 This lecture is part of an online algebraic geometry course on schemes and we'll be about the relation between Cartier devisors and Vey devisors So you remember last lecture we showed there was a homomorphism from Cartier devisors to Vey devisors And we want to ask, you know, when is this an isomorphism or when is it injective or when is it surjective? Let's start with an example when this is not injective So if there are Cartier devisors can't always be distinguished by their zeros and for this We're going to take our ring to be the spectrum of k t squared t cubed So this ring is all power series of the form a naught plus a 2 t squared plus a 3 t cubed and so on Where you notice there's there's no term a 1 t to the 1 And This is a local ring and its spectrum has two points and it has one closed point Which which corresponds to the Ideal generated by t squared t cubed and it has an Open points that I'm going to draw as a sort of cusp because this is actually a Sort of germ of a cusp like if we put y equals t squared and Let's put y equals and we want y squared x cubed so y equals t cubed and x equals t squared you find that y squared is t cubed Which is the cusp here And we're sort of looking at this tiny piece of it by by working with formal power series Anyway, this is a local ring so the Cartier devisors are easy to work out so the Cartier devisors Will just be the units of the function field k which is the ring of Laurent power series Should be t to the minus one And we quotient out by the units of R Okay, so why aren't we taking an open cover of this? The reason is if we take any open cover of this one of the sets of the open cover must be the whole space So we may as well Just take the single open set consisting of the whole space and then Cartier devisors look like that Well the units of K Can be written as well that all none zero elements of k so you can write them in the form t to the n times Sorry, let's put a t to the n times 1 plus a 1 t plus a 2 t squared and so on units of R on the other hand Just look like things of the form a times 1 plus a 2 t squared and so on where here a is units of K So you see there's no a1 term there And So we see the quotient of the units of k by the units of R star is just isomorphic to Z times the field k where z comes from this n here and K comes from this coefficient there And on the other hand the very devisors are just isomorphic to Z because there's only one co-dimension one closed set so the map from the Cartier devisors to vey devisors has a kernel consisting of Corresponding to K and you can easily check that it's actually isomorphic to the additive group of K. I mean what else could it be? We notice that the element 1 plus a t has no zeros Or poles in some sense But is not a unit of R So this is different from the case we had when we The example we had last week when we were looking at the last lecture when we look at the affine line With its function field K X you notice that any localization of this If an element in the localization has no zeros or poles and it must actually be a unit and this this difference is why the Picard group to the vague group was injective in this case, but not in this case So let's have a look at a general condition when the map is Injective so suppose Let's take X to be notarian and integral Notarian is to avoid lots of tires and technical complications and integral is basically put in because I'm feeling kind of lazy So this means that X as a function field K which makes devices easy to think about then if X is normal then Cartier devisors are a subset of vey devisors. Well They're not a subset of a devisors what I mean by this is the natural map from Cartier devisors to vey devisors is Injective, but I'm going to be sloppy and say that means Cartier devisors are a subset of vey devisors So what does normal mean well normal means all local rings are Integrally closed in in in their field of quotients and Since we are assuming the scheme was integral all the local rings are domain So we don't have to have a headache one wondering about what this means Now what we do is We recall the basic theorem from commutative algebra Which says that if a ring R is normal then R is the intersection of all co-dimension Intersection over all co-dimension one primes of RP. So RP is the local ring at P And so what we do is we look at some any open affine of the form spectrum of R in X and Suppose a Cartier deviser is represented locally on X by G over H on on this open set X If its image is Nought in the vey devisors this means that G over H is contained in All local rings are over P and similarly H over G is also contained in all local rings are over P So using this theorem about normal rings. We see that G over H is a unit of R so is trivial as A Cartier deviser So what we're saying is that if a Cartier deviser has image zero in the vey devisors and The scheme is normal then this implies the Cartier deviser must be trivial everywhere. So The the kernel of the map of the map from Cartier devisors to vey devisors is Just zero And for example, let's look at the example we had we had this Ring k t squared t cubed and we noticed that this is not normal because t is in the in its integral closure T is obviously a root of X squared minus t squared equals naught which is an integral equation and Furthermore, we see that There's only one prime and RP is just k The ring of all formal power series in T. So R is not equal to the intersection over all P of RP so that shows where The example we had where where the map from Cartier to vey groups is not injective Fails Because it's not normal We can for this example The We can also ask about whether the Cartier deviser classes is contained in the group of vey deviser classes So in the previous example Although the Cartier devisors are not contained in the vey devisors every Cartier deviser is in fact At least the ones of zero degree are in fact principal So that doesn't quite give a counter example What we can do is we can modify the example slightly and find an example Where the Cartier deviser classes are not contained in the vey deviser classes and for this We just take the spectrum of k t squared t cubed This differs from the previous example because we're now just looking at polynomials not power series So this is just the coordinate ring of the cusp y squared equals x cubed with y Was t cubed x equals t squared Now the vey devisors are the same as for k t and you can easily check that the vey deviser classes are just zero More or less the same as for that there you notice there's a natural map from this Ring to this ring. So there's a natural map from the affine line to this ring with a cusp and Now that's sort of more or less identifies vey devisors of both But There's a Cartier deviser Um But sorry the following cart of Cartier deviser is not principal So I'm going to take the following Cartier deviser. I'm going to take the Cartier deviser to be given by the function 1 plus a t on the open set Where t is not equal to a to minus 1 here a is in k and I'm going to take it to be 1 on the open set Where t is non zero So here we've defined a Cartier deviser by covering the spectrum with two open sets and using different Meromorphic functions on these two open sets But it's not in the image of the function field k which is Rational functions of t Um, um, so This Cartier deviser is Not even principal so so in fact the group of Cartier devisors contains a Subgroup Isomorphic to the additive group of k in fact It is Isomorphic to the whole of k plus which I'll just leave as a fairly easy exercise Um, the next question we can ask is when is the map from the Cartier devisors to the very devisors and Isomorphism So we found conditions under which it need not be injective And we can also have an example when it's not surjective So here we're going to take Well, in fact, I'm going to give an example where the Cartier deviser classes to the very deviser classes is not on to and for this we're going to take the Um cone xy equals z squared so it's a sort of Draw it as a cone like this and We're going to take the deviser Look like this. So here's the deviser and this is the deviser y equals zero and This is a very deviser and we're going to show that it's not the image of a Cartier deviser Even near the point zero zero zero And so we just have to show I mean it's enough to show that it's not locally a Cartier deviser Near the point zero because and then we don't need to worry about what happens elsewhere And what we do is we look at the local ring of the variety x so x is big x is going to be the This double cone and we're going to look at the local ring at nought nought nought and it's got a maximal ideal Generated by x y and z rather obviously So let's call this m and we notice that m over m squared has Dimension three and this gen span by x y and z you notice this dimension is bigger than the dimension of the cone so it's a singular point And If we take the Sorry, so this isn't the idea why it was mortis. It's Generated by y equals Z equals zero So the ideal P Generated by y and z Has image y z In m over m squared which is Is two-dimensional two-dimensional vector space so it cannot be generated by one element So this ideal P is not principle Even in the local ring at zero. So so this So the very deviser Is not the image of Cartier deviser On the other hand we notice that if we take p squared, this is the ideal y squared z squared y z Which is equal y squared x y y z Which is just the ideal y so so P squared is a Cartier deviser Or at least it's the image of a Cartier deviser So geometrically what's going on is this here the Deviser y equals zero is A sort of double line Which is the Intersection of a tangent plane with the cone and if you sort of think about it You'll see the tangent plane meets the cone on a sort of double line And if we look at x equals zero It's again a sort of double line like this And if we look at the divisor of z equals zero It sort of looks like this and now you can see that the divisor of z squared is Equal to the divisor of x plus the divisor of y Which is not very surprising if you remember that z squared is equal to x y You can also Observe that it's sort of visibly not a unique factorization domain because this gives Is two factorizations of the element z squared you can either write the z squared or as x y and there's no way to Find a Write z as a product of Of other things so so so unique factorization doesn't hold in the local ring at this point Well, the reason I'm emphasizing unique factorization is we have the following rather simple theorem so suppose x is notarian and Again, I'm going to assume it integral just for safety then if all local rings of x are unique factorization domains then the map from Cartier divisors to V divisors is An isomorphism Well unique factorization domains imply that it's normal So we already know the map from Cartier devices to V devices is injective So we just need to show The map is onto and so We may as well pick some One dimension so Irreducible subset D of co-dimension one and It's enough to show that Cartier devices map onto this because the V divisors are generated by these things and what we do is we Pick any point X So here's our divisor D, which may or may not contain X And The local ring at X is a unique factorization domain and this implies that D is locally of the form FX near X In other words, it's it's at least locally the zeros of some function So we can find an open set U X containing X such that the V divisor is given by F of X And now what we do is we cover X by a finite number of Such open sets So we might have you know U X you why And you Z where is the divisor D? Here's X and here's Y here Z There's no reason why Z should actually lie on the divisor and on each of these open sets. We've got a function f X f Y and f C so that F X equals D on you X F Y Equals D on you Y and so on so the The the sets F F star you star define a Cartier divisor because The quotient of FX and FY on the intersection of you X and you Y must be a unit because And the the the only common zero is on DX and and the zeros cancel out and The All the local rings are normal because they're unique factorization domains So so if something has no zeros or poles, it must be a unit So we've constructed a Cartier divisor mapping on to this divisor D So to summarize the map from V divisors So the map from Cartier divisors to V divisors is Injective if X is normal and an isomorphism if X is also locally You unique factorization domain or its local rings a unique factorization here. We're assuming that X is notarian and integral So it's also worth remarking that if X is regular Meaning that all its local rings are regular local rings in other words X has no singular points then this implies X is locally Unique factorization domain So Cartier divisors at the same as V divisors So that's why all the examples we had have singular points if if a scheme doesn't have singular points Then Cartier divisors are essentially the same as V divisors Okay, next lecture. We're going to study the relation between Cartier divisors and the Pickard group of invertible sheaves