Yoshida - Non-abelian Lubin-Tate 5

Some remarks.

• Let $f:X \to Y$ be a morphism of (integral) schemes. Then we say that $f$ is a Galois covering if $K(X)/K(Y)$ is a Galois extension. $A_0$ is clearly a domain and $A_m$ is a regular local ring, then by Matsumura, Theorem 14.3, we know that $A_m$ is a domain. Therefore $\mathrm{Spec}~A_m \to \mathrm{Spec}~A_0$ is a morphism of integral schemes.
• According to the post in mathSE, $Y$ becomes a quotient of $X$ by the group of deck transformations of $f$ which turns out to be the Galois group. 
• According to the Wiki article, discrete valuation ring is a regular ring with dimension 1. The converse is also true. Therefore $A_0:=W[[X_1, \ldots, X_{n-1}]]$ is a regular ring with dimension $n$.
• Review on representable functor
• A functor $\mathcal{F}: \mathcal{C} \to (\mathrm{Set})$ is called representable if there exists $A \in \mathcal{C}$ such that $\mathcal{F}$ is naturally isomorphic to $Hom(A,-)$ (or $Hom(-,A)$ for contravariant functor).
• A pair $(A, \Phi)$ is called a representation if $\Phi$ is the natural isomorphism between $Hom(A,-)$ and $\mathcal{F}$.
• By Yoneda, we know that $\Phi$ corresponds to an element $u \in \mathcal{F}(A)$.
• Such element corresponds one-to-one with representation if $(A,u)$ is a universal element, i.e. for every pair $(B,v)$ with $v \in \mathcal{F}(B)$, there exists a unique morphism $f: A \to B$ such that $\mathcal{F}(f)(u) = v$.
• Consider the level zero deformation functor $\mathcal{F}_0: \mathcal{C} \to (\mathrm{Set})$. This is represented by $A_0:=W[[X_1, \ldots, X_n]]$ with some natural transformation. The corresponding element would in $\mathcal{F}(A)$ would be the universal deformation.
• We must ask how morphisms interact. Given $f: A \to B$, the deformation $\Sigma$ over $B$ becomes a deformation of $A$ by applying $f$, i.e. if $f=id_A$, then $\mathcal{F}(id_A) = id_{\mathcal{F}(A)}$ as we all know (as we need to prove that $\mathcal{F}$ is actually a functor).
• Let $(A, \Phi)$ and $(A, \Phi')$ be representations, i.e. both $(A, u)$ and $(A, v)$ be universal objects with $u:=\Phi_A(id_A)$ and $v:=\Phi'_A(id_A)$. Let $f: A \to A$ and $g: A \to A$ with $\mathcal{F}(f)(u) = v$ and $\mathcal{F}(g)(v) = u$. By the universal property, $f$ and $g$ are inverses of each other, i.e. $f$ is an automorphism.
• The problem boils down to $f: A \to A$ automorphism, then $\Sigma \cong \Sigma \otimes_{A,f} A$.
• If my understanding is correct, then given $F$ an formal $\mathcal{O}_K$-module over $A$,
  $$F(X,Y) = \sum a_{ij} X^i Y^j$$
  and if $G = F \otimes_A B$ with the $A$-algebra structure morphism $f:A \to B$
  $$G(X, Y) = \sum f(a_{ij}) X^i Y^j \in B[[X,Y]]$$
• Suppose $f$ is an isomorphism with $g$ as its inverse. Can you construct a morphism $H$ and $H^{-1}$ such that $H : F \to G$ is an isomorphism of formal groups?