We will construct the maximal abelian extension of a local field $K$ in this section. It gives us the solution of the Hilbert's twelfth problem for local fields, note that by using this result, we can solve the Hilbert's twelfth problem for $\mathbb{Q}$, which is a corollary of the global Kronecker-Weber's theorem. We will see that the global Kronecker-Weber's theorem can be deduced from the local Kronecker-Weber's theorem.
As in the previous section, we will fix a local field $K$, and its Lubin-Tate's tower
$$K\subset K_1\subset ...\subset K_n\subset K_\infty$$
The goal if this section is to prove that $K^{\text{ab}}=K_\infty K^{\text{un}}$, where $K^{\text{ab}}$ is the maximal abelian extension of $K$, $K^{\text{un}}$ is the maximal unramified extension of $K$. We first need to get familiar with (un)ramified extensions for compositum of fields.
Lemma 4.1. Let $L$ be a finite unraimfied extension of $K_\infty$, then $L=K_\infty L'$ for some $L'$: finite unramified extension of $K_n$, for some $n$.
Proof. We recall that a finite extension $M/F$ is an unramified extension of fields if $e(M|F)=1$, and $M$ is separable over $F$. Hence, $L/K_\infty$ is unramified implies that $L/K_\infty$ is separable. Because $L/K_\infty$ is finite, this yields $L=K_\infty(\alpha)$, by the primitive element theorem. Let $f(X)=a_0+a_1X+...+X^n$ is a minimal polynomial of $\alpha$. This yields $\overline{f(X)}$ is irreducible and separable in $k_{K_\infty}$-the residue field of $K_\infty$. Also, because $K_\infty=\cup_{n\ge 1}K_n$, there exists some $n$ such that all $a_i\in K_n$. Let $L'=K_n(\alpha)$, one has $\overline{f(X)}$ is also irreducible in $k_{K_n}$, and it is separable there, since $k_{K_n}$ is finite, and hence, perfect. From this, one can see $L'/K_n$ is unramified. And $L=K_\infty L'$. (Q.E.D)
Lemma 4.2. Let $L'$ be a finite unramified extension of $K_n$, then $L'=K_nL''$, for some unramified extension $L''$ over $K$.
Proof. Because $K_n$ is also a local field, and because $K_n/K$ is totally ramified, we have $k_K=k_{K_n}$. Also $L'/K_n$ is an unramified extension implies that $L'=K_n(\zeta)$, where $\zeta$ is a $q^m-1$-th root of unity in $K_n$. This yiels $\zeta$ is also $q^m-1$-th root of unity in $K$. Hence $K(\alpha)/K$ is unramified. Let $L'':=K(\alpha)$, we have $L'=K_nL''$ (Q.E.D)
By the two lemmas above, we obtain
Corollary 4.3. Let $L$ be a finite unramified extension of $K_\infty$, then $L\subset K_\infty K^{\text{un}}$.
Proof. By Lemma 4.1, we have $L\subset K_\infty L'$, for some $L'/K_n$ is unramified extension, for some $n$. By Lemma 4.2, $L=K_nL''$, for some $L''/K$ is unramified extension. This yields $L\subset K_\infty K^{\text{un}}$. (Q.E.D)
We will need more about Galois groups (extensions) of compositum of fields.
Lemma 4.4. Let $F$ be a field, $L/F, K/F$ is finite Galois extensions, with $L,K\subset \overline{F}$, then so is $L\cap K/F, LK/F$.
Proof. Obvious.
Lemma 4.5. Let $F, L,K$ be as above, then
$$Gal(KL/F)=\{(\sigma,\psi)\in Gal(L/F)\times Gal(K/F)|\sigma_{L\cap K}=\psi_{L\cap K}\}$$
Proof. It can be seen that the map $Gal(KL/F)\to Gal(K/F)\times Gal(L/F)$ defined by
$$\sigma\mapsto (\sigma_{|K}, \sigma_{|L})\subset \{(\sigma,\psi)\in Gal(L/F)\times Gal(K/F)|\sigma_{L\cap K}=\psi_{L\cap K}\}$$
is an injective map. We will prove that this map is in fact an isomorphism, by deducing they have the same number of elements.
Let $L\cap K=M$, $k:=\#Gal(KL:K), l:=\#Gal(KL/L), n:=\#Gal(M/F)$ and $A:=Gal(KL/K), B:=Gal(KL/L)$, then it can be seen that $A\cap B=id_M, (KL)^{AB}=M$. And hence, $Gal(KL/M)\cong A\times B$. This yields $\#Gal(KL/M)=kl$. And from this, one obtains $[L:M]=k, [K:M]=l, \#Gal(KL/F)=nkl$.
Due to Lemma 4.4, and the primitive element, we can write $M=F(\alpha), K=F(\alpha, \beta), L=F(\alpha,\gamma)$. Let $\alpha_1:=\alpha,...,\alpha_n$ be all Galois conjugates of $\alpha$, similarly for $\beta_1,...,\beta_l$ and $\gamma_1,...,\gamma_k$. From this, the group in the LHS of the statement is
$$\{(\sigma,\psi)|\sigma(\alpha)=\alpha_i, \sigma(\beta)=\beta_j, \psi(\alpha)=\alpha_i,\psi(\gamma)=\gamma_k\}$$
So, in particular, we have this group has $nkl$ elements. This yields the injective map in the beginning of the proof is an isomorphism. (Q.E.D)
We are now ready for the second corollary.
Corollary 4.6. Let $L/K_\infty$ be a finite abelian extension of exponent $m$, then $L\subset L_tK_m$, for $K_m/K_\infty$: an unramified extension of degree $m$, and $L_t/K_\infty$ is a totally ramified abelian extension.
Proof. Consider $G:=Gal(LK_m/K_\infty)$, by the previous lemma, this group is also finite, abelian. Let $\sigma$ be an element in this group, then $((\sigma|_L)^m=(\sigma)|_{K_m})^m=id$, so we get $G$ is an abelian group of exponent $m$.
Let $\tau\in G$ such that $\tau|_{K_m}$ is the Frobenius element, then $\tau^m=1$. And hence, we can express $G=\langle \tau\rangle \times H$. Let $L_t:=(LK_m)^{\langle \tau \rangle}$, then $Gal(LK_m/L_t)=\langle \tau \rangle$. Also, $Gal(LK_m/K_m)=H$, since $Gal(K_m/K_\infty)=\langle \tau \rangle$.
Hence, by the previous lemma, $L_t\cap K_m=K_\infty$. Let $L'$ be an intermediate field between $L_t$ and $K_\infty$, such that $L_t/L'$ is totally ramified and $L'/K_\infty$ is unramified. From the hypothesis, one gets $Gal(L'/K)$ is a cyclic group of exponent $m$, i.e. $L'$ is also a subfield of $K_m$. This yields $L'=K_\infty$. And hence, $L_t/K_\infty$ is an abelian totally ramified extension, and $L\subset L_tK_m$. (Q.E.D)
The following result is more difficult to prove, because it is quite long, and require results of ramification groups (See for example Milne's note on CFT (Lemma 4.9)).
Fact 4.7. Let $L$ be an abelian totally ramified extension of $K$, if $K_\infty \subset L$, then $K_\infty = L$.
This result yields $K_\infty$ is a maximal abelian totally ramified extension of $K$. We are now ready for the main result
Theorem 4.8. $K^{\text{ab}}=K_\infty K^{\text{un}}$. And $Gal(K^{\text{ab}}/K)\cong \mathscr{O}_K^\times \times \hat{\mathbb{Z}}$.
Proof. Let $K'$ be any a finite abelian extension of $K$. Then $L:=K'K_{\infty}$ is a finite abelian extension of $K_{\infty}$. By Corollary 4.6, $L\subset L_tK_m$, for some $L_t/K_\infty$ is a finite abelian totally ramified extension, and $K_m/K_{\infty}$ is a finite unramified extension. By Fact 4.7, $L_t=K_{\infty}$, and by Corollary 4.3, $K_m\subset K_{\infty} K^{\text{ab}}$. This yields $K'\subset K_{\infty} K^{\text{un}}$. Hence, $K^{\text{ab}}\subseteq K_{\infty} K^{\text{un}}$. The other inclusion is clear. Hence, $K^{\text{ab}}=K_{\infty} K^{\text{un}}$.
For the second statement, we can see that $K^{\text{un}}\cap K_{\infty}=K$, hence, the statement about Galois group follows from (infinite version) of Lemma 4.5.
(Q.E.D)
Showing posts with label Class Field Theory. Show all posts
Showing posts with label Class Field Theory. Show all posts
Tuesday, December 26, 2017
Sunday, December 24, 2017
[Local Class Field Theory 3] Lubin-Tate's Towers
We assume the notations in Section 2, with $\overline{K}$ the algebraic closure of K. Let $M:=\{z\in \overline{K}|v_K(z)>0\}$ the maximal ideal of $\overline{K}$. Recall that for any $z_1,...,z_n\in M$, and any power series $F(X_1,...,X_n)\in A[[X_1,...,X_n]]$, $F(z_1,...,z_n)$ converges in $\overline{K}$, since norms of terms go to $0$, when the indexes go to infinity.
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.
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.
Thursday, December 21, 2017
[Local Class Field Theory 2] Lubin-Tate Formal Group Law
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 $[a]_{f,f}=[a]_f$.
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 $[a]_{f,f}=[a]_f$.
Wednesday, December 20, 2017
[Local Class Field Theory 1] Formal Group Laws
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)
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)
Tuesday, December 19, 2017
Local Class Field Theory: Some Words before Beginning
The story begins with Hilbert's twelfth problem, that is: how to construct all abelian extensions of a number fields? It is somewhat equivalent to the question: let $K$ be a number field, construct the maximal abelian extension $K^{\text{ab}}$ of $K$. And this question is a starting point of class field theory. In local fields, one can ask for a similar question, let $K$ be a local field, construct the maximal abelian extension $K^{\text{ab}}$ of $K$. This question can be done completely by Lubin-Tate's theory.
Lubin-Tate's theory is based on formal groups, which is a vast and beautiful topic with many applications. Let $R$ be a (commutative) ring, and $F\in R[[X,Y]]$ be a series, then $F$ is said to be a commutative formal group if
1. $F(X,0)=X, F(0,Y)=Y$.
2. $F(X,F(Y,Z))=F(F(X,Y),Z)$ for all $Z\in R[[X,Y]]$.
3. $F(X,Y)=F(Y,X)$.
If one remember about the group structures of elliptic curves over a base field $k$, then addition law on an elliptic curve around the point $\infty$ is exactly a formal group (Silverman, Chapter IV.1). And one can define homomorphism between two formal group. In elliptic curves case, we obtain an isogeny represented around the two infinity points.
Lubin-Tate's formal group is also a specific example of formal groups. It is defined for Forbenius series over a local field. And by attaching suitable things, we obtain an abelian maximal totally ramified extension $K^{\text{ab.tot}}$ of $K$. We also obtain that $Gal(K^{\text{ab.tot}}/K)\cong \mathscr{O}_K^\times$.
Recall that any extension of local field can be decomposed into two parts: totally ramified part and unramified part. For the unramified part, it is somewhat easy, since to obtain an unramified extension of degree $n$, we just need to add $(q^n-1)$ root of unity, where $q$ is the cardinality of the residue field of $K$. And adjoining them all, we obtain $K^{\text{un}}$-the maximal unramified part of $K$. And from this, one has $Gal(K^{\text{un}}/K)\cong \hat{\mathbb{Z}}:=\varprojlim \mathbb{Z}/n\mathbb{Z}$
And now, $K^{\text{ab}}=K^{\text{ab.tot}}K^{\text{un}}$, with $Gal(K^{\text{ab}}/K)=\mathscr{O}_K^\times\times \hat{\mathbb{Z}}$. More than that, with the Artin map, we can also classify abelian extensions of $K$ with given degree.
Lubin-Tate's theory is based on formal groups, which is a vast and beautiful topic with many applications. Let $R$ be a (commutative) ring, and $F\in R[[X,Y]]$ be a series, then $F$ is said to be a commutative formal group if
1. $F(X,0)=X, F(0,Y)=Y$.
2. $F(X,F(Y,Z))=F(F(X,Y),Z)$ for all $Z\in R[[X,Y]]$.
3. $F(X,Y)=F(Y,X)$.
If one remember about the group structures of elliptic curves over a base field $k$, then addition law on an elliptic curve around the point $\infty$ is exactly a formal group (Silverman, Chapter IV.1). And one can define homomorphism between two formal group. In elliptic curves case, we obtain an isogeny represented around the two infinity points.
Lubin-Tate's formal group is also a specific example of formal groups. It is defined for Forbenius series over a local field. And by attaching suitable things, we obtain an abelian maximal totally ramified extension $K^{\text{ab.tot}}$ of $K$. We also obtain that $Gal(K^{\text{ab.tot}}/K)\cong \mathscr{O}_K^\times$.
Recall that any extension of local field can be decomposed into two parts: totally ramified part and unramified part. For the unramified part, it is somewhat easy, since to obtain an unramified extension of degree $n$, we just need to add $(q^n-1)$ root of unity, where $q$ is the cardinality of the residue field of $K$. And adjoining them all, we obtain $K^{\text{un}}$-the maximal unramified part of $K$. And from this, one has $Gal(K^{\text{un}}/K)\cong \hat{\mathbb{Z}}:=\varprojlim \mathbb{Z}/n\mathbb{Z}$
And now, $K^{\text{ab}}=K^{\text{ab.tot}}K^{\text{un}}$, with $Gal(K^{\text{ab}}/K)=\mathscr{O}_K^\times\times \hat{\mathbb{Z}}$. More than that, with the Artin map, we can also classify abelian extensions of $K$ with given degree.
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