1. Formal group law and their homomorphisms. We always fix $A$ a commutative ring, let us warm up with the useful statement for the ring of power series of one variable $A[[X]]$.
Lemma 1.1. Let $f=a_1X+a_2X^2+...\in A[[X]]$, then there exists $g(X)\in XA[[X]]$ such that $f\circ g(X) = X$ iff $a_1\in A^\times$. Also, if such $g$ exists, then it is unique, and $f\circ g(X)=g\circ f(X) = X$.
Proof. Let $g(X) = b_1X+b_2X^2+...\in A[[X]]$. We then have
$$f(g(X))=a_1(b_1X+b_2X^2+...)+a_2(b_1X+b_2X^2+...)^2+...=$$
$$=(a_1b_1)X+(a_1b_2+a_2b_1)X^2+...$$
Then $f(g(X))=X$ iff $a_1b_1=1, a_1b_2+a_2b_1=0,...$. Hence, $f\circ g(X)=X$ iff $a_1\in A^\times$. The uniqueness of $g(X)$ directly follows from this.
Assume that $a_1\in A^\times$, and $g(X)$ is constructed as above. Because $b_1=(a_1)^{-1}\in A^\times$, we can construct $h(X)\in XA[[X]]$ such that $g\circ h(X)= X$, and hence
$$h(X) = (f\circ g)\circ h(X)=f\circ(g\circ h)(X)=f(X)$$
Hence, $f\circ g(X)=g\circ f(X) = X$. (Q.E.D)
We are now ready for the definition of formal group.
Definition. Let $A[[X,Y]]$ be the ring of formal power series ring of two variables, $F(X,Y)\in A[[X,Y]]$, then $F$ is said to be a commutative formal group law if:
(i) $F(X,Y)=X+Y+$(terms of degree $\ge 2$).
(ii) $F(X,F(Y,Z))=F(F(X,Y),Z)$.
(iii) There exists only one $i_F\in A[[X]]$ such that $F(X,i_F(X))=F(Y,i_F(Y))=0$.
(iv) $F(X,Y)=F(Y,X)$.
For convenience, we will often denote "terms of degree $\ge n$" as $\mod \deg n$. Assume that $F$ is a commutative formal group law, denote $f(X):=F(X,0)=X\mod \deg 2$, then by the definition, we have
$$F(0,F(X,0))=F(F(X,0),0)=f\circ f(X)$$
And $F(F(X,0),0)=F(X, F(0,0))=F(X,0)=f(X)$, we obtain $f\circ f=f$. This follows from Lemma 1 that there exists a unique $g\in XA[[X]]$ such that $f\circ g=g\circ f=X$. Hence,
$$f(X)=f\circ (f\circ g(X))=(f\circ f)\circ g(X)=(f\circ g)(X)=X$$
And this yields $F(X,0)=X$. By symmetry, we also have $F(0,Y)=Y$. And this yields any commutative formal group law is of the form
$$F(X,Y)=X+Y+\sum_{1\le i,j}a_ib_jX^iY^j$$
We hence have proved that, in fact, the formula above is equivalent to the condition $F(X,0)=X, F(Y,0)=Y$ for any commutative formal group law. Also, by comparing coefficients, we can see that (i) and (ii) also imply (iii). That means, such a power series $i_F$ is unique.
Example 1.2. Let $F(X,Y):=X+Y$, then it can be easily checked that $F$ defines a commutative formal group law. Similarly, $F(X,Y):=X+Y+XY$ also defines a commutative formal group law.
Important remark 1.3. Let $f,g,h\in XA[[X]]$, then the conditions of commutative formal group law imply that
(i) $F(f,0)=F(0,f)=f$.
(ii) $F(f,F(g,h))=F(F(f,g),h)$.
(iii) $F(f, i_F(f))=0$.
(iv) $F(f,g)=F(g,f)$.
That means, for all $f,g\in XA[[X]]$, $f+_Fg:=F(f,g)$ defines a group law on $XA[[X]]$.
We are now ready to define homomorphisms between formal group laws.
Definition. Let $F, G\in A[[X,Y]]$ be two formal group laws, then a power series $h(X)\in A[[X]]$ is said to be a homomorphism from $F$ to $G$ (say, a homomorphism $h: F\to G$) if
$$f(F(X,Y))=G(f(X),f(Y))$$
$h$ is said to be an isomorphism if there exists a homomorphism $h': G\to F$ and $h\circ h'(X)=h'\circ h(X) = X$.
Based on Lemma 1.1, there is a useful characterization of isomorphisms between formal group laws.
Lemma 1.4. Let $F, G, h$ be as above, then $h$ is an isomorphism iff $h(X)=a_1X\mod\deg 2$, with $a_1\in A^\times$.
Proof. One can see by Lemma 1.1 that there exists $h'\in A[[X]]$ such that $h\circ h'=h'\circ h=X$ iff $a_1\in A^\times$. And in this case, we have
$$h'(G(X,Y))=h'(G(h\circ h'(X), h\circ h'(Y)))=(h'\circ h)\circ F(h'(X), h'(Y))=F(h'(X),h'(Y))$$
This proves that $h':G\to F$ is also a homomorphism. And this finishes our proof. (Q.E.D)
Example 1.5. Let $F(X,Y):=X+Y+XY=(1+X)(1+Y)-1$ be a commutative formal group law on Example 1.2, for a prime number $p$, we can define $h(X)=(1+X)^p-1$, then
$$h(F(X,Y))=(1+F(X,Y))^p-1=(1+X)^p(1+Y)^p-1$$
And
$$F(h(X),h(Y))=F((1+X)^p-1,(1+Y)^p-1)=(1+X)^p(1+Y)^p-1$$
This yields $h:F\to F$ is a homomorphism.
The example above is very important for your later development of Lubin-Tate's theory. We will conclude this section by the following about the endomorphism ring of formal group law
Proposition 1.6. Let $F$ be a commutative formal group law, then
$$End(F)=\{f:F\to F|f \text{ is a homomorphism}\}$$
forms a ring, with addition $+_F, \circ_F$ is defined as $f+_Fg:=F(f,g)$, and $f\circ_F g:=f\circ g$.
Proof. The proof of the proposition above is not difficult, but slightly long, with repetition steps.
Step 0. We easily see that $id:F\to F$ defined as $id\circ F=F$, and $0: F\to F$ defined as $0\circ F=0$ are certainly in $End(F)$.
Step 1. Let $f, g\in End(F)$, we have
$$f\circ g\circ F(X,Y)=F(f\circ g(X),f\circ g(Y))$$
This yields $f\circ g\in End(F)$, with $f\circ id=id\circ f=f$.
Step 2. Let $f,g,h\in End(E)$, or more generally, with $f,g,h\in XA[[X]]$, we can easily see that $(f\circ g)\circ h=f\circ (g\circ h)$.
Step 3. We first let $Z:=F(i_F(X),i_F(Y))$, we have
$$F(Y,Z)=F(Y,F(i_F(X),i_F(Y)))=F(F(Y,i_F(Y)),i_F(X))=F(i_F(X),0)=i_F(X)$$
And from this,
$$F(F(X,Y),Z)=F(X,F(Y,Z))=F(X,i_F(X))=0$$
Also, we have $F(F(X,Y),i_F(F(X,Y)))=0$. Because of the uniqueness of $i_F$, we get
$$i_F(F(X,Y))=F(i_F(X),i_F(Y))$$
This follows that $i_F\in End(F)$.
Step 4. Let $f,g\in End(F)$, we can define $h(X):=F(f(X),g(X))=f+_Fg$. Then,
$$h(F(X,Y))=F(f(F(X,Y)),g(F(X,Y)))=F(F(f(X),f(Y)),F(g(X),g(Y)))$$
Similar to Step 3, we can interchange terms and get
$$h(F(X,Y))=F(F(f(X),g(X)),F(f(Y),g(Y)))=F(h(X),h(Y))$$
And this yields $h\in End(F)$, and it is easy to check that $f+_F0=0+_Ff=f$ and $f+_Fg=g+_Ff$.
Step 5. One can see, by Step 1 and Step 3, $-f:=i_F\circ f\in End(F)$. Also, similar to Step 4, we can see $f+_F(-f)=(-f)+_Ff=0$.
Step 6. We have, for all $f,g,h\in End(F)$
$$e(X):=f\circ(g+_Fh)=f(F(g(X),h(X)))=F(f\circ g,f\circ h)=(f\circ g)+_F(f\circ h)$$
And similarly, $(g+_Fh)\circ f=(g\circ f)+_F(h\circ f)$.
We can now conclude that $End(F)$ is a ring with addition and multiplication laws defined as above. (Q.E.D)
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