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 is a complete non-archimedean field with a non-discret valuation of rank such that the residue field is of characteristic and the Frobenius map is surjective.
Here several points are to be clarified: the topology on is induced by the valuation . Non-discret valuation here means that is non-discret in the topology of the latter is the Euclidean topology.
Classical examples for perfectoid spaces are the completions of , , , the valuation of the first and the third fields is induced by that of and the valuation of the second is induced by the -adic valuation.
Next we want to define a tilting process for a perfectoid field . We choose an element with and take . Then we have
(1) has the discret topology;(2) is a perfect ring.
The first point is not hard. Note that consists of elements of valuation . And this is an open set in , thus in 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 with the inverse limit topology and we have that
(3)there is a multiplicative homeomorphism, given by projection;(4) the right-hand side is independent of ;(5)there is thus a multiplicative map .
We first show the point (5). Suppose that , and choose any such that is a lift of , then we define . This point is a simple calculation(here we shall use the assumption that is complete to ensure the existence of the limit). As for the point (3), we set . It is easy to see that this map is inverse to the projection map . And thus the point (4) is clear, since is independent of . By the way, it is not hard to see that these two maps are homeomorphisms.
(6)there exists an element such that ;(7)if we set equipped with the -adic topology, then we have a multiplicative homeomorphism , and there is a map .
For the point (6), we have to notice that it does not say that we can find an image of in , it simply says that there exists an element in it with the same norm as . Note that is not discrete in , which has the following consequence:
Lemma: is divisible.
Proof(of the lemma): since , we have that is generated by elements such that where . For any such , since the Frobenius map is surjective, there is an element such that . Yet , thus . Thus is divisible.
So, in this way, we can find a such that . Using the fact that the Frobenius map is surjective, we can find an element . It is then easy to see that , and thus proving the point (6). As for point (7), we first observe that is multiplicative, so extends to . Note that the topology on is the same as the -adic topology since the valuation is of rank , and in the same way, the topology on is the same as the -adic topology, so is . This in turn shows that the topology on is the same as the topology induced by the valuation , which is a valuation of rank . So, in this way, we give a valuation . The rest part of (7) is easy. Moreover, we see that .
(8) is a perfectoid field of characteristic , and . If are the maximal ideals of ,then ;(9)if is of characteristic , then .
It is easy to see that is complete since is so(the projective limit of a system of discrete spaces). It is clear that is of characteristic , then so is its residue field. Moreover, , so the Frobenius map on is surjective, this proves that is indeed a perfectoid field. Moreover, the rest part of (8) is easy, so point (8) is done. For (9), note that if for , we have that , then . This is fairly straight-forward, and this shows that , and thus , finishing (9).
In this way, we call the tilt of . In the next posts, we will use almost ring theory to show that the category of algebras and the category of algebras are canonically isomorphic.