Let us first fix some notations: a local field $K$, with its residue field $k_K$, and $\#k_K=q$, and $p$ is the characteristic of $k_K$. Its ring of integers $A:=\mathscr{O}_K$ is a D.V.R with its unique maximal ideal $M_K$ generated by $\pi:=\pi_K$.
Definition. Let $f\in A[[X]]$ be a formal power series, then $f$ is said to be a Frobenius series if
(i) $f(X)=\pi X\mod\deg 2$.
(ii) $f(X)\equiv X^q\mod \pi$.
Example 2.1. $f(X):=\pi X+X^q$ is a Frobenius series. Also, when $K=\mathbb{Q}_p$, and $\pi=p$, then $f(x):=(1+X)^p-1$ is a Frobenius series.
Let us warm-up with the following
Lemma 2.2. Let $f,g$ be two Frobenius series, and $F(X)\in F[[X_1,...,X_n]]$ be a formal power series in $n$-variables, then $f\circ F\equiv F(g,...,g)\mod \pi$.
Proof. We have $f\circ F(X_1,...,X_n)\equiv F(X_1,...,X_n)^q\mod \pi$, and $F(g(X_1),...,g(X_n))\equiv F(X_1^q,...,X_n^q)\mod \pi$. And we can easily see that $F(X_1,...,X_n)^q\equiv F(X_1^q,...,X_n^q)\mod \pi$. (Q.E.D)
Using this, we can prove the key lemma for this section
Lemma 2.3. Let $f,g$ be two Frobenius series, and $\psi(X_1,...,X_n):=a_1X_1+...+a_nX_n$ a linear form in $A[X_1,...,X_n]$. Then there exists a unique $F\in A[X_1,...,X_n]$ such that
(i) $F=F_1\mod\deg 2$
(ii) $f\circ F=F(g,...,g)$
Proof. We will construct $F$ from polynomials in $A[X_1,...,X_n]$ by reduction with these conditions for all $r\ge 0$
(1)$F_r\in A[X_1,...,X_n]$ is a polynomial of degree r.
(2) $f\circ F_r=F_r(g,...,g)\mod \deg r+1$.
(3) $F_{r+1}=F_r+E_{r+1}$, where $E_{r+1}$ is a homogeneous polynomial of degree $r+1$ in $A[X_1,...,X_n]$.
Assume that such $F_r$ are constructed, we let $F:=F_r+E_{r+1}+E_{r+2}+...$. Then it can be seen for all $r$
$$f(F(X_1,...,X_n))=f(F_r+\text{terms of degree}\ge r+1)=f(F_r)\mod \deg (r+1)$$
And because of condition (2) and (1), we have
$$f\circ F=F_r(g,...,g)\mod\deg (r+1)=F(g,...,g)\mod \deg (r+1)$$
So, we get $f\circ F=F(g,...,g)$. Hence, it is sufficient for us to construct $F_r$. First, one can see that $F_1$ is exactly $\psi$, due to the condition (i), and the condition (ii) is also satisfied since
$$f(\psi(X_1,...,X_n))=\pi(a_1X_1+...+a_nX_n)\mod\deg 2$$
And also
$$\psi(g(X_1),...,g(X_n))=a_1g(X_1)+...+a_ng(X_n)\equiv \pi(a_1X_1+...+a_nX_n)\mod \deg 2$$
And this follows that $F_1=\psi$. Now, assume that we already constructed $F_r$, and we want to construct $F_{r+1}$. Let $F_{r+1}=F_r+E_{r+1}$. We have
$$f(F_r+E_{r+1})=f(F_r)+\pi E_{r+1}\mod \deg (r+2)$$
And
$$F_{r+1}(g,...,g)=F_r(g,...,g)+E_{r+1}(g,...,g)=$$
$$=F_r(g,...,g)+E_{r+1}(\pi X_1,...,\pi X_n)\mod \deg (r+2)$$
$$=F_r(g,...,g)+\pi^{r+1}E_{r+1}(X_1,...,X_n)\mod \deg (r+2)$$
The last equality follows since $E_{r+1}$ is homogeneous of degree $r+1$. The condition (2) for $F_r$ implies that $f\circ F_r=F_r(g,...,g)$. And we want
$$f\circ F_{r+1}=F_{r+1}(g,...,g)$$
And this is equivalent to say
$$E_{r+1}=\frac{F_r(g,...,g)-f\circ F_r}{\pi(1-\pi^r)}$$
But in Lemma 2.2, we have prove that $\pi | (F_r(g,...,g)-f\circ F_r)$, and $(1-\pi^r)\in A^\times$. So, we can construct $E_{r+1}$ by this formula, and hence $F_{r+1}$. Now, the uniqueness of $F$ follows easily from this construction. (Q.E.D)
The latter development will be applications of Lemma 2.3. The first one is
Corollary 2.4. Let $f$ be a Frobenius series, the there exists a unique commutative formal group law $F_f$ such that $f\in End(F_f)$.
Proof. Based on Lemma 2.3, there exists a unique formal power series $F\in A[[X,Y]]$ such that
(i)$F(X,Y)=X+Y\mod\deg 2$.
(ii)$f\circ F=F(f,f)$
And we need to check that in fact $F$ is a commutative formal group law. Because the axiom about inverse is automatically true if the two axioms of associativity and $F(X,Y)=X+Y\mod \deg 2$. So we just need to check two things.
(1) $F(X,Y)=F(Y,X)$. Let $G_1=F(X,Y), G_2=F(Y,X)$, then $G_i=X+Y\mod\deg 2$, and also $f\circ G_i=G_i(f, f)$. So by the uniqueness in Lemma 2.3, we get $G_1=G_2$.
(2) (Associativity) Let $G_1(X,Y,Z)=F(X,F(Y,Z)), G_2(X,Y,Z)=F(F(X,Y),Z)$, then $G_i=X+Y+Z\mod\deg 2$, and $f\circ G_i=G_i(f,f,f)$. So, by the uniqueness of Lemma 2.3 again, we get $G_1=G_2$.
This follows directly that $F$ is commutative formal group law. (Q.E.D)
Definition. Let $f$ be a Frobenius series, then such a commutative formal group law $F_f$ in Corollary 2.4 is called a Lubin-Tate's formal group law.
Example 2.5. Let $K=\mathbb{Q}_p, \pi=p, f(X)=(1+X)^p-1$ is a Frobenius series, then $F(X,Y):=X+Y+XY=(1+X)(1+Y)-1$ is a Lubin-Tate formal group law of $f$, as in Example 1.5 presented.
With the help of Lemma 2.3, we can now easily construct homomorphism between two Lubin-Tate's formal group laws. Let $f,g$ be two Frobenius series, then by Lemma 2.3, there exists a unique $[a]_{g,f}\in A[[X]]$ such that $[a]_{g,f}=aX\mod\deg 2$ and $g\circ [a]_{g,f}=[a]_{g,f}\circ f$.
Proposition 2.6. Such an $[a]_{g,f}$ defined above is a homomorphism from $F_f$ to $F_g$.
Proof. Let $h:=[a]_{g,f}=aX\mod\deg 2$, then we want to show $h\circ F_f=F_g(h,h)$. Let $H_1:=h\circ F_f$, and $H_2:=F_g(h,h)$. Then one can see both $H_1,H_2$ have linear term as $aX+aY$. Also,
$$g\circ H_1=g\circ h\circ F_f=h\circ f\circ F_f=h\circ F(f,f)=H_1(f,f)$$
And
$$g\circ H_2=g\circ F_g(h,h)=F_g(g\circ h,g\circ h)=F_g(h\circ f,h\circ f)=H_2(f,f)$$
So, by the uniqueness of Lemma 2.3, we get $H_1=H_2$. This yields $[a]_{g,f}$ is a homomorphism from $F_f$ to $F_g$. (Q.E.D)
Here is a nice corollary of the proposition above.
Corollary 2.7. Let $f,g$ be two Frobenius series, then $F_f\cong F_g$.
Proof. One can see $a\in A^\times$, then $[a]_{g,f}=aX\mod\deg 2$ defines an isomorphism between $F_f$ and $F_g$, as presented in Lemma 1.4. (Q.E.D)
Proposition 2.8. Let $f$ be a Frobenius series, $a,b\in A$, then
$$[ab]_{f,f}=[a]_{f,f}[b]_{f,f}=[ba]_{f,f}=[b]_{f,f}[a]_{f,f}$$
Proof. We can see that the four element above (in the ring $End(F_f)$) have the same linear term $abX$. We have $f\circ [ab]_{f,f}=[ab]_{f,f}\circ f$, and
$$f\circ [a]_{f,f}\circ [b]_{f,f}=[a]_{f,f}\circ f\circ [b]_{f,f}=[a]_{f,f}\circ [b]_{f,f}\circ f$$
Also, $f\circ [ba]_{f,f}=[ba]_{f,f}\circ f$. So, by the uniqueness of Lemma 2.3, we have $[ab]_{f,f}=[a]_{f,f}b_{f,f}=[ba]_{f,f}=[b]_{f,f}[a]_{f,f}$. The last equality follows by interchanging $a,b$ in the first equality. (Q.E.D)
We are now ready for the Lubin-Tate theory. For convenience, from now on, we can denote $[a]_{f,f}=[a]_f$.
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