Now in this post we will use Witt vectors to show that the category of perfectoid -algebras is equivalent to the category of perfectoid -algebras(which are defined in this post).
Suppose that is a perfectoid -algebra, which means that the Frobenius is surjective. Then just like the process of passing to , we can construct where is an element with the same valuation as in .
Next we show how can we get back from .
We first note that any element has a unique representative such that . Note that we can write explicitly as where is any lift of in . In fact, this is just the Teichmüller representative of in .
We then take the Witt vector ring of , and set . We can show that:
Proposition-1: is a surjective ring homomorphism.
If this is true, then in some sense, after passing to localization, we get a functor from the category of perfectoid -algebras to the category of perfectoid -algebras. Combining with the first step from to , we get an equivalence between these two categories.
Proof(of Proposition-1):We first show that is a ring homomorphism. We can show is a morphism modulo for all , which implies that is a morphism. The definition of shows that we can consider just the elements of of the form , that is, we can consider just . We can show that this can factorize through , that is, there are two maps and such that . Indeed, if we set , , then . Note that it is easy to see if there is a morphism of rings , then there is an induced morphism of rings . So if we consider the projection , then we get the induced morphism , and the composed map of and is just . Note that this time is a morphism of rings. Indeed, look at what the addition and multiplication means here: just takes an element to each of their -th component . And since the addition and multiplication of is just component for component, so this shows that is indeed a morphism of rings. On the other hand, we can show that depends only on elements modulo . That is to say, . Yet for , we have that in , the coefficient is . If , then , thus . If , then we still have that . So all in all, we showed that . So, this means that can factorize through . Moreover, is a morphism of rings, and is too, this shows that , their composition is also a morphism of rings.
Next we show that is surjective. This is simple as we can proceed as follows: if we can find for each an element such that , then we are done. Indeed, if this is so, then consider , we can again find a such that . This means that (here we can choose such that ) since is a morphism of rings. So we can iterate this process to infinity and get a sequence . It is not hard to see that using the topology of , we can show that converges in . So this means that there is an element in which is sent to for any , thus showing that is surjective, which finishes the whole proof.
Now as we have remarked, we first construct a functor from to , and then another functor , which sends each perfectoid -algebra to . The problem with this construction is that, we should know that corresponds to some in . To get rid of this restriction, we show the following:
Proposition-2: is a principal ideal of .
Proof(of Proposition-2): Now for any such that , this means that , so for some . For any , since , this means that since is integrally closed in . So, in this way, we find an element which is in such that . If we define two elements , then for , we have that , and for , . So, this means that . We can take a factor from such that . And note that this time , so we can redo the above process with , and we shall get some such that , then again we proceed to infinity and finally we get, using the completeness of , that , this shows that , an ideal generated by .
Note that this description of gives an intrinsic definition of using the kernel of , that is .
So in this way we have these two functors the definitions of which depend only on the tilting .
One word about the correspondence of , a perfectoid -algebra and (the same for -algebras). Note that, since is a Banach -algebra, so for any ,there is some such that , so . Note that for any , we have that , so this shows that . So, in we see that there is a one-to-one correspondence between the category of perfectoid -algebras and the category of -algebras.
Combining all the above results, we arrive at proving that the category of perfectoid -algebras and the category of perfectoid -algebras are equivalent.