Perfectoid spaces-7

Now in this post we will use Witt vectors to show that the category of perfectoid $K$-algebras is equivalent to the category of perfectoid $K^{\flat}$-algebras(which are defined in this post).

Suppose that $A$ is a perfectoid $K$-algebra, which means that the Frobenius $F: A^o/p\rightarrow A^o/p$ is surjective. Then just like the process of passing $K^o$ to $K^{\flat o}$, we can construct $A^{\flat}=\varprojlim_F A^o/p[1/p']$ where $p'\in \varprojlim_F K^o/p$ is an element with the same valuation as $p$ in $K$.

Next we show how can we get back from $A^{\flat}$.

We first note that any element $x^{\flat}=(x_0,x_1,...)\in A^{\flat o}$ has a unique representative $(x^{(0)},x^{(1)},...)\in (A^o)^{\mathbb{N}}$ such that $x^{(n)}=x_n(\text{mod} p),(x^{(n)})^p=x^{(n-1)},\forall n$. Note that we can write $x^{(n)}$ explicitly as $x^{(n)}=\lim_m(x_{n+m}')^{p^m}$ where $x_{n+m}'$ is any lift of $x_{n+m}$ in $A^o$. In fact, this is just the Teichmüller representative of $x_n$ in $A^o$.

We then take $W(A^{\flat o})$ the Witt vector ring of $A^{\flat o}$, and set $T: W(A^{\flat o})\rightarrow A^o, a=(a_0,a_1,...)\mapsto \sum_n p^na_n^{(n)}$. We can show that:

Proposition-1:$T$ is a surjective ring homomorphism.

If this is true, then in some sense, after passing to localization, we get a functor from the category $K^{\flat}-Perf$ of perfectoid $K^{\flat}$-algebras to the category $K-Perf$ of perfectoid $K$-algebras. Combining with the first step from $K-Perf$ to $K^{\flat}-Perf$, we get an equivalence between these two categories.

Proof(of Proposition-1):We first show that $T$ is a ring homomorphism. We can show $T$ is a morphism modulo $p^n$ for all $n>0$, which implies that $T$ is a morphism. The definition of $T$ shows that we can consider just the elements of $W(A^{\flat o})$ of the form $(a_0,a_1,...,a_{n-1},0,0,...)$, that is, we can consider just $T_n:W_n(A^{\flat o})\rightarrow A^o/p^nA^o$. We can show that this $T_n$ can factorize through $W_n(A^{\flat o}/pA^{\flat o})$, that is, there are two maps $f:W_n(A^{\flat o})\rightarrow W_n(A^{\flat o}/p^nA^{\flat o})$ and $S_n:W_n(A^{\flat o}/p^nA^{\flat o})\rightarrow A^o$ such that $S_n\circ f=T_n$. Indeed, if we set $f(a_0,a_1,...,a_{n-1})=(a_0^{(n)},a_1^{(n)},...,a_{n-1}^{(n)})(\text{mod}p^n)$, $S_n(y_0,y_1,...,y_{n-1})=\sum_ip^i y_i^{p^{n-i}}$, then $S_n\circ f(a_0,...,a_{n-1})=\sum_ip^i (a_i^{(n)})^{p^{n-i}}=\sum_ip^i a_i^{(i)}=T_n(a_0,a_1,...,a_{n-1})$. Note that it is easy to see if there is a morphism of rings $s: R\rightarrow R'$, then there is an induced morphism of rings $s':W_n(R)\rightarrow W_n(R')$. So if we consider the projection $\pi:A^{\flat o}/p^n\rightarrow A^{\flat o}/pA^{\flat o}$, then we get the induced morphism $\pi':W_n(A^{\flat o}/p^nA^{\flat o})\rightarrow W_n(A^{\flat o}/pA^{\flat o})$, and the composed map of $f$ and $\pi'$ is just $\pi'\circ f: W_n(A^{\flat o})\rightarrow W_n(A^{\flat o}/pA^{\flat o}),(a_0,...,a_{n-1})\mapsto (a_0^{(n)},...,a_{n-1}^{(n)})(\text{mod}p)$. Note that this time $\pi'\circ f$ is a morphism of rings. Indeed, look at what the addition and multiplication means here: $\pi'\circ f$ just takes an element $a=(a_0,...,a_{n-1})$ to each of their $n$-th component $(a_{0,n},...,a_{n-1,n})$. And since the addition and multiplication of $W_n(A^{\flat o})$ is just component for component, so this shows that $\pi'\circ f$ is indeed a morphism of rings. On the other hand, we can show that $S_n$ depends only on elements modulo $p$. That is to say, $S_n((y_0+p z_0,...,y_{n-1}+p z_{n-1}))=\sum_i p^i (y_i+p z_i)^{p^{n-i}}$. Yet for $j>0$, we have that in $p^i{p^{n-i}\choose j}y_i^{p^{n-i}-j}(p z_i)^{j}$, the coefficient is $p^{i+j}{p^{n-i}\choose j}$. If $j, then $p^{n-i}|{p^{n-i}\choose j}$, thus $p^n|p^{i+j}{p^{n-i}\choose j}$. If $j=p^{n-i}$, then we still have that $p^n|p^{i+j}{p^{n-i}\choose j}$. So all in all, we showed that $S_n((y_0+p z_0,...,y_{n-1}+p z_{n-1}))\simeq S_n((y_0,...,y_{n-1}))(\text{mod}p)$. So, this means that $T_n$ can factorize through $W_n(A^{\flat o}/pA^{\flat o})$. Moreover, $\pi'\circ f$ is a morphism of rings, and $S_n$ is too, this shows that $T_n$, their composition is also a morphism of rings.

Next we show that $T$ is surjective. This is simple as we can proceed as follows: if we can find for each $a'\in A^{\flat}$ an element $a\in W(A^{\flat o})$ such that $T(a)=a'(\text{mod}p)$, then we are done. Indeed, if this is so, then consider $(a'-T(a))/p\in A^{\flat}$, we can again find a $b\in W(A^{\flat})$ such that $T(b)=(a'-T(a))/p(\text{mod}p)$. This means that $a'=T(a+p' b)(\text{mod}p^2)$ (here we can choose $p'\in W(A^{\flat o})$ such that $T(p')=p(\text{mod} p^2)$) since $T$ is a morphism of rings. So we can iterate this process to infinity and get a sequence $a,b,c,...\in W(A^{\flat o})$. It is not hard to see that using the topology of $W(A^{\flat o})$, we can show that $a+p' b+p'^2 c+...$ converges in $W(A^{\flat o})$. So this means that there is an element in $W(A^{\flat o})$ which is sent to $a'$ for any $a'\in A^{\flat}$, thus showing that $T$ is surjective, which finishes the whole proof.

Now as we have remarked, we first construct a functor $S$ from $K-Perf$ to $K^{\flat}-Perf$, and then another functor $T:K^{\flat}-Perf\rightarrow K-Perf$, which sends each perfectoid $K^{\flat}$-algebra $A^{\flat}$ to $T(W(A^{\flat o}))[1/p]=A$. The problem with this construction is that, we should know that $A^{\flat}$ corresponds to some $A$ in $K-Perf$. To get rid of this restriction, we show the following:

Proposition-2: $Ker(T)$ is a principal ideal of $W(A^{\flat o})$.

Proof(of Proposition-2): Now for any $a\in W(A^{\flat o})$ such that $T(a)=\sum_n p^n a_n^{(n)}=0$, this means that $a_0^{(0)}=0(\text{mod}p)$, so $a_0^{(0)}=p^{(0)} b_0$ for some $b_0\in A^o$. For any $n>0$, since $(a_0^{(n)})^{p^n}=a_0^{(0)}$, this means that $b_n=a_0^{(n)}/p'^{(n)}\in A^o$ since $A^o$ is integrally closed in $A$. So, in this way, we find an element $b'=(b_0,b_1,...)$ which is in $A^{\flat o}$ such that $a_0=p' b'$. If we define two elements $b=(b',0,0,...),\tilde{p}=(p',0,0,...)\in W(A^{\flat o})$, then for $c=a-\tilde{p}b$, we have that $c_0=a_0-p' b'=0$, and for $n>0$, $c_n=a_0^{p^n}-(p'b')^{p^n}+p(...)=0(\text{mod}p)$. So, this means that $c=a-\tilde{p}b\in pW(A^{\flat o})$. We can take a factor from $W(A^{\flat o})$ such that $a-(\tilde{p}-p)b\in pW(A^{\flat o})$. And note that this time $T(a-(\tilde{p}-p)b)=T(a)-T(\tilde{p}-p)T(b)=-T(b)(T(\tilde{p})-T(p))=-T(b)(p^{(0)}-p)=0$, so we can redo the above process with $(a-(\tilde{p}-p)b)/p\in W(A^{\flat o})$, and we shall get some $c\in W(A^{\flat o})$ such that $(a-(\tilde{p}-p)b)/p-(\tilde{p}-p)c\in pW(A^{\flat o})$, then again we proceed to infinity and finally we get, using the completeness of $W(A^{\flat o})$, that $a=(\tilde{p}-p)(b+p c+p^2 d+...)\in (\tilde{p}-p)W(A^{\flat o})$, this shows that $Ker(T)=(\tilde{p}-p)$, an ideal generated by $\tilde{p}-p$.

Note that this description of $Ker(T)$ gives an intrinsic definition of $T(W(A^{\flat o}))$ using the kernel of $T$, that is $A=W(A^{\flat o})/(\tilde{p}-p)$.

So in this way we have these two functors $S,T$ the definitions of which depend only on the tilting $K\rightarrow K^{\flat}$.

One word about the correspondence of $A$, a perfectoid $K$-algebra and $A^{o}$(the same for $K^{\flat}$-algebras). Note that, since $A$ is a Banach $K$-algebra, so for any $a\in A$,there is some $k\in K^*$ such that $||k a||<1$, so $k a\in A^o$. Note that for any $k\in K-K^o$, we have that $K=K^o[k]$, so this shows that $A=A^o[1/(1/k)]=A^o[k]$. So, in we see that there is a one-to-one correspondence between the category of perfectoid $K$-algebras and the category of $K^o$-algebras.

Combining all the above results, we arrive at proving that the category of perfectoid $K$-algebras and the category of perfectoid $K^{\flat}$-algebras are equivalent.