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.