Matrix calculation used to determine local deformation ring

Question

Let $C$ be a local ring with maximal ideal $m_C$ and residue field isomorphic to a finite extension of $\mathbb{F}_l$. Suppose $q$ is a power of a prime $p \neq l$ and suppose $A,B,P,Q,R,S \in m_C$ and $\alpha,\beta \in C^{\times}$ such that $\alpha/\beta \not\equiv 1,q,q^{-1} \pmod{m_C}$ and $$\begin{pmatrix} \alpha + A & 0 \\ 0 & \beta + B\end{pmatrix}\begin{pmatrix} 1 + P& R \\ S & 1 + Q\end{pmatrix}\begin{pmatrix} \alpha + A & 0 \\ 0 & \beta + B\end{pmatrix}^{-1} = \begin{pmatrix} 1 + P& R \\ S & 1 + Q\end{pmatrix}^q.$$ I want to show that $R = S = 0$. This is part of a calculation used in Jack Shotton's paper "Local deformation rings and a Breuil-Mézard conjecture when $\ell \neq p$" to explicitly describe a local deformation ring. While skimming the paper I assumed it would be trivial, but after looking at it closely I'm actually not sure how it works. The paper says to just "look at the top-right/bottom-left entries". This is easy on the left-hand side and we just get $\gamma R$ and $\gamma^{-1} S$ where $\gamma = (\alpha + A)/(\beta + B)$, but on the right-hand we'll get some awful expressions in $P,R,S,Q$. And the ring $C$ is not necessarily an integral domain, so even if this results in an equation of the form $\gamma R = cR$ with $c \neq \gamma$, for instance we can't conclude that $R = 0$ from this.

Answer 1

I agree this is not entirely obvious! If I'm not mistaken, the point is that the RHS has the form $(1 + X)^q$ with $X \in Mat_{2 \times 2}(\mathfrak{m})$, so it is $1 + q X \bmod \mathfrak{m}^2$. So, for the top-right term, you get $\gamma R$ on the LHS, which is $\tfrac{\alpha}{\beta} R \bmod \mathfrak{m}^2$; and on the RHS, you get something which is $qR \bmod \mathfrak{m}^2$. So you have $(q-\tfrac{\alpha}{\beta})R \in \mathfrak{m}^2$; but $q-\tfrac{\alpha}{\beta}$ is a unit in $C$, so $R \in \mathfrak{m}^2$.

You can now repeat the argument to deduce inductively that $R \in \mathfrak{m}^k$ for every $k$; and because $C$ is $\mathfrak{m}$-adically complete, we deduce $R = 0$. The argument for the bottom left corner is similar, using the assumption $\tfrac{\alpha}{\beta} \ne q^{-1}$.