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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 $\cap M_n=0$.

### Invertible Sheaves and Picard Groups

This is a blog post on invertible sheaves, which form elements (over a fixed algebraic variety) of the Picard Group. The group operation here is the tensor product. We will closely follow the developments in Victor I. Piercey’s paper.

We will develop invertible sheaves on algebraic varieties. However, instead of studying sheaves over varieties, we will be studying the algebraic analogues of these geometric entities- we’ll be studying modules over coordinate rings.

First we discuss what it means for a module to be invertible over a ring. Over a ring $A$, a module $I$ is invertible if it is finitely generated and if for any prime ideal $p\subset A$, we have $I_p\simeq A_p$ as $A_p$-modules. Here $A_p$ is the localization of the ring $A$ with respect to the prime ideal $p$, and $I_p$ is just the ideal over the localized ring $A_p$. What does the expression $I_p\simeq A_p$ mean? One way that this condition is easily seen to be satisfied is that $I$ is generated by a single element over $A$. I can’t think of any other ways right now. It is perhaps fitting that the article says next that this condition implies that $I_p$ is locally free of rank $1$.

The reason that the notation $I$ is chosen for an invertible module is that we shall soon see that every invertible module is isomorphic to an invertible ideal. How does one see that? An ideal of a ring is definitely a module over that ring. Assuming that the ideal is a principal ideal and the module under consideration is also generated by a single element, all we need to do is to map the generator of the module to the generator of the ideal. The reason we can assume that the ideal is principal and that the module is generated by a single element is that we want both the modules to be locally of rank $1$, and this is the easiest way of doing so.

We will now discuss an ideal of a module that is locally free, but not principal. Let $A=\Bbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$. It is easy to see that this ideal is not principal. Also, $A/I=F_2$. Hence, $I$ is maximal in $A$. Now if $I\not\subset p$, where $p$ is the prime ideal under consideration, then $I\cap A\setminus p\neq\emptyset$. Hence, $I_p=A_p$. This is because there is an element of $I$ which has been inverted, which causes the ideal to be equal to the ring. We therefore assume that $I\subset p$. As $I$ is maximal, we conclude that $I=p$. We observe that $3$ is not in $I$, and hence invertible in $A_p$ (which can now be written as $A_I$). Now $2$, which is one of the generators of $I$, is written as an element of $I_p$ (it is written as $\{(1+\sqrt{-5})(1-\sqrt{-5})\}/3$). This shows that the ideal can be generated by a single element in $A_p$, which makes it isomorphic to $A_p$.

The isomorphism classes of invertible modules over the ring $A$ form the Picard group. The identity element is the isomorphism class of $A$ over itself. Given an invertible module $I$, its inverse is the module $I^*=\text{Hom}(I,A)$. Why is this the inverse element? This is because there is a natural map $I^*\otimes I\to A$, which is defined as $\psi\otimes a=\psi(a)$. As the isomorphism class of $A$ is the identity element, this is a map of the product of two elements to the identity, which makes one the inverse of the other. What about $I\otimes I^*$? Shouldn’t we have a two-sided inverse? Remember that in general, for any two modules $M$ and $N, M\otimes N\simeq N\otimes M$. Hence, we can define $a\otimes \psi$ to be the same as $\psi\otimes a$, and get away with it.

Theorem 1: If $I$ is an $A$-module, then $I$ is invertible if and only if the natural map $\mu:I^*\otimes I\to A$ is an isomorphism.

The proof and subsequent theorems in the paper will be discussed in a later blog post.