### Filtrations and Gradings

This is going to be a blog post on Filtrations and Gradings. We’re going to closely follow the development in Local Algebra by Serre.

A filtered ring is a ring with the set of ideals such that , , and . An example would be , where is the ideal generated by in .

Similarly, a filtered module over a filtered ring is defined as a module with a set of submodules such that , , and . Why not just have ? This is because multiplication between elements of a module may not be defined. An example would be the module generated by by the element over , where .

Filtered modules form an additive category with morphisms such that . A trivial example is , defined using the grading above, and the map being defined as .

If is a submodule, then the induced filtration is defined as . Is every a submodule of ? Yes, because every is by definition a submodule of , and the intersection of two submodules ( and in particular) is always a submodule. Simialrly, the quotient filtration is also defined. As the quotient of two modules, the meaning of is clear. However, what about the filtration of ? Turns out the filtration of is defined the following way: . We need to have as the object under consideration because it is not necessary that .

An important example of filtration is the -adic filtration. Let be an ideal of , and let the filtration of be defined as . Similarly, for a module over , the -adic filtration of is defined by .

Now we shall discuss the topology defined by filtration. If is a filtered module over the filtered ring, then form a basis for neighbourhoods around . This obviously is a nested set of neighbourhoods, and surely enough the intersection of a finite number of neighbourhoods is also a neighbourhood, and so is the union of any set of neighbourhoods. Hence, the usual topological requirements for a basis is satisfied. But why ?

**Proposition**: Let be a submodule of a filtered module . Then the closure of of is defined as . How does this work? If one were to hand wave a bit, we are essentially finding the intersection of all neighbourhoods of . Remember that each is a neighbourhood of . We’re translating each such neighbourhood by , which is another way of saying we’re now considering all neighbourhoods of . And then we find the intersection of all such neighbourhoods to find the smallest closed set containing . There is an analogous concept in metric spaces- the intersection of all open sets containing , for instance, is the closed set . The analogy is not perfect, as the intersection of all neighbourhoods of is itself, which is not a closed set. But hey. We at least have something to go by.

**Corollary**: is Hausdorff if and only if .