[proofplan]
We apply the Lifting Lemma twice. First, with $n = 2$ and $L(X, Y) = X + Y$, the Lifting Lemma produces a unique $F_e(X, Y)$ satisfying $F_e \equiv X + Y \pmod{(X, Y)^2}$ and $e(F_e(X, Y)) = F_e(e(X), e(Y))$. We then verify the formal group axioms (associativity, commutativity, identity) using the uniqueness clause. Second, applying the Lifting Lemma with $n = 1$ and $L(X) = aX$ yields $[a]_{e_1, e_2}$. The formal $\mathcal{O}_K$-module structure follows from uniqueness: the identities $[a + b]_e = F_e([a]_e, [b]_e)$, $[ab]_e = [a]_e \circ [b]_e$, and $[1]_e = X$ all hold because both sides satisfy the same lifting problem with the same linear part.
[/proofplan]
[step:Construct $F_e(X, Y)$ via the Lifting Lemma]
Apply the [Lifting Lemma for Lubin--Tate Series](/theorems/2391) with $e_1 = e_2 = e$, $n = 2$, and $L(X, Y) = X + Y$. The lemma produces a unique power series $F_e(X, Y) \in \mathcal{O}_K[[X, Y]]$ satisfying
\begin{align*}
F_e(X, Y) &\equiv X + Y \pmod{(X, Y)^2}, \\
e(F_e(X, Y)) &= F_e(e(X), e(Y)).
\end{align*}
[/step]
[step:Verify associativity and commutativity using uniqueness]
**Commutativity.** The power series $F_e(Y, X)$ satisfies $F_e(Y, X) \equiv Y + X = X + Y \pmod{(X, Y)^2}$ and
\begin{align*}
e(F_e(Y, X)) = F_e(e(Y), e(X)),
\end{align*}
which is the same intertwining relation (with arguments permuted). By uniqueness in the Lifting Lemma, $F_e(Y, X) = F_e(X, Y)$.
**Associativity.** Consider $A(X, Y, Z) = F_e(F_e(X, Y), Z)$ and $B(X, Y, Z) = F_e(X, F_e(Y, Z))$. Both have linear part $X + Y + Z$ modulo degree $2$. For the intertwining relation, using $e(F_e(X, Y)) = F_e(e(X), e(Y))$ twice:
\begin{align*}
e(A) &= e(F_e(F_e(X, Y), Z)) = F_e(e(F_e(X, Y)), e(Z)) = F_e(F_e(e(X), e(Y)), e(Z)) = A(e(X), e(Y), e(Z)).
\end{align*}
Similarly $e(B) = B(e(X), e(Y), e(Z))$. Since $A$ and $B$ have the same linear part $X + Y + Z$ and satisfy the same functional equation $e_1 \circ G = G \circ (e_2, e_2, e_2)$ with $e_1 = e_2 = e$, uniqueness in the Lifting Lemma (applied with $n = 3$) gives $A = B$.
[guided]
The strategy is always the same: to prove two power series are equal, verify that both satisfy the same lifting problem — same linear part, same intertwining relation — and invoke uniqueness. This is the central technique in the Lubin--Tate construction, and it will appear repeatedly.
For commutativity, the point is that swapping $X$ and $Y$ in $F_e$ does not change the linear part ($X + Y$ is symmetric) and does not break the intertwining relation (since $e$ is applied to each variable independently). Uniqueness forces $F_e(X, Y) = F_e(Y, X)$.
For associativity, both $F_e(F_e(X, Y), Z)$ and $F_e(X, F_e(Y, Z))$ have linear part $X + Y + Z$. The intertwining computation for $A = F_e(F_e(X, Y), Z)$ uses the intertwining property of $F_e$ twice:
\begin{align*}
e(A) &= e(F_e(F_e(X, Y), Z)) \\
&= F_e(e(F_e(X, Y)), e(Z)) \quad \text{(intertwining applied to the outer } F_e\text{)} \\
&= F_e(F_e(e(X), e(Y)), e(Z)) \quad \text{(intertwining applied to the inner } F_e\text{)} \\
&= A(e(X), e(Y), e(Z)).
\end{align*}
The identical computation works for $B$. The Lifting Lemma with $n = 3$ and $L(X, Y, Z) = X + Y + Z$ gives uniqueness, so $A = B$.
[/guided]
[/step]
[step:Construct $[a]_{e_1, e_2}$ for each $a \in \mathcal{O}_K$ via the Lifting Lemma]
For $e_1, e_2 \in \mathcal{E}_\pi$ and $a \in \mathcal{O}_K$, apply the [Lifting Lemma](/theorems/2391) with $n = 1$ and $L(X) = aX$. This yields a unique $[a]_{e_1, e_2}(X) \in \mathcal{O}_K[[X]]$ satisfying
\begin{align*}
[a]_{e_1, e_2}(X) &\equiv aX \pmod{X^2}, \\
e_1([a]_{e_1, e_2}(X)) &= [a]_{e_1, e_2}(e_2(X)).
\end{align*}
When $e_1 = e_2 = e$, write $[a]_e = [a]_{e, e}$.
[/step]
[step:Verify the formal $\mathcal{O}_K$-module axioms using uniqueness]
We must show that $a \mapsto [a]_e$ is a ring homomorphism $\mathcal{O}_K \to \operatorname{End}(F_e)$.
**Additivity: $[a + b]_e = F_e([a]_e, [b]_e)$.** Both sides have linear part $(a + b)X$ modulo $X^2$. For the intertwining relation:
\begin{align*}
e(F_e([a]_e(X), [b]_e(X))) &= F_e(e([a]_e(X)), e([b]_e(X))) \\
&= F_e([a]_e(e(X)), [b]_e(e(X))).
\end{align*}
This shows $G(X) := F_e([a]_e(X), [b]_e(X))$ satisfies $e(G(X)) = G(e(X))$ with $G(X) \equiv (a+b)X \pmod{X^2}$. By uniqueness, $G = [a + b]_e$.
**Multiplicativity: $[ab]_e = [a]_e \circ [b]_e$.** The composition $[a]_e([b]_e(X))$ has linear part $a \cdot bX = abX \pmod{X^2}$. For intertwining:
\begin{align*}
e([a]_e([b]_e(X))) = [a]_e(e([b]_e(X))) = [a]_e([b]_e(e(X))).
\end{align*}
Both sides of the proposed identity $[ab]_e = [a]_e \circ [b]_e$ satisfy the same lifting problem with linear part $abX$, so they agree by uniqueness.
**Unit: $[1]_e = X$.** The identity $X$ has linear part $X$ and satisfies $e(X) = e(X)$, which is the intertwining relation for $[1]_e$. By uniqueness, $[1]_e(X) = X$.
**Endomorphism property: $[a]_e(F_e(X, Y)) = F_e([a]_e(X), [a]_e(Y))$.** Both sides, viewed as elements of $\mathcal{O}_K[[X, Y]]$, have linear part $a(X + Y)$ and satisfy the intertwining relation with $e$ (verified by the same chain-of-substitutions argument). By uniqueness in the Lifting Lemma with $n = 2$, they are equal. This confirms $[a]_e \in \operatorname{End}(F_e)$.
Together, these identities show that $a \mapsto [a]_e$ is a ring homomorphism $\mathcal{O}_K \to \operatorname{End}(F_e)$, making $F_e$ a formal $\mathcal{O}_K$-module.
[guided]
The remarkable feature of the Lubin--Tate construction is that every algebraic identity one needs follows from a single principle: the uniqueness clause of the Lifting Lemma. To show two power series $P$ and $Q$ are equal, one checks:
1. They have the same linear part.
2. They both satisfy the intertwining relation $e \circ G = G \circ e$.
Let us verify the endomorphism property in detail, since it is the least immediate. We need $[a]_e(F_e(X, Y)) = F_e([a]_e(X), [a]_e(Y))$ in $\mathcal{O}_K[[X, Y]]$. Set $P(X, Y) = [a]_e(F_e(X, Y))$ and $Q(X, Y) = F_e([a]_e(X), [a]_e(Y))$.
**Linear parts:** $P(X, Y) \equiv a(X + Y) \pmod{(X,Y)^2}$ since $[a]_e(Z) \equiv aZ$ and $F_e(X,Y) \equiv X + Y$. Similarly $Q(X, Y) \equiv F_e(aX, aY) \equiv aX + aY \pmod{(X,Y)^2}$. So both have linear part $a(X + Y)$.
**Intertwining for $P$:** $e(P) = e([a]_e(F_e(X,Y))) = [a]_e(e(F_e(X,Y))) = [a]_e(F_e(e(X), e(Y))) = P(e(X), e(Y))$.
**Intertwining for $Q$:** $e(Q) = e(F_e([a]_e(X), [a]_e(Y))) = F_e(e([a]_e(X)), e([a]_e(Y))) = F_e([a]_e(e(X)), [a]_e(e(Y))) = Q(e(X), e(Y))$.
Both $P$ and $Q$ satisfy the lifting problem with $n = 2$ and $L(X,Y) = a(X+Y)$. By uniqueness, $P = Q$.
[/guided]
[/step]
