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This is from p. 459 of Wiles's Fermat paper.

Theorem: If $D_{p}$ is a decomposition group at $p$, $A$ is an Artinian local ring with maximal ideal $\mathfrak{m}$ and finite residue field $k$ of characteristic $p$, $\pi:D_{p}\longrightarrow\operatorname{GL}_{2}(A)$ a continuous representation so that $\pi\cong\left(\begin{matrix}\chi_{1}\epsilon & *\\0 & \chi_{2}\end{matrix}\right)$ then the residue representation $\chi_{1}\neq\chi_{2}$ both unramified characters and residue representation $\overline{\pi}:D_{p}\longrightarrow\operatorname{GL}_{2}(k)$ comes from a finite flat group scheme.

The proof seems straightforward to me, it's just a few points that I don't understand. Wiles (or Diamond) writes: One can replace $\pi$ by $\pi\otimes\chi_{2}^{-1}$, and set $\varphi:=\chi_{1}\chi_{2}^{-1}$. Then $\pi\cong\left(\begin{matrix}\varphi\epsilon & t\\0 & 1\end{matrix}\right)$ for a cocycle $t:D_{p}\longrightarrow M(1)$. QUESTION: Why $M(1)$? What does the (1) stand for? To me, $M$ makes more sense here than $M(1)$.

$t$ defines a cohomology class $u\in H^{1}(D_{p},M(1))$ with image $u_{0}\in H^{1}(D_{p},M_{0}(1))$ where $M_{0}=M/\mathfrak{m}$. Let $G=\ker(\varphi)$ with fixed field $F$, which is an unramified extension of $\mathbb{Q}_{p}$ since $\varphi$ is unramified.

$A$ is Artinian, so there exists $n\in\mathbb{N}:p^{n}A=0$. So far, so good. Wiles claims that $H^{2}(G,\mu_{p^{r}})\longrightarrow H^{2}(G,\mu_{p^{s}})$ is an injection for $r\leq s$. Why is that so? Is that some sort of consequence of Hilbert 90? He also makes the claim that this implies that $H^{1}(G,\mu_{p^{n}})\otimes_{\mathbb{Z}_{p}}M\longrightarrow H^{1}(G,M(1))$ is an isomorphism. Why is that the case?

FINALLY, AND MOST IMPORTANTLY: Wiles proves that the elements of $M^{D_{p}}$ map to $0$ in $M_{0}=M/\mathfrak{m}M$ and therefore come from elements of $\mathsf{O}^{\times}_{F}/(\mathsf{O}^{\times}_{F})^{n}\otimes_{\mathbb{F}_{p}}M_{0}\subseteq H^{1}(G,M_{0}(1))$. From this he concludes that $\operatorname{res}u_{0}$ lies in $\mathsf{O}^{\times}_{F}/(\mathsf{O}^{\times}_{F})^{n}\otimes_{\mathbb{F}_{p}}M_{0}\subseteq H^{1}(G,M_{0}(1))$ as well, where $u$ is the cohomology class in $H^{1}(D_{p},M(1))$ represented by the cocycle $t$ and $\operatorname{res}u_{0}$ is the image of $u$ unter the composite map \begin{equation*} H^{1}(D_{p},M(1))\longrightarrow H^{1}(D_{p},M_{0}(1))\longrightarrow H^{1}(G,M_{0}(1))\text{.} \end{equation*} But this means that $u$ maps into $H^{1}(G,M(1))^{D_{p}}$. Why is that the case?

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    $\begingroup$ (1) means Tate twist. Means tensoring with $\varepsilon$. Note that the group acts on the module $M$ by $\varphi$. $\endgroup$
    – tau
    Commented Jan 30, 2022 at 0:15
  • $\begingroup$ In my opinion it acts via $\varphi\epsilon$. Which is exactly why we don't need the twist. What am I getting wrong? $\endgroup$
    – The Thin Whistler
    Commented Jan 30, 2022 at 2:18
  • $\begingroup$ On line 3 of the proof he defines what M is, and it is a free $A$-module of rank $1$ on which $D_p$ acts by $\varphi$. From this it follows that $M(1)$ is a free $A$-module on which $D_p$ acts by $\varphi\varepsilon$, which matches your matrix description, as the entry 12 in the matrix gives the cocycle. $\endgroup$
    – tau
    Commented Jan 30, 2022 at 11:08
  • $\begingroup$ If you look at the l.e.s. in chomology associated to s.e.s. $1\rightarrow \mu_{p^r}\rightarrow \mu_{p^s} \rightarrow \mu_{p^{s-r}}\rightarrow 1$ then you want to show that the map $H^1(\mu_{p^s})\rightarrow H^1(\mu_{p^{s-r})$ is surjective. This would follow from Hilbert 90th as you write. $\endgroup$
    – tau
    Commented Jan 30, 2022 at 11:33
  • $\begingroup$ Partial answer here: mathoverflow.net/questions/350713/… $\endgroup$
    – The Thin Whistler
    Commented Jan 31, 2022 at 0:02

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