# perfectoid spaces-2

The main references for this series of posts on perfectoid spaces are Scholze’s papers of this aspect.

This post will say something on almost mathematics, or more precisely almost ring theory.

Through out this post, we will assume that $K$ is a perfectoid field with a residue of characteristic $p$, and $K^o$ is the subring of elements whose valuations are $\leq 1$ and $m$ the maximal ideal of $K^o$ consisting of elements whose valuations are $<1$.

For $M$ a $K^o$-module, $x\in M$ is an element, we say that $x$ is almost zero if $mx=0$. $M$ is said to be an almost $K^o$-module if all its elements are almost zero.

We can show that the full subcategory of almost $K^o$-modules is a Serre subcategory of the category $K^o-mod$ of $K^o$-modules, that is

For an exact sequence $0\rightarrow M_1\rightarrow M_2\rightarrow M_2\rightarrow 0$ in $K^o-mod$, then $M_1,M_3$ are almost zero if and only if $M_2$ is almost zero.

Before proving this proposition, we need a lemma

Lemma: $m^2=m$.

Since $Im(v)$ is $p$-divisible, so for any $k\in m$, there exists $k'\in K$ such that $v(k')^p=v(k)<1$. This means that $v(k')<1$ and thus $k'\in m$, moreover $k',k'^{p-1}\in m$ showing that $m\subset m^2$, thus proving the lemma(the other direction is evident).

Return to the proof to the proposition. If $M_1,M_3$ are almost zero, then $mM_1=0=mM_3$. So for any $x\in M_2$, and any $k\in m$, $kx$ is zero in $M_3$, so $kx\in M_1$, thus $mkx=0$, yet using the lemma shows that $mx=0$. If $M_2$ is almost zero, then the proof is evident, thus the proposition.

Then using the fact that this subcategory is a Serre subcategory, we can define a quotient category, that is, the $K^{o a}-mod$ category, the objects of which are the objects of $K^o-mod$, and the morphisms are defined to be $Hom_{K^{o a}}(M^a,N^a)=\varprojlim_{(M',N')}Hom_{K^o}(M',N')$ where the inverse limit is taken over the system of pairs $(M',N')$ with $M'$ being a sub-$K^o$-mod of $M$, $N'$ being a sub-$K^o$-mod of $N$ and $M/M',N'$ being almost zero. Here $M^a$ is the image of $M$ in $K^{o a}-mod$.

A little remark on categories is in order. We can show that, this quotient category, $K^{o a}-mod$ is the localization of $K^o-mod$ with respect to some set $S$ of morphisms of $K^o-mod$. More precisely, we set $S=\{f\in Mor(K^o-mod)|Ker(f),Coker(f)\in Obj(K^o-mod)\}$. Then we can show that $S^{-1}(K^o-mod)=K^{o a}-mod$(we shall denote this functor as $F^a$, that is $F^a(M)=M^a$). So, we sometimes say that the functor from $K^o-mod$ to $K^{o a}-mod$ constructed as above is a localization functor. In this description, objects of $S^{-1}(K^o-mod)$ are the same as $K^o-mod$, yet a morphism from $M$ to $N$ is of the form $M\xrightarrow{f}P\xleftarrow{g}N$ such that $g\in S$, that is $Ker(g),Coker(g)\in Obj(K^o-mod)$ with $P$ a $K^o$-module.

Next we will give the left and right adjoint functors of $F^a$. That is:

For two $K^o$modules(which are automatically $K^{o a}$-modules), we have that $Hom_{K^{o}}(m\bigotimes_{K^o}M,N)=Hom_{K^{o a}}(M,N^a)$.

In other words, the functor $T:K^{o a}-mod\rightarrow K^o-mod, M\mapsto m\bigotimes M$ is left-adjoint to the functor $F^a$, i.e. $T\vdash F^a$. We have to define how $T$ acts on the morphisms. This time we shall use the localization description of $K^{o a}-mod$. That is to say, given two objects $M,N$ in $K^{o a}-mod$ and a morphism $M\xrightarrow{f}P\xleftarrow{g}N$, then we know that $Ker(g),Coker(g)$ are almost zero. We need a lemma to finish the definition of $T$:

Lemma: $m\bigotimes_{K^o}m=m$

This is not difficult to show. We simply set $f:m\bigotimes m\rightarrow m,k\bigotimes k'\mapsto kk'$. For its inverse, we use the fact that $m=m^2$.

Now we define $T(f,g):m\bigotimes M=m\bigotimes m\bigotimes M\rightarrow m\bigotimes N, x\bigotimes x'\bigotimes k\mapsto x\bigotimes g^{-1}(x'f(k))$. The first demanding problem is to show that $T(f,g)$ is well-defined. Note that $Coker(g)=P/g(N)$ is almost zero, so $f(k)$ is almost zero in $Coker(g)$, this means that $x'f(k)\in g(N)$. Now suppose that there are two elements $k',k''\in N$ such that $g(k')=g(k'')=x'f(k)$, then $k'-k''\in Ker(g)$, thus $k'-k''$ is almost zero. And we can show that $x\bigotimes(k'-k'')=0$ by writing $x=\sum x_ix_i'\in m^2$. So, $x\bigotimes g^{-1}(x'f(k))$ is well-defined. It is not difficult to show that $T(f,g)$ is a $K^o$-morphism, thus finishing the proof that $T$ is a well-defined functor.

Now we need to show the isomorphism $Hom_{K^{o}}(m\bigotimes_{K^o}M,N)=Hom_{K^{o a}}(M,N^a)$. Given a morphism $f:m\bigotimes M\rightarrow N$, we want to construct a corresponding one in $Hom_{K^{o a}(M,N^a)}$. Note that there is a natural morphism $g:m\bigotimes M\rightarrow M, x\bigotimes k\mapsto xk$. Now we take the push-out of the pair $M\xleftarrow{g}m\bigotimes M \xrightarrow{f}N$, and we get $M\xrightarrow{f'}P\xleftarrow{g'}N$. Note that $Ker(g),Coker(g)$ are almost zero modules, and we can verify that the corresponding $Ker(g'),Coker(g')$ are also almost zero(in some sense, concretely $P$ is just $M\bigoplus N$ modulo something). In this way, we get a morphism $M\xrightarrow{f'}P\xleftarrow{g'}N$ in $Hom_{K^{o a}}(M,N^a)$. For the converse, given $M\xrightarrow{f}P\xleftarrow{g}N$ with $P$ in $S$, we have to find a $K^o$-morphism from $M$ to $N$. This is very similar to proving that $T$ is well-defined. Another short-cut is to use the composition of morphism, that is, $T(f,g):m\bigotimes M\rightarrow m\bigotimes N$ composed with $m\bigotimes N\rightarrow N$ gives the desired morphism from $m\bigotimes M$ to $N$. It is not too difficult to show that these two processes are inverse one to the other. For one direction, given $f:m\bigotimes M\rightarrow N$, with the canonical $g:m\bigotimes M\rightarrow M$, we have $M\xrightarrow{f'}P\xleftarrow{g'}N$ where $P=M\bigoplus N/R$ with $R$ generated by elements of the form $(g(x\bigotimes k),-f(x\bigotimes k))=(xk,-f(x\bigotimes k))$ where $x$ runs through $m$ and $k$ runs through $M$. And in turn, from $(f',g')$, we get $f'':m\bigotimes M=m\bigotimes m\bigotimes M\rightarrow N, x'\bigotimes x\bigotimes k\mapsto x'g'^{-1}(xf'(k))$. Note that $g'^{-1}(xf'(k))=g'^{-1}(\overline{(xk,0)})$. We have to show that this element is the same as that through $f$, that is $f(x'\bigotimes x\bigotimes k)=f(x'x\bigotimes k)=x'f(x\bigotimes k)$, or equivalently, we should show that $g'(f(x\bigotimes k))=\overline{(xk,0)}$. Yet, the left-hand side is $\overline{(0,f(x\bigotimes k))}$, this means that $g'(f(x\bigotimes k))-\overline{(xk,0)}=0\in P$. The other direction is not difficult, either.

So, in this way, for any two objects $M,N$ in $K^{o a}-mod$, we have that $Hom_{K^{o a}}(M,N)$ has a natural $K^o$-module structure. Moreover, for any $f\in Hom_{K^o}(m\bigotimes M,N)$ and any $x\in m$, we have that $xf(x'\bigotimes k)=f(xx'\bigotimes k)$, so $mf=0$ means that $f=0$ since $m=m^2$. This shows that $Hom_{K^{o a}}(M,N)$, as a $K^o$-module, has no almost zero element at all. Thus we write $alHom(M,N)=(Hom_{K^{o a}}(M,N))^a$.

We have the following proposition on the category $K^{o a}-mod$:

$K^{o a}-mod$ is an abelian tensor category(i.e. we can define kernel, co-kernel, tensor product compatible with the functor $F^a$).

For example, we can define the tensor product of two $K^{o a}$-modules $M^a,N^a$ as $(M\bigotimes_{K^o}N)^a$. We can also define $K^{o a}$-algebras, and the modules on these algebras.

Next we give a right-adjoint functor of $F^a$, that is:

We set the functor $T':K^{o a}-mod\rightarrow K^o-mod, M\mapsto M_*=Hom_{K^{o a}}(K^{o a},M)$, then $F^a\vdash T'$.

Of course, we have to precise how $T'$ acts on the morphisms. This is fairly obvious. For any two objects $M,N$ in $K^{o a}-mod$ and a morphism $M\xrightarrow{f}P\xleftarrow{g}N$, we construct $T'(f,g)$ as follows: for any $K^o\xrightarrow{i}P'\xleftarrow{j}M\in alHom(K^o,M)$, we take the push-out of $P'\xleftarrow{j}M\xrightarrow{f}P$, and we get $P'\xrightarrow{f'}P''\xleftarrow{j'}P$, note that this morphism is in $K^{o a}-mod$; then using composition of morphisms, we get that $K^o\xrightarrow{f'\circ i}P''\xleftarrow{j'\circ g}N$, and since $j$ is in $S$(defined far above), so is $j'$, and since $g$ is also in $S$, so is their composition $j'\circ g$, so this shows that $K^o\xrightarrow{f'\circ i}P''\xleftarrow{j'\circ g}N$ is in $alHom(K^o,N)$. This finishes the construction of $T'(f,g)$.

Then we have to show that

$Hom_{K^{o a}}(M^a,N^a)=Hom_{K^o}(M,T'(N))$

in order to show that $F^a\vdash T'$. This is simpler than the process of the first pair of adjoint functors above. Given a morphism $M\xrightarrow{f}P\xleftarrow{g}N$, we construct a morphism which sends each element in $M$ to an almost morphism from $K^{o a}$ to $N$. This is very obvious, for each $x\in M$, we define an almost morphism $H(f,g)(x)$ as $K^o\xrightarrow{f'}P\xleftarrow{g}N$where $f'$ is defined as $f'(k)=kf(x)$. This is clearly an element in $alHom(K^o,N)$ and it is easy to see that $H(f,g)$ is a morphism from $M$ to $T'(N)$. Conversely, for $H\in Hom_{K^o}(M,T'(N))$, this means that $H(x)$($\forall x\in M$) is an almost morphism, $K^o\xrightarrow{f_x}P_x\xleftarrow{g_x}N$. How to piece these morphisms together? We can proceed as follows: we set $P=\bigoplus_{x\in M}P_x/P'$ where $P'$ is generated by elements of the form $f_{kx}(1)-kf_x(1)$ or $g_x(y)-g_{x'}(y)$ or $g_{kx}(y)-kg_x(y)$ where $x,x'$ runs through $M$, $y$ runs through $N$ and $k$ runs through $K^o$. Now we define $f:M\rightarrow P, x\mapsto f_x(1)$ and $g:N\rightarrow P,y\mapsto g_x(y)$. We verify easily that $f,g$ are well-defined and are indeed $K^o$-morphisms. Moreover, if $g(y)=0$ in $P$, this means that $g_x(y)=\sum (f_{k_ix_i}(1)-k_if_{x_i}(1)+g_{k_ix_i}(y_i)-k_ig_{x_i}(y_i))+\sum_j(g_{z_j}(w_j)-g_{z'_j}(w_j))\in P'$ for a fixed $x$. Using this, we can show that $Ker(g)$ is almost zero. It is also easy to show that $Coker(g)$ is almost zero, too(a good reference for this is Gabber and Ramero’s book, ‘almost ring theory'(2003), propositions 2.2.*).

Having constructed the morphisms in both directions, we can prove that this is indeed an isomorphism(another method of doing this is to use the isomorphism $Hom_{K^{o a}}(L,alHom(M,N))=Hom_{K^{o a}}(L\bigotimes M,N)$ where $L,M,N$ are objects in $K^{o a}-mod$).

Moreover, we have that:

If $M$ is an object in $K^{o a}-mod$, then the adjunction morphism $(M_*)^a\mapsto M$ is an isomorphism; if $M$ is an object in $K^o-mod$, then $(M^a)_*=Hom_{K^o}(m,M)$.

The adjunction morphism can be written out explicitly, and this is not difficult. The second identity comes from the first pair of adjoint functors.

In the next post, we will extend some notions in commutative algebra to $K^{o a}$-algebras.