用户: Aripriner/Lubin-Tate Theory

We always assume that all rings are commutative and have the unit 1. My main reference book is [J.S.Milne 2020]. [J.W.S.Cassels–A.Frohlich 1967] is also one classic book I use.

1Local Class Field Theory

We know that the composite of two abelian extensions of a field K is abelian. So we can talk about the maximal abelian extension $K^{ab}$ of K, which is the union of all abelian extensions of K in a fixed algebraic closure $K^{al}$. It is easy to see that its Galois group is the quotient of $Gal(K^{al}/K)$ by the closure of its commutator subgroup.

An example of abelian extension is unramified extension. If L is an unramified extension over K, the Galois group $Gal(L/K)\cong Gal(l/k)$ is a cyclic group, hence abelian. So $Gal(L/K)$ is generated by the Frobenius element $\sigma$ such that $\sigma (x)\equiv x\mathrm{mod}{\mathfrak{m}}_L$ for any $x\in {\mathcal{O}}_L$.

Now we can give the statement of the main theorems of local class field theory:

定理 1.1 (Local Reciprocity Law). For every nonarchimedian local field K, we have a unique homomorphism $$\begin{array}{r}{\varphi }_K:K^{\times }\to Gal(K^{ab}/K)\end{array}$$with the following property:

1.	For every uniformizer $\pi$ of K and any finite unramified extension L over K, ${\varphi }_K(\pi )$ is just the Frobenius element $Fro{b}_{L/K}$ in $Gal(L/K)$.
2.	For any finite abelian extension L over K, the restriction of the image of $N{m}_{L/K}(L^{\times })$ in $Gal(L/K)$ is trival. And we have the isomorphism $$\begin{array}{r}{\varphi }_{L/K}:K^{\times }/N{m}_{L/K}(L^{\times })\to Gal(L/K)\end{array}$$

We call ${\varphi }_K$ and ${\varphi }_{L/K}$ the local Artin maps for K and L/K. And the subgroups of $K^{\times }$ of the form $Nm(L^{\times })$ with L is an abelian finite extension over K are called the norm groups in $K^{\times }$.

定理 1.2 (Local Existence Theorem). These norm subgroups of abelian extensions in $K^{\times }$ are exactly the open subgroups of finite index in $K^{\times }$.

In the view of philosophy, the local class field theory tells us that the structure of all abelian extensions is someway decided by the topological structure of the field itself.

Let us assume first that the theorems are right. I list some property of the norm groups without proofs:

命题 1.3. Let $K$ be a nonarchimedian local field. Then:

1.	The map $L\to Nm(L^{\times })$ gives the bijection from the set of abelian finite extensions over K to the set of the norm groups of K.
2.	$L\subseteq L^{\prime }\iff Nm(L^{\times })\supseteq Nm(L^{\prime \times })$.
3.	$Nm((L\cdot L^{\prime }{)}^{\times })=Nm(L^{\times })\cap Nm(L^{\prime \times })$.
4.	$Nm((L\cap L^{\prime }{)}^{\times })=Nm(L^{\times })Nm(L^{\prime \times })$
5.	Every subgroups of $K$ containing a norm subgroup is a norm subgroup.
6.	For any extension L over K, if $Nm(L^{\times })$ is of finite index in $K^{\times }$, then it is open. Specially, $Nm(L)$ is open if $L$ is an abelian finite extension over $K$.

By the local Artin maps, we get the isomorphisms $$\begin{array}{r}{\varphi }_{L/K}:K^{\times }/N{m}_{L/K}(L^{\times })\to Gal(L/K)\end{array}$$These form an inverse system as L runs through all the abelian extensions over K. Passing to the limit, we get the isomorphism $$\begin{array}{r}{\hat{\varphi }}_K:\widehat{K^{\times }}\to Gal(K^{ab}/K)\end{array}$$We can also know that $$\begin{array}{r}\widehat{K^{\times }}\cong U_K\times \hat{\mathbb{Z}}\end{array}$$Let $\pi$ an uniformizer of K. Thus we can have a decomposition$$\begin{array}{r}{K}^{ab}=K_{\pi }\cdot K^{un}\end{array}$$where $K_{\pi }$ is the subfield fixed by ${\hat{\varphi }}_K(\hat{\mathbb{Z}})$ and $K^{un}$ is fixed by ${\hat{\varphi }}_K(U_K)$. In another way, $K_{\pi }$ is the union of all finite abelian extensions L over K such that $\pi \in Nm(L^{\times })$ and $K^{un}$ is the union of all finite abelian extensions L over K such that $U_K\subseteq Nm(L^{\times })$.

In the reminder of this section, I will follow [J.S.Milne 2020] to give the outline of the proof of the theorems(the second method in [J.S.Milne 2020])

First, I will show that if the local existence theorem 1.2 holds, the uniqueness of the local Artin map is then proved.

命题 1.4. Assume Local Existence Theorem 1.2, then the there exist at most one homomorphism satisfying the conditions in Theorem 1.1.

证明. For every uniformizer $\pi$, by proposition 1.3 and theorem 1.2, we have the extension $K_{\pi ,n}$ such that $Nm(K_{\pi ,n})=U_n\cdot (\pi )$, and $K_{\pi }$ in the decomposition 1 is exactly ${\cup }_{n\ge 0}{K}_{\pi ,n}$. So $\varphi (\pi )$ in $Gal(K^{ab}/K)=Gal(K_{\pi }\cdot K^{un}/K)$ is uniquely decied by the unique decomposition in 1. Because the group $K^{\times }$ is generated by these uniformizers, the local Artin map is hence uniquely decided.

$\square$

In [J.S.Milne 2020],Chapter 3, Theorem 3.4, the author uses group cohomology to prove the existence of the local Artin maps $\varphi$. It remains to prove the local existence theorem. Next we will show how to choose suitable abelian finite extensions to prove Theorem 1.2(Hence finish all Theorem 1.1).

命题 1.5. If we have a finite abelian extension $K_{\pi ,n}$ of K for every positve integer n and uniformizer $\pi$, and a homomorphism $$\begin{array}{r}{\varphi }_{\pi }:K^{\times }\to Gal((K_{\pi }\cdot K^{un}/K))\end{array}$$where $K_{\pi }={\cup }_{n\ge 1}{K}_{\pi ,n}$, which has the following properties:

1.	$K_{\pi }\cdot K^{un}$ and ${\varphi }_{\pi }$ are independent of the choice of $\pi$
2.	$[K_{\pi ,n}:K]=(q-1)q^{n-1}$
3.	for all n, $\pi$ is in the norm of $K_{\pi ,n}$
4.	${\varphi }_{\pi }(\pi )|K^{un}=Fro{b}_K$
5.	for all m and n, ${\varphi }_{\pi }(a)|(K_{\pi ,n}\cdot K_m)=id$ for $a\in U_n\cdot ({\pi }^m)$, here $K_m$ is the unramified extension of K with degree m.

Then Theorem 1.2 holds.

Let $K^{\prime }=K_{\pi }\cdot K^{un}$, ${\varphi }^{\prime }={\varphi }_{\pi }$. We first prove two lemmas.

引理 1.6. For all $a\in K^{\times }$, $\varphi (a)|K^{\prime }={\varphi }^{\prime }(a)$

证明. Trivial.

$\square$

引理 1.7 (Local Kronecker-Weber Theorem). For every uniformizer $\pi$, $$\begin{array}{r}{K}^{ab}=K_{\pi }\cdot K^{un}\end{array}$$

证明. Denote $K_{n,m}=K_{\pi ,n}\cdot K_m$. By the property of the local Artin map, for any $a\in U_n\cdot ({\pi }^m)$ $\varphi (a)$ trivially acts in $K_{\pi ,n}$ and $K_m$, thus also trivially in $K_{n,m}$. So $U_n\cdot ({\pi }^m)\subseteq Nm(K_{n,m}^{\times })$. Denote $U_{n,m}=U_n\cdot ({\pi }^m)$, we have $$\begin{array}{rl}[K^{\times }:U_{n,m}]& =(U:U_n)\cdot ((\pi ):({\pi }^m))\\ & =(q-1)q^{n-1}\cdot m\\ & =[K_{\pi ,n}:K][K_m:K]\\ & =[K_{n,m}:K]\\ & =Gal(K_{n,m}/K)\\ & =[K^{\times }:Nm(K_{n,m})]\end{array}$$Hence $U_{n,m}=Nm(K_{n,m})$. Now for any $L$ which is an abelian finite extension over K, by proposition 1.3, $Nm(L^{\times })$ is open in $K^{\times }$. So it must contain some $U_{n,m}=Nm(K^{n,m})$. Then again proposition 1.3, $L\subseteq K_{n,m}$. It follows that $K^{ab}=K_{\pi }\cdot K^{un}$.

$\square$

The proof of proposition 1.5. We just need to prove that if $H$ is an open subgroup with finite index then $H$ is a norm group. But $H$ must contain some $U_{n,m}$ which is a norm group. Then by the proposition 1.3, $H$ is a norm group.

$\square$

In the following sections, I will introduce the basic of Lubin-Tate theory to explicitly construct these extensions we want.

2Lubin-Tate Formal Group Laws

The fundamental tool in Lubin-Tate theory is Power Series. This definition makes the maximal ideal in the ring of integers a Galois module, which is very different from the usual additive group structure. Thus we can get very useful representations from it.

定义 2.1. Let A be a ring. A power series with coefficient in A is an infinite sequence $$\begin{array}{r}f=(a_0,a_1,\cdots ,a_n,\cdots ),a_i\in A,i\in \mathbb{N}\end{array}$$with addition and multiplication structure given by $$\begin{array}{rl}(a_0,a_1,a_2,\cdots )+(b_0,b_1,b_2,\cdots )& =(a_0+b_0,a_1+b_1,a_2+b_2,\cdots )\\ (a_0,a_1,a_2,\cdots )\cdot (b_0,b_1,b_2,\cdots )& =(a_0{b}_0,a_1{b}_0+a_0{b}_1,a_2{b}_0+a_1{b}_1+a_0{b}_2,\cdots )\end{array}$$This formula can be seen as a formal power series $$\begin{array}{r}f=\sum _{i\ge 0}{a}_i{X}^i\end{array}$$We can see that the power series consist a commutative ring, which we denote by $A[[X]]$.

For any two elements $f,g\in A[[X]]$, we want to define their composition $f\circ g=f(g)$ as usual expansion. But we notice that it exists only when $g\in XA[[X]]$. We list some simple properties of $A[[X]]$ without proofs.

命题 2.2.

1.	For $f\in A[[X]]$ and $g,h\in XA[[X]]$, the composition has association law: $$\begin{array}{r}f\circ (g\circ h)=(f\circ g)\circ h\end{array}$$
2.	For $f={\sum }_{i\ge 1}{a}_i{X}^i\in XA[[X]]$, there exists an element $g\in XA[[X]]$ such that $f\circ g=X$ iff $a_1$ is a unit in A, in which g is unique and also $g\circ f=X$

Now we will use power series to create a group!

定义 2.3 (Formal Group Law). Let A be a commutative ring. A one-parameter commutative formal group law is a power series $F\in A[[X,Y]]$ such that

1.	$F(X,Y)=X+Y+\text{ terms of degree}\ge 2$
2.	$F(X,F(Y,Z))=F(F(X,Y),Z)$
3.	there exists a unique $i_F(X)\in XA[[X]]$ such that $F(X,i_F(X))=0$
4.	$F(X,Y)=F(Y,X)$

注 2.4. Taking $Y=Z=0$, the first equality implies that $f(X)=F(X,0)$ has an inverse, and the second implies $f\circ f=f$. Thus $f=X$. Similarly $F(0,Y)=Y$. Hence $$\begin{array}{r}F(X,Y)=X+Y+\sum _{i,j\ge 1}{a}_{i,j}{X}^i{Y}^j\end{array}$$

定义 2.5. Let $F(X,Y)$ and $G(X,Y)$ be formal group laws. A homomorphism $F\to G$ is a power series $h\in XA[[X]]$ such that $$\begin{array}{r}h(F(X,Y))=G(h(X,Y))\end{array}$$When there exists a homomorphism $h^{\prime }:G\to F$ such that $$\begin{array}{r}h\circ h^{\prime }=X=h^{\prime }\circ h\end{array}$$$h$ is called an isomorphism. We also say $h$ has an inverse. A homomorphism $h:F\to F$ is called an endomorphism of $F$.

For a formal group law $G$ and any $f,g\in XA[[X]]$, we denote $f{+}_{G}g=G(f(X),g(X))$.

We can check that for any formal group law $F$ and $G$, the set $Hom(F,G)$ has a natural abelian group structure $f{+}_{G}g$ and $End(F)$ has a natural ring structure with multiplication $f\circ g$.

We now try to construct one kind of special formal group law for a local field. We let $A={\mathcal{O}}_K$ the ring of integers in a nonarchimedian local field $K$, and we choose a uniformizer $\pi$. Denote $q$ as the cardinality of the residue field of ${\mathcal{O}}_K$.

定义 2.6. Let ${\mathcal{F}}_{\pi }$ be the set of $f(x)\in A[[X]]$ such that

1.	$f(X)=\pi X+\text{terms of degree}\ge 2$
2.	$f(X)\equiv X^q\mathrm{mod}\pi$

引理 2.7. Let $f,g\in {\mathcal{F}}_{\pi }$, and let ${\varphi }_1(X_1,\cdots ,X_n)$ be a linear form with coefficients in $A$. There is a unique $\varphi \in A[[X_1,\cdots ,X_n]]$ such that

1.	$\varphi (X_1,\cdots ,X_n)={\varphi }_1+\text{terms of degree}\ge 2$
2.	$f({\varphi }_r(X_1,\cdots ,X_n))=\varphi (g(X_1),\cdots ,g(X_n))$

证明. We prove by induction on r that there exists a unique polynomial ${\varphi }_r(X_1,\cdots ,X_n)$ of degree $r$ such that

1.	${\varphi }_r(X_1,\cdots ,X_n)={\varphi }_1+\text{terms of degree}\ge 2$
2.	$f({\varphi }_r(X_1,\cdots ,X_n))={\varphi }_r(g(X_1),\cdots ,g(X_n))+\text{terms of degree}\ge r+1$

For $r=1$, clearly we can only choose ${\varphi }_1$ as our first polynomial.

Suppose we have defined ${\varphi }_r$ for some $r\ge 1$. In fact, because ${\varphi }_r$ is unique, ${\varphi }_{r+1}$ must be the form ${\varphi }_r+Q$, where $Q$ is a homogeneous polynomial with degree in $A[X_1,\cdots ,X_n]$. We need $$\begin{array}{r}f({\varphi }_{r+1}(X_1,\cdots ,X_n))={\varphi }_{r+1}(g(X_1),\cdots ,g(X_n))+\text{terms of degree}\ge r+2\end{array}$$The left hand side is $$\begin{array}{r}f({\varphi }_r(X_1,\cdots ,X_n))+\pi Q(X_1,\cdots ,X_n)+\text{terms of degree}\ge r+2\end{array}$$And the right hand side is $$\begin{array}{r}{\varphi }_r(g(X_1),\cdots ,g(X_n))+{\pi }^{r+1}Q(X_1,\cdots ,X_n)+\text{terms of degree}\ge r+2\end{array}$$So we need $$\begin{array}{r}({\pi }^{r+1}-\pi )Q(X_1,\cdots ,X_n)=f({\varphi }_r(X_1,\cdots ,X_n))-{\varphi }_r(g(X_1),\cdots ,g(X_n))+\text{terms of degree}\ge r+2\end{array}$$Thus if $$\begin{array}{r}\frac{f({\varphi }_r(X_1,\cdots ,X_n))-{\varphi }_r(g(X_1),\cdots ,g(X_n))}{{\pi }^{r+1}-\pi }\in A[[X]]\end{array}$$then we just make the terms of degree of $r+1$ the polynomial $Q$ we want. This follows from $$\begin{array}{r}f\circ {\varphi }_r(X_1,\cdots ,X_n)-{\varphi }_r(g(X_1),\cdots ,g(X_n))\equiv ({\varphi }_r(X_1,\cdots ,X_n){)}^q-{\varphi }_r(X_1^q,\cdots ,X_n^q)\equiv 0\mathrm{mod}\pi \end{array}$$and ${\pi }^r-1$ is a unit in $A$.

$\square$

命题 2.8. For every $f\in {\mathcal{F}}_{\pi }$, there is a unique formal group law $F_f$ with coefficients in A admitting $f$ as an endomorphism.

证明. According to lemma 2.7, there is a unique power series $F_f(X,Y)$ such that the linear part of $F_f$ is $X+Y$ and it commutes with $f$. The commutativity and associativity then also follows from the uniqueness.

$\square$

命题 2.9. For $f,g\in {\mathcal{F}}_{\pi }$ and $a\in A$, let $[a{]}_{g,f}$ be the unique element of $A[[X]]$ such that

1.	$[a{]}_{g,f}(X)=aX+\text{terms of degree}\ge 2$
2.	$g\circ [a{]}_{g,f}=[a{]}_{g,f}\circ f$

Then $[a{]}_{g,f}$ is a homomorphism $F_f\to F_g$.

证明. The existence of $h=[a{]}_{g,f}$ follows form lemma 2.7. To show it’s a homomorphism, we need to show $$\begin{array}{r}h(F_f(X,Y))=F_g(h(X),h(Y))\end{array}$$Clealy the linear part of both side is $aX+aY$, and $$\begin{array}{r}h(F_f(f(x),f(Y)))=(h\circ f)(F_f(X,Y))=g(h(F_f(X,Y)))\end{array}$$$$\begin{array}{r}{F}_g(h(f(X)),h(f(Y)))=F_g(g(h(X)),g(h(Y)))=g(F_g(h(X),h(Y)))\end{array}$$Thus they are equal from the uniqueness in lemma 2.7

$\square$

命题 2.10. For any $a,b\in A$, $$\begin{array}{r}[a+b{]}_{g,f}=[a{]}_{g,f}{+}_{F_g}[b{]}_{g,f}\end{array}$$and $$\begin{array}{r}[ab{]}_{h,f}=[a{]}_{h,g}\circ b[g,f]\end{array}$$

证明. Because both side has the same linear terms, this proposition follows from the uniqueness of lemma 2.7.

$\square$

推论 2.11. $[\pi {]}_f=f$ by checking the definition.

推论 2.12. For any $f,g\in {\mathcal{F}}_{\pi }$, $F_f\cong F_g$

In fact, we can choose $[1{]}_{g,f}$ which is the unique isomorphism $h:F_f\to F_g$ such that $h(X)=X+\cdots$

推论 2.13. For each $a\in A$, there is a unique endomorphism $[a{]}_f:F_f\to F_f$ such that $[a{]}_f=aX+\text{terms of degree}\ge 2$ and $[a{]}_f$ commutes with $f$. The map $$\begin{array}{r}a\mapsto [a{]}_f:A\hookrightarrow End(F_f)\end{array}$$is an injective ring homomorphism.

We conclude what we have done in this section. For any $f\in {\mathcal{F}}_{\pi }$, there exists a unique formal group law $F_f$ admitting $f$ as an endomorphism. And there is a unique $A-$module structure $a\mapsto [a{]}_f:A\to End(F_f)$ on $F_f$ such that

1.	$[a{]}_f$ has linear terms $aX$.
2.	$[a{]}_f$ commutes with $f$.

We also have $[\pi {]}_f=f$ and for $g\in {\mathcal{F}}_{\pi }$, $F_f\cong F_g$

3The Construction of the Extensions

In this section, $A={\mathcal{O}}_K$, where $K$ is a nonarchimedian local field with finite residue field $k=A/m$ having $q$(a power of $p$) elemnets. We fix a uniformizer $\pi$. I will use formal group laws to construct the extensions $K_{\pi ,n}$.

First, we can uniquely extend the absolute value on $K$ to the algebraic closure $K^{al}$. Let $f\in {\mathcal{F}}_{\pi }$. For any $\alpha ,\beta \in K^{al}$ with $|\alpha |,|\beta |<1$ and $a\in A$, the series $F_f(\alpha ,\beta )$ and $[a{]}_f(\alpha )$ converge. Thus we can define the $A-$module ${\Lambda }_f$ as:$$\begin{array}{rl}{\Lambda }_f& =\{\alpha \in K^{al}||\alpha |<1\}\\ \alpha {+}_{{\Lambda }_f}\beta & =\alpha {+}_{F_f}\beta =F_f(\alpha ,\beta )\\ a\ast \alpha & =[a{]}_f(\alpha )\end{array}$$We define ${\Lambda }_n$ to be submodule of ${\Lambda }_f$ of elements annihilated by $[\pi {]}_f^n$. Because $F_f$ and $F_g$ are isomorphic, the module ${\Lambda }_f$ and ${\Lambda }_g$ are isomorphic.

Note that $A$ is in fact a principal ideal domain, thus every finitely generated $A-$module $M$ can be written to the form $$\begin{array}{r}M=A^r\oplus A/({\pi }^{r_1})\oplus \cdots \oplus A/({\pi }^{r_n}),r_1\le \cdots \le r_n\end{array}$$

Now we prove a lemma:

引理 3.1. Let $M$ be an $A-$ module, and let $M_n=Ker({\pi }^n:M\to M)$. Assume

1.	$|M_1|=q=|A/(\pi )|$
2.	$\pi :M\to M$ is surjective

Then $M_n\cong A/({\pi }^n)$. In particular, it has $q^n$ elements.

证明. We use induction on n. When $n=1$, by $|M_1|=q$ and the formula 3, $M_1$ must be $A/(\pi )$.

For common case, consider the exact sequence

The exactness follows from condition 2. Then $|M_n|=q^n$ and ${\pi }^{n-1}{M}_n=M_1=A/(\pi )$. Again by formula 3, $M_n\cong A/({\pi }^n)$.

$\square$

We can prove an interesting result now.

命题 3.2. The $A-$module ${\Lambda }_n$ is isomorphic to $A/({\pi }^n)$. Hence $En{d}_A({\Lambda }_n)\cong A/({\pi }^n)$ and $Au{t}_A(\Lambda )\cong (A/({\pi }^n){)}^{\times }$.

证明. By what we discuss before, we can choose $f\in {\mathcal{F}}_{\pi }$ be a polynomial of the form $\pi X+\cdots +X^q$. This is an Eisenstein polynomial. So it has $q$ distinct roots with all abusolte values smaller than 1([J.S.Milne 2020],p.129). Because $[\pi {]}_f(X)=f(X)$, ${\Lambda }_n$ is indeed the set of the roots of $f^{(n)}=f\circ f\circ \cdots \circ f$. Thus ${\Lambda }_1$ has $q$ elements. And for any $\alpha \in K^{al}$ with its absolute value smaller than 1, we also know that the roots of $$\begin{array}{r}f(X)-\alpha =X^q+\cdots +\pi X-\alpha \end{array}$$have absolute value smaller than 1, hence belong to ${\Lambda }_f$. By the lemma 3.1, ${\Lambda }_n\cong A/({\pi }^n)$.

$\square$

引理 3.3. Let $L$ be a finite Galois extension of a local field $K$, with Galois group $G$. For every $F\in {\mathcal{O}}_K[[X_1,\cdots ,X_n]]$ and ${\alpha }_1,\cdots ,{\alpha }_n\in {\mathfrak{m}}_L$, $$\begin{array}{r}F(\tau {\alpha }_1,\cdots ,\tau {\alpha }_n)=\tau F({\alpha }_1,\cdots ,{\alpha }_n),\forall \tau \in G\end{array}$$

证明. Because the absolute value in L extended by the absolute value in K is unique, it is preserved by the Galois action. Hence every element in $G$ is continuous and commutes with the limit.

$\square$

Now we construct the fields we want.

定理 3.4. Let $K_{\pi ,n}$ denote the subfield $K[{\Lambda }_n]$ of $K^{al}$.

1.	For each $n$, $K_{\pi ,n}/K$ is totally ramified of degree $(q-1)q^{n-1}$.
2.	The action of $A$ on ${\Lambda }_n$ defines an isomorphism $$\begin{array}{r}(A/{\mathfrak{m}}^n{)}^{\times }\to Gal(K_{\pi ,n}/K)\end{array}$$In particular, $K_{\pi ,n}/K$ is abelian.
3.	For each $n$, $\pi$ is a norm from $K^{\pi ,n}$.

证明. We still asuume that the element $f$ in ${\mathcal{F}}_{\pi }$ is $\pi X+\cdots +X^q$.

For (1) and (2), we choose a nonzero root ${\pi }_1$ of $f(X)$ and inductively a root of ${\pi }_n$ of $f(X)-{\pi }_{n-1}$. Consider the sequence of fields $$\begin{array}{r}K[{\Lambda }_n]\supset K[{\pi }_n]\stackrel{q}{\supset }K[{\pi }_{n-1}]\stackrel{q}{\supset }\cdots \stackrel{q}{\supset }K[{\pi }_1]\stackrel{q-1}{\supset }K\end{array}$$Each extension is Eisenstein with degree indicated above. Therefore $K[{\pi }_n]$ is totally ramified over $K$ of degree $q^{n-1}(q-1)$([J.S.Milne 2020],7.55).

Because $K[{\Lambda }_n]$ is in fact the splitting fiels of $f^{(n)}$, the Galois group $Gal(K[{\Lambda }_n]/K)$ can be seen a subgroup of the permutation group of ${\Lambda }_n$. And the lemma 3.3 ensures that these elements of the Galois group are $A-$isomorphism. Thus $$\begin{array}{r}|Aut({\Lambda }_n)|=(q-1)q^{n-1}\ge |Gal(K[{\Lambda }_n]/K)|=[K[{\Lambda }_n]:K]\ge [K[{\pi }_n]:K]=(q-1)q^{n-1}\end{array}$$So we must have all equalities. Thus $Gal(K[{\Lambda }_n]:K)\cong (A/{\mathfrak{m}}^n{)}^{\times }$ and $K[{\Lambda }_n]=K[{\pi }_n]$.

For (3), let $f^{[n]}(X)=(f/X)\circ f\circ \cdots \circ f$(all n terms), so that $$\begin{array}{r}{f}^{[n]}(X)=\pi +\cdots +X^{(q-1)q^{n-1}}\end{array}$$Then $f^{[n]}({\pi }_n)=f^{[n-1]}({\pi }_{n-1})=\cdots =f({\pi }_1)=0$. Because $f^{[n]}$ is monic with degree $(q-1)q^n$, it must be the minimal polynomial of ${\pi }_n$ over $K$. Therefore, $$\begin{array}{rl}N{m}_{K[{\Lambda }_n]/K}({\pi }_n)& =(-1{)}^{(q-1)q^{n-1}}\pi \\ & =\pi \text{ unless }q=2,n=1\end{array}$$But when $q=2,n=1$, $K[{\Lambda }_1]=K$, $\pi$ is clearly a norm.

$\square$

In fact, this theorem tells us how to add reasonable elements to get these totally ramified extensions.

From the proof, we can see that the action $a\ast \lambda =[a{]}_f(\lambda )$ of $a\in A,\lambda \in {\Lambda }_n$ induces the isomorphism $$\begin{array}{r}(A/{\mathfrak{m}}^n{)}^{\times }\to Gal(K_{\pi ,n}/K)\end{array}$$Passing to the inverse limit, we get the isomorphism $$\begin{array}{r}{A}^{\times }\to Gal(K_{\pi }/K)\end{array}$$

4The Construction of the Map

Now we have the extensions $K_{\pi ,n}$. We still need to construct the proper homomorphism ${\varphi }_{\pi }$.

Let $a\in K^{\times }$. Because $K_{\pi }\cap K^{un}=K$, we just need to describe the action of ${\varphi }_{\pi }(a)$ in $K^{un}$ and $K_{\pi }$ separately. Assume $a=u{\pi }^m,u\in U$. We let ${\varphi }_{\pi }(a)$ act on $K^{un}$ as $Fro{b}^m$, and on $K_{\pi }$ following the rule $$\begin{array}{r}{\varphi }_{\pi }(a)(\lambda )=[u^{-1}{]}_f(\lambda ),\forall \lambda \in \bigcup {\Lambda }_n\end{array}$$

Now we prove

定理 4.1. Both $K_{\pi }\cdot K^{un}$ and ${\varphi }_{\pi }$ are independent of the choice of $\pi$.

We know that $K^{al}$ is not complete([J.S.Milne 2020],exercise 7-7). In fact, even $K^{un}$ is not complete. We write $\widehat{K^{un}}$ for its completion and $B$ for the valuation ring of $\widehat{K^{un}}$. We write $\sigma$ as the Frobenius $Fro{b}_K$ of $K^{un}/K$, and also for its extension to $\widehat{K^{un}}$. For a power series $\theta (X)=\sum b_i{X}^i\in B[[X]]$, we define $(\sigma \theta )(X)$ to be the power series $\sum (\sigma b_i)X^i$.

命题 4.2. Let $F_f$ and $F_g$ be the formal group laws defined by $f\in {\mathcal{F}}_{\pi }$ and $g\in {\mathcal{F}}_{\omega }$, where $\pi$ and $\omega =u\pi$ are two uniformizers of $K$. Then $F_f$ and $F_g$ become $A-$isomorphic over $B$. More precisely, there exists an $ϵ\in B^{\times }$ such that $\sigma ϵ=ϵu$, and a power series $\theta (X)\in B[[X]]$ such that

1.	$\theta (X)=ϵX+\text{terms of degree}\ge 2$.
2.	$\sigma \theta =\theta \circ [u{]}_f$
3.	$\theta (F_f(X,Y))=F_g(\theta (X),\theta (Y))$
4.	$\theta \circ [a{]}_f=[a{]}_g\circ \theta$

The last two conditions say that $\theta$ is a homomorphism $F_f\to F_g$ commuting with the actions of of $A$, and the first conditon implies that $\theta$ is an isomorphism.

引理 4.3. The homomorphisms $$\begin{array}{r}b\mapsto \sigma b-b:B\to B,\\ b\mapsto \sigma b/b:B^{\times }\to B^{\times }\end{array}$$are surjective with kernels $A$ and $A^{times}$ respectively.

证明. Denote $R$ be the valuation ring in $K^{un}$, and $\mathfrak{n}$ be its maximal ideal. Then $B=\underleftarrow{\lim }R/{\mathfrak{n}}^n$. We prove the sequence

is exact by induction:

For $n=1$, the sequence is

This is obviously right. Assume the sequence is exact of $n-1$, and consider the diagram

By snake lemma, the homomorphism $\sigma -1:R/{\mathfrak{n}}^n\to R/{\mathfrak{n}}^n$ is surjective and that its kernel has $q^n$ elements. Because $A/{\mathfrak{m}}_K^{}$ has $q^n$ elements and is contained is the kernel, it must be the kernel. Thus the sequence is exact for $n$. Now pass to the inverse limit, we get the exact sequence

The proof for $B^{\times }$ is similar.

$\square$

We prove the proposition 4.2 by four steps:

Step 1	There exists $\theta (X)\in B[[X]]$ satisfy the condition (1) and(2).
Step 2	The $\theta$ in Step 1 can chosen so that $g=\sigma \theta \circ f\circ {\theta }^{-1}$.
Step 3	The power series $\theta (F_f({\theta }^{-1}(X),{\theta }^{-1}(Y)))$ has the property characterizing $F_g(X,Y)$, hence equals to it.
Step 4	The power series $\theta \circ [a{]}_f\circ {\theta }^{-1}$ has the property characterizing $[a{]}_g$, and therefore equals to it.

Proof of Step 1. Choose an $ϵ\in B^{\times }$ such that $\sigma ϵ=ϵu$ (Lemma 4.3). Starting with ${\theta }_1(X)=ϵX$, we will construct a sequence of polynomials ${\theta }_r$ such that $$\begin{array}{r}{\theta }_r(X)={\theta }_{r-1}(X)+b{X}^r,\text{for some }b\in B\\ \sigma {\theta }_r={\theta }_r\circ [u{]}_f+\text{terms of degree}\ge r+1\end{array}$$Suppose ${\theta }_r$ has been found, and we wish to find the $b$ we want. Write $b=a{ϵ}^{r+1},a\in B$. Then the second equation becomes $$\begin{array}{r}(\sigma {\theta }_r)(X)+(\sigma a)(\sigma ϵ{)}^{r+1}{X}^{r+1}={\theta }_r([u{]}_f(X))+a{ϵ}^{r+1}(uX{)}^{r+1}+\text{terms of degree}\ge r+2\end{array}$$Thus we need $(\sigma a-a)(ϵu{)}^{r+1}$ to be the coefficient of $X^{r+1}$ in ${\theta }_r\circ [u{]}_f-\sigma {\theta }_r$. This is obvious by Lemma 4.3.

$\square$

Proof of Step 2. Define $$\begin{array}{r}h=\sigma \theta \circ f\circ {\theta }^{-1}=\theta \circ [u{]}_f\circ f\circ {\theta }^{-1}=\theta \circ f\circ [u{]}_f\circ {\theta }^{-1}\end{array}$$Because $f$ and $[u{]}_f$ have coefficients in $A$, $$\begin{array}{r}\sigma h=\sigma \theta \circ f\circ [u{]}_f\circ \sigma {\theta }^{-1}=\sigma \theta \circ f\circ {\theta }^{-1}=h\end{array}$$The second equality follows from $\theta \circ [u{]}_f\circ \sigma {\theta }^{-1}=X$. Thus $h\in A[[X]]$. Moreover, $$\begin{array}{r}h(X)=\sigma ϵ\cdot \pi \cdot {ϵ}^{-1}X+\cdots =\omega X+\cdots \end{array}$$and $$\begin{array}{r}h(X)\equiv \sigma \theta \circ ({\theta }^{-1}{)}^q\equiv \sigma \theta (\sigma {\theta }^{-1}(X^q))\equiv X^q\mathrm{mod}{\mathfrak{m}}_K\end{array}$$Thus $h\in {\mathcal{F}}_{\omega }$. Let ${\theta }^{\prime }=[1{]}_{g,h}\theta$. Then ${\theta }^{\prime }$ obviously satisfies the condition (1) of the proposition, and it also satisfies the conditon (b) because $[1{]}_{g,h}\in A[[X]]$. Moreover, $$\begin{array}{r}\sigma {\theta }^{\prime }\circ f\circ {\theta }^{\prime -1}=[1{]}_{g,h}\circ h\circ [1{]}_{g,h}^{-1}=g\end{array}$$

$\square$

Proof of Step 3 and 4. Just compute the equation then use the Lemma 2.7.

$\square$

Now we finish the proof of Theorem 4.1. First, we prove a lemma:

引理 4.4. Every subfield $E$ of $K^{al}$ containing $K$ is closed.

证明. Let $H=Gal(K^{al}/E)$. Then $H$ fixes every element of $E$, hence by continuity, $H$ fixes every element in the closure of $E$. By Galois correspondence, the closure of $E$ is just $E$. Thus $E$ is closed.

$\square$

Proof for ${\mathbf{K}}_{\mathbf{\pi }}\cdot {\mathbf{K}}^{\mathbf{un}}$ is independent of $\mathbf{\pi }$.

Let $\pi$ and $\omega$ be two uniformizers of $K$. From Propositon 4.2 we find that $$\begin{array}{r}(\sigma \theta )\circ [\pi {]}_f=\theta \circ [u{]}_f\circ [\pi {]}_f=\theta \circ [\omega {]}_f=[\omega {]}_g\circ \theta \end{array}$$so $$\begin{array}{r}(\sigma \theta )(f(X))=g(\theta (X))\end{array}$$

Therefore, for any $\alpha \in K^{al}$, $$\begin{array}{r}f(\alpha )=0\to g(\theta (\alpha ))=0\end{array}$$Similarly, $$\begin{array}{r}g(\alpha )=0\to f({\theta }^{-1}(\alpha ))=0\end{array}$$Thus $\theta$ defines a bijection ${\Lambda }_{f,1}\to {\Lambda }_{g,1}$, and so $$\begin{array}{r}\widehat{K^{un}}[{\Lambda }_{g,1}]=\widehat{K^{un}}[\theta ({\Lambda }_{f,1})]\subseteq \widehat{K^{un}}[{\Lambda }_{f,1}]=\widehat{K^{un}}[{\theta }^{-1}({\Lambda }_{g,1})]\subseteq \widehat{K^{un}}[{\Lambda }_{g,1}]\end{array}$$Hence $$\begin{array}{r}\widehat{K^{un}}[{\Lambda }_{g,1}]=\widehat{K^{un}}[{\Lambda }_{f,1}]\end{array}$$By Lemma 4.4, $$\begin{array}{r}\widehat{K^{un}}[{\Lambda }_{g,1}]\cap K^{al}=K^{un}[{\Lambda }_{g,1}],\widehat{K^{un}}[{\Lambda }_{f,1}]\cap K^{al}=K^{un}[{\Lambda }_{f,1}],\end{array}$$So $K^{un}[{\Lambda }_{g,1}]=K^{un}[{\Lambda }_{f,1}]$. Simliarly, $K^{un}[{\Lambda }_{g,n}]=K^{un}[{\Lambda }_{f,n}]$ for all $n$. Hence $K^{un}\cdot K_{\omega }=K^{un}\cdot K_{\pi }$.

$\square$

Proof for ${\mathbf{\varphi }}_{\mathbf{\pi }}$ is independent of $\mathbf{\pi }$. We need to show that for any twq uniformizers $\pi$ and $\omega$ $$\begin{array}{r}{\varphi }_{\pi }(\omega )={\varphi }_{\omega }(\omega )\end{array}$$Then for any uniformizers $\pi$, ${\pi }^{\prime }$, $\omega$, we have $$\begin{array}{r}{\varphi }_{\pi }(\omega )={\varphi }_{\omega }(\omega )={\varphi }_{({\pi }^{\prime })}(\omega )\end{array}$$then obviously ${\varphi }_{\pi }$ and ${\varphi }_{{\pi }^{\prime }}$ are the same.

On $K^{un}$, both of them induce the Frobenius. It remains for $K_{\omega }$.

Let $\theta$ be the isomorphism $F_f\to F_g$ over $\widehat{K^{un}}$ as in Proposition 4.2. It induces an isomorphism ${\Lambda }_{f,n}\to {\Lambda }_{g,n}$ for all $n$. By definition, ${\varphi }_{\omega }(\omega )$ is trivial on $K_{\omega }$, and since $K_{\omega ,n}$ is generated over $K$ by the elements $\theta (\lambda )$ for $\lambda \in {\Lambda }_{f,n}$, we need to show that $$\begin{array}{r}{\varphi }_{\pi }(\theta (\lambda ))=\theta (\lambda ),\forall \lambda \in {\Lambda }_{f,n}\end{array}$$Write $\omega =u\pi$, then ${\varphi }_{\pi }(\omega )={\varphi }_{\pi }(u)\cdot {\varphi }_{\pi }(\pi )$. We know that $\tau ={\varphi }_{\pi }(u)$ is trivial in $K^{un}$ and $[u{]}_f^{-1}$ on $\lambda$, $\sigma ={\varphi }_{\pi }(\pi )$ is the Frobenius in $K^{un}$ and trivial on $\lambda$. So we have $$\begin{array}{r}{\varphi }_{\pi }(\omega )(\theta (\lambda ))=\tau \sigma (\theta (\lambda ))=(\sigma \theta )(\tau \lambda )=(\sigma \theta )([u^{-1}{]}_f(\lambda ))=\theta (\lambda )\end{array}$$

$\square$

J.S.Milne (2020). “Class field theory”. (web)

J.S.Milne (2020). “Algebraic number theory”. (web)

J.W.S.Cassels, A.Frohlich (1967). “Algebraic number theory: proceedings of an instructional conference”.