perfectoid spaces-4

Suppose that K is a perfectoid field. In this post, we will introduce algebras over K as follows:

A perfectoid K-algebra is a Banach K algebra R such that the subset R^o of R consisting of power-bounded elements(as defined in the previous post) is open and bounded, and the Frobenius morphism F:R^o/\omega\rightarrow R^o/\omega is surjective(here the element \omega is defined in the first post of this series). Morphisms between perfectoid K-algebras are the continuous morphism of K-algebras. A perfectoid K^{o a}-algebra is a \omega-adically complete flat K^{o a}-algebra R on which the Frobenius morphism F induces an isomorphism F:R/\omega^{1/p}\rightarrow R/\omega(here \omega-adically complete means that R\simeq \varprojlim R/\omega^n); morphisms between perfectoid K^{o a}-algebras are morphisms of K^{o a}-algebras. A perfectoid K^{o a}/\omega-algebra is a \omega-adically complete flat K^{o a}-algebra R on which the Frobenius morphism induces an isomorphism F:R/\omega^{1/p}\rightarrow R; morphisms between perfectoid K^{o a}/\omega-algebras are morphisms of K^{o a}/\omega-algebras.

We will use Scholze’s notations, that is, we write K-perf denotes the category of perfectoid K-algebras and K^{o a}-perf denotes the category of perfectoid K^{o a}-algebras, and so on.

The following proposition shows that perfectoid K-algebras and perfectoid K^{o a}-algebras are closely related:

If R is a perfectoid K-algebra, then the Frobenius morphism F induces an isomorphism R^o/\omega^{1/p}\simeq R^o/\omega, and R^{o a} is a perfectoid K^{o a}-algebra.

Proof: By definition, F is surjective. Suppose that x\in R is sent to \omega R^o, that is x^p\in \omega R^o, say x^p=\omega y(y\in R^o). Since \omega y is power bounded, so is x, so x\in R^o. Moreover, it is clear that x\in \omega^{1/p}R^o, so F induces an isomorphism. Next we need to show that R^o is \omega-adically complete. This is not difficult. We define f:R^o\rightarrow \varprojlim R^o/\omega^n, r\mapsto (r+(\omega^n)), and g:\varprojlim R^o/\omega^n\rightarrow R^o,(x_n+(\omega^n))\mapsto \lim_n x_n. It is necessary to verify that g is well-defined. Indeed, note that, according to the definition, x_n-x_{n+k}\in (\omega^n) for k>0, thus (x_n)_n is a Cauchy sequence in R^o, as R is complete, so is R^o, thus \lim_n x_n exists. Moreover, if \lim_n x_n=0, this means that x_n-0=\lim_k x_n-x_{n+k}\in (\omega^n) for all n, so (x_n+(\omega^n))=(0+(\omega^n)). So, g is indeed well-defined. And it is not difficult to show that f and g are inverse one to the other, so we showed that R^o is \omega-adically complete. One last thing is to show that R^o is a flat K^{o a}-module. This is not difficult. To finish the proof, we need the following results:

If N is a flat K^o-module, then N^a is a flat K^{o a}-module.

Conversely, we have that

If A is a K^{o a}-algebra, and let R=A_*[1/\omega]. If we give R the Banach K-algebra structure making A_* open and bounded in R, then A_*=R^o is the set of power-bounded elements, and R is perfectoid K-algebra, and the Frobenius morphism F:A_*/\omega^{1/p}\simeq A_*/\omega.


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