_\chi^\mathrm{loc}[[\underline{x}]]} S(\chi, \infty) = \Spec R_\chi^\mathrm{loc}[[\underline{x}]], \quad \operatorname{supp} R_{\chi,\emptyset}^\mathrm{univ}(S(\chi, \emptyset)) = \Spec R_{\chi,\emptyset}^\mathrm{univ}. $$ Unfortunately, we are not done because $R_1^\mathrm{loc}$ is not irreducible.
Fact 5. If $\chi_v \neq 1$ then for all $v \in R$ we have $R_\chi^\mathrm{loc}$ is irreducible.