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Divisors, Invertible Sheaves, and Their Relationship (With Examples)

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Divisors, Invertible Sheaves, and Their Relationship (With Examples) DIVISORS, INVERTIBLE SHEAVES, AND THEIR RELATIONSHIP (WITH EXAMPLES) EOIN MACKALL Notation and Conventions. For this talk a variety is going to be an integral scheme, separated of finite type over a field k. We make two assumptions throughout this talk to make some parts simpler: k is going to be a fixed algebraically closed base field and by variety I mean only the closed points. 0. introduction Let's start with some recollections on terminology. Definition 0.1. A vector bundle V over a variety X is a variety together with a map π : V ! X such that over a covering of X by open subvarieties fUigi there is a commuting diagram n ∼ −1 A × Ui π (Ui) πjUi Ui with the top (horizontal) arrow an isomorphism. If n = 1, then V is called a line bundle. Definition 0.2. A finite rank locally free sheaf L on X is a sheaf so that over an open cover fUigi of X there are isomorphisms Lj =∼ O⊕n where O is the structure sheaf of U . If n = 1, then L Ui Ui Ui i is called an invertible sheaf. The next remark is purely for clarification { we won't use it at all. We're going to, in this talk, stay strictly with invertible sheaves. Remark 0.3. There is an equivalence between the category of finite rank locally free sheaves and vector bundles on a variety X. More precisely, if V is a vector bundle on X then there is a functor LX which assigns to V a finite rank locally free sheaf on X with sections over an open subvariety U of X: −1 LX (V )(U) := fσ : U ! π (U): π ◦ σ = idjU g: There is also a functor SpecX (Sym(−)) that assigns to a finite rank locally free sheaf L a vector bundle over X. On an open U of X, the variety SpecX (Sym(L)) is defined to be Spec(Sym(L(U))), the spectrum of the symmetric algebra of the OU -module L(U). These varieties glue together over the open sets U to give a variety over X. The composition L 7! SpecX (Sym(L)) 7! LX (SpecX (Sym(L))) is canonically isomorphic with the dual, taking L to Hom(L; OX ). Sometimes, to avoid this duality, one takes the dual of the bundle they're working with before taking the associated vector bundle. Date: October 21, 2018. 1 Now fix a variety X, which we're going to assume is projective (meaning it has a morphism into a projective space of some dimension that realizes X as a closed subvariety). A question one might ask is the following: in how many ways can I realize X as a subvariety of projective space? Another n question one might ask: what is the minimal dimension n so that I can embed X into P ? These questions can be answered, if one has a specific variety X in mind, using invertible sheaves. First, let's give an example. 1 n Example 0.4. Let X = P . Then X admits morphisms into P for any n ≥ 1 by taking [x : y] to [x : y : 0 : ··· : 0] for example. But there are other morphisms one can choose as well. If n = 2 then X can be realized as a closed subvariety using the map [x : y] 7! [x2 : xy : y2]: 2 2 In this case X is isomorphic with the subvariety of P given by the vanishing of x0x2 = x1. There's 2 also the map [x : y] 7! [x : y : 0] that realizes X as a linear subvariety of P . If n = 3 then X can be realized as a closed subvariety using the map [x : y] 7! [x3 : x2y : xy2 : y3]: 3 In this case X is isomorphic with the subvariety of P given by the vanishing of the polynomials 2 2 x0x3 = x2x1, x1 = x0x2, and x3x1 = x2. This subvariety is called a twisted cubic. 1 There are clearly a lot of ways to embed even P into various projective spaces. However, for 2 3 embeddings into P and P , these are essentially the only cases (allowing for automorphisms of the target space which could change the coordinates for example). The main purpose of this talk is to elucidate the above concept. In the first section of these notes, we make precise the relationship between invertible sheaves and morphisms into projective 2 space. In the second section, we begin to answer the questions posed above Example 0.4. More precisely, we introduce the Divisor class group, the Picard group, and their relationship (along with the relationships between these objects and other well-known objects like certain cohomology groups). A number of examples are included in each section although they are limited by the knowledge of the author (i.e. they might not be the simplest but they're all I got!). Remark 0.5. A notable absence from these notes is the concept of ample invertible sheaves. These objects are incredibly useful { especially in examples and in areas like the theory of curves, for instance. However, this talk itself aims to cover a much more foundational area of algebraic geometry which is still interesting in its own right. I would suggest the interested reader take a look at this material, e.g. in [Har77, Chapter 2, Sections 5-6; Chapters 3-5] or [Laz04, Chapter 1]. 1. maps to projective space Roughly, this is how we should continue: given an invertible sheaf L on a variety X and n + 1 n global sections of s0; :::; sn L, one can try to define a morphism to P by sending a point x to the point [s0(x): ··· : sn(x)], where we use s0(x); :::; sn(x) to denote the images of these sections in the residue field ∼ Lx = OX;x k(x): There are some immediate problems one might have with this method at first glance though. E.g. ∼ \Why is this well-defined? We had to use a choice of an isomorphism Lx = OX;x that relies on an open set about x that trivializes L!" and \Even if this was well-defined, is it ever the case that these sections are all 0 and so do not define a point of projective space?" Both of which we'll try to answer. First, let's set up the morphisms we want to talk about in precise terms. We still fix a variety X over k, an invertible sheaf L, and we assume s0; :::; sn are global sections of L, possibly with repetitions and possibly with some being the 0 section. But, to solve the second question above, we add some assumptions. We'll need to assume that, for any point x, the values s0(x); :::; sn(x) are not all 0. So in the definition x 7! [s0(x): ··· : sn(x)], the target is always a point of projective space (even if the map is not well-defined for other reasons). Notice also that it's not necessarily true these assumptions can be met for our particular L but, we'll assume they hold anyway (these conditions are very frequently satisfied). Of course the map we defined above turns out to be well-defined. By definition, L comes with a ∼ collection of open sets fUigi covering X and isomorphisms 'i : LjUi = OUi . In the morphism we're ∼ defining, we first take the isomorphism 'i and restrict it to an isomorphism on stalks Lx = OX;x. It does not have to be the case, if x were in two open sets Ui;Uj, that s0(x); :::; sn(x) have the same images in k(x) under the two different isomorphisms 'i;'j. However, if we take two such isomorphisms, over open sets Ui;Uj, and consider the morphism −1 ' jU \U 'Ui jUi\Uj Uj i j OUi\Uj −−−−−−!LjUi\Uj −−−−−−!OUi\Uj of OUi\Uj -modules then this composition is completely determined by where the section correspond- ing to 1 in OUi\Uj (Ui \ Uj) goes. Being an isomorphism, the 1-section must be sent to an invertible element t of OUi\Uj (Ui \ Uj). The difference in the two maps over our two different choices of open set Ui;Uj is then given by multiplication by t x 7! [s0(x): ··· : sn(x)] and x 7! [ts0(x): ··· : tsn(x)]: Due to the definition of projective space, these are actually the same point showing this map is well-defined. Altogether we've sketched out a proof of the following statement. 3 Lemma 1.1. Let X be a variety over k and L an invertible sheaf on X. Assume L has global sections s0; :::; sn such that for every point x in X the values s0(x); :::; sn(x) are not simultaneously n 0, then there is a well-defined morphism X ! P . Proof. All that's left is show that there is a morphism # f : OPn ! f∗OX n where f : X ! P is the map defined (on topological spaces) in the previous paragraphs. Let fUigi be a collection of open subsets of X that trivialize L and cover X, i.e. we assume there ∼ are isomorphisms 'i : LjUi = OUi on these opens. Fix one such Ui. Since not all of the sj are simultaneously vanishing, we can assume (by possibly making Ui smaller) that sj is not 0 in k(x) n for any point x in Ui. Now the image of Ui under f lands in the open affine of P where xj is # invertible (by the definition of f).
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