Recall that if $f$ is a Frobenius series, and $F_f$ its corresponding Lubin-Tate's formal group law, we can equip $M$ with the addition defined as $a+_{F_f}b=F_f(a,b)(\forall a,b\in M)$. This turns out $(M,+_{F_f})$ is an abelian group. We can further equip $M$ with an $A$-module structure as $(a,z):=[a]_f(z)$ for all $z\in M$. This is well-defined, since
i. $[1]_f(z)=z$.
ii. $[a]_f(z_1+z_2)=[a]_fF_f(z_1,z_2)=F_f([a]_f(z_1),[a]_f(z_2))=[a]_f(z_1)+_{F_f}[a]_f(z_2)$.
iii. $[ab]_f(z)=[a]_f\circ [b]_f(z)$. This follows from Proposition 2.8.
Let $g$ be another Frobenius series, recall that for all $a\in A$, we can construct the map $[a]_{g,f}:F_f\to F_g$. This induces a homomorphism of abelian group
$$[a]_{g,f}: (M,+_{F_f})\to (M,+_{F_g})$$
Remark 3.1. The homomorphism $[a]_{g,f}$ defined above is also an $A$-module homomorphism.
Proof. It is sufficient to prove that $[a]_{g,f}\circ [b]_f(z)=[b]_g\circ [a]_{g,f}$. The both power series have the same linear term $abX$. Also,
$$g\circ [a]_{g,f}\circ [b]_f=[a]_{g,f}\circ f\circ [b]_f=[a]_{g,f}\circ [b]_f\circ f$$
And furthermore,
$$g\circ [b]_g\circ[a]_{g,f}=[b]_g\circ g\circ [a]_{g,f}=[b]_g\circ [a]_{g,f}\circ f$$
So, the statement now follows by the uniqueness of Lemma 2.3. (Q.E.D)
Via this remark, one can see for any $a\in A, [a]_f:(M,+_{F_f})\to (M,+_{F_f})$ is an $A$-module homomorphism. And hence $\ker [a]_f$ is an $A$-submodule of $(M,+_{F_f})$. Let $a:=\pi^n$, we obtain $\mathscr{F}_n:=\ker [\pi^n]_f=\{z\in M|[\pi^n]_f(z)=0\}$. Using the uniqueness of Lemma 2.3 again, we can see that $[\pi]_f=f, [\pi^2]_f=f\circ f,...,[\pi^n]_f=f\circ f\circ...\circ f \text{ (n terms)}$.
Remark 3.2. $\mathscr{F}_n$ has a structure of $\mathscr{O}_K/\pi^n\mathscr{O}_K$-module, and we have a increasing sequence of $A$-modules
$$\mathscr{F}_1\subset \mathscr{F}_2\subset...\subset \mathscr{F}_n$$
And hence, the increasing of field extensions
$$K\subset K_1:=K(\mathscr{F}_1)\subset ...\subset K_n:=K(\mathscr{F}_n)\subset...\subset K_{\infty}:= \cup_{n\ge 1}K_n$$
Such sequence of field extensions is called Lubin-Tate's tower.
Proof. Assume that $a=b+c\pi^n$, for $a,b,c\in A$, we have for all $z\in \mathscr{F}_n$,
$$[a]_f(z)=[b+c\pi^n]_f(z)=[b]_f(z)+_{F_f}[c\pi^n]_f(z)=[b]_f(z)+_{F_f}[c]_f\circ [\pi^n]_f(z)=[b]_f(z)$$
And this yields $\mathscr{F}_n$ has a $\mathscr{O}_K/\pi^n\mathscr{O}_K$-module structure. The increasing sequences are easily obtain by the fact $f\in XA[[X]]$ and
$$[\pi^n]_f=f\circ f...\circ f \text{ (n terms)}$$
(Q.E.D)
From now on, we will deduce properties of Lubin-Tate tower. One can see that $\mathscr{F}_n$ is obviously dependent on the choice of $f$, but we will prove that it is not the case for $K_n$.
Lemma 3.3. Let $g$ be another Frobenius series with $\mathscr{F'}_n=\ker [\pi^n]_g$, then $K(\mathscr{F}_n)=K(\mathscr{F'}_n)$.
Proof. Choose any $u\in A^\times$ with $[u]_{g,f}:F_f\xrightarrow{\sim} F_g$ is an isomorphism. This induces an isomorphism of $A$-module $(M,+_{F_f})\xrightarrow{\sim} F_g$. And also, it induces an $A/\pi^nA$-isomorphism $\mathscr{F}_n\xrightarrow{\sim}\mathscr{F'}_n$. Hence, in particular, we get $z\in \mathscr{F}_n$ iff $[u]_{g,f}(z)\in \mathscr{F'}_n$. But then, since $z\in M$, $[u]_{g,f}(z)$ converges in $K(z)$, we obtain $K(\mathscr{F'}_n)\subset K(\mathscr{F}_n)$. By symmetry, we obtain $K(\mathscr{F}_n)=K(\mathscr{F'}_n)$. (Q.E.D)
Via this proof, we can see the explore the algebraic properties of the $A/\pi^nA$-module $\mathscr{F}_n$, it is sufficient for us to choose a simple Frobenis series, $f(X):=\pi X+X^q$.
Lemma 3.3. With $f$ is chosen as above, the map $[\pi]_f: \mathscr{F}_n\to \mathscr{F}_{n-1}$ sending $z\mapsto [\pi]_f(z)$ is a surjective homomorphism of $A$-module, and its kernel is $\mathscr{F}_1$.
Proof. One can see easily that $[\pi]_f$ is a well-defined homomorphism. Take any $z_{n-1}\in \mathscr{F}_{n-1}$, we want to find $z_n\in \mathscr{F}_n$, such that $[\pi]_f(z_n)=z_{n-1}$. One can see the equation $\pi X+X^q=z_{n-1}$ always has solutions in $\overline{K}$, and since $v_K(z_{n-1})>0$, such a solution $z_n$ also lie in $M$. And we have $[\pi]_f(z_n)=z_{n-1}$. This yields $[\pi^n]_f(z_n)=0$, i.e. $z_n\in \mathscr{F}_n$. And hence, $[\pi]_f$ is a surjective homomorphism. The kernel of $[\pi]_f$ now directly follows. (Q.E.D)
For the main results of this section, we need the following
Lemma 3.4. Let $z\in M$, then the polynomial $g(X)=z+\pi X+X^q$ has distinct roots in $\overline{K}$.
Proof. We have $g'(X)=\pi+qX^{q-1}$, which is $\pi$ when $char(K) = p$. Hence, the statement is obviously true when $char(K)>0$. Assume for now $char(K)=0$, and that there exists some $x\in \overline{K}$, with $g(x)=g'(x)=0$, then $x^{q-1}=-\pi/q$. This yields $|x| \ge 1$, because $|\pi/q|\ge 1$. This yields $|\pi x|<|x|\le |x|^q$, which yields $|z| = |\pi x+x^q|\ge 1$. It is a contradiction, since $v_p(z)>0$. (Q.E.D)
We are now ready the for the an important result
Proposition 3.5. $\mathscr{F}_n$ is a free $A/\pi^nA$-module of rank $1$. This implies $Aut_{A/\pi^nA}(\mathscr{F}_n)\cong (A/\pi^nA)^\times$.
Proof. We will prove this fact the induction. When $n=1$, the equation $f(X)=\pi X+X^q=X(\pi + X^{q-1})$ has $q$-distinct roots in $\overline{K}$ (Lemma 3.4), and $\mathscr{F}_1$ has a structure of $A/\pi A$-vector space structure. This yields $\mathscr{F}_1$ is a 1-dimensional $A/\pi A$-vector space.
Assume that the statement holds to $n-1(n\ge 2)$, then there exists $z_{n-1}$, the generator of $\mathscr{F}_{n-1}$, and an isomorphism $\phi_{n-1}: A/\pi^{n-1}A\xrightarrow{\sim} \mathscr{F}_{n-1}$ defined as $a\mapsto [a]_f(z_{n-1})$. By using Lemma 3.3, there exists $z_n\in \mathscr{F}_n$, with $[\pi]_f(z_n)=z_{n-1}$. Also, the map $\phi_n: A/\pi^nA\xrightarrow F_{n}$ defined as $a\mapsto [a]_f(z_n)$ making the following diagram commute:
where rows are exact, with the first and the last vertical arrows are isomorphisms. This yields the arrow in the middle is an isomorphism, too. Hence, $\mathscr{F}_n$ is a free $A/\pi^nA$-module of rank 1. The later statement is now clear. (Q.E.D)
Because $\#A/\pi^nA=q^n$, we have $\#\mathscr{F}_n=q^n$. And hence $[K_n:K]<+\infty$. And we conclude this section by the following
Theorem 3.6. The Lubin-Tate's tower
$$K\subset K_1\subset...\subset K_n$$
is a tower of totally ramified Galois extension, with $[K_n:K]=q^{n-1}(q-1)$. Moreover, if $z_n$ is a generator for $\mathscr{F}_n$ as $A/\pi^nA$-module, then $z_n$ is a uniformizer for $K_n$. And that $Gal(K_{\infty}/K)\cong A^\times$.
Proof. One can see that $K_1=K(\mathscr{F}_1)$, i.e. $K_1$ is obtained by adjoining roots of the polynomial $f(X)=\pi X+X^q$, which is separable. Hence, $K_1/K$ is Galois of degree $q-1$. If $z_1\ne 0$ is a root of $f(X)$, we can see that $z_1$ is a root of $g(X):=\pi+X^{q-1}$, which is an Eisentein polynomial. Hence, $z_1$ is a uniformizer for $K_1$. For $n\ge 2$, assume that the statements hold for $n-1$, we can see $K(\mathscr{F}_n)$ is an extension of $K(\mathscr{F}_{n-1})$ by adjoining roots of the polynomial $\pi X+X^q=z_{n-1}$, for all $z_{n-1}$: generator of $\mathscr{F}_{n-1}$ as $A/\pi^{n-1}A$-module. For such $z_{n-1}$, the polynomial $g(X):=-z_{n-1}+\pi X+X^q$ is Eisentein of degree $q$ over $K_{n-1}$ (since $z_{n-1}$ is a uniformizer for $K_{n-1}$), and by Lemma 3.4, $g(X)$ is separable. This implies $K_n/K_{n-1}$ is totally ramified Galois extension of degree at least $q$. Hence, one obtains $[K_n:K]\ge q^{n-1}(q-1)$.
It is the time to see how Galois actions play a role. For any $\sigma \in Gal(\overline{K}/K)$, because $\sigma$ acts as identity map in $K$, $\sigma$ acts on $(M,+_{F_f})$ as an $A$-module isomorphism, since for all $z, z_1,z_2\in M$, $\sigma([a]_f(z))=[a]_f(\sigma(z))$, and also $\sigma F_f(z_1,z_2)=F_f(\sigma(z_1),\sigma(z_2))$. From this, $\sigma$ induces an $A/\pi^nA$-module automorphism on $\mathscr{F}_n$. This yields by Proposition 3.5, for each $\sigma$, there exists only one $\phi_\sigma\in Aut_{A/\pi^nA}(\mathscr{F}_n)$, such that $\sigma(z)=\phi_\sigma(z)$, for all $z\in \mathscr{F}_n$. And hence, one obtain an embedding from $Gal(K_n/K)$ to $Aut_{A/\pi^nA}(\mathscr{F}_n)$.
By Proposition 3.5 again, this yields $\#Gal(K_n/K)\le q^{n-1}(q-1)$. It follows by our previous argument that $Gal(K_n/K)\cong (A/\pi^nA)^\times$, and that $[K_n:K]=q^{n-1}(q-1)$, and that $K_n/K_{n-1}$ is totally ramified, and $z_n$ is a uniformizer of $K_n$ since the polynomial $g(X)$ defined above is Eisentein. With this result at hand, we obtain
$$Gal(K_{\infty}/K)=\varprojlim Gal(K_n/K)\cong \varprojlim (A/\pi^nA)^\times \cong A^\times$$
(Q.E.D)
Remark 3.7. One can see that Lubin-Tate construction basically gives us the 1-dimensional representation of the absolute Galois group. It is very similar to the 2-dimensional representation obtained by using Tate modules on elliptic curves.
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