# 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 $F:A_*/\omega^{1/p}\simeq A_*/\omega$.