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<v(\omega)<1 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<v(k)<1 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.


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