 This lecture is part of an online algebraic geometry course on the schemes and will be about the relation between Cartier divisors and the Picard group of invertible sheaves. So you recall that last lecture we defined a map from Cartier divisors to very divisors which was an isomorphism for reasonably well-behaved schemes. Similarly, we have a map between the Cartier divisor classes and the Bay divisor classes. And what we're going to do this lecture is construct a homomorphism from Cartier divisor classes to the Picard group of invertible sheaves or more or less equivalently line bundles. And this map here will in practice usually be an isomorphism. So let's just see how this construction works for Riemann surfaces. So suppose C is a Riemann surface, which we may as well take to be compact, and it has a divisor D consisting of some of points with multiplicities. And we're going to make this into an invertible sub-sheaf of the sheaf of meromorphic functions. Let's call this sheaf K of meromorphic functions. So how do we do this? Well, it just corresponds to the sheaf. This sheaf will be called L of D. And sections of L of D over an open set are just going to correspond to meromorphic functions such that F plus D is greater than or equal to zero. So I'll just remind you what this means. And if we've got a divisor sum of NiPi being greater than or equal to zero, this just means that all Ni are greater than or equal to zero. So these are the positive divisors. F is the divisor of zeros of F, whereas usual poles count as negative zeros. So what does this mean? It says that if D is sum of NiPi, then F is allowed a pole of order at most Ni at Pi, or zero of order at least minus Ni if Ni is less than zero. First of all, this is kind of obviously a line bundle because locally it's isomorphic with the sheaf of coordinate functions because multiplication by the meromorphic function F gives a local isomorphism. Now what we want to do is to copy this for schemes, for schemes X. So K is going to be the sheaf of total quotient rings, which you remember for X integral is more or less the constant sheaf of the function field of X, which most of the time is what we will do. So what we do is we want to take Cartier divisor and we want to turn this Cartier divisor into an invertible sub sheaf of K. And the Cartier divisor means we cover X by open subsets Ui, so you can think of Ui, Uj, Uk and so on. And on each of these we have a function Fi, so we've got Fi on Ui and Fj and Uj and Fk and Uk. So Fk is a section of the total quotient sheaf of Ui, so you can think of it as being something like a meromorphic function on Ui. And Fi over Fj has to be a unit of KuI intersection Uj. And we take the sub sheaf, so L of U is going to be the sub sheaf of K of functions G given by functions G of K and Ui, so that's G of Fi is regular on Ui. So if Fi is a zero of order three say this means roughly that G is allowed to have a pole of order three. And again Fi is an isomorphism from L of D to the sheaf of regular functions on Ui. So locally the sheaf is isomorphic to the sheaf of regular functions, so it's an invertible sheaf. So this gives a map from Cartier divisors to the Picard group. And there are some easy things to check which I'm not going to go through in detail. First of all it's a homomorphism. Second, it's injective from Cartier divisor classes to the Picard group. So took a Cartier divisor as image zero in the Picard group if and only if it's a principal Cartier divisor. And thirdly the image is the sub group of the Picard group of invertible sheaves that are isomorphic to a sub sheaf of the sheaf of total quotient rings. It's usually an isomorphism from CACL on to the Picard group. There are some counter examples where there are elements of the Picard group that aren't in the image of a Cartier divisor but they tend to be a bit weird. For instance, one example found by Klein which starts off with Hirunaka's example of a non-projective three-dimensional manifold we discussed earlier and fiddles around with it a bit and produces an example where this map isn't on to. So you're sort of unlikely to encounter examples where this map isn't an isomorphism unless you're searching very hard for counter examples. In particular, if x is integral, then the map from CACL to the Picard group is an isomorphism and this isn't difficult to see. Since x is integral, the sheaf k is just the constant sheaf of the field of functions of x, which I'll also denote by k. I mean, sometimes people denote the sheaf by a calligraphic k and because I'm not very good at drawing calligraphic letters so I won't bother, I just use k for both of them. And suppose L is invertible, so suppose it's an invertible sheaf, then this means we can cover x with open sets ui so that L on ui is isomorphic to O on ui. This is the definition of being invertible. And this means that k tensored with L is isomorphic to k on each ui. So k tensored with L is constant on each ui. This means it's a constant sheaf, which means on each subset of ui it has the same value except of course for the empty subset. So this implies it is constant on x as x is irreducible. So what's going on here is if we pick any open subset u of x and any smaller open subset v, then the values of this constant sheaf on u is the same as the value of the constant sheaf on v. So at least for sets of the open cover. So as it's a sheaf, it takes the same value on every open set of u. So L tensored with k is actually just isomorphic with k on the whole of x because it has the same value. You can identify its value on each open set. So this gives a map from L to L tensored k where you just map L to L tensored one. So L is a sub sheaf of k because this is isomorphic to the constant sheaf k. So in order to illustrate a bit of this, let's calculate the Picard group of projective space. So what we've seen is the same as the Cartier divisive classes of projective space and is the same as the v divisive classes of projective space and we'll see that it's actually isomorphic to z. So you remember we've actually found line bundles o of n for n in z. And all we want to do is to check that every invertible sheaf on the Picard group is isomorphic to one of these line bundles or invertible sheaves or whatever you want to call them. So first of all, let's take a look at finding the Picard group of any scheme x. Let's cover x by open sets ui. And suppose L is an invertible sheaf and suppose it's trivial on each ui. That means it's isomorphic to the sheaf of coordinate functions on each ui. Well, then we can define L by giving the isomorphism between the regular functions on here and the regular functions on here for each open set i. So L is defined by elements bij in o of ui intersection uj star. So we pick a unit of the regular functions on for each intersection. And this gives an isomorphism from o of ui restricted to ui intersection uj to o of uj restricted to ui intersection uj. And L is kind of formed by gluing all these things together using using these units bij. So bij, the set of numbers called bij is called a one co-chain. And there are some none zero. So there are some compatibility conditions on them because if we go from ui to uj and then from uj to uk and then from uk to ui, we must get back to where we started. So bij, bijk, bi must be equal to one in order for all these gluings to be compatible. And this condition says that bij is a one co-cycle, more or less the definition of a one co-cycle. So this line bundle might be trivial. So suppose we pick some element ai on each ui. So we pick an element ai, which is going to be a unit of the regular functions on ui. And then if we multiply the something in the coordinate functions of ai by ai, we don't actually change the invertible sheaf up to isomorphism. So if bij is equal to aij to minus one, then the corresponding line bundle is trivial. So if bi has this property, we say that bij is a one co-boundary. So think of this as being the co-boundary of these elements ai. So we find that invertible sheaves trivial on all the ui is just isomorphic to the one co-cycles over the one co-boundaries. And now to find all invertible sheaves is a sort of direct limit of this over all coverings. We won't actually need to take this direct limit because we'll actually start with a covering on which all invertible sheaves are trivial. But what we've actually defined is the first check co-homology group. So the first check co-homology group arises very naturally if you try and classify line bundles over a scheme or for that matter over any ringed space. Line bundles or invertible sheaves are classified exactly by the elements of this direct limit, which is more or less by definition the check co-homology group. So now let's get back to the Picard group of Pn. Well, we know that Pn is covered by n plus one copies of the affine space An. And we notice that the Picard group of An is just zero. And the reason for this is that we know the Picard group of An is isomorphic to the Baye divisor class group of An by what we've said before because An is locally factorial and notarian and integral and all the rest of it. And we worked out this, this is isomorphic to Z because the coordinate ring is Kx1 up to xn and this is a unique factorization domain. So we show that for any unique factorization domain, the vial, the group of Baye divisor classes was just trivial. Sorry, that's not a Z, that's zero. And similarly, if you've got any unique factorization domain you can see the Picard group of that unique factorization domain vanishes. So now we can cover Pn by these An's and if L is a line bundle then L on each An is trivial because all line bundles on affine space are trivial. So we don't need to go around taking a direct limit or anything, we can just calculate the one co-cycles modulo, the one co-boundaries for this particular covering by affine spaces. So let's take a look at what's going on. So we've got spaces ui, uj and uk and we need to remember what the ring of coordinate functions on ui is. Well, all of ui can be identified with Kx0 up to xn and then we invert xi and then we take the degree zero pieces of it. But this is actually just another polynomial ring but we're writing it in this funny form because this makes it easier to go between these different coverings. And we want the elements bij to be in the units of ui intersection uj and this is just given by the units of Kx0 up to xn. Now we have to invert xi and xj and take the degree zero pieces. The units of this are all of the form c times xi over xj to the nij, cij, where cij is a non-zero element of our field and nij is just an integer. It's because these are the only degree zero units of this ring. Bij is given by these numbers here. Now we apply the one co-cycle condition which says that bij, bjk, bk, i equals one and we find xi over xj, bnij times xj over xk to the njk times xk over xi. The nk i times, there were some c, cij, cjk, ck i is equal to one. And there's an immediate consequence of this, you can see that nij equals njk equals nk i. So all the nijs are equal to say some number n. I guess I'm using that for the dimension of projected space as well, which I shouldn't have done. Now we've still got these constants cij to get rid of, but those are easy to get rid of. So here we've got these numbers u1, i, uj, and here we've got a number cij, cjk and so on. And what we do is we're going to eliminate the cijs by showing that they're actually co-boundary. So what we do is we set a0 equals one and we set aj equals cj0, where cj0 equals c0j minus one of course. And then the fact that cij, cjk, ck i equals one just implies that cij is equal to aij to minus one. So cij is a one co-boundary and you can just remove it because multiplying a one co-cycle by one co-boundary doesn't affect its line bundle. So the line bundle L is given by the one co-cycle bij equals xi over xj to the n for some integer n in z. So the Picard group of p, I shouldn't really call it pn because I've overused n, pr, is just isomorphic to z. So this one co-cycle obviously corresponds either to the sheaf o, n, or possibly to the sheaf o minus n. And I can never remember which way around they go and it's easy to make a sign error with this depending on which way around you put these xi's and j's so I'm not going to bother. Okay, next lecture, we will look at the Picard group and the various divisor class groups for curves and dedikind domains.