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Let $V$ be an integral domain with an ideal $m\subset V$ and put $K = S^{-1}V$ where $S = {1}\cup m$ (a multiplicatively closed subset). Is it true that the category of almost $(V,m)$-modules is equivalent to the category of $K$-modules via $-\otimes_V K$? By the category of almost $(V,m)$-modules I mean the category whose objects are $V$-modules and morphisms are defined by $$Hom(M,N) = \varinjlim Hom(M',N/N')$$ where the direct limit is taken over $M'\subset M$ and $N'\subset N$ with $M/M'$ and $N'$ both $m$-torsion (a module $M$ is $m$-torsion if for every $x\in M$, there is some $u\in m$ such that $ux=0$). This is the Serre quotient of the category of $V$-modules by the subcategory of $m$-torsion modules.


Graham: yes, thank you. Suppose that $m$ is the maximal ideal of a valuation ring (and remove zero!).

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Surely $S$ is not what it is made to be in the second sentence, for then it contains zero. – Mariano Suárez-Alvarez May 16 '11 at 3:16
Probably supposed to be $S = 1 + m$, so that $K = V_m$. – Hurkyl May 16 '11 at 5:16
up vote 6 down vote accepted

At least in Faltings's setting, $m$ is the maximal ideal of a non-Noetherian valuation domain $V$. If we let $S$ be the non-zero elements of $m$, then the localization of $V$ at $S$ will be the field of fractions $K$ of $V$. The category of $K$-modules (i.e. $K$-vector spaces) is obtained as a Serre quotient of $V$-modules, but one quotients out modules which are killed by some non-zero element of $V$, while the category of amost modules is obtained by quotienting out by a much smaller category, namely the modules which are killed by every element of $m$.

In other words, I think you have misinterpreted the meaning of $m$-torsion, at least in so far it is used in the context of Faltings's "almost commutative algebra".

Added in response to the comment below: Let me not use the word $m$-torsion anymore, since it can be interpreted in different ways, and seems to be causing confusion.

Let $\mathcal M$ be the category of $V$-modules. Here are two Serre subcategories of $\mathcal M$:

  1. $\mathcal C$, the category of almost zero modules, i.e. modules for which $m M = 0$. This is a Serre subcategory because $m^2 = m$ in the context of Faltings's theory.

  2. $\mathcal C'$, the category of torsion modules, i.e. modules such that $x M = 0$ for some non-zero $x \in m$. This is a Serre category just because $V$ is a domain.

Clearly $\mathcal C$ is contained in $\mathcal C'$, but they are far from equal!

The quotient $\mathcal M/\mathcal C$ is the category of almost modules. The quotient $\mathcal M/\mathcal C'$ is equivalent to the category of $K$-vector spaces (the equivalence being given by applying the functor $K\otimes\text{--}$, where $K$ is the fraction field of $V$).

In the statement of the question, you (mistakenly) define almost modules to be $\mathcal M/\mathcal C'$, and (correctly) conclude that this is equivalent to $K$-vector spaces. Once you replace this mistaken definition with the correct definition of almost modules (i.e. $\mathcal M/\mathcal C$), you will see that the resulting category is not equivalent to $K$-vector spaces.

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A module is almost zero when $mM = 0$. What is the Serre subcategory that gives rise to $K$-vector spaces? – Nicolás May 16 '11 at 0:59
Dear unknown, Yes, I agree with your comment defining almost zero modules, and the category of almost modules is obtained by taking the quotient by the Serre subcatgory of almost zero modules. However, in your question, you wrote something different (and incorrect), namely that almost modules are obtained by quotienting out by the Serre subcategory consisting of modules annihilated by some element of $m$, which is not the same thing. This latter quotient category is not the category of almost modules; rather, it is the category of $K$-vector spaces. Regards, Matthew – Emerton May 16 '11 at 3:04
Question edited to try to clear up confusion. – Emerton May 16 '11 at 3:11
Thanks for your answer, Matt. – Nicolás May 16 '11 at 3:26

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