perfectoid spaces-3

In this post, we will say something on K^{o a}-algebras. At first sight, this notion is a bit mysterious, because we do not know whether the category K^{o a}-mod has any ring or field serving as base ring or base field, just like R-mod where R is a ring or a field. In fact, the essential point in defining an algebra A is a morphism, F:A\times A\rightarrow A, to serve as the multiplication operation on latex $A$(of course, if the algebra has a unit, more conditions should be posed). This F is bi-additive(just like the bilinear forms), so this inspires us to redefine the multiplication operation as f:A\bigotimes A\rightarrow A. So, here the tensor product appears naturally, and as we have said, K^{o a}-mod is a tensor category, that is, the notion of tensor product makes sense in this category. As a result, we can safely talk about K^{o a}-algebras.

Suppose that A is a K^{o a}-algebra, then as a K^{o a}-module, A_*=Hom_{K^{o a}}(K^{o a},A)=Hom_{K^o}(m,A). Surprisingly, we can make it into an K^o-algebra. The point is, suppose f,g:m\rightarrow A two K^o-morphisms, then we can define the product of f,g as fg:m=m\bigotimes m\rightarrow A\bigotimes A\rightarrow A, x\bigotimes y\mapsto f(x)g(y). Note that this is indeed a K^o-morphism, and the product thus defined indeed gives A_* a structure of K^o-algebra. Using the last proposition of the precedent post, we see that A_*^a=A, which means that any K^{o a}-algebra comes from a K^o-algebra. Similarly, for any A-module M, we have that M_* is an A_*-module, and M_*^a=M shows that A-modules come from A_*-modules. We can also show that the category A_mod is again an abelian tensor category, so again we can define A-algebras, and things the like.

Now we give several definitions stemmed from commutative algebra:

Suppose A is a K^{o a}-algebra, then(1) an A-module M is flat if the functor N\mapsto M\bigotimes N on A-mod is exact;(2) an A-module M is projective if the functor N\mapsto alHom(M,N) on A-mod is exact;(3)if R is a K^o-algebra and N is an R-module, then N^0 is an almost finitely generated(resp. almost finitely presented) R^a-module if and only if for all x\in N^o, there is a finitely generated(resp. finitely presented) R-module M_x with a morphism f_x:N_x\rightarrow N^o such that Ker(f_x),Coker(f_x) are annihilated by x. We say that N^o is uniformly almost finitely generated if there is some integer n such that N_x can be chosen to be generated by n elements for all x.

Here are some easy consequences of these definitions:

If R is a K^o-algebra and N is an R-module, then the R^a-module N^a is flat if and only if Tor_i^R(N,X) is almost zero for all R-module X and for all i>0; N^a is almost projective if and only if Ext^i_R(N,X) is almost zero for all R-module X and for all i>0. For their proofs, see Gabber and Ramero’s book, ‘almost ring theory’ section 2.3 and 2.4.

We consider the following example: take K to be the completion of \mathbb{Q}_p(p^{1/p^{\infty}}) and L=K(p^{1/2}). Then K^o is the completion of \mathbb{Z}_p[p^{1/p^{\infty}}] and its maximal ideal m consists of those elements of K^o whose valuations are <1, that is, m is generated by p^{1/p^{\infty}}. As for L, we know that L^o=K^o[p^{1/2}]. For any x=p^{1/p^{\infty}}\in m, we can find an injection f_x: L_x=K^o\bigoplus x^{1/2}K^o\rightarrow L^o=K^o\bigoplus p^{1/2}K^o, then we see easily that xL^o\subset L_x, thus x Coker(f_x)=0, so this means that L^o, as a K^o is uniformly almost finitely generated.

There are other concepts to be introduced later.

In the next post, we will introduce the perfectoid K-algebras and show some equivalences of categories.

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