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 $K$.
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