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 \{A_n\}_{n\in\Bbb{Z}} such that A_0=A, A_{n+1}\subset A_n, and A_pA_q=A_{p+q}. An example would be A_n=(2^n), where (2^n) is the ideal generated by 2^n in \Bbb{Z}.

Similarly, a filtered module M over a filtered ring A is defined as a module with a set of submodules \{M_n\}_{n\in\Bbb{N}} such that M_0=M, M_{n+1}\subset M_n, and A_pM_q\subset M_{p+q}. Why not just have M_pM_q\subset M_{p+q}? This is because multiplication between elements of a module may not be defined. An example would be the module generated by by the element v over \Bbb{Z}, where M_n=2^n M.

Filtered modules form an additive category F_A with morphisms u:M\to N such that u(M_n)\subset N_n. A trivial example is \Bbb{Z}\to\Bbb{Z}, defined using the grading above, and the map being defined as x\to -x.

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

An important example of filtration is the m-adic filtration. Let m be an ideal of A, and let the filtration of A be defined as A_n=m^n. Similarly, for a module M over A, the m-adic filtration of M is defined by M_n=m^nM.

Now we shall discuss the topology defined by filtration. If M is a filtered module over the filtered ring, then M_n form a basis for neighbourhoods around 0. 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 0?

Proposition: Let N be a submodule of a filtered module M. Then the closure of \overline{N} of N is defined as \bigcap(N+M_n). How does this work? If one were to hand wave a bit, we are essentially finding the intersection of all neighbourhoods of N. Remember that each M_n is a neighbourhood of 0. We’re translating each such neighbourhood by N, which is another way of saying we’re now considering all neighbourhoods of N. And then we find the intersection of all such neighbourhoods to find the smallest closed set containing N. There is an analogous concept in metric spaces- the intersection of all open sets containing [0,1], for instance, is the closed set [0,1]. The analogy is not perfect, as the intersection of all neighbourhoods of (0,1) is (0,1) itself, which is not a closed set. But hey. We at least have something to go by.

Corollary: M is Hausdorff if and only if \cap M_n=0.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush

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