# perfectoid spaces-1

This series of posts will focus on perfectoid spaces.

Perfectoid spaces are a kind of spaces over perfectoid fields both of which introduced by Peter Scholze.

First let’s see what is a perfectoid field.

A perfectoid field $K$ is a complete non-archimedean field with a non-discret valuation $v=| |:K\rightarrow \mathbb{R}_{>0}\bigcup {0}$ of rank $1$ such that the residue field $\kappa=K^o/m=\{k\in K|v(k)\leq 1\}/ \{k\in K|v(k)<1\}$ is of characteristic $p\neq 0$ and the Frobenius map $F:K^o/p\rightarrow K^o/p, k\mapsto k^p$ is surjective.

Here several points are to be clarified: the topology on $K$ is induced by the valuation $v$. Non-discret valuation $v$ here means that $Im(v)$ is non-discret in $\mathbb{R}_{>0}$ the topology of the latter is the Euclidean topology.

Classical examples for perfectoid spaces are the completions of $\mathbb{Q}_p(p^{1/p^{\infty}})$, $\mathbb{F}_p[[t]]({t^{1/p^{\infty}}})$, $\overline{\mathbb{Q}_p}$, the valuation of the first and the third fields is induced by that of $\mathbb{Q}_p$ and the valuation of the second is induced by the $t$-adic valuation.

Next we want to define a tilting process for a perfectoid field $K$. We choose an element $\omega \in K$ with $1/p and take $\varprojlim_F K^{o}/\omega$. Then we have

(1)$K^o/\omega$ has the discret topology;(2)$\varprojlim_F K^{o}/\omega$ is a perfect ring.

The first point is not hard. Note that $\omega K^o$ consists of elements of valuation $\leq v(\omega)$. And this is an open set in $K$, thus in $K^o$ according to the definition of the topology induced by this valuation. I will not prove (2) here since we will not use it later.

We equip $\varprojlim_F K^{o}/\omega$ with the inverse limit topology and we have that

(3)there is a multiplicative homeomorphism, $\varprojlim_{k\mapsto k^p}K^o\rightarrow \varprojlim_FK^o/\omega$ given by projection;(4) the right-hand side is independent of $\omega$;(5)there is thus a multiplicative map $\varprojlim_F K^o/\omega \rightarrow K^o, k\mapsto k^\sharp$.

We first show the point (5). Suppose that $k=(\overline{k_0},\overline{k_1},...)\in \varprojlim_F K^o/\omega$, and choose any $k_n\in K^o$ such that $k_n$ is a lift of $\overline{k_n}$, then we define $k^{\sharp}=\lim_nk_n^{p^n}$. This point is a simple calculation(here we shall use the assumption that $K$ is complete to ensure the existence of the limit). As for the point (3), we set $G:\varprojlim_FK^o/\omega\rightarrow \varprojlim_{k\mapsto k^p}K^o,k\mapsto (k^{\sharp},(k^{1/p})^\sharp,...)$. It is easy to see that this map is inverse to the projection map $\varprojlim_{k\mapsto k^p}K^o\rightarrow \varprojlim_FK^o/\omega$. And thus the point (4) is clear, since $\varprojlim_{k\mapsto k^p}K^o$ is independent of $\omega$. By the way, it is not hard to see that these two maps are homeomorphisms.

(6)there exists an element $\omega^\flat\in\varprojlim_FK^o/\omega$ such that $v((\omega^\flat)^\sharp)=v(\omega)$;(7)if we set $K^\flat=(\varprojlim_FK^o/\omega)[(\omega^\flat)^{-1}]$ equipped with the $(\omega^\flat)^{-1}$-adic topology, then we have a multiplicative homeomorphism $K^\flat=\varprojlim_{k\mapsto k^p}K$, and there is a map $K^\flat\rightarrow K,k\mapsto k^\sharp$.

For the point (6), we have to notice that it does not say that we can find an image of $\omega$ in $\varprojlim_FK^o/\omega$, it simply says that there exists an element in it with the same norm as $\omega$. Note that $Im(v)$ is not discrete in $\mathbb{R}$, which has the following consequence:

Lemma: $Im(v)$ is $p-$divisible.

Proof(of the lemma): since $Im(v)\neq p^{\mathbb{Z}}$, we have that $Im(v)$ is generated by elements $v(k)$ such that $1/p where $k\in K$. For any such $k$, since the Frobenius map $F$ is surjective, there is an element $k'\in K$ such that $v(k-k'^p)\leq1/p$. Yet $v(k-k'^p)\neq v(k)$, thus $v(k'^p)=v(k'^p-k+k)=\max(v(k'^p-k),v(k))=v(k)$. Thus $Im(v)$ is $p-$divisible.

So, in this way, we can find a $\omega'$ such that $v(\omega')^p=v(\omega)$. Using the fact that the Frobenius map is surjective, we can find an element $\omega^\flat=(\omega'^p,\omega',k_2,...)\in \varprojlim_FK^o/\omega$. It is then easy to see that $v((\omega^\flat)^\sharp)=v(\omega)$, and thus proving the point (6). As for point (7), we first observe that $G$ is multiplicative, so $G$ extends to $G:K^\flat=\varprojlim_{F}(K^o/\omega)[(\omega^\sharp)^{-1}]\rightarrow \varprojlim_{k\mapsto k^p}K$. Note that the topology on $K$ is the same as the $\omega K^o$-adic topology since the valuation $v$ is of rank $1$, and in the same way, the topology on $\varprojlim_FK^o/\omega$ is the same as the $\omega^\flat(\varprojlim_FK^o/\omega)$-adic topology, so is $K^\flat$. This in turn shows that the topology on $K^\flat$ is the same as the topology induced by the valuation $v^\flat(k)=v(G(k))$, which is a valuation of rank $1$. So, in this way, we give $K^\flat$ a valuation $v^\flat$. The rest part of (7) is easy. Moreover, we see that $v^\flat(K^\flat-0)=v(K-0)$.

(8)$K^\flat$ is a perfectoid field of characteristic $p$, and $K^{\flat o}/\omega^\flat= K^o/\omega$. If $m,m^\flat$ are the maximal ideals of $K^o,K^{\flat o}$,then $K^{\flat o}/m^\flat=K^o/m$;(9)if $K$ is of characteristic $p$, then $K^\flat=K$.

It is easy to see that $K^\flat$ is complete since $\varprojlim_FK^o/\omega$ is so(the projective limit of a system of discrete spaces). It is clear that $K^\flat$ is of characteristic $p$, then so is its residue field. Moreover, $K^{\flat o}=\varprojlim_FK^o/\omega$, so the Frobenius map on $K^{\flat o}=K^{\flat o}/p$ is surjective, this proves that $K^\flat$ is indeed a perfectoid field. Moreover, the rest part of (8) is easy, so point (8) is done. For (9), note that if for $k,k'\in K$, we have that $k^p=k'^p$, then $k=k'$. This is fairly straight-forward, and this shows that $\varprojlim_{k\mapsto k^p}K=K$, and thus $K^\flat=K$, finishing (9).

In this way, we call $K^\flat$ the tilt of $K$. In the next posts, we will use almost ring theory to show that the category of $K-$algebras and the category of $K^\flat-$algebras are canonically isomorphic.