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<p^{n-i}, 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.


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