The tilting correspondence should be my first step towards $p$-adic Hodge theory. The results from tilting and un-tilting process give us truly beautiful and surprising results about the connection between char. 0 and char. $p$. We already saw how to construct the tilt of a perfectoid field. Via this process, we obtain a complete, perfect, immediate field between $\widehat{L_\infty}^\flat$ and $\mathbb{C}_p^\flat$. Later, we will see how to un-tilt a complete, perfect, immediate field between $\widehat{L_\infty}^\flat$ and $\mathbb{C}_p$. We will again obtain a perfectoid, immediate field between $\widehat{L}_\infty$ and $\mathbb{C}_p$.
Let me explain how it works. The reduction from char. 0 to char. $p$ is often easy, because one just need to take $\mod p$. To do the converse, we need Witt vectors to lift from char. $p$ to char. 0 (For example: $W(\mathbb{F}_p)=\mathbb{Z}_p$). By using Witt vectors, the un-tilting process can be done. This gives us a bijective map between perfectoid fields (of char. 0) and perfect, complete field (of char. $p$). This would be the first step. This step also leads to some relations between the field of norm $E_L$ (it is basically the field $k_L((X))$) and $\mathbb{C}_p^\flat$, and $\widehat{L_\infty}^\flat$.
For the second step, we even go further, to generalize the result of Fontaine as mentioned in Part I. Because we are dealing with metric spaces basically, we need to equip Witt vectors a suitable topology, and it turns out that the weak topology on Witt vectors are natural. By using this, and the field of norm $E_L$, we deduce the topological isomorphism $Gal(\overline{\mathbb{Q}_p}/L_\infty)\cong Gal(k_L((X))^\text{sep}/k_L((X)))$.
The last step is even more interesting, we deduce the correspondence between finite (Galois) extensions of perfectoid fields and finite (Galois) extensions of their tilts. For me, it is a very amazing result, because one actually needs tools from field extensions of char. $p$ to deduce facts of field extension of char. 0. Both, in my mind, before I read these results, have no relation, but it turns out that they have very close relations, and as elementary arithmetic, char. $p$ is very useful, and somewhat easier than char. 0.
The notes will come very soon...
Sunday, February 11, 2018
Tuesday, January 30, 2018
[Tilts and Field of Norms I] Perfectoid Fields and Tilts
Let us fix these notations, $L/\mathbb{Q}_P$ a finite extension, with $\mathscr{O}:=\mathscr{O}_L$ is its ring of integers, $\pi:=\pi_L$ is a uniformizer, with $q:=\#k_L$, where $k_L$ is the residue field of $L$. $\overline{\mathbb{Q}_p}$ denotes the algebraic closure of $\mathbb{Q}_p$, and $\mathbb{C}_p$ its completion. Let $L_\infty$ be the totally ramified extension of $L$ as in the theory of Lubin-Tate.
When $L:=\mathbb{Q}_p$, $L_\infty=\mathbb{Q}_p(\zeta_{p^\infty})$, and a theorem of Fontaine yields $Gal(\overline{\mathbb{Q}_p}/\mathbb{Q}_p(\zeta_{p^\infty}))$ is isomorphic (as topological group) to $Gal(\mathbb{F}_p((t))^{\text{sep}}/\mathbb{F}_p((t)))$. This helps us classify all $p$-adic Galois representations, by the classification of Galois representations in char. $p$. Fontaine dealt with this by the theory of field of norms. It was generalized by Peter Scholze by the notions of perfectoid fields and their tilts. The goal of this part is to understand the construction of Peter Scholze.
Definition. Let $L\subset K\subset \mathbb{C}_p$ be an immediate field, then $K$ is said to be a perfectoid field if
i) $K$ is complete.
(ii) $|K^\times|$ is dense in $\mathbb{R}^\times_{>0}$.
(iii) The map $\mathscr{O}_K/p\mathscr{O}_K\to \mathscr{O}_K/p\mathscr{O}_K$ defined by $x\mapsto x^p$ is surjective.
Lemma 1.1. Let $K$ be a perfectoid field and $a\in K^\times$, then there exists $b\in K^\times$, such that $|a|=|b|^p$.
Proof. Because $K^\times$ is dense in $\mathbb{R}^\times_{>0}$, there exists some $\omega\in K^\times$ such that $|p|<|\omega|\le 1$, and some $m\in \mathbb{Z}$, such that $|\omega|^{m+1}\le |a|\le |\omega|^m$. From this, $|\omega|\le |a\omega^{-m}|\le 1$, and hence $|p|<|a\omega^{-m}|\le 1$, and $|a|=|\omega^m||a\omega^{-m}|$. And it is sufficient to prove that whenever $|p|<|a|\le 1$, there exists $b\in K^\times$, such that $|a|=|b|^p$. The condition (iii) of the definition above yields there exists some $b\in K^\times$, such that $|a-b^p|<|p|$. If $|a|\ne |b^p|$, then $|a-b^p|=\max\{|a|,|b^p|\}\ge |a|>|p|$, a contradiction. Hence, $|a|=|b|^p$. (Q.E.D)
Let us fix some $\omega\in K^\times$, where $K$ is a perfectoid fields, such that $|\omega|\ge |\pi|$ (so that $\omega \mathscr{O}_K\supset \pi \mathscr{O}_K\supset p\mathscr{O}_K$. We consider the following projective limit
$$\mathscr{O}_{K^\flat}:=\varprojlim (...\xrightarrow{(.)^q}\mathscr{O}_K/\omega \mathscr{O}_K\xrightarrow{(.)^q}\mathscr{O}_K/\omega \mathscr{O}_K\xrightarrow{(.)^q}\mathscr{O}_K/\omega\mathscr{O}_K)=$$
$$=\{(...,\alpha_i,...,\alpha_1,\alpha_0)|\alpha_i\in \mathscr{O}_K/\omega\mathscr{O}_K,\alpha_{i+1}^q=\alpha_i\}$$
Lemma 1.2. $\mathscr{O}_{K^\flat}$ is a perfect $k_L$-algebra.
Proof. There is a map from $(\mathscr{O}\mod \pi \mathscr{O})$ to $\mathscr{O}_{K^\flat}$ defined as
$$(a\mod \pi \mathscr{O}) \mapsto (...,a\mod \omega \mathscr{O}_K,...,a\mod \omega \mathscr{O}_K)$$
Because $a^q\equiv a \mod \pi \mathscr{O}$ for all $a\in \mathscr{O}$, we have $a^q\equiv a\mod \omega\mathscr{O}_K$, this yields a well-defined map from $k_L$ to $\mathscr{O}_K/\omega \mathscr{O}_K$. It is easy to check that this map is a ring homomorphism. Hence, $\mathscr{O}_{K^\flat}$ is a $k_L$-algebra.
Let us consider the map $\mathscr{O}_{K^\flat}\to \mathscr{O}_{K^\flat}$ defined as $\alpha \mapsto \alpha^q$. Assume that $\alpha^q:=(...,\alpha_i^q,...,\alpha_1^q,\alpha_0^q)=0$. The fact that $\alpha_{i+1}^q=\alpha_i$ yields $\alpha_i=0$, for all $i$, and hence, $\alpha=0$. Also, it is easy to see that $(...,\alpha_i,...,\alpha_1,\alpha_0)=(...,\alpha_i,...,\alpha_1)^q$. So, the map is also surjective. This yields $\mathscr{O}_{K^\flat}$ is perfect. (Q.E.D)
Now, for any $\alpha=(...,\alpha_i,...,\alpha_1,\alpha_0)$, we can lift $\alpha_i$ to $a_i\in \mathscr{O}_{K}$, such that $a_i\mod \omega \mathscr{O}_{K}=\alpha_i$, and we have $a_{i+1}^q=a_i\mod \omega \mathscr{O}_{K}$. And this yields
$$a_{i+1}^{q^{i+1}}\equiv a_i^{q^i}\mod \omega\mathscr{O}_{K}$$
so that the sequence $(a_i^{q^i})$ converges in $\mathscr{O}_{K}$. It can be checked easily that the limit of this sequence does not depend on the choice of $a_i$. And we denote this limit as $\alpha^\sharp$.
Lemma 1.3. The map
$$\varprojlim_{(.)^q}\mathscr{O}_K\xrightarrow{\psi} \mathscr{O}_{K^\flat}$$
defined by
$$(..., a_i,...,a_1,a_0)\mapsto (...,a_i\mod \omega\mathscr{O}_{K},...,a_1\mod \omega \mathscr{O}_{K},a_0\mod \omega \mathscr{O}_{K})$$
and the map
$$\mathscr{O}_{K^\flat}\xrightarrow{\theta} \varprojlim_{(.)^q}\mathscr{O}_{K}$$
defined by $\alpha\mapsto (...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)$ are multiplicative inverse of each other.
Proof. We can see that $\psi$ is well-defined. Also, if we denote $\alpha:=(...,\alpha_i,...,\alpha_1,\alpha_0)$, then it can be seen that $\alpha^{1/q^i}=(...,\alpha_{i+1},\alpha_i)$, and
$$(\alpha^{1/q^i})^\sharp=\lim_{j\to \infty}(a_{i+j}^{q^j})=(\lim_{k\to \infty}(a_k)^{q^k})^{1/q^i}=(\alpha^\sharp)^{1/q^i}$$
when we change variables $k=i+j$, and $a_i$ are lifts of $\alpha_j$. And this yields $\theta$ is also well-defined. Now, if we begin with $(...,a_i,...,a_1,...,a_0)\in \varprojlim_{(.)^q} \mathscr{O}_{K}$, then
$$\theta\psi(...,a_i,...,a_1,a_0)=\theta(...,a_i\mod \omega \mathscr{O}_{K},...,a_0\mod \omega \mathscr{O}_{K})=$$
$$=\theta (...,\alpha_i,...,\alpha_1,\alpha_0)=(...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp, \alpha^\sharp)$$
where $\alpha_i=a_i\mod \omega \mathscr{O}_{K}$, and $\alpha=(...,\alpha_i,...,\alpha_0)$. We have $(\alpha^{1/q^i})^\sharp=\lim_{j\to\infty}(a_{i+j}^{q^j})$. Because $a_{i+1}^q=a_i$, we have $a_j^{q^{i+j}}=a_i$. And hence $(\alpha^{1/q^i})^\sharp=a_i$. And hence, $\theta\circ \psi$ is just the identity map.
Now, if we begin with $\alpha=(...,\alpha_i,...,\alpha_1,\alpha_0)\in \mathscr{O}_{K^\flat}$, then we first note that $\alpha^\sharp\equiv \alpha_0\mod \omega \mathscr{O}_{K}$, hence
$$\psi\circ\theta(\alpha)=\psi(...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)=\alpha$$
So, this yields $\psi$ and $\theta$ are inverse of each other. The multiplicative properties are easy to check. (Q.E.D)
Our net goal is to prove that in fact $\mathscr{O}_{K^\flat}$ is an integral domain of char. $p$, and that it is complete. We first introduce the following norm on $\mathscr{O}_{K^\flat}$
$$||_\flat: \mathscr{O}_{K^\flat} \to \mathbb{R}$$
defined as $|\alpha|_\flat:=|\alpha^\sharp|$.
Proposition 1.4.
(i) The norm $|.|_\flat$ on $\mathscr{O}_{K^\flat}$ is non-archimedean.
(ii) $|\mathscr{O}_{K^\flat}|=|\mathscr{O}_{K}|$
(iii) For $\alpha,\beta\in \mathscr{O}_{K^\flat}$, $\alpha\mathscr{O}_{K^\flat}\subset \beta \mathscr{O}_{K^\flat}$ iff $|\alpha|_\flat\le |\beta|_\flat$.
(iv) $\mathscr{O}_{K^\flat}$ is a local domain of char. $p$, with the unique maximal ideal $\mathfrak{m}_{K^\flat}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat<1\}$.
(v) $\mathscr{O}_{K^\flat}/\mathfrak{m}_{K_\flat}\cong \mathscr{O}_{K}/\mathfrak{m}_K$.
(vi) Let $\omega^\flat\in \mathscr{O}_{K^\flat}$, such that $|\omega^\flat|_\flat=|\omega|$, then the map $\mathscr{O}_{K^\flat}/\omega^\flat\mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\omega \mathscr{O}_{K}$ defined as $\alpha\mapsto \alpha^\sharp \mod \omega \mathscr{O}_{K}$ is an isomorphism of rings.
Proof. We first fix $\alpha:=(...,\alpha_i,...,\alpha_1,\alpha_0), \beta:=(...,\beta_i,...,\beta_1,\beta_0)$ in $\mathscr{O}_{K^\flat}$, and $a_i:=(\alpha^{1/q^i})^\sharp, b_i:=(\beta^{1/q^i})^\sharp$, we know that $b_{i+1}^q=b_i, a_{i+1}^q=a_i$.
(i) We have
$$|\alpha+\beta|_\flat=|(\alpha+\beta)^\sharp|=|\lim_{i\to \infty}(a_i+b_i)^{q^i}|=\lim_{i\to \infty}|(a_i+b_i)^{q^i}|$$
$$\le \lim_{i\to \infty}\max\{|a^{q^i}|, |b^{q^i}|\}=\lim_{i\to \infty}\{|a_0|,b_0\}=\max\{|\alpha|^\sharp|,|\beta^\sharp|\}=\max\{|\alpha|_\flat,|\beta|_\flat\}$$
Also, assume that $|\alpha|_\flat=0$, this yields $\alpha^\sharp=a_0=0$, and this yields $\alpha=0$. The multiplicative property of $|.|_\flat$ is easy to check. So, it is a non-archimedean norm on $\mathscr{O}_{K^\flat}$.
(ii) From the definition, we have $|\mathscr{O}_{K^\flat}|_\flat\subseteq |\mathscr{O}_{K}|$. Take any $a\in \mathscr{O}_{K}$, we know that there exists some $b$, such that $|\omega|<|b|\le 1$, and $|a|=|b|^{q^m}$. We can find $\alpha\in \mathscr{O}_{K^\flat}$ such that $\alpha_0\equiv b\mod \omega \mathscr{O}_{K}$. This yields $\alpha^\sharp\equiv b\mod \omega \mathscr{O}_{K}$, and $|\beta^\sharp - b|\le |\omega|$. It follows that $|\beta^\sharp|=|b|$. So, we get $|\alpha|_\flat=|b|$, and $|a|=|\beta^{q^m}|_\flat$. So $\mathscr{O}_{K^\flat}=\mathscr{O}_{K}$.
(iii) Assume that $\alpha \mathscr{O}_{K^\flat}\subseteq \beta \mathscr{O}_{K^\flat}$. Then there exists some $\gamma\in \mathscr{O}_{K^\flat}$, such that $\alpha=\beta\gamma$, and this yields $|\alpha|_\flat\le |\beta|_\flat$. Conversely, assume $|\alpha|_\flat\le |\beta|_\flat$, which yields $|(\alpha^{1/q^i})^\sharp|\le |(\beta^{1/q^i})^\sharp|$, because $|\alpha^{1/q^i}|_\flat\le |\beta^{1/q^i}|_\flat$. And this yields $|a_i|\le |b_i|$, and there exists some $c_i\in \mathscr{O}_{K}$, such that $c_ia_i=b_i$. It follows directly that $c_{i+1}^q=c_i$. And hence, $\gamma:=(...,c_i\mod \omega \mathscr{O}_{K},...,c_1 \mod \omega \mathscr{O}_{K}, c_0\mod \omega \mathscr{O}_{K})$ defines an element in $\mathscr{O}_{K^\flat}$. And it is clear that $\alpha\gamma=\beta$, and $\alpha \mathscr{O}_{K^\flat}\subseteq \beta \mathscr{O}_{K^\flat}$.
(iv) Now, if we take any element $\gamma\in \mathscr{O}_{K^\flat}\setminus\mathfrak{m}_{K^\flat}$, then we can see by our recent argument that $\gamma \mathscr{O}_{K^\flat}=\mathscr{O}_{K^\flat}$, i.e. $\gamma$ is invertible. This yields $\mathscr{O}_{K^\flat}$ is local with maximal ideal $\mathfrak{m}_{K^\flat}$. Assume for now, $\alpha\beta=0$, this yields $|\alpha\beta|_\flat=|a_0b_0|=0$, and hence, $a_0=0$ or $b_0=0$. From this $\alpha=0$ or $\beta=0$. This implies $\mathscr{O}_{K^\flat}$ is a domain.
(v) Let us consider the map $\psi: \mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\mathfrak{m}_K$ defined by $\psi(\alpha)=\alpha^\sharp\mod \mathfrak{m}_K$. We can see easily that $\psi(\alpha\beta)=\psi(\alpha)\psi(\beta)$. Also,
$$\psi(\alpha+\beta)\equiv(\alpha+\beta)^\sharp\equiv a_0+b_0\mod \omega \mathscr{O}_{K^\flat}\equiv a_0+b_0\mod \mathfrak{m}_K$$
So, $\psi$ is a ring homomorphism. Take any $a_0\in \mathscr{O}_{K}$, we can find $a_1\in \mathscr{O}_{K}$ such that $a_1^q\equiv a_0\mod p\mathscr{O}_{K}$. It follows $a_1^q\equiv a_0\mod \mathfrak{m}_K$ and $a_1^q\equiv a_0\mod \omega \mathscr{O}_{K}$. Continuing this process, we get $\alpha:=(...,a_i\mod \omega \mathscr{O}_{K},...,a_0\mod \omega \mathscr{O}_{K})\in \mathscr{O}_{K^\flat}$, and $\alpha^\sharp\equiv a_0\mod \omega \mathscr{O}_{K}\equiv a_0\mod \mathfrak{m}_K$. And $\psi$ is surjective. From (iii), we have
$$\ker \psi=\{\alpha\in \mathscr{O}_{K^\flat}|\alpha^\sharp\in \mathfrak{m}_K\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha^\sharp|<1\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat<1\}=\mathfrak{m}_{K^\flat}$$
And this yields $\mathscr{O}_{K^\flat}/\mathfrak{m}_{K^\flat}=\mathscr{O}_{K}/\mathfrak{m}_K$.
(vi) It follows from (v) that the map $\theta: \mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\mathfrak{m}_K$ is surjective, with
$$\ker\theta=\{\alpha\in \mathscr{O}_{K^\flat}|\alpha^\sharp|\le |\omega|\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat\le |\omega^\flat|_\flat\}=\omega^\flat\mathscr{O}_{K^\flat}$$
So, $\mathscr{O}_{K^\flat}/\omega^\flat \mathscr{O}_{K^\flat}\cong \mathscr{O}_{K}/\omega \mathscr{O}_{K}$. (Q.E.D)
We will conclude this section by the following
Proposition 1.5. $\mathscr{O}_{K^\flat}$ is complete with respect to the norm $|.|_\flat$.
Proof. We have $\mathscr{O}_{K^\flat}=\varprojlim_{(.)^q}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$, and we can equip each $\mathscr{O}_{K}/\omega \mathscr{O}_{K}$ the discrete topology, and $\prod_{\mathbb{N}}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$ the product topology, and $\mathscr{O}_{K^\flat}$ is a topological subgroup of $\prod_{\mathbb{N}}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$, which has a fundamental system of open neighborhoods around $0$ defined as
$$U_m:=\{(...,a_{m+1},0,...,0)\} (m\ge 1)$$
and $U_1\supset U_{2}\supset ...$ forms a filtration. We will prove that with this topology, $\mathscr{O}_{K^\flat}$ is complete, and it coincides with the topology defined by $|.|_\flat$. For the first statement, it is sufficient to prove any Cauchy sequence converges in $\mathscr{O}_{K^\flat}$.
Let $(x_n)_n$ be a Cauchy sequence in $\mathscr{O}_{K^\flat}$. We can represent each $x_n$ as $(...,x_{n,i},...,x_{n,1}, x_{n,0})$, with $x_{n,i+1}^q=x_{n,i}$. And for all $k\ge 0$, there exists some $m_k$ such that $\forall m,n\ge m_k$, $x_m-x_n\in U_{k+1}$, and $m_{k+1}>m_k$. This yields $x_{m,i}-x_{n,i}=0(\forall 0\le i\le k+1; m,n\ge m_k)$.
Let $x:=(...,x_{m_i,i},...,x_{m_1,1},x_{m_0,0})$. We can see that $x_{m_i,i+1}=x_{m_{i+1},i+1}$, and hence $x_{m_i,i+1}^q=x_{m_i,i}=x_{m_{i+1},i+1}^q$. So, $x\in \mathscr{O}_{K^\flat}$. Now, for any $k\ge 0$, $n\ge m_k$, we have $x-x_n=(x-x_{m_k})-(x_n-x_{m_k})$. It can be seen that for any $0\le i\le k+1$, we have $x_{m_k,i}-x_{m_i,i}=0$, so $x-x_{m_k}\in U_{k+1}$, and $x_n-x_{m_k}\in U_{k+1}$. And this yields $(x_n)_n$ converges to $x$. Hence, with this topology, $\mathscr{O}_{K^\flat}$ is complete. On the other hand, we have
$$U_m=\{\alpha\in \mathscr{O}_{K^\flat}|(\alpha^{1/q^m})^\sharp\in \omega\mathscr{O}_{K}\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha^{1/q^m}|_\flat\le |\omega^\flat|_\flat\}=(\omega^\flat)^{q^m}\mathscr{O}_{K^\flat}$$
And hence $\{U_m\}_{m\ge 1}$ also forms a fundamental system of open neighborhoods around $0$ with the topology induced by $|.|_\flat$. It follows that the two topology coincide. And this yields $\mathscr{O}_{K^\flat}$ is complete. (Q.E.D)
For now, it makes sense to talk about $K^\flat$, the fraction field of $\mathscr{O}_{K^\flat}$. It is a field of characteristic $p$. By extending the norm $|.|_\flat$ to $K^\flat$, it is complete and non-archimedean. Also, the inverse map of $\psi$ in Lemma 1.3 can be extended to a multiplicative bijection
$$K^\flat \xrightarrow{\cong} \varprojlim_{(.)^q}K$$
defined by $\alpha \mapsto (...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)$.
Definition. $K^\flat$ is called the tilt of $K$.
When $L:=\mathbb{Q}_p$, $L_\infty=\mathbb{Q}_p(\zeta_{p^\infty})$, and a theorem of Fontaine yields $Gal(\overline{\mathbb{Q}_p}/\mathbb{Q}_p(\zeta_{p^\infty}))$ is isomorphic (as topological group) to $Gal(\mathbb{F}_p((t))^{\text{sep}}/\mathbb{F}_p((t)))$. This helps us classify all $p$-adic Galois representations, by the classification of Galois representations in char. $p$. Fontaine dealt with this by the theory of field of norms. It was generalized by Peter Scholze by the notions of perfectoid fields and their tilts. The goal of this part is to understand the construction of Peter Scholze.
Definition. Let $L\subset K\subset \mathbb{C}_p$ be an immediate field, then $K$ is said to be a perfectoid field if
i) $K$ is complete.
(ii) $|K^\times|$ is dense in $\mathbb{R}^\times_{>0}$.
(iii) The map $\mathscr{O}_K/p\mathscr{O}_K\to \mathscr{O}_K/p\mathscr{O}_K$ defined by $x\mapsto x^p$ is surjective.
Lemma 1.1. Let $K$ be a perfectoid field and $a\in K^\times$, then there exists $b\in K^\times$, such that $|a|=|b|^p$.
Proof. Because $K^\times$ is dense in $\mathbb{R}^\times_{>0}$, there exists some $\omega\in K^\times$ such that $|p|<|\omega|\le 1$, and some $m\in \mathbb{Z}$, such that $|\omega|^{m+1}\le |a|\le |\omega|^m$. From this, $|\omega|\le |a\omega^{-m}|\le 1$, and hence $|p|<|a\omega^{-m}|\le 1$, and $|a|=|\omega^m||a\omega^{-m}|$. And it is sufficient to prove that whenever $|p|<|a|\le 1$, there exists $b\in K^\times$, such that $|a|=|b|^p$. The condition (iii) of the definition above yields there exists some $b\in K^\times$, such that $|a-b^p|<|p|$. If $|a|\ne |b^p|$, then $|a-b^p|=\max\{|a|,|b^p|\}\ge |a|>|p|$, a contradiction. Hence, $|a|=|b|^p$. (Q.E.D)
Let us fix some $\omega\in K^\times$, where $K$ is a perfectoid fields, such that $|\omega|\ge |\pi|$ (so that $\omega \mathscr{O}_K\supset \pi \mathscr{O}_K\supset p\mathscr{O}_K$. We consider the following projective limit
$$\mathscr{O}_{K^\flat}:=\varprojlim (...\xrightarrow{(.)^q}\mathscr{O}_K/\omega \mathscr{O}_K\xrightarrow{(.)^q}\mathscr{O}_K/\omega \mathscr{O}_K\xrightarrow{(.)^q}\mathscr{O}_K/\omega\mathscr{O}_K)=$$
$$=\{(...,\alpha_i,...,\alpha_1,\alpha_0)|\alpha_i\in \mathscr{O}_K/\omega\mathscr{O}_K,\alpha_{i+1}^q=\alpha_i\}$$
Lemma 1.2. $\mathscr{O}_{K^\flat}$ is a perfect $k_L$-algebra.
Proof. There is a map from $(\mathscr{O}\mod \pi \mathscr{O})$ to $\mathscr{O}_{K^\flat}$ defined as
$$(a\mod \pi \mathscr{O}) \mapsto (...,a\mod \omega \mathscr{O}_K,...,a\mod \omega \mathscr{O}_K)$$
Because $a^q\equiv a \mod \pi \mathscr{O}$ for all $a\in \mathscr{O}$, we have $a^q\equiv a\mod \omega\mathscr{O}_K$, this yields a well-defined map from $k_L$ to $\mathscr{O}_K/\omega \mathscr{O}_K$. It is easy to check that this map is a ring homomorphism. Hence, $\mathscr{O}_{K^\flat}$ is a $k_L$-algebra.
Let us consider the map $\mathscr{O}_{K^\flat}\to \mathscr{O}_{K^\flat}$ defined as $\alpha \mapsto \alpha^q$. Assume that $\alpha^q:=(...,\alpha_i^q,...,\alpha_1^q,\alpha_0^q)=0$. The fact that $\alpha_{i+1}^q=\alpha_i$ yields $\alpha_i=0$, for all $i$, and hence, $\alpha=0$. Also, it is easy to see that $(...,\alpha_i,...,\alpha_1,\alpha_0)=(...,\alpha_i,...,\alpha_1)^q$. So, the map is also surjective. This yields $\mathscr{O}_{K^\flat}$ is perfect. (Q.E.D)
Now, for any $\alpha=(...,\alpha_i,...,\alpha_1,\alpha_0)$, we can lift $\alpha_i$ to $a_i\in \mathscr{O}_{K}$, such that $a_i\mod \omega \mathscr{O}_{K}=\alpha_i$, and we have $a_{i+1}^q=a_i\mod \omega \mathscr{O}_{K}$. And this yields
$$a_{i+1}^{q^{i+1}}\equiv a_i^{q^i}\mod \omega\mathscr{O}_{K}$$
so that the sequence $(a_i^{q^i})$ converges in $\mathscr{O}_{K}$. It can be checked easily that the limit of this sequence does not depend on the choice of $a_i$. And we denote this limit as $\alpha^\sharp$.
Lemma 1.3. The map
$$\varprojlim_{(.)^q}\mathscr{O}_K\xrightarrow{\psi} \mathscr{O}_{K^\flat}$$
defined by
$$(..., a_i,...,a_1,a_0)\mapsto (...,a_i\mod \omega\mathscr{O}_{K},...,a_1\mod \omega \mathscr{O}_{K},a_0\mod \omega \mathscr{O}_{K})$$
and the map
$$\mathscr{O}_{K^\flat}\xrightarrow{\theta} \varprojlim_{(.)^q}\mathscr{O}_{K}$$
defined by $\alpha\mapsto (...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)$ are multiplicative inverse of each other.
Proof. We can see that $\psi$ is well-defined. Also, if we denote $\alpha:=(...,\alpha_i,...,\alpha_1,\alpha_0)$, then it can be seen that $\alpha^{1/q^i}=(...,\alpha_{i+1},\alpha_i)$, and
$$(\alpha^{1/q^i})^\sharp=\lim_{j\to \infty}(a_{i+j}^{q^j})=(\lim_{k\to \infty}(a_k)^{q^k})^{1/q^i}=(\alpha^\sharp)^{1/q^i}$$
when we change variables $k=i+j$, and $a_i$ are lifts of $\alpha_j$. And this yields $\theta$ is also well-defined. Now, if we begin with $(...,a_i,...,a_1,...,a_0)\in \varprojlim_{(.)^q} \mathscr{O}_{K}$, then
$$\theta\psi(...,a_i,...,a_1,a_0)=\theta(...,a_i\mod \omega \mathscr{O}_{K},...,a_0\mod \omega \mathscr{O}_{K})=$$
$$=\theta (...,\alpha_i,...,\alpha_1,\alpha_0)=(...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp, \alpha^\sharp)$$
where $\alpha_i=a_i\mod \omega \mathscr{O}_{K}$, and $\alpha=(...,\alpha_i,...,\alpha_0)$. We have $(\alpha^{1/q^i})^\sharp=\lim_{j\to\infty}(a_{i+j}^{q^j})$. Because $a_{i+1}^q=a_i$, we have $a_j^{q^{i+j}}=a_i$. And hence $(\alpha^{1/q^i})^\sharp=a_i$. And hence, $\theta\circ \psi$ is just the identity map.
Now, if we begin with $\alpha=(...,\alpha_i,...,\alpha_1,\alpha_0)\in \mathscr{O}_{K^\flat}$, then we first note that $\alpha^\sharp\equiv \alpha_0\mod \omega \mathscr{O}_{K}$, hence
$$\psi\circ\theta(\alpha)=\psi(...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)=\alpha$$
So, this yields $\psi$ and $\theta$ are inverse of each other. The multiplicative properties are easy to check. (Q.E.D)
Our net goal is to prove that in fact $\mathscr{O}_{K^\flat}$ is an integral domain of char. $p$, and that it is complete. We first introduce the following norm on $\mathscr{O}_{K^\flat}$
$$||_\flat: \mathscr{O}_{K^\flat} \to \mathbb{R}$$
defined as $|\alpha|_\flat:=|\alpha^\sharp|$.
Proposition 1.4.
(i) The norm $|.|_\flat$ on $\mathscr{O}_{K^\flat}$ is non-archimedean.
(ii) $|\mathscr{O}_{K^\flat}|=|\mathscr{O}_{K}|$
(iii) For $\alpha,\beta\in \mathscr{O}_{K^\flat}$, $\alpha\mathscr{O}_{K^\flat}\subset \beta \mathscr{O}_{K^\flat}$ iff $|\alpha|_\flat\le |\beta|_\flat$.
(iv) $\mathscr{O}_{K^\flat}$ is a local domain of char. $p$, with the unique maximal ideal $\mathfrak{m}_{K^\flat}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat<1\}$.
(v) $\mathscr{O}_{K^\flat}/\mathfrak{m}_{K_\flat}\cong \mathscr{O}_{K}/\mathfrak{m}_K$.
(vi) Let $\omega^\flat\in \mathscr{O}_{K^\flat}$, such that $|\omega^\flat|_\flat=|\omega|$, then the map $\mathscr{O}_{K^\flat}/\omega^\flat\mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\omega \mathscr{O}_{K}$ defined as $\alpha\mapsto \alpha^\sharp \mod \omega \mathscr{O}_{K}$ is an isomorphism of rings.
Proof. We first fix $\alpha:=(...,\alpha_i,...,\alpha_1,\alpha_0), \beta:=(...,\beta_i,...,\beta_1,\beta_0)$ in $\mathscr{O}_{K^\flat}$, and $a_i:=(\alpha^{1/q^i})^\sharp, b_i:=(\beta^{1/q^i})^\sharp$, we know that $b_{i+1}^q=b_i, a_{i+1}^q=a_i$.
(i) We have
$$|\alpha+\beta|_\flat=|(\alpha+\beta)^\sharp|=|\lim_{i\to \infty}(a_i+b_i)^{q^i}|=\lim_{i\to \infty}|(a_i+b_i)^{q^i}|$$
$$\le \lim_{i\to \infty}\max\{|a^{q^i}|, |b^{q^i}|\}=\lim_{i\to \infty}\{|a_0|,b_0\}=\max\{|\alpha|^\sharp|,|\beta^\sharp|\}=\max\{|\alpha|_\flat,|\beta|_\flat\}$$
Also, assume that $|\alpha|_\flat=0$, this yields $\alpha^\sharp=a_0=0$, and this yields $\alpha=0$. The multiplicative property of $|.|_\flat$ is easy to check. So, it is a non-archimedean norm on $\mathscr{O}_{K^\flat}$.
(ii) From the definition, we have $|\mathscr{O}_{K^\flat}|_\flat\subseteq |\mathscr{O}_{K}|$. Take any $a\in \mathscr{O}_{K}$, we know that there exists some $b$, such that $|\omega|<|b|\le 1$, and $|a|=|b|^{q^m}$. We can find $\alpha\in \mathscr{O}_{K^\flat}$ such that $\alpha_0\equiv b\mod \omega \mathscr{O}_{K}$. This yields $\alpha^\sharp\equiv b\mod \omega \mathscr{O}_{K}$, and $|\beta^\sharp - b|\le |\omega|$. It follows that $|\beta^\sharp|=|b|$. So, we get $|\alpha|_\flat=|b|$, and $|a|=|\beta^{q^m}|_\flat$. So $\mathscr{O}_{K^\flat}=\mathscr{O}_{K}$.
(iii) Assume that $\alpha \mathscr{O}_{K^\flat}\subseteq \beta \mathscr{O}_{K^\flat}$. Then there exists some $\gamma\in \mathscr{O}_{K^\flat}$, such that $\alpha=\beta\gamma$, and this yields $|\alpha|_\flat\le |\beta|_\flat$. Conversely, assume $|\alpha|_\flat\le |\beta|_\flat$, which yields $|(\alpha^{1/q^i})^\sharp|\le |(\beta^{1/q^i})^\sharp|$, because $|\alpha^{1/q^i}|_\flat\le |\beta^{1/q^i}|_\flat$. And this yields $|a_i|\le |b_i|$, and there exists some $c_i\in \mathscr{O}_{K}$, such that $c_ia_i=b_i$. It follows directly that $c_{i+1}^q=c_i$. And hence, $\gamma:=(...,c_i\mod \omega \mathscr{O}_{K},...,c_1 \mod \omega \mathscr{O}_{K}, c_0\mod \omega \mathscr{O}_{K})$ defines an element in $\mathscr{O}_{K^\flat}$. And it is clear that $\alpha\gamma=\beta$, and $\alpha \mathscr{O}_{K^\flat}\subseteq \beta \mathscr{O}_{K^\flat}$.
(iv) Now, if we take any element $\gamma\in \mathscr{O}_{K^\flat}\setminus\mathfrak{m}_{K^\flat}$, then we can see by our recent argument that $\gamma \mathscr{O}_{K^\flat}=\mathscr{O}_{K^\flat}$, i.e. $\gamma$ is invertible. This yields $\mathscr{O}_{K^\flat}$ is local with maximal ideal $\mathfrak{m}_{K^\flat}$. Assume for now, $\alpha\beta=0$, this yields $|\alpha\beta|_\flat=|a_0b_0|=0$, and hence, $a_0=0$ or $b_0=0$. From this $\alpha=0$ or $\beta=0$. This implies $\mathscr{O}_{K^\flat}$ is a domain.
(v) Let us consider the map $\psi: \mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\mathfrak{m}_K$ defined by $\psi(\alpha)=\alpha^\sharp\mod \mathfrak{m}_K$. We can see easily that $\psi(\alpha\beta)=\psi(\alpha)\psi(\beta)$. Also,
$$\psi(\alpha+\beta)\equiv(\alpha+\beta)^\sharp\equiv a_0+b_0\mod \omega \mathscr{O}_{K^\flat}\equiv a_0+b_0\mod \mathfrak{m}_K$$
So, $\psi$ is a ring homomorphism. Take any $a_0\in \mathscr{O}_{K}$, we can find $a_1\in \mathscr{O}_{K}$ such that $a_1^q\equiv a_0\mod p\mathscr{O}_{K}$. It follows $a_1^q\equiv a_0\mod \mathfrak{m}_K$ and $a_1^q\equiv a_0\mod \omega \mathscr{O}_{K}$. Continuing this process, we get $\alpha:=(...,a_i\mod \omega \mathscr{O}_{K},...,a_0\mod \omega \mathscr{O}_{K})\in \mathscr{O}_{K^\flat}$, and $\alpha^\sharp\equiv a_0\mod \omega \mathscr{O}_{K}\equiv a_0\mod \mathfrak{m}_K$. And $\psi$ is surjective. From (iii), we have
$$\ker \psi=\{\alpha\in \mathscr{O}_{K^\flat}|\alpha^\sharp\in \mathfrak{m}_K\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha^\sharp|<1\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat<1\}=\mathfrak{m}_{K^\flat}$$
And this yields $\mathscr{O}_{K^\flat}/\mathfrak{m}_{K^\flat}=\mathscr{O}_{K}/\mathfrak{m}_K$.
(vi) It follows from (v) that the map $\theta: \mathscr{O}_{K^\flat}\to \mathscr{O}_{K}/\mathfrak{m}_K$ is surjective, with
$$\ker\theta=\{\alpha\in \mathscr{O}_{K^\flat}|\alpha^\sharp|\le |\omega|\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha|_\flat\le |\omega^\flat|_\flat\}=\omega^\flat\mathscr{O}_{K^\flat}$$
So, $\mathscr{O}_{K^\flat}/\omega^\flat \mathscr{O}_{K^\flat}\cong \mathscr{O}_{K}/\omega \mathscr{O}_{K}$. (Q.E.D)
We will conclude this section by the following
Proposition 1.5. $\mathscr{O}_{K^\flat}$ is complete with respect to the norm $|.|_\flat$.
Proof. We have $\mathscr{O}_{K^\flat}=\varprojlim_{(.)^q}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$, and we can equip each $\mathscr{O}_{K}/\omega \mathscr{O}_{K}$ the discrete topology, and $\prod_{\mathbb{N}}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$ the product topology, and $\mathscr{O}_{K^\flat}$ is a topological subgroup of $\prod_{\mathbb{N}}\mathscr{O}_{K}/\omega \mathscr{O}_{K}$, which has a fundamental system of open neighborhoods around $0$ defined as
$$U_m:=\{(...,a_{m+1},0,...,0)\} (m\ge 1)$$
and $U_1\supset U_{2}\supset ...$ forms a filtration. We will prove that with this topology, $\mathscr{O}_{K^\flat}$ is complete, and it coincides with the topology defined by $|.|_\flat$. For the first statement, it is sufficient to prove any Cauchy sequence converges in $\mathscr{O}_{K^\flat}$.
Let $(x_n)_n$ be a Cauchy sequence in $\mathscr{O}_{K^\flat}$. We can represent each $x_n$ as $(...,x_{n,i},...,x_{n,1}, x_{n,0})$, with $x_{n,i+1}^q=x_{n,i}$. And for all $k\ge 0$, there exists some $m_k$ such that $\forall m,n\ge m_k$, $x_m-x_n\in U_{k+1}$, and $m_{k+1}>m_k$. This yields $x_{m,i}-x_{n,i}=0(\forall 0\le i\le k+1; m,n\ge m_k)$.
Let $x:=(...,x_{m_i,i},...,x_{m_1,1},x_{m_0,0})$. We can see that $x_{m_i,i+1}=x_{m_{i+1},i+1}$, and hence $x_{m_i,i+1}^q=x_{m_i,i}=x_{m_{i+1},i+1}^q$. So, $x\in \mathscr{O}_{K^\flat}$. Now, for any $k\ge 0$, $n\ge m_k$, we have $x-x_n=(x-x_{m_k})-(x_n-x_{m_k})$. It can be seen that for any $0\le i\le k+1$, we have $x_{m_k,i}-x_{m_i,i}=0$, so $x-x_{m_k}\in U_{k+1}$, and $x_n-x_{m_k}\in U_{k+1}$. And this yields $(x_n)_n$ converges to $x$. Hence, with this topology, $\mathscr{O}_{K^\flat}$ is complete. On the other hand, we have
$$U_m=\{\alpha\in \mathscr{O}_{K^\flat}|(\alpha^{1/q^m})^\sharp\in \omega\mathscr{O}_{K}\}=\{\alpha\in \mathscr{O}_{K^\flat}||\alpha^{1/q^m}|_\flat\le |\omega^\flat|_\flat\}=(\omega^\flat)^{q^m}\mathscr{O}_{K^\flat}$$
And hence $\{U_m\}_{m\ge 1}$ also forms a fundamental system of open neighborhoods around $0$ with the topology induced by $|.|_\flat$. It follows that the two topology coincide. And this yields $\mathscr{O}_{K^\flat}$ is complete. (Q.E.D)
For now, it makes sense to talk about $K^\flat$, the fraction field of $\mathscr{O}_{K^\flat}$. It is a field of characteristic $p$. By extending the norm $|.|_\flat$ to $K^\flat$, it is complete and non-archimedean. Also, the inverse map of $\psi$ in Lemma 1.3 can be extended to a multiplicative bijection
$$K^\flat \xrightarrow{\cong} \varprojlim_{(.)^q}K$$
defined by $\alpha \mapsto (...,(\alpha^{1/q^i})^\sharp,...,(\alpha^{1/q})^\sharp,\alpha^\sharp)$.
Definition. $K^\flat$ is called the tilt of $K$.
Wednesday, January 24, 2018
[Topology on Witt Vectors III] Weak Topology on Witt Vectors
We still fix these notations, $L$ is a non-archimedean local field, $k:=k_L$ is a residue field of $L$, $\mathscr{O}:=\mathscr{O}_L$ is its ring of integers with $\pi:=\pi_L$ is a fixed uniformizer, $q=\#k_L$, $B$ is a perfect topological $k_L$-algebra, and $W(B)$ is the ring of Witt vectors with coefficients in $L$. For simplicity, we will denote the addition and multiplication on $W(B)$ as usual, instead of $\boxplus, \boxdot$.
For any open ideal $\mathfrak{a}$ of $B$, we define
$$V_{\mathfrak{a}, m}:=\ker(W(B)\xrightarrow{pr}W_m(B)\xrightarrow{W(pr)}W_m(B/\mathfrak{a}))=$$
$$=\{(b_0,...,b_{m-1})\in W(B)|b_0,...,b_{m-1}\in \mathfrak{a}\}$$
We can see that $V_{\mathfrak{a},m}$ is an ideal of $W(B)$, and
$$V_{\mathfrak{a}\cap \mathfrak{b}, \max{m,n}}\subset V_{\mathfrak{a},m}\cap V_{\mathfrak{b},n}$$
For any open ideal $\mathfrak{b}$ of $B$. And hence, by Proposition 1.6, there exists a unique topological structure on $W(B)$ such that $W(B)$ is a topological ring and that such $V_{\mathfrak{a},m}$ become a fundamental system of open neighborhoods around $0$. If we consider
$$W_m:=\pi^mW(B)=\{(0,...,0,b_m,...)\in W(B)|b_m,b_{m+1},...\in B\}$$
then for any $V_{\mathfrak{a},m}$, we always have $W_m\subset V_{\mathfrak{a},m}$. So, the topology on $W(B)$ we have equipped is weaker than the $\pi$-adic topology on $W(B)$. We call it the weak topology on $W(B)$.
Lemma 3.1. For any $a=(a_0,a_1,...)\in W(B)$, we have
$$a + W_{\mathfrak{a},m}=\{(b_0,b_1,...)\in W(B)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\}$$
Hence, the weak topology on $W(B)$ coincides with the product topology on $B\times B\times ...$
Proof. Take any $(c_0,c_1,...)\in V_{\mathfrak{a},m}$, i.e. $c_0,...,c_{m-1}\in \mathfrak{a}$, we have
$$(a_0,a_1,...) + (c_0,c_1,...)=(a_0+c_0,...)=:(b_0,b_1,...)$$
We can see that $b_0=a_0+c_0\equiv a_0\mod \mathfrak{a}$. Assume that $b_i\equiv a_i\mod \mathfrak{a}$ holds to $n-1$ where $1\le n\le m-1$, we will prove that this holds for $n$. By the addition formula for Witt vectors, we have
$$\Phi_n(a_0,...,a_n)+\Phi_n(c_0,...,c_n)=\Phi_n(b_0,...,b_n)$$
Assume that $b_i=a_i+d_i$ for $d_i\in \mathfrak{a}$, $0\le i\le n-1$, we deduce from the definition of Witt polynomials that
$$\Phi_{n-1}(a_0^q,...,a_{n-1}^q)+\Phi_{n-1}(c_0^q,...,c_{n-1}^q)+\pi^n(a_n+c_n)=\Phi_{n-1}(b_0^q,...,b_{n-1}^q)+\pi^nb_n$$
And this yields
$$\frac{\Phi_{n-1}(a_0^1,...,a_{n-1}^q)+\Phi_{n-1}(c_0^q,...,c_{n-1}^q)-\Phi_{n-1}(b_0^q,...,b_{n-1}^q)}{\pi^n}+a_n+c_n=b_n (*)$$
And we know that
$$\Phi_{n-1}(b_0^q,...,b_{n-1}^q)=\Phi_{n-1}((a_0+d_0)^q,...,(a_{n-1}+d_{n-1})^q)=$$
$$=\Phi_{n-1}(a_0^q+d_0^q,...,a_{n-1}^q+d_{n-1}^q)=(a_0^q+d_0^q)^{q^{n-1}}+...+\pi^{n-1}(a_{n-1}^q+d_{n-1}^q)$$
$$=\Phi_{n-1}(a_0^q,...,a_{n-1}^q)+\Phi_{n-1}(d_0^q,...,d_{n-1}^q)$$
And from $(*)$, we get
$$d+a_n+c_n=b$$
for some $d\in \mathfrak{a}$, and this yields $b_n\equiv a_n\mod \mathfrak{a}$, since $c_n$ is also in $\mathfrak{a}$. We then get
$$a + V_{\mathfrak{a},m}\subset \{(b_0,...,b_{m-1},...)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\} $$
For the converse direction, with the same argument, we deduce that for $b_i\equiv a_i\mod \mathfrak{a}$, for all $0\le i\le m-1$,
$$(b_0,...,b_{m-1},...) - (a_0,...,a_{m-1},...)=(c_0,...,c_{m-1},...)$$
with $c_i\in \mathfrak{a}$, for $0\le i\le m-1$. For the second statement, we can see by the first statement that the set
$$a+V_{\mathfrak{a},m}=\{(b_0,...,b_{m-1})\in W(B)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\}$$
forms a fundamental system of open neighborhoods around $a$. And this follows directly that the weak topology on $W(B)$ is the same as the product topology $B\times B\times ...$. (Q.E.D)
Via this lemma, we can prove
Proposition 3.2. If $B$ is Hausdorff (complete), then $W(B)$ is Hausdorff (complete, resp.).
Proof. It follows easily that if $B$ is Hausdorff then the product topology $B\times B\times...$ is also Hausdorff. Now, assume that $B$ is complete. In this case, the canonical map
$$\phi: B\to \varprojlim_{\mathfrak{a}} B/\mathfrak{a} $$
is surjective. Let $\mathfrak{c}$ be its kernel, we have $B/\mathfrak{c}\cong \varprojlim_{\mathfrak{a}} B/\mathfrak{a}$. And this yields
$$W_m(B/\mathfrak{c})\cong W_m(\varprojlim_{\mathfrak{a}} B/\mathfrak{a})\cong \varprojlim_{\mathfrak{a}} W_m(B/\mathfrak{a})=\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
where the second isomorphism comes from the functorial properties of Witt vectors, and the last isomorphism follows from the fact that the map $W(B)\xrightarrow{W(pr)} W(B/\mathfrak{a})\xrightarrow{pr} W_m(B/\mathfrak{a})$ has kernel $V_{\mathfrak{a},m}$.
Now, it follows from our (W.4) in the chapter about Witt vectors that
$$W(B/\mathfrak{c})\cong \varprojlim_{m}W_m(B/\mathfrak{c})\cong \varprojlim_m\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
And so, we obtain the following surjective map
$$W(B)\xrightarrow{W(pr)} W(B/\mathfrak{c})\cong \varprojlim_m\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
And this yields $W(B)$ is complete. (Q.E.D)
Remark 3.3. In the case $B$ is complete and Hausdorff, we can equip the topological structure on the ring $W_m(B)$, with $m$ is fixed, by the method in the proof, such that $W_m(B)$ is Hausdorff and complete.
Proof. In this case, the canonical map $B\to \varprojlim_{\mathfrak{a}}B/\mathfrak{a}$ is bijective, and this yields by the previous proof that
$$W(B)/V_m(B)=W_m(B)\cong \varprojlim_{\mathfrak{a}}W_m(B/\mathfrak{a})\cong$$
$$\cong \varprojlim_{\mathfrak{a}} W(B)/V_{\mathfrak{a},m}\cong \varprojlim_{\mathfrak{a}}(W(B)/V_m(B))/(V_{\mathfrak{a},m}/V_m(B))$$
From this, there exists a unique topological structure on $W_m(B)$, such that $\{V_{\mathfrak{a},m}/V_m(B)|\mathfrak{a}\subset B: \text{ open ideal }\}$ becomes a fundamental system of open neighborhood around $0$. And $W_m(B)$ is also Hausdorff, and complete. (Q.E.D)
We will be mainly interested in the case $B:=\mathscr{O}_F$, where $F$ is a complete, non-archimedean, perfect field containing $k$. In this case, we get $W(B)$ is Hausdorff, complete, and is a subring of $W(F)$.
Lemma 3.4. Let $\mathscr{O}_F$ be as above, then an ideal $\mathfrak{a}$ of $\mathscr{O}_F$ is open iff $\mathfrak{a}$ is non-zero.
Proof. Assume that $\mathfrak{a}$ is open, then it is obvious that $\mathfrak{a}$ is non-zero. Now, let $\mathfrak{a}\subset \mathscr{O}_F$ be any non-zero ideal. Take $0\ne x\in \mathfrak{a}$, it is sufficient to prove that $(x)$-the ideal generated by $x$ is open in $\mathscr{O}_F$. We can see that
$$(x) =\{y\in \mathscr{O}_F||y|\le |x|)\}$$
Let us take any $z\in \mathscr{O}_F$, such that $|y-z|<|x-y|$. This yields $|y-z|< \max\{x,y\}\le |x|$. From this, we have $|z|\le |x|$, and $z\in (x)$. This yields $(x)$ is open, and hence, $\mathfrak{a}$ is open. (Q.E.D)
We can define for any open ideal $\mathfrak{a}$ of $\mathscr{O}_F$, and any $m\ge 1$ an $\mathscr{O}_F$-submodule
$$U_{\mathfrak{a},m}:=V_{\mathfrak{a},m} + \pi^mW(F):=\{(b_0,...,b_{m-1},...)\in W(F)|b_0,...,b_{m-1}\in \mathfrak{a}\}$$
We note that $U_{\mathfrak{a},m}$ are not ideals of $W(F)$, and we again have
$$U_{\mathfrak{a}\cap \mathfrak{b},\max\{m,n\}}\subset U_{\mathfrak{a},m}\cap U_{\mathfrak{b},n}$$
And these subgroups satisfy the conditions of Proposition 1.3. So, there exists a unique topology on $W(F)$, such that $W(F)$ is a topological group, and $U_{\mathfrak{a},m}$ forms a fundamental system of neighborhoods around $0$. We recall that because $F$ is an extension of $k$, and perfect, $W(F)$ is a D.V.R, with maximal ideal generated by $\pi$. Again, the topology we have equipped for $W(F)$ is weaker that the $\pi$-adic topology. We can call it the weak topology on $W(F)$. We actually want to prove that this topology actually defines a topological ring on $W(F)$, and that when $\mathscr{O}_F$ admits a filtered fundamental system, then $W(F)$ is complete.
We will need the multiplicative property of Teichmuller's representative.
Lemma 3.5.
(i) Let $a_1,...,a_r\in W(F)$, then there exists $0\ne \alpha\in \mathscr{O}_F$, such that
$$\tau(\alpha)a_1,...,\tau(\alpha)a_r\in U_{\mathscr{O}_F,m}$$
(ii) Let $\mathfrak{a}$ be an open ideal of $\mathscr{O}_F$, then for any $0\ne \alpha\in \mathscr{O}_F$, and $m\ge 1$, we have
$$\tau(\alpha^{-1})U_{\alpha^{q^{m-1}}\mathfrak{a},m}\subset U_{\mathfrak{a},m}$$
Proof.
(i) By Proposition 4.6 in the section about Witt vectors, we can represent
$$a_i=\sum_{j\ge 0}\tau(a_{i,j})\pi^j$$
And from this
$$\tau(\alpha)a_i=\sum_{j\ge 0}\tau(\alpha a_{i,j})\pi^j=(\alpha a_{i,0},\alpha a_{i,1},...)$$
And we can choose $\alpha$ such that $\alpha a_{i,j}\in \mathscr{O}_F$, for all $1\le i\le r, 0\le j\le m-1$.
(ii) Take $a=(a_0,a_1,...)\in \alpha^{q^{m-1}}\mathfrak{a}$, we can represent
$$(\alpha_0,\alpha_1,...)=\sum_{i\ge 0}\tau(a_i^{1/q^i})\pi^i$$
Hence
$$\tau(\alpha^{-1})a=\sum_{i\ge 0}\tau(\alpha^{-1}a_i^{1/q^i})\pi^i=\sum_{i\ge 0}(\alpha^{-q^i}a_i)$$
And hence, $\alpha^{-q^i}a_i\in U_{\mathfrak{a},m}$, for all $0\le i\le m-1$. (Q.E.D)
We are now ready for the main result of this section.
Proposition 3.7.
(i) $W(F)$ is a Hausdorff topological ring.
(ii) If $\mathscr{O}_F$ admits a filtered fundamental system, then so is $W(F)$, and in this case, $W(F)$ is complete.
Proof.
(i) We will prove that the multiplication map
$$W(F)\times W(F)\to W(F)$$
is continuous. Take any $a,b\in W(F)$, and an open neighborhood of $ab+U_{\mathfrak{a},m}$, for some open ideal $\mathfrak{a}$ of $\mathscr{O}_F$, and $m\ge 1$. By Lemma 3.6, one can find $0\ne \alpha\in \mathscr{O}_F$ such that $\tau(\alpha)a,\tau(\alpha)b\in U_{\mathscr{O}_F,m}$, which is equivalent to $a,b\in \tau(\alpha^{-1})U_{\mathscr{O}_F,m}$. By Lemma 3.5 again, we have
$$(a+U_{\alpha^{q^{m-1}}\mathfrak{a},m})(b+U_{\alpha^{q^{m-1}}\mathfrak{a},m})\subset ab+U_{\mathscr{O}_F,m}U_{\mathfrak{a},m}+U_{\mathfrak{a},m}\subset ab+ U_{\mathfrak{a},m}$$
And by Lemma 3.4, $U_{\alpha^{q^{m-1}}\mathfrak{a},m}$ is open. Hence, $W(F)$ is a topological ring. Moreover, one get easily that
$$\cap_{\mathfrak{a},m}U_{\mathfrak{a},m}=\cap_{\mathfrak{a},m}\{(b_0,...,b_{m-1},...)\in W(F)|b_i\in \mathfrak{a}\}=0$$
since $\mathscr{O}_F$ is Hausdorff, and the intersection of all open ideals is just 0.
(ii) Let $I_1\supset I_2\supset ...$ be a filtered fundamental system of $\mathscr{O}_F$. Let
$$U_{n,m}:=\{(b_0,...,b_{m-1},...)\in W(F)|b_0,...,b_{m-1}\in I_n\}$$
And $U_m:=U_{m,m}$. Then it can be seen easily that $\{U_{n,m}\}$ forms a fundamental system of open neighborhood around $0$ in $W(F)$, because for any open ideal $\mathfrak{a}\in \mathscr{O}_F$, there exists $n$ such that $I_n\subset \mathfrak{a}$. And one has $U_{n,m}\subset U_{\min{n,m}}, U_m=U_{m,m}$. So $\{U_m\}$ also forms a fundamental system around $0$ in $W(F)$. And this yields $W(F)$ admits a filtration.
Now, in the case $\mathscr{O}_F$ is complete, and admits a filtered fundamental system, it is sufficient to prove any Cauchy sequence converges in $W(F)$. The main ideal of the proof is that we will use Lemma 3.5 to reduce the induced Cauchy sequence to $W_m(\mathscr{O}_F)$, which is complete, by Remark 3.3. And then, with the completeness of the $\pi$-adic topology on $W(F)$, we will prove that our sequence converges in $W(F)$.
Take any $(a_n)_n$ is a Cauchy sequence in $W(F)$. Fix an integer $m\ge 1$, then for any $\mathfrak{a}$: open ideal in $\mathscr{O}_F$, there exists an integer $n_{\mathfrak{a}}$ such that for all $n,n'\ge n_\mathfrak{a}$, $a_n-a_{n'}\in U_{\mathfrak{a},m}$. Then by Lemma 3.5, we can choose $0\ne \alpha\in \mathscr{O}_F$, such that
$$\tau(\alpha) a_1,...,\tau(\alpha) a_{n_\mathfrak{a}}\in U_{\mathscr{O}_F,m}$$
And hence, for all $n\ge n_\mathfrak{a}$, we have
$$\tau(\alpha)(a_n-a_{n_\mathfrak{a}})\in \tau(\alpha)U_{\mathscr{O}_F,m}\subset U_{\mathscr{O}_F,m}$$
And hence $(\tau(\alpha)a_n)_n\in U_{\mathscr{O}_F,m}$, and for $n,n'\ge n_{\mathfrak{a}}$, we have
$$\tau(\alpha)(a_n-a_{n'})\in \tau(\alpha)U_{\mathfrak{a},m}\subset U_{\mathfrak{a},m}$$
Take $(b_n)_n\in W(\mathscr{O}_F)$ such that $\tau(\alpha)-b_n\in \pi^mW(F)$. We then have
$$b_n-b_{n'}\in (\tau(\alpha)(a_m-a_n)+\pi^mW(F))\cap W(\mathscr{O}_F)\subset (U_{\mathfrak{a},m}+\pi^m W(F))\cap W(\mathscr{O}_F)=V_{\mathfrak{a},m}$$
for all $n,n'\ge n_{\alpha}$. This yields the sequence $(b_n\mod V_m(\mathscr{O}_F))_n$ is Cauchy, and hence, converges to some $b\mod V_m(\mathscr{O}_F)$ in $W_m(\mathscr{O}_F)$. Hence, for any open ideal $\mathfrak{b}\subset \mathscr{O}_F$, there exists $n_{\mathfrak{b}}$ such that for all $n\ge n_{\mathfrak{b}}$, we have $b-b_n\in V_{\alpha^{q^{m-1}}\mathfrak{b},m}$. Let us denote $a(m):=\tau(\alpha^{-1})b$, then
$$a(m)-a_n=\tau(\alpha^{-1})b-a_n= \tau(\alpha^{-1})(b-\tau(\alpha)a_n)=\tau(\alpha^{-1})(b-b_n+b_n-\tau(\alpha)a_n)$$
$$\subset \tau(\alpha^{-1})(V_{\alpha^{q^{m-1}}\mathfrak{b},m}+\pi^mW(F))\subset U_{\mathfrak{b},m}+\pi^mW(F)\subset U_{\mathfrak{a},m}$$
for all $n\ge n_{\mathfrak{a}}$. Now, if we vary $m$, then we will get
$$a(m+1)-a(m)=(a(m+1)-a_n)-(a(m)-a_n)\in U_{\mathfrak{b},m}$$
for $n$ is sufficiently large. That means
$$a_{m+1}-a_m\in \{(b_0,...,b_{m-1})\in W(B)|b_0,...,b_{m-1}\in \mathfrak{b}, \forall \mathfrak{b}\subset \mathscr{O}_F: \text{ open}\}$$
And this yields $a(m+1)-a(m)\in \pi^mW(F)$. Now, this yields $(a(m))_m$ is a Cauchy sequence with respect to the $\pi$-adic topology, and hence, a Cauchy sequence in the weak topology. Let $a$ be the convergent value of $(a(m))_m$ in the $\pi$-adic topology, we will prove that $a$ is also the convergent value of $(a_n)_n$. For any ideal $\mathfrak{a}\subset \mathscr{O}_F$: open, and any $m$, we have $\pi^m W(F)\subset U_{\mathfrak{a},m}$, and there exists some $n'$ such that $a-a(n')\in \pi^mW(F)$ and $a(n')-a_n\in U_{\mathfrak{a},m}$, for some $n\ge n_\mathfrak{a}$. Hence, $(a-a_n)\in U_{\mathfrak{a},m}$. This yields $a$ is the convergent value of $W(F)$, and $W(F)$ is complete. (Q.E.D)
For any open ideal $\mathfrak{a}$ of $B$, we define
$$V_{\mathfrak{a}, m}:=\ker(W(B)\xrightarrow{pr}W_m(B)\xrightarrow{W(pr)}W_m(B/\mathfrak{a}))=$$
$$=\{(b_0,...,b_{m-1})\in W(B)|b_0,...,b_{m-1}\in \mathfrak{a}\}$$
We can see that $V_{\mathfrak{a},m}$ is an ideal of $W(B)$, and
$$V_{\mathfrak{a}\cap \mathfrak{b}, \max{m,n}}\subset V_{\mathfrak{a},m}\cap V_{\mathfrak{b},n}$$
For any open ideal $\mathfrak{b}$ of $B$. And hence, by Proposition 1.6, there exists a unique topological structure on $W(B)$ such that $W(B)$ is a topological ring and that such $V_{\mathfrak{a},m}$ become a fundamental system of open neighborhoods around $0$. If we consider
$$W_m:=\pi^mW(B)=\{(0,...,0,b_m,...)\in W(B)|b_m,b_{m+1},...\in B\}$$
then for any $V_{\mathfrak{a},m}$, we always have $W_m\subset V_{\mathfrak{a},m}$. So, the topology on $W(B)$ we have equipped is weaker than the $\pi$-adic topology on $W(B)$. We call it the weak topology on $W(B)$.
Lemma 3.1. For any $a=(a_0,a_1,...)\in W(B)$, we have
$$a + W_{\mathfrak{a},m}=\{(b_0,b_1,...)\in W(B)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\}$$
Hence, the weak topology on $W(B)$ coincides with the product topology on $B\times B\times ...$
Proof. Take any $(c_0,c_1,...)\in V_{\mathfrak{a},m}$, i.e. $c_0,...,c_{m-1}\in \mathfrak{a}$, we have
$$(a_0,a_1,...) + (c_0,c_1,...)=(a_0+c_0,...)=:(b_0,b_1,...)$$
We can see that $b_0=a_0+c_0\equiv a_0\mod \mathfrak{a}$. Assume that $b_i\equiv a_i\mod \mathfrak{a}$ holds to $n-1$ where $1\le n\le m-1$, we will prove that this holds for $n$. By the addition formula for Witt vectors, we have
$$\Phi_n(a_0,...,a_n)+\Phi_n(c_0,...,c_n)=\Phi_n(b_0,...,b_n)$$
Assume that $b_i=a_i+d_i$ for $d_i\in \mathfrak{a}$, $0\le i\le n-1$, we deduce from the definition of Witt polynomials that
$$\Phi_{n-1}(a_0^q,...,a_{n-1}^q)+\Phi_{n-1}(c_0^q,...,c_{n-1}^q)+\pi^n(a_n+c_n)=\Phi_{n-1}(b_0^q,...,b_{n-1}^q)+\pi^nb_n$$
And this yields
$$\frac{\Phi_{n-1}(a_0^1,...,a_{n-1}^q)+\Phi_{n-1}(c_0^q,...,c_{n-1}^q)-\Phi_{n-1}(b_0^q,...,b_{n-1}^q)}{\pi^n}+a_n+c_n=b_n (*)$$
And we know that
$$\Phi_{n-1}(b_0^q,...,b_{n-1}^q)=\Phi_{n-1}((a_0+d_0)^q,...,(a_{n-1}+d_{n-1})^q)=$$
$$=\Phi_{n-1}(a_0^q+d_0^q,...,a_{n-1}^q+d_{n-1}^q)=(a_0^q+d_0^q)^{q^{n-1}}+...+\pi^{n-1}(a_{n-1}^q+d_{n-1}^q)$$
$$=\Phi_{n-1}(a_0^q,...,a_{n-1}^q)+\Phi_{n-1}(d_0^q,...,d_{n-1}^q)$$
And from $(*)$, we get
$$d+a_n+c_n=b$$
for some $d\in \mathfrak{a}$, and this yields $b_n\equiv a_n\mod \mathfrak{a}$, since $c_n$ is also in $\mathfrak{a}$. We then get
$$a + V_{\mathfrak{a},m}\subset \{(b_0,...,b_{m-1},...)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\} $$
For the converse direction, with the same argument, we deduce that for $b_i\equiv a_i\mod \mathfrak{a}$, for all $0\le i\le m-1$,
$$(b_0,...,b_{m-1},...) - (a_0,...,a_{m-1},...)=(c_0,...,c_{m-1},...)$$
with $c_i\in \mathfrak{a}$, for $0\le i\le m-1$. For the second statement, we can see by the first statement that the set
$$a+V_{\mathfrak{a},m}=\{(b_0,...,b_{m-1})\in W(B)|b_i\equiv a_i\mod \mathfrak{a}, 0\le i\le m-1\}$$
forms a fundamental system of open neighborhoods around $a$. And this follows directly that the weak topology on $W(B)$ is the same as the product topology $B\times B\times ...$. (Q.E.D)
Via this lemma, we can prove
Proposition 3.2. If $B$ is Hausdorff (complete), then $W(B)$ is Hausdorff (complete, resp.).
Proof. It follows easily that if $B$ is Hausdorff then the product topology $B\times B\times...$ is also Hausdorff. Now, assume that $B$ is complete. In this case, the canonical map
$$\phi: B\to \varprojlim_{\mathfrak{a}} B/\mathfrak{a} $$
is surjective. Let $\mathfrak{c}$ be its kernel, we have $B/\mathfrak{c}\cong \varprojlim_{\mathfrak{a}} B/\mathfrak{a}$. And this yields
$$W_m(B/\mathfrak{c})\cong W_m(\varprojlim_{\mathfrak{a}} B/\mathfrak{a})\cong \varprojlim_{\mathfrak{a}} W_m(B/\mathfrak{a})=\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
where the second isomorphism comes from the functorial properties of Witt vectors, and the last isomorphism follows from the fact that the map $W(B)\xrightarrow{W(pr)} W(B/\mathfrak{a})\xrightarrow{pr} W_m(B/\mathfrak{a})$ has kernel $V_{\mathfrak{a},m}$.
Now, it follows from our (W.4) in the chapter about Witt vectors that
$$W(B/\mathfrak{c})\cong \varprojlim_{m}W_m(B/\mathfrak{c})\cong \varprojlim_m\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
And so, we obtain the following surjective map
$$W(B)\xrightarrow{W(pr)} W(B/\mathfrak{c})\cong \varprojlim_m\varprojlim_{\mathfrak{a}}W(B)/V_{\mathfrak{a},m}$$
And this yields $W(B)$ is complete. (Q.E.D)
Remark 3.3. In the case $B$ is complete and Hausdorff, we can equip the topological structure on the ring $W_m(B)$, with $m$ is fixed, by the method in the proof, such that $W_m(B)$ is Hausdorff and complete.
Proof. In this case, the canonical map $B\to \varprojlim_{\mathfrak{a}}B/\mathfrak{a}$ is bijective, and this yields by the previous proof that
$$W(B)/V_m(B)=W_m(B)\cong \varprojlim_{\mathfrak{a}}W_m(B/\mathfrak{a})\cong$$
$$\cong \varprojlim_{\mathfrak{a}} W(B)/V_{\mathfrak{a},m}\cong \varprojlim_{\mathfrak{a}}(W(B)/V_m(B))/(V_{\mathfrak{a},m}/V_m(B))$$
From this, there exists a unique topological structure on $W_m(B)$, such that $\{V_{\mathfrak{a},m}/V_m(B)|\mathfrak{a}\subset B: \text{ open ideal }\}$ becomes a fundamental system of open neighborhood around $0$. And $W_m(B)$ is also Hausdorff, and complete. (Q.E.D)
We will be mainly interested in the case $B:=\mathscr{O}_F$, where $F$ is a complete, non-archimedean, perfect field containing $k$. In this case, we get $W(B)$ is Hausdorff, complete, and is a subring of $W(F)$.
Lemma 3.4. Let $\mathscr{O}_F$ be as above, then an ideal $\mathfrak{a}$ of $\mathscr{O}_F$ is open iff $\mathfrak{a}$ is non-zero.
Proof. Assume that $\mathfrak{a}$ is open, then it is obvious that $\mathfrak{a}$ is non-zero. Now, let $\mathfrak{a}\subset \mathscr{O}_F$ be any non-zero ideal. Take $0\ne x\in \mathfrak{a}$, it is sufficient to prove that $(x)$-the ideal generated by $x$ is open in $\mathscr{O}_F$. We can see that
$$(x) =\{y\in \mathscr{O}_F||y|\le |x|)\}$$
Let us take any $z\in \mathscr{O}_F$, such that $|y-z|<|x-y|$. This yields $|y-z|< \max\{x,y\}\le |x|$. From this, we have $|z|\le |x|$, and $z\in (x)$. This yields $(x)$ is open, and hence, $\mathfrak{a}$ is open. (Q.E.D)
We can define for any open ideal $\mathfrak{a}$ of $\mathscr{O}_F$, and any $m\ge 1$ an $\mathscr{O}_F$-submodule
$$U_{\mathfrak{a},m}:=V_{\mathfrak{a},m} + \pi^mW(F):=\{(b_0,...,b_{m-1},...)\in W(F)|b_0,...,b_{m-1}\in \mathfrak{a}\}$$
We note that $U_{\mathfrak{a},m}$ are not ideals of $W(F)$, and we again have
$$U_{\mathfrak{a}\cap \mathfrak{b},\max\{m,n\}}\subset U_{\mathfrak{a},m}\cap U_{\mathfrak{b},n}$$
And these subgroups satisfy the conditions of Proposition 1.3. So, there exists a unique topology on $W(F)$, such that $W(F)$ is a topological group, and $U_{\mathfrak{a},m}$ forms a fundamental system of neighborhoods around $0$. We recall that because $F$ is an extension of $k$, and perfect, $W(F)$ is a D.V.R, with maximal ideal generated by $\pi$. Again, the topology we have equipped for $W(F)$ is weaker that the $\pi$-adic topology. We can call it the weak topology on $W(F)$. We actually want to prove that this topology actually defines a topological ring on $W(F)$, and that when $\mathscr{O}_F$ admits a filtered fundamental system, then $W(F)$ is complete.
We will need the multiplicative property of Teichmuller's representative.
Lemma 3.5.
(i) Let $a_1,...,a_r\in W(F)$, then there exists $0\ne \alpha\in \mathscr{O}_F$, such that
$$\tau(\alpha)a_1,...,\tau(\alpha)a_r\in U_{\mathscr{O}_F,m}$$
(ii) Let $\mathfrak{a}$ be an open ideal of $\mathscr{O}_F$, then for any $0\ne \alpha\in \mathscr{O}_F$, and $m\ge 1$, we have
$$\tau(\alpha^{-1})U_{\alpha^{q^{m-1}}\mathfrak{a},m}\subset U_{\mathfrak{a},m}$$
Proof.
(i) By Proposition 4.6 in the section about Witt vectors, we can represent
$$a_i=\sum_{j\ge 0}\tau(a_{i,j})\pi^j$$
And from this
$$\tau(\alpha)a_i=\sum_{j\ge 0}\tau(\alpha a_{i,j})\pi^j=(\alpha a_{i,0},\alpha a_{i,1},...)$$
And we can choose $\alpha$ such that $\alpha a_{i,j}\in \mathscr{O}_F$, for all $1\le i\le r, 0\le j\le m-1$.
(ii) Take $a=(a_0,a_1,...)\in \alpha^{q^{m-1}}\mathfrak{a}$, we can represent
$$(\alpha_0,\alpha_1,...)=\sum_{i\ge 0}\tau(a_i^{1/q^i})\pi^i$$
Hence
$$\tau(\alpha^{-1})a=\sum_{i\ge 0}\tau(\alpha^{-1}a_i^{1/q^i})\pi^i=\sum_{i\ge 0}(\alpha^{-q^i}a_i)$$
And hence, $\alpha^{-q^i}a_i\in U_{\mathfrak{a},m}$, for all $0\le i\le m-1$. (Q.E.D)
We are now ready for the main result of this section.
Proposition 3.7.
(i) $W(F)$ is a Hausdorff topological ring.
(ii) If $\mathscr{O}_F$ admits a filtered fundamental system, then so is $W(F)$, and in this case, $W(F)$ is complete.
Proof.
(i) We will prove that the multiplication map
$$W(F)\times W(F)\to W(F)$$
is continuous. Take any $a,b\in W(F)$, and an open neighborhood of $ab+U_{\mathfrak{a},m}$, for some open ideal $\mathfrak{a}$ of $\mathscr{O}_F$, and $m\ge 1$. By Lemma 3.6, one can find $0\ne \alpha\in \mathscr{O}_F$ such that $\tau(\alpha)a,\tau(\alpha)b\in U_{\mathscr{O}_F,m}$, which is equivalent to $a,b\in \tau(\alpha^{-1})U_{\mathscr{O}_F,m}$. By Lemma 3.5 again, we have
$$(a+U_{\alpha^{q^{m-1}}\mathfrak{a},m})(b+U_{\alpha^{q^{m-1}}\mathfrak{a},m})\subset ab+U_{\mathscr{O}_F,m}U_{\mathfrak{a},m}+U_{\mathfrak{a},m}\subset ab+ U_{\mathfrak{a},m}$$
And by Lemma 3.4, $U_{\alpha^{q^{m-1}}\mathfrak{a},m}$ is open. Hence, $W(F)$ is a topological ring. Moreover, one get easily that
$$\cap_{\mathfrak{a},m}U_{\mathfrak{a},m}=\cap_{\mathfrak{a},m}\{(b_0,...,b_{m-1},...)\in W(F)|b_i\in \mathfrak{a}\}=0$$
since $\mathscr{O}_F$ is Hausdorff, and the intersection of all open ideals is just 0.
(ii) Let $I_1\supset I_2\supset ...$ be a filtered fundamental system of $\mathscr{O}_F$. Let
$$U_{n,m}:=\{(b_0,...,b_{m-1},...)\in W(F)|b_0,...,b_{m-1}\in I_n\}$$
And $U_m:=U_{m,m}$. Then it can be seen easily that $\{U_{n,m}\}$ forms a fundamental system of open neighborhood around $0$ in $W(F)$, because for any open ideal $\mathfrak{a}\in \mathscr{O}_F$, there exists $n$ such that $I_n\subset \mathfrak{a}$. And one has $U_{n,m}\subset U_{\min{n,m}}, U_m=U_{m,m}$. So $\{U_m\}$ also forms a fundamental system around $0$ in $W(F)$. And this yields $W(F)$ admits a filtration.
Now, in the case $\mathscr{O}_F$ is complete, and admits a filtered fundamental system, it is sufficient to prove any Cauchy sequence converges in $W(F)$. The main ideal of the proof is that we will use Lemma 3.5 to reduce the induced Cauchy sequence to $W_m(\mathscr{O}_F)$, which is complete, by Remark 3.3. And then, with the completeness of the $\pi$-adic topology on $W(F)$, we will prove that our sequence converges in $W(F)$.
Take any $(a_n)_n$ is a Cauchy sequence in $W(F)$. Fix an integer $m\ge 1$, then for any $\mathfrak{a}$: open ideal in $\mathscr{O}_F$, there exists an integer $n_{\mathfrak{a}}$ such that for all $n,n'\ge n_\mathfrak{a}$, $a_n-a_{n'}\in U_{\mathfrak{a},m}$. Then by Lemma 3.5, we can choose $0\ne \alpha\in \mathscr{O}_F$, such that
$$\tau(\alpha) a_1,...,\tau(\alpha) a_{n_\mathfrak{a}}\in U_{\mathscr{O}_F,m}$$
And hence, for all $n\ge n_\mathfrak{a}$, we have
$$\tau(\alpha)(a_n-a_{n_\mathfrak{a}})\in \tau(\alpha)U_{\mathscr{O}_F,m}\subset U_{\mathscr{O}_F,m}$$
And hence $(\tau(\alpha)a_n)_n\in U_{\mathscr{O}_F,m}$, and for $n,n'\ge n_{\mathfrak{a}}$, we have
$$\tau(\alpha)(a_n-a_{n'})\in \tau(\alpha)U_{\mathfrak{a},m}\subset U_{\mathfrak{a},m}$$
Take $(b_n)_n\in W(\mathscr{O}_F)$ such that $\tau(\alpha)-b_n\in \pi^mW(F)$. We then have
$$b_n-b_{n'}\in (\tau(\alpha)(a_m-a_n)+\pi^mW(F))\cap W(\mathscr{O}_F)\subset (U_{\mathfrak{a},m}+\pi^m W(F))\cap W(\mathscr{O}_F)=V_{\mathfrak{a},m}$$
for all $n,n'\ge n_{\alpha}$. This yields the sequence $(b_n\mod V_m(\mathscr{O}_F))_n$ is Cauchy, and hence, converges to some $b\mod V_m(\mathscr{O}_F)$ in $W_m(\mathscr{O}_F)$. Hence, for any open ideal $\mathfrak{b}\subset \mathscr{O}_F$, there exists $n_{\mathfrak{b}}$ such that for all $n\ge n_{\mathfrak{b}}$, we have $b-b_n\in V_{\alpha^{q^{m-1}}\mathfrak{b},m}$. Let us denote $a(m):=\tau(\alpha^{-1})b$, then
$$a(m)-a_n=\tau(\alpha^{-1})b-a_n= \tau(\alpha^{-1})(b-\tau(\alpha)a_n)=\tau(\alpha^{-1})(b-b_n+b_n-\tau(\alpha)a_n)$$
$$\subset \tau(\alpha^{-1})(V_{\alpha^{q^{m-1}}\mathfrak{b},m}+\pi^mW(F))\subset U_{\mathfrak{b},m}+\pi^mW(F)\subset U_{\mathfrak{a},m}$$
for all $n\ge n_{\mathfrak{a}}$. Now, if we vary $m$, then we will get
$$a(m+1)-a(m)=(a(m+1)-a_n)-(a(m)-a_n)\in U_{\mathfrak{b},m}$$
for $n$ is sufficiently large. That means
$$a_{m+1}-a_m\in \{(b_0,...,b_{m-1})\in W(B)|b_0,...,b_{m-1}\in \mathfrak{b}, \forall \mathfrak{b}\subset \mathscr{O}_F: \text{ open}\}$$
And this yields $a(m+1)-a(m)\in \pi^mW(F)$. Now, this yields $(a(m))_m$ is a Cauchy sequence with respect to the $\pi$-adic topology, and hence, a Cauchy sequence in the weak topology. Let $a$ be the convergent value of $(a(m))_m$ in the $\pi$-adic topology, we will prove that $a$ is also the convergent value of $(a_n)_n$. For any ideal $\mathfrak{a}\subset \mathscr{O}_F$: open, and any $m$, we have $\pi^m W(F)\subset U_{\mathfrak{a},m}$, and there exists some $n'$ such that $a-a(n')\in \pi^mW(F)$ and $a(n')-a_n\in U_{\mathfrak{a},m}$, for some $n\ge n_\mathfrak{a}$. Hence, $(a-a_n)\in U_{\mathfrak{a},m}$. This yields $a$ is the convergent value of $W(F)$, and $W(F)$ is complete. (Q.E.D)
Sunday, January 21, 2018
[Topology on Witt Vectors II] Linearly Topologized Rings
For our further purposes, we often use inverse limit to define topology on a commutative ring $A$.
Definition. Let $A$ be a topological group (ring), we say $A$ is linearly topologized group (ring, resp.) if $0$ has a fundamental system of open neighborhoods $\mathscr{N}$ consisting of subgroups (ideals, resp.) of $A$. Such a fundamental system is called linearly topologized fundamental system. And we denote $(A, \mathscr{N})$ for the topological group (ring) $A$ with linearly topologized fundamental system $\mathscr{N}$.
The following lemma is quite useful to detect when are two linearly topoligized fundamental system generate the same topology.
Lemma 2.1. If $A$ is an abelian group (commutative ring) and $\mathscr{N}, \mathscr{M}$ are two sets containing subgroups of $A$ satisfying conditions of Proposition 1.3 (Proposition 1.6, respectively), then they induces the same topology on $A$ iff for any $I\in \mathscr{N}$, there exists some $J\in \mathscr{M}$ such that $J\subset I$, and vice versa.
Proof. It follows directly from our construction in Proposition 1.3, since they generate the same base for the topology on $A$. (Q.E.D)
By the following lemma, it makes sense to give the definition of the completion of a linearly topologized ring $A$.
Lemma 2.2. Let $A$ be an abelian group (commutative ring), $\mathscr{M}, \mathscr{N}$ are sets of subgroups of $A$, such that for every $G\in \mathscr{M}$, there exists $H_G\in \mathscr{N}$, such that $G\subset H_G$, and for every $H\in \mathscr{N}$, there exists $G_H\in \mathscr{M}$ such that $G_H\subset H$. Then there is a canonical isomorphism
$$\varprojlim_{G\in \mathscr{M}}A/G\cong \varprojlim_{H\in \mathscr{N}}A/H$$
Proof. For any $G\subset H$: subgroups of $A$, we denote $\psi^G_H$ the canonical map $A/G\to A/H$. Let $(a_G\mod G)_{G\in \mathscr{M}}\in \varprojlim_{G\in \mathscr{M}} A/G$, for $a_G\in A$, due to the assumption, for any $H\in \mathscr{N}$, there is $G_H\in \mathscr{M}$ such that $G_H\subset H$, and we can define a map
$$\psi: \varprojlim_{G\in \mathscr{M}}A/G\to \varprojlim_{H\in \mathscr{N}}A/H$$
by $\psi((a_G\mod G)_G)=(\psi^{G_H}_H(a_{G_H}))_H=(a_{G_H}\mod H)_H$. It is well-defined, since $(\psi^{G_H}_H)$ is compatible with the inverse system $\varprojlim_{H\in \mathscr{M}}A/H$.
Conversely, for any $G\in \mathscr{N}$, there exists $H_G\in \mathscr{M}$, such that $H_G\subset G$, and one can also define a map
$$\theta: \varprojlim_{H\in \mathscr{N}}A/H\to \varprojlim_{G\in \mathscr{M}}A/G$$
as follows $\theta(b_H\mod H)_H=(b_{H_G}\mod G)_{G\in \mathscr{N}}$, and we have the composition $\theta\circ \psi$ would be
$$(a_G\mod G)_G\mapsto (a_{G_H}\mod H)_H\mapsto (a_{G_{H_G}}\mod G)_G$$
We have $G_{H_G}\subset H_G\subset G$, and because we are in inverse systems, we have
$$a_{G_{H_G}}\mod G_{H_G}=a_G\mod G_{H_G}$$
And that means $a_{G_{H_G}}-a_G\in G_{H_G}\subset G$. So, $a_{G_{H_G}}\mod G=a_G\mod G$, and $\theta\circ\psi$ is just the identity map. It is similar to show that $\psi\circ\theta$ is also the identity map. And this yields the desired isomorphism. (Q.E.D)
Via Lemma 2.1 and Lemma 2.2 we have this
Definition. Let $(A, \mathscr{N})$ be a topological group (ring), we say that $A$ is complete if the canonical map $A\to \lim_{I\in \mathscr{N}}A/I$ is surjective.
We can also obtain an algebraic characterization of Hausdorff on a linearly topologized ring $A$.
Proposition 2.3. Let $(A, \mathscr{N})$ be a topological group (ring), then $A$ is Hausdorff iff the canonical map
$$\phi: A\to \varprojlim_{I\in \mathscr{N}} A/I$$
is injective. Also, $A$ is Hausdorff iff $\cap_{I\in \mathscr{N}} I=0$.
Proof. Assume $\phi$ is injective, but $A$ is not Hausdorff, i.e. there exists $0\ne a\in A$, such that all open neighborhoods at $0$ contain $a$. But then, it is a contradiction, since $\phi(a)=0$, and $\phi$ is not injective.
Conversely, assume that $A$ is Hausdorff, and $\phi$ is not injective, that means, there exists $0\ne a\in A$ and $a\in I(\forall I\in \mathscr{N})$. Due to the definition of a fundamental system, if we take any open neighborhood $U$ of 0, then there exists $I\in \mathscr{N}$, and $I\subset U$, and $I$ contains both $0$ and $a$, and hence, so does $U$. That means, $A$ is not Hausdorff, a contradiction.
For the second statement, assume that
$$\phi: A\to \varprojlim_{I\in \mathscr{N}}A/I$$
is injective, and $0\ne a\in \cap_{I\in \mathscr{N}}I$, then $\phi(a)=0$, so $\phi$ is not injective, a contradiction. Conversely, assume that $\cap _{I\in \mathscr{N}}I=0$, and $\phi$ is not injective, i.e. there exists $0\ne a\in A$ such that $\phi(a)=0$. That means $a\in I(\forall I\in \mathscr{N})$, a contradiction. We hence obtain the statements. (Q.E.D)
Remark 2.4. This follows directly from the definition that if $(A, \mathscr{N})$ is Hausdorff then if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergence is unique.
Proof. Assume that $A$ is Hausdorff, and $(x_n)_n$ is a Cauchy sequence converges in $A$ to $x, x'$, then it is clear that for all $I\in \mathscr{N}$, $x-x'\in I$. But then, since $\cap_{I\in \mathscr{N}}I=0$, we have $x=x'$. (Q.E.D)
For the converse of the remark above, we need to have a more special linearly topologized fundamental system. Let $A$ be an abelian group (commutative ring), and $I_1 \supset I_2\supset ...$ is a chain of ideals in $A$. We can see that $\mathscr{N}:=\{I_1,I_2,...\}$ satisfies conditions Proposition 1.6. Hence, there exists a unique topology on $A$ such that $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. We say such $\mathscr{N}$ a (countable) filtered fundamental system.
Remark 2.5. The converse of Remark 2.4 holds for a topological group (ring) $(A, \mathscr{N})$, where $\mathscr{N}$ is a filtered fundamental system. That means, in this case, $(A, \mathscr{N})$ is Hausdorff if and only if for any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergent value is unique.
Proof. For the converse, assume that $A$ is not Hausdorff, i.e. there exists $0\ne a\in \cap_{I\in \mathscr{N}}I$, then we can choose a sequence $(x_n)_n$ with $x_n\in I_n$, where $I_n\in \mathscr{N}$. This sequence is Cauchy, since we always have $x_{n+1}-x_n\in I_n$, for all $n$. This follows that both $a\ne -a$ are convergent values of $(x_n)_n$, a contradiction. (Q.E.D)
Lemma 2.5. Let $(A, \mathscr{N})$ be a topological group (ring), where $\mathscr{N}:=\{(I_n)_n\}$ is a filtered fundamental system, then the canonical map
$$\phi: A\to \varprojlim_n A/I_n$$
is surjective iff $A$ is complete with respect to $\mathscr{N}$.
Proof. Assume that $\phi: A\to \varprojlim_n A/I_n$ is surjective, and $(x_n)_n$ is a Cauchy sequence. Then for any $k\ge 1$, there exists $n_k$ such that $n_{k+1}> n_{k}$ and for all $m,n\ge n_k$, we have $x_m-x_n\in I_k$. We can define $y_k:=x_{n_k}$, then $(y_k)_k$ is an element in $\varprojlim_k A/I_k$, because $y_{k+1}-y_k=x_{n_{k+1}}-x_{n_k}\in I_k$. By the surjectivity of $\phi$, there exists $y\in A$ such that $\phi(y)=(y_k)_k$, i.e. $y\mod I_k=y_k, \forall k$ and we can see easily that $y$ is a convergent value of $(y_k)_k$. This also follows that $y$ is a convergent value of $(x_n)_n$ because for any $k$, and for all $n\ge n_k$, $y-x_n=y-x_{n_k}+x_{n_k}-x_n\in I_k$.
Conversely, let $(x_n)_n\in \varprojlim_n A/I_n$ be any element, we can see that $x_{n+1}-x_n\in I_n$, and $(x_n)_n$ is a Cauchy sequence in $A$, and there exists $x\in A$ is a convergent value of $(x_n)_n$. Due to the definition, for every $n$, and there exists some $m_n$ such that for all $m\ge m_n$, $x-x_m\in I_n$. We can choose $m\ge \max\{n,m_n\}$, and $x-x_n=x-x_m+x_m-x_n\in I_n$. And this yields, for all $n$, $x-x_n\in I_n$. Hence, $\phi(x)=(x_n)_n$, and $\phi$ is surjective. (Q.E.D)
Via this lemma, for the case $A$ has a filtered fundamental system, we obtain an equivalence between the notions of sequential complete and complete. We conclude this section by a
Proposition 2.7. Let $A$ be an topological group (ring), and $I_1\supset I_2\supset ...$ a linearly topologized fundamental system around $0$. Then $A$ is Hausdorff and complete iff the canonical map
$$\phi: A\to \varprojlim_n A/I_n$$
is bijective.
Definition. Let $A$ be a topological group (ring), we say $A$ is linearly topologized group (ring, resp.) if $0$ has a fundamental system of open neighborhoods $\mathscr{N}$ consisting of subgroups (ideals, resp.) of $A$. Such a fundamental system is called linearly topologized fundamental system. And we denote $(A, \mathscr{N})$ for the topological group (ring) $A$ with linearly topologized fundamental system $\mathscr{N}$.
The following lemma is quite useful to detect when are two linearly topoligized fundamental system generate the same topology.
Lemma 2.1. If $A$ is an abelian group (commutative ring) and $\mathscr{N}, \mathscr{M}$ are two sets containing subgroups of $A$ satisfying conditions of Proposition 1.3 (Proposition 1.6, respectively), then they induces the same topology on $A$ iff for any $I\in \mathscr{N}$, there exists some $J\in \mathscr{M}$ such that $J\subset I$, and vice versa.
Proof. It follows directly from our construction in Proposition 1.3, since they generate the same base for the topology on $A$. (Q.E.D)
By the following lemma, it makes sense to give the definition of the completion of a linearly topologized ring $A$.
Lemma 2.2. Let $A$ be an abelian group (commutative ring), $\mathscr{M}, \mathscr{N}$ are sets of subgroups of $A$, such that for every $G\in \mathscr{M}$, there exists $H_G\in \mathscr{N}$, such that $G\subset H_G$, and for every $H\in \mathscr{N}$, there exists $G_H\in \mathscr{M}$ such that $G_H\subset H$. Then there is a canonical isomorphism
$$\varprojlim_{G\in \mathscr{M}}A/G\cong \varprojlim_{H\in \mathscr{N}}A/H$$
Proof. For any $G\subset H$: subgroups of $A$, we denote $\psi^G_H$ the canonical map $A/G\to A/H$. Let $(a_G\mod G)_{G\in \mathscr{M}}\in \varprojlim_{G\in \mathscr{M}} A/G$, for $a_G\in A$, due to the assumption, for any $H\in \mathscr{N}$, there is $G_H\in \mathscr{M}$ such that $G_H\subset H$, and we can define a map
$$\psi: \varprojlim_{G\in \mathscr{M}}A/G\to \varprojlim_{H\in \mathscr{N}}A/H$$
by $\psi((a_G\mod G)_G)=(\psi^{G_H}_H(a_{G_H}))_H=(a_{G_H}\mod H)_H$. It is well-defined, since $(\psi^{G_H}_H)$ is compatible with the inverse system $\varprojlim_{H\in \mathscr{M}}A/H$.
Conversely, for any $G\in \mathscr{N}$, there exists $H_G\in \mathscr{M}$, such that $H_G\subset G$, and one can also define a map
$$\theta: \varprojlim_{H\in \mathscr{N}}A/H\to \varprojlim_{G\in \mathscr{M}}A/G$$
as follows $\theta(b_H\mod H)_H=(b_{H_G}\mod G)_{G\in \mathscr{N}}$, and we have the composition $\theta\circ \psi$ would be
$$(a_G\mod G)_G\mapsto (a_{G_H}\mod H)_H\mapsto (a_{G_{H_G}}\mod G)_G$$
We have $G_{H_G}\subset H_G\subset G$, and because we are in inverse systems, we have
$$a_{G_{H_G}}\mod G_{H_G}=a_G\mod G_{H_G}$$
And that means $a_{G_{H_G}}-a_G\in G_{H_G}\subset G$. So, $a_{G_{H_G}}\mod G=a_G\mod G$, and $\theta\circ\psi$ is just the identity map. It is similar to show that $\psi\circ\theta$ is also the identity map. And this yields the desired isomorphism. (Q.E.D)
Via Lemma 2.1 and Lemma 2.2 we have this
Definition. Let $(A, \mathscr{N})$ be a topological group (ring), we say that $A$ is complete if the canonical map $A\to \lim_{I\in \mathscr{N}}A/I$ is surjective.
We can also obtain an algebraic characterization of Hausdorff on a linearly topologized ring $A$.
Proposition 2.3. Let $(A, \mathscr{N})$ be a topological group (ring), then $A$ is Hausdorff iff the canonical map
$$\phi: A\to \varprojlim_{I\in \mathscr{N}} A/I$$
is injective. Also, $A$ is Hausdorff iff $\cap_{I\in \mathscr{N}} I=0$.
Proof. Assume $\phi$ is injective, but $A$ is not Hausdorff, i.e. there exists $0\ne a\in A$, such that all open neighborhoods at $0$ contain $a$. But then, it is a contradiction, since $\phi(a)=0$, and $\phi$ is not injective.
Conversely, assume that $A$ is Hausdorff, and $\phi$ is not injective, that means, there exists $0\ne a\in A$ and $a\in I(\forall I\in \mathscr{N})$. Due to the definition of a fundamental system, if we take any open neighborhood $U$ of 0, then there exists $I\in \mathscr{N}$, and $I\subset U$, and $I$ contains both $0$ and $a$, and hence, so does $U$. That means, $A$ is not Hausdorff, a contradiction.
For the second statement, assume that
$$\phi: A\to \varprojlim_{I\in \mathscr{N}}A/I$$
is injective, and $0\ne a\in \cap_{I\in \mathscr{N}}I$, then $\phi(a)=0$, so $\phi$ is not injective, a contradiction. Conversely, assume that $\cap _{I\in \mathscr{N}}I=0$, and $\phi$ is not injective, i.e. there exists $0\ne a\in A$ such that $\phi(a)=0$. That means $a\in I(\forall I\in \mathscr{N})$, a contradiction. We hence obtain the statements. (Q.E.D)
Remark 2.4. This follows directly from the definition that if $(A, \mathscr{N})$ is Hausdorff then if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergence is unique.
Proof. Assume that $A$ is Hausdorff, and $(x_n)_n$ is a Cauchy sequence converges in $A$ to $x, x'$, then it is clear that for all $I\in \mathscr{N}$, $x-x'\in I$. But then, since $\cap_{I\in \mathscr{N}}I=0$, we have $x=x'$. (Q.E.D)
For the converse of the remark above, we need to have a more special linearly topologized fundamental system. Let $A$ be an abelian group (commutative ring), and $I_1 \supset I_2\supset ...$ is a chain of ideals in $A$. We can see that $\mathscr{N}:=\{I_1,I_2,...\}$ satisfies conditions Proposition 1.6. Hence, there exists a unique topology on $A$ such that $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. We say such $\mathscr{N}$ a (countable) filtered fundamental system.
Remark 2.5. The converse of Remark 2.4 holds for a topological group (ring) $(A, \mathscr{N})$, where $\mathscr{N}$ is a filtered fundamental system. That means, in this case, $(A, \mathscr{N})$ is Hausdorff if and only if for any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergent value is unique.
Proof. For the converse, assume that $A$ is not Hausdorff, i.e. there exists $0\ne a\in \cap_{I\in \mathscr{N}}I$, then we can choose a sequence $(x_n)_n$ with $x_n\in I_n$, where $I_n\in \mathscr{N}$. This sequence is Cauchy, since we always have $x_{n+1}-x_n\in I_n$, for all $n$. This follows that both $a\ne -a$ are convergent values of $(x_n)_n$, a contradiction. (Q.E.D)
Lemma 2.5. Let $(A, \mathscr{N})$ be a topological group (ring), where $\mathscr{N}:=\{(I_n)_n\}$ is a filtered fundamental system, then the canonical map
$$\phi: A\to \varprojlim_n A/I_n$$
is surjective iff $A$ is complete with respect to $\mathscr{N}$.
Proof. Assume that $\phi: A\to \varprojlim_n A/I_n$ is surjective, and $(x_n)_n$ is a Cauchy sequence. Then for any $k\ge 1$, there exists $n_k$ such that $n_{k+1}> n_{k}$ and for all $m,n\ge n_k$, we have $x_m-x_n\in I_k$. We can define $y_k:=x_{n_k}$, then $(y_k)_k$ is an element in $\varprojlim_k A/I_k$, because $y_{k+1}-y_k=x_{n_{k+1}}-x_{n_k}\in I_k$. By the surjectivity of $\phi$, there exists $y\in A$ such that $\phi(y)=(y_k)_k$, i.e. $y\mod I_k=y_k, \forall k$ and we can see easily that $y$ is a convergent value of $(y_k)_k$. This also follows that $y$ is a convergent value of $(x_n)_n$ because for any $k$, and for all $n\ge n_k$, $y-x_n=y-x_{n_k}+x_{n_k}-x_n\in I_k$.
Conversely, let $(x_n)_n\in \varprojlim_n A/I_n$ be any element, we can see that $x_{n+1}-x_n\in I_n$, and $(x_n)_n$ is a Cauchy sequence in $A$, and there exists $x\in A$ is a convergent value of $(x_n)_n$. Due to the definition, for every $n$, and there exists some $m_n$ such that for all $m\ge m_n$, $x-x_m\in I_n$. We can choose $m\ge \max\{n,m_n\}$, and $x-x_n=x-x_m+x_m-x_n\in I_n$. And this yields, for all $n$, $x-x_n\in I_n$. Hence, $\phi(x)=(x_n)_n$, and $\phi$ is surjective. (Q.E.D)
Via this lemma, for the case $A$ has a filtered fundamental system, we obtain an equivalence between the notions of sequential complete and complete. We conclude this section by a
Proposition 2.7. Let $A$ be an topological group (ring), and $I_1\supset I_2\supset ...$ a linearly topologized fundamental system around $0$. Then $A$ is Hausdorff and complete iff the canonical map
$$\phi: A\to \varprojlim_n A/I_n$$
is bijective.
[Topology on Witt Vectors I] Topological Groups and Rings
Definition. Let $A$ be an abelian group, and $A$ is also a topological space. Then $A$ is said to be a topological group if the addition map $add: A\times A\to A$, defined by $add(a,b)=a+b$, and the inverse map $inv: A\to A$ defined by $inv(a)=-a$ are continuous, where $A\times A$ is equipped with the product topology.
We begin with
Lemma 1.1. For all $c\in A$, the map $\tau_c: A\to A$ defined by $\tau_c(a)=c+a$ is a homeomorphism.
Proof. One can see that the restriction of the topology on $A\times A$ to $\{c\}\times A$ makes $\{c\}\times A$ homeomorphic to $A$ via the map $pr(c,b) = b$. Also, the map $add|_{\{c\}\times A}$ is continuous and bijective. This yields $\tau_c:=add|_{\{c\}\times A}\circ pr^{-1}: A\to A$ is also continuous and bijective. Its inverse map $\tau_{-c}$, by similar arguments, is also continuous. Hence $\tau_c$ is a homeomorphism (Q.E.D)
For most of cases, we want to equip topology for a given abelian group $A$. Here is a way to do this.
Definition. Let $A$ be a topological group, $x\in A$ be a point, then a fundamental system of open neighborhoods around $x$ is a set $\mathscr{N}_x$ containing open neighborhoods of $x$, such that for all $x\in U$, where $U\subset A$ open, there exists $V\in \mathscr{N}_x$ such that $V\subset U$. We denote a fundamental system of open neighborhoods around 0 as $\mathscr{N}$, instead of $\mathscr{N}_0$.
Lemma 1.2. The following holds for $\mathscr{N}$:
(T.G.1) $\forall U\in \mathscr{N}, \forall c\in U$, there exists $V\in \mathscr{N}$, such that $c+V\subset U$.
(T.G.2) $\forall U\subset \mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V+V\subset U$.
Let $\mathscr{B}:=\{a+U|a\in A, U\in \mathscr{N}\}$, then $\mathscr{B}$ forms a base for the topology on $A$.
Proof. For (T.G.1), by Lemma 1.1, we have $U-c$ is also an open neighborhood of $0$, and hence, due to the definition of $\mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V\subset U-c$, i.e. $V+c\subset U$.
For (T.G.2), let us look at $add^{-1}(U)$, it is an open neighborhood of $(0,0)$, and hence, there exists $V_1,V_2$: open neighborhoods of $0$ such that $add(V_1\times V_2)\subset U$. Take $V\in \mathscr{N}$, and $V\subset V_1\cap V_2$, we get $add(V+V)\subset U$.
Now, it follows by Lemma 1.1. that if $\mathscr{N}$ is a fundamental system around $0$, then $\{x+U|U\in \mathscr{N}\}$ is a fundamental system around $x$. Let $x\in A$ be any element, and $U_x$ is an open neighborhood of $x$, then $U_x-x$ is an open neighborhood of $0$, and by (T.G.1), there exists $V\in \mathscr{N}$, such that $V\subset U_x-x$, i.e. $x+V\subset U_x$, and $x+V\in \mathscr{B}$, and $x+V$ is an open neighborhood of $x$. This yields, any open subset of $A$ is a union of elements in $\mathscr{B}$. Hence, $B$ is a base for the topology on $A$. (Q.E.D)
We are now ready to see how to define a suitable topology for an abelian group so that it becomes a topological group.
Proposition 1.3. Let $\mathscr{N}$ be a set containing subsets of $A$, which contain $0$, such that $\mathscr{N}$ satisfies (T.G.1), (T.G.2) and
(T.G.3) $\forall U\in \mathscr{N}$, $-U\in \mathscr{N}$.
(T.G.4) $\forall U,V\in \mathscr{N}$, there exists $W\in \mathscr{N}$, such that $W\subset U\cap V$
then there exists a unique topology on $A$ that makes $A$ become a topological group, and $\mathscr{N}$ a fundamental system around 0.
Proof. We first define a topology on $A$ as follows. $U\subset A$ is open if either $U$ is empty, or for all $x\in U$, there exists $V\in \mathscr{N}$, such that $x+V\subset U$. With this kind of open sets, we now prove $A$ is a topological space.
i. $\emptyset, A$ are open.
ii. Let $U_i(i\in I)$ be open subsets of $A$, then it is obvious to see that $\cup_{i\in I} U_i$ is also open.
iii. Let $U_1,...,U_n$ be open subsets of $A$, we will prove that $U_1\cap ...\cap U_n$ is also open. By induction, it is sufficient for us to prove the case $n=2$. If $U_1\cap U_2=\emptyset$, then there is nothing to prove. Otherwise, let $x\in U_1\cap U_2$, then there exists $V_i(i=1,2)$, such that $V_i\in \mathscr{N}$, and $x+V_i\subset U_i(i=1,2)$. By (T.G.4), we can choose $V\subset V_1\cap V_2$ and $V\in \mathscr{N}$, and this yields $x+V\subset U_1\cap U_2$. Hence, $U_1\cap U_2$ is also open.
We now prove that with this kind of topology, $A$ becomes a topological group, that means, we have to prove $add$ and $inv$ are continuous. First, let $U\subset A, U\ne \emptyset$ be an open subset, then $inv^{-1}(U)=-U$. Take any $x\in -U$, we have $-x\in U$, and there exists $V\in \mathscr{N}$, such that $-x+V\subset U$, i.e. $x-V\subset U$. By (T.G.3), $-V$ is also in $\mathscr{N}$. This yields $-U$ is open, and $inv$ is continuous.
Next, we will prove that if $U\subset A, U\ne \emptyset$ is an open subset, then for any $c\in A$, $c+U$ is also open. Taking any $b\in c+U$, i.e. $b-c\in U$, and there exists $V\in \mathscr{N}$, such that $b-c+V\subset U$, i.e. $b+V\subset c+U$. And this yields $c+U$ is open.
Now, let $U\ne \emptyset, U\subset A$ be an open subset, consider $(c,d)\in add^{-1}(U)$, i.e. $c+d\in U$, then there exists $V\in \mathscr{N}$, such that $c+d+V\subset U$. By (T.G.2), there exists $W\in \mathscr{N}$ such that $W+W\subset V$, and this yields $(c+W)+(d+W)\subset U$. And by our previous argument, both $c+W, d+W$ are open and $(c+W)\times (d+W)\subset add^{-1}(U)$. Hence, $add$ is also continuous. Hence, $A$ is a topological group with this topology.
By (T.G.1), all elements in $\mathscr{N}$ are open subsets of $A$, and they form a fundamental neighborhood of $0$, since for any $U\subset A$: open, containing 0; there exists $V\in \mathscr{N}$, such that $0+V=V\subset U$. By Lemma 1.2, $\mathscr{B}:=\{a+U|a\in A,U\in \mathscr{N}\}$ forms a base for the topology on $A$, and this yields the uniqueness part. (Q.E.D)
We now come to the definition of topological rings.
Definition. Let $(A,+,.)$ be a commutative ring, then $A$ is said to be a topological ring if $A$ is also a topological space, and
i. (A,+) is a topological group.
ii. The map $mult: A\times A\to A$ defined by $mult(a,b)=ab$ is continuous.
Lemma 1.4. Let $A$ be a topological ring, and $a\in A$, then the map $m_c:A\to A$ defined by $m_c(a)=ac$ is continuous.
Proof. We have $\{c\}\times A$ is homeomorphic to $A$ via the map $pr(c,b)=b$, and the restriction of $mult|_{\{c\}\times A}$ is continuous, and hence, the composition $m_c=mult|_{\{c\}\times A}\circ pr^{-1}$ is continuous. (Q.E.D)
Similarly to earlier argument, we obtain
Lemma 1.5. Let $\mathscr{N}$ be a fundamental system around 0 of a topological ring $A$, then
(T.R.1) $\forall U\in \mathscr{N}, \forall c\in A$, there exists $V\in \mathscr{N}$, such that $cV\subset U$.
(T.R.2) $\forall U\in \mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V.V:={uv|u,v\in V}\subset U$.
Proof. For (T.R.1), using Lemma 1.4, the map $m_c:A\to A$ is continuous, and this yields $m_c^{-1}(U)\subset A$ is also open, and contains $0$. So, there exists $V\in \mathscr{N}$, such that $V\subset m_c^{-1}(U)$, i.e. $cV\subset U$.
For (T.R.2), the map $mult: A\times A\to A$ is continuous, and hence, $mult^{-1}(A)$ is open in $A\times A$. Let $V_1\times V_2\subset mult^{-1}(U)$, and $V_1, V_2$ are open neighborhoods of $0$, we can now choose $v\in \mathscr{N}$ and $V\subset V_1\cap V_2$, and $V.V=mult(V\times V)\subset U$. (Q.E.D)
And similar to commutative group, here is a way to put a topology on a commutative ring $A$ so that it becomes a topological ring.
Proposition 1.6. Let $\mathscr{N}$ be a set containing subsets of $A$, which contain $0$, such that $\mathscr{N}$ satisfies (T.G.1)-(T.G.4) and (T.R.1), (T.R.2), then there exists a unique way to equip $A$ a topology such that $\mathscr{N}$ becomes a fundamental system around $0$.
Proof. By Proposition 1.3, there is a unique way to equip $A$ a topology such that $A$ becomes a topological group. So, it is sufficient to prove that, with this kind of topology, $A$ now becomes a topological ring. That means we have to check that the map $mult$ is continuous.
First, let $m_c: A\to A$ be defined as above, then for any $U\subset A$: open, non-empty, and $x\in m_c^{-1}(U)$, i.e. $cx\in U$. By (T.G.1), there exists $V\in \mathscr{N}$, such that $cx+V\subset U$, and by (T.R.1), there exists $W\in \mathscr{N}$, such that $cW\subset V$, and hence, $cx+cW\subset U$, that means, $x+W\subset m_c^{-1}(U)$. And this yields $m_c$ is continuous.
We will now prove that $mult$ is continuous. Let $U$ be as above, take $(c,d)\in mult^{-1}(U)$, i.e. $cd\in U$. By (T.G.1), there exists $V\in \mathscr{N}$ such that $cd+V\subset U$. And by (T.R.2), there exists $W\in \mathscr{N}$ such that $W.W\subset V$. Now, by (T.R.1), we can choose $W',W''\in \mathscr{N}$ such that $dW'\subset V, cW''\subset V$. By (T.G.4), there exists $W_c, W_d\in \mathscr{N}$ such that $W_c\subset W'\cap W, W_d\subset W''\cap W$, then
$$(c+W_c)(d+W_d)=cd+dW_c+cW_d+W_cW_d\subset U$$
Hence, $mult$ is a continuous map. This yields $A$ is a topological ring. (Q.E.D)
We now discuss about Cauchy sequence on a topological ring $A$.
Definition. Let $A$ be a topological ring with $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$, and $(x_n)_n$ a sequence of elements in $A$, then $(x_n)_n$ is said to be a Cauchy sequence with respect to $\mathscr{N}$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$ such that for all $m,n\ge n_U$, we have $x_m-x_n\in U$. We say that $x$ is a convergent value of $(x_n)_n$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$, such that for all $m\ge n_U$, we have $x-x_m\in U$. We say $(A, \mathscr{N})$ is sequential complete with respect to $\mathscr{N}$ if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$.
Proposition 1.7. Let $A$ be a topological ring, and $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. If $\mathscr{M}$ is another fundamental system of open neighborhoods around $0$, then $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$ iff $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{M}$. Hence, the definition of Cauchy sequences, and sequential complete do not depend on the choice of fundamental systems.
Proof. Assume that $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$, then by definition, for any $V\in \mathscr{M}$, there exists $U\in \mathscr{N}$, such that $U\subset V$, then if we choose $m,n\ge n_U$, we get $x_m-x_n\in U\subset V$. Hence, $(x_n)_n$ is also a Cauchy sequence with respect to $V$. The converse is the same.
Assume that a Cauchy sequence $(x_n)_n$ with respect to $\mathscr{N}$ converges to $x\in A$, then it is clear from our earlier argument that $(x_n)_n$ also converges to $x$ with respect to $\mathscr{M}$. Hence, the definition of completion also does not depend on the choice of fundamental systems.
(Q.E.D)
We now discuss about Cauchy sequence on a topological ring $A$.
Definition. Let $A$ be a topological ring with $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$, and $(x_n)_n$ a sequence of elements in $A$, then $(x_n)_n$ is said to be a Cauchy sequence with respect to $\mathscr{N}$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$ such that for all $m,n\ge n_U$, we have $x_m-x_n\in U$. We say that $x$ is a convergent value of $(x_n)_n$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$, such that for all $m\ge n_U$, we have $x-x_m\in U$. We say $(A, \mathscr{N})$ is sequential complete with respect to $\mathscr{N}$ if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$.
Proposition 1.7. Let $A$ be a topological ring, and $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. If $\mathscr{M}$ is another fundamental system of open neighborhoods around $0$, then $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$ iff $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{M}$. Hence, the definition of Cauchy sequences, and sequential complete do not depend on the choice of fundamental systems.
Proof. Assume that $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$, then by definition, for any $V\in \mathscr{M}$, there exists $U\in \mathscr{N}$, such that $U\subset V$, then if we choose $m,n\ge n_U$, we get $x_m-x_n\in U\subset V$. Hence, $(x_n)_n$ is also a Cauchy sequence with respect to $V$. The converse is the same.
Assume that a Cauchy sequence $(x_n)_n$ with respect to $\mathscr{N}$ converges to $x\in A$, then it is clear from our earlier argument that $(x_n)_n$ also converges to $x$ with respect to $\mathscr{M}$. Hence, the definition of completion also does not depend on the choice of fundamental systems.
(Q.E.D)
Witt Vectors: A Complete Introduction
Witt vectors are my old friends. I had a very hard time to digest it in the book of Jacobson at the first time we met. However, things are easier now when I read the book of Schneider about Galois representation. His treatment about Witt vectors is general for local fields. The main thing is that for Witt's vectors, Frobenius acts in a nice way. We don't have it in the ring of integer of a local field of characteristic $0$, though they are isomorphic, if we construct Witt vectors from the residue field with coefficients in our given local field.
Tuesday, December 26, 2017
[Local Class Field Theory 4] The Maximal Abelian Extensions of Local Fields
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)
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)
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