We had a very unramified representation $\bar{r} \colon G_{\mathbb{Q},S} \to \mathrm{GL}_ 2(\mathbb{F}_ \ell)$ where $S = \lbrace 2, \ell \rbrace$ with its universal deformation $(R^\mathrm{univ}, r^\mathrm{univ})$ that is similarly unramified. We proved that $R^\mathrm{univ} \ge 1$ and we were trying to show that $R^\mathrm{univ}$ is finite over $\mathcal{O}$.
Let $F/\mathbb{Q}$ be a finite Galois extension unramified at $\ell$ satisfying $F \cap \bar{\mathbb{Q}}^{\ker \bar{r}} = \mathbb{Q}$. There is then a similar universal deformation ring $R_F^\mathrm{univ}$ of $r^\mathrm{univ} \vert_{G_F}$, and a ring homomorphism $R_F^\mathrm{univ} \to R^\mathrm{univ}$.
Claim 1. The ring homomorphism $R_F^\mathrm{univ} \to R^\mathrm{univ}$ is finite.
To show this, it suffices to prove by Nakayama’s lemma that $R^\mathrm{univ} / \mathfrak{m}_ F$ is finite over $\mathbb{F}$, where $\mathfrak{m}_ F \subseteq R_F^\mathrm{univ}$ is the maximal ideal. But what is this parametrizing? These are parametrizing Galois representations factoring as $$ G_\mathbb{Q} \to \Gal(\bar{\mathbb{Q}}^{\ker \bar{r}} F / \mathbb{Q}) \to \GL_2(R^\mathrm{univ} / \mathfrak{m}_ F). $$ Now this is a deformation theory of representations of finite groups, but this is still not so easy.
Lemma 2. Suppose $R \supseteq S$ is in $\mathcal{C}_ \mathcal{O}$, and let $I \subseteq R$ be an ideal. Let $\Gamma$ be a group and $r \colon \Gamma \to \GL_n(R)$ is such that $\bar{r} = r \bmod{\mathfrak{m}_ R}$ is absolutely irreducible and $\tr r$ is valued in $S$. If $r \bmod{I}$ is valued in $\GL_n(S/I \cap S)$ then there exists a matrix $A \in 1 + M_{n \times n}(I)$ with $A r A^{-1}$ valued in $\GL_n(S)$.
Proof.
We will reduce to the case of $\mathbb{F}[\epsilon] \supseteq \mathbb{F}$, and do this step in a clever way. For the reduction step, we consider the set $$ \mathfrak{X} = \lbrace (J, A) : I \subseteq J \subseteq R, A \in 1 + M_{n \times n}(I/J), A (r \bmod{J}) A^{-1} \in \mathrm{GL}_ n(S/J \cap S) \rbrace. $$ We can use Zorn’s lemma where we say $(J, A) \le (J^\prime, A^\prime)$ if $J \subseteq J^\prime$ and $A \bmod{J^\prime} = A^\prime$. Chains do have lower bounds, and so there is a minimal element.
If $J \neq (0)$, we have $J^\prime \subseteq J$ with $\dim_ \mathbb{F} J / J^\prime = 1$. We may replace $r$ by $r^\mathrm{new} = \tilde{A} r \tilde{A}^{-1}$ and $S$ by $S^\mathrm{new} = S / S \cap J^\prime$ and $R$ by $R^\mathrm{new} = R/J^\prime \times_{R/J} S / S \cap J$. If the lemma is true for $R^\mathrm{new}, S^\mathrm{new}, r^\mathrm{new}$, then there exists a matrix $B \in 1 + M_{n \times n}(J/J^\prime)$ and then $B \tilde{A} r \tilde{A}^{-1} B^{-1}$ is valued in $\GL_n(S/J^\prime \cap S)$ and so we would get a contradiction.
So we are reduced to the case when $R/I = S/S \cap I$ and $\dim_\mathbb{F} I = 1$. If $I \subseteq S$ then $R = S$ and so we may assume $R = S \oplus I$ with the natural multiplication $(s, a) (s^\prime, a^\prime) = (s s^\prime, s a^\prime + a s^\prime)$. It suffices to treat the case of $S/\mathfrak{m}_ S \oplus I \supseteq S/\mathfrak{m}_ S$ because then we will get some $A \in 1 + M_{n \times n}(I)$ and this will do the job for $S \oplus I$ as well.
Now we are really in this explicit case of $R = \mathbb{F}[\epsilon]$ and $S = \mathbb{F}$. We now have ring homomorphisms $$ \mathbb{F}[\Gamma] \xrightarrow{r} M_{n \times n}(\mathbb{F}[\epsilon]) \twoheadrightarrow M_{n \times n}(\mathbb{F}) $$ where the composition is surjective because $\bar{r}$ is absolutely irreducible. We also know that $\tr r \in \mathbb{F}$. For every $x \in \ker \bar{r}$ and $y \in \mathbb{F}[\Gamma]$ we have $$ 0 = \tr \bar{r}(xy) = \tr r(xy) = \tr\Bigl( \frac{r(x)}{\epsilon} y \Bigr). $$ Because $\bar{r} \colon \mathbb{F}[\Gamma] \to M_{n \times n}(\mathbb{F})$ is surjective, we have $\tr((r(x)/\epsilon) A) = 0$ for all $A \in M_{n \times n}(\mathbb{F})$, and because $\tr$ is a non-degenerate pairing, we have $r(x)/\epsilon = 0$. That is $r(x) = 0$ for every $x \in \ker \bar{r}$.
Now we are in the situation where we have $$ M_{n \times n}(\mathbb{F}) \xrightarrow{r} M_{n \times n}(\mathbb{F}[\epsilon]) \twoheadrightarrow M_{n \times n}(\mathbb{F}) $$ composes to the identity, and this satisfies $\tr r \subseteq \mathbb{F}$. Write $r(x) = x + \phi(x) \epsilon$ where $\phi$ is $\mathbb{F}$-linear and $\tr \phi = 0$. This satisfies the relation $$ \phi(xy) = x \phi(y) + \phi(x) y. $$ We now want to show that there exists a $A \in M_{n \times n}(\mathbb{F})$ such that $\phi(x) = Ax - xA$. We simply check that $$ A = \sum_{j=1}^n \phi(e_{j1}) e_{1j} $$ works.
Corollary 3. The ring $R_r^\mathrm{univ}$ is topologically generated over $\mathcal{O}$ by the traces $\tr(r^\mathrm{univ}(\sigma))$ for $\sigma \in G_{\mathbb{Q},S}$.
Proof.
If we let $S \subseteq R^\mathrm{univ}$ be the subalgebra generated by $\tr(r^\mathrm{univ}(\sigma))$ for $\sigma \in G_{\mathbb{Q},S}$, then we can conjugate $r^\mathrm{univ}$ so that it is valued in $S$. Then we have a ring homomorphism $R^\mathrm{univ} \to S \subseteq R^\mathrm{univ}$ that is the identity.
Anyways, we wanted to prove that $R^\mathrm{univ} / \mathfrak{m}_ F R^\mathrm{univ}$ is finite. By the lemma above it is topologically generated by these $r^\mathrm{univ}(\sigma)$ for $\sigma \in \Gal(E/\mathbb{Q})$, where $E = F \bar{\mathbb{Q}}^{\ker \bar{r}}$. On the other hand, for $m = [E:\mathbb{Q}]$ we have that $r^\mathrm{univ}(\sigma)^m = 1$, so then we can show that $\tr r^\mathrm{univ}(\sigma)$ is integral over $\mathbb{F}$. It follows that $R^\mathrm{univ} / \mathfrak{m}_ F R^\mathrm{univ}$ is finite.