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.