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 is a perfectoid field with a residue of characteristic , and is the subring of elements whose valuations are and the maximal ideal of consisting of elements whose valuations are .

**For a -module, is an element, we say that is almost zero if . is said to be an almost -module if all its elements are almost zero.**

We can show that the full subcategory of almost -modules is a Serre subcategory of the category of -modules, that is

**For an exact sequence in , then are almost zero if and only if is almost zero.**

Before proving this proposition, we need a lemma

**Lemma: .**

Since is -divisible, so for any , there exists such that . This means that and thus , moreover showing that , thus proving the lemma(the other direction is evident).

Return to the proof to the proposition. If are almost zero, then . So for any , and any , is zero in , so , thus , yet using the lemma shows that . If 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 category, the objects of which are the objects of , and the morphisms are defined to be where the inverse limit is taken over the system of pairs with being a sub--mod of , being a sub--mod of and being almost zero. Here is the image of in .

A little remark on categories is in order. We can show that, this quotient category, is the localization of with respect to some set of morphisms of . More precisely, we set . Then we can show that (we shall denote this functor as , that is ). So, we sometimes say that the functor from to constructed as above is a localization functor. In this description, objects of are the same as , yet a morphism from to is of the form such that , that is with a -module.

Next we will give the left and right adjoint functors of . That is:

**For two modules(which are automatically -modules), we have that .**

In other words, the functor is left-adjoint to the functor , i.e. . We have to define how acts on the morphisms. This time we shall use the localization description of . That is to say, given two objects in and a morphism , then we know that are almost zero. We need a lemma to finish the definition of :

**Lemma: **

This is not difficult to show. We simply set . For its inverse, we use the fact that .

Now we define . The first demanding problem is to show that is well-defined. Note that is almost zero, so is almost zero in , this means that . Now suppose that there are two elements such that , then , thus is almost zero. And we can show that by writing . So, is well-defined. It is not difficult to show that is a -morphism, thus finishing the proof that is a well-defined functor.

Now we need to show the isomorphism . Given a morphism , we want to construct a corresponding one in . Note that there is a natural morphism . Now we take the push-out of the pair , and we get . Note that are almost zero modules, and we can verify that the corresponding are also almost zero(in some sense, concretely is just modulo something). In this way, we get a morphism in . For the converse, given with in , we have to find a -morphism from to . This is very similar to proving that is well-defined. Another short-cut is to use the composition of morphism, that is, composed with gives the desired morphism from to . It is not too difficult to show that these two processes are inverse one to the other. For one direction, given , with the canonical , we have where with generated by elements of the form where runs through and runs through . And in turn, from , we get . Note that . We have to show that this element is the same as that through , that is , or equivalently, we should show that . Yet, the left-hand side is , this means that . The other direction is not difficult, either.

So, in this way, for any two objects in , we have that has a natural -module structure. Moreover, for any and any , we have that , so means that since . This shows that , as a -module, has no almost zero element at all. Thus we write .

We have the following proposition on the category :

** is an abelian tensor category(i.e. we can define kernel, co-kernel, tensor product compatible with the functor ).**

For example, we can define the tensor product of two -modules as . We can also define -algebras, and the modules on these algebras.

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

**We set the functor , then .**

Of course, we have to precise how acts on the morphisms. This is fairly obvious. For any two objects in and a morphism , we construct as follows: for any , we take the push-out of , and we get , note that this morphism is in ; then using composition of morphisms, we get that , and since is in (defined far above), so is , and since is also in , so is their composition , so this shows that is in . This finishes the construction of .

Then we have to show that

in order to show that . This is simpler than the process of the first pair of adjoint functors above. Given a morphism , we construct a morphism which sends each element in to an almost morphism from to . This is very obvious, for each , we define an almost morphism as where is defined as . This is clearly an element in and it is easy to see that is a morphism from to . Conversely, for , this means that () is an almost morphism, . How to piece these morphisms together? We can proceed as follows: we set where is generated by elements of the form or or where runs through , runs through and runs through . Now we define and . We verify easily that are well-defined and are indeed -morphisms. Moreover, if in , this means that for a fixed . Using this, we can show that is almost zero. It is also easy to show that 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 where are objects in ).

Moreover, we have that:

**If is an object in , then the adjunction morphism is an isomorphism; if is an object in , then .**

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 -algebras.