 So this lecture is part of an online algebraic geometry course on schemes and will be examples of Picard group of a scheme X So you remember that for reasonable schemes the Picard group is the same as the group of Cartier divisors Which is the same as the group of vay divisors. So we want X to be I guess notarian integral and Locally factorial Which will automatically be true because we're just going to take X to be non singular So what I'm going to do is give a sort of survey without proofs of some examples of the Picard group Most of these examples that I'm about to give take a fair amount of effort to actually prove Picard group of something if you're given something you can usually work out its Picard group But it usually takes a fair amount of effort So some of the earliest examples were due to Gauss in his book disquisitione is arithmetic I so what Gauss did was he worked out things like the Picard group of Picard group of I guess the spectrum of our but our is the integers of an imaginary quadratic field Actually Gauss didn't state his results in terms of imaginary quadratic fields or real quadratic fields He stated them in terms of binary quadratic forms, but there's a sort of easy dictionary that allows you to go from one to the other So for example, if we take R to be Z root minus five in the Picard group We've sort of more or less seen was is of order two and It's quite difficult Understanding how the Picard group behaves for example Gauss found that if R is Z one plus root minus one six three over two then the Picard group is just trivial and Gauss asked if the if this was the largest Imaginary quadratic field which had Picard number zero or I guess Gauss didn't actually talk about Picard number because Picard only came a century later Gauss talked about the class number So this is the famous class number one problem Are there imaginary quadratic fields beyond this with class number one and this is finally solved by Stark and Haigner and Baker There's a very complicated history about this that Haigner sort of almost solved it But there was a minor gap in his proof So his his paper was ignored for several years until Stark pointed out it was easily repairable. So that's You know understanding the Picard group of imaginary quadratic fields is quite tricky There's a if you talk about real quadratic fields again, you can work out the Picard group of these and Again, there are some basic open problems. So we can ask for real quadratic fields. Are there infinitely many with Picard group of order one and As far as I know, this is an unsolved question. In fact, it's an unsolved question about whether there are an infinite number of Imaginary quadratic fields of any degree with class number one There's theoretical and numerical evidence that sort of suggests the answer is yes, but no one's been able to prove it yet So The work of CUMMA on Fermat's last theorem and reciprocity laws also is essentially about Picard groups so CUMMA worked out the Picard groups of Of the spectrum of R where R is a cyclotomic field. So Z of zeta Where zeta is a root of X to the p minus one over X minus one So these are the famous cyclotomic fields and What CUMMA one reason CUMMA was interested in them is that he showed that if P does not divide the order of the Picard group of the cyclotomic field with some prime Then Fermat's last theorem is true for the exponent P And Yeah, until Wiles came along. This was the basis of the The best results we had about Fermat's last theorem For example, he showed that the Picard group of R Is equal to one if P is less than 23 but the Picard group of R is Ciclic order three For P equals 23 So this is the first case when it's non-trivial and there are cases When P does divide the order of the Picard group So he found the first case when this happens is the Picard group of R was Z over 37 Z for P equals 37 And the Picard group can get quite spectacularly large for example if P is 257 Then the order of the Picard group of cyclotomic fields Is given by the following number you get five four five two four eight five oh two three Four one nine and I have to continue two three oh eight seven three two two three eight two two six two five five five And I haven't finished yet nine six four four six one four seven six four two two eight five four six six two one six eight Three two one okay, so it's got about well over 50 digits. So so There's a theorem of Minkowski that says the Picard group of The ring of integers of an algebraic number field is always finite but as you see from this finite doesn't mean small Next we can look at Picard groups of Curves C so C is going to be a projective non-singular curve Over the complex numbers so non-singular Projective curves over the complex numbers are more or less the same as compact Riemann surfaces So analysts work with Riemann surfaces and geometers work with non-singular projective curves and the Picard group Of this we've worked out in two cases. First of all the Picard group of the projective line We've seen as just z and we saw that the Picard group of an elliptic curve Fits into an extension nought goes to the elliptic curve goes to the Picard group goes to z Goes to zero where this map is the degree And this is the connected component of the of the Picard group So in general the Picard group is described by the Arbel Jacobi theorem And it depends on the genus Of c and you remember the genus of c is the number of handles So c is a sphere the genus is zero And if c looks like a torus the genus is one and if c There's two handles like this. It is genus two and so on So the projective line and elliptic curves are the case of genus zero and one So for higher genus And it looks like this first of all the space of holomorphic one forms As dimension g As a complex vector space a holomorphic one form is something that looks in local coordinates like f of z D z where f is a is a holomorphic function And what we can do is we can pick some point c nought in c And then we try and define a map from the curve c to the dual Of the space of one forms So one forms might be denoted by omega and what we can try doing is we can map a point c to the integral from c nought to c of omega So this is linear in omega and here we're just integrating from c nought to c. Well There's a little bit of a problem because there may be more than one path from c nought to c. Well, if the paths are homotopic then that's okay because by Cauchy's theorem We can just move them around But there may be lots of paths from c nought to c that are not homotopic to each other So this is not actually well defined map Instead what we get Is a map from the curve to the dual Of the space of holomorphic one forms Except we've got to quotient it out by The first homology group H1 of c with coefficients in z So the first homology group kind of describes the different homotopy Group of paths So these you can think of as being essentially described paths From c nought to c up to homotopy or some homology or something like that Um So, um This thing here This group here on the top Is a complex vector space of complex dimension g On the other hand this um first homology group Isomorphic to z to the two g And the quotient is something called a complex torus So as a topological space Is isomorphic to a two g dimensional vector space quotient it out by two g um Free z module so it's isomorphic to a product of two g copies of a circle um so um So we we get a well-defined map from c To this space here, um, which depends on The the choice Of c nought and we can extend it to divisors So we get a map from um divisors To this space um Where we take one forms Dual and quotient it out by h one Um, and we can restrict it to degree zero divisors And it turns out to be well-defined modulo principle divisors And this map here is an isomorphism This is the arbel jacobi theorem. So arbel proved It's injective And jacobi Proved that it's onto So, um, this gives us Um a description of the picard group. So this group here is the famous jacobian of the curve And the arbel jacobi theorem just says we've got a map taking nought to the to j which is the jacobian To the picard group to z To zero where this map here is just the degree um, so in the special case of The projective line the jacobian is just zero because there aren't any Holographic differentials on the projective line apart from zero and in the special case of elliptic curves. This is just the map from the Elliptic curve to the Picard variety and in the case of an elliptic curve it's isomorphic to its own jacobian Um for curves of genus greater than one the jacobian has dimension greater than one So what we do is we end up with the curve as a sub variety of the jacobian um so, um, this works for um Curves over the complex numbers Um, you see where we're doing complex contour integration in order to define this map um, they extended um to curves over fields Of characteristic p greater than zero So he managed to find an analog of the jacobian variety only he had to do it algebraically and his his original constructions rather complicated In fact, it was so complicated. It wasn't even obvious that Waze jacobian was a projective variety. Although this was later proved by chow um, so These jacobians turn out to be projective varieties. This isn't quite obvious from the construction here, but it's not too difficult to show And projective varieties that are also abelian groups are called abelian varieties and are Notoriously difficult to handle um So next we're going to look at some examples of picard groups of surfaces um, so we've already done the projective plane That's just got a picard group, which is the integers. So let's look at a degree two Quadric in p3 so it might be defined by w x equals y z and if we set um um So so in in in real space, this ends up looking like a sort of hyperboloid of revolution So we get surfaces that kind of look a bit like this um Now these surfaces are easy to deal with because they're isomorphic to product of two copies of p1 And you can see this because if you take coordinates ab and the first p1 And cd and second p1 you can map this to a c ad bc bd in p3 and if this is W x y z you can check that w z equals x y. So I suppose it's slightly different from that but never mind um, and the picard group of this Turns out to be just z times z and we get um, you can sort of see this Geometrically because if you think of this as being a product of the first p1 and the second p1 and the first um p1 gives you a device which gives you the first c and the second p1 gives you the Another device in which which gives you the second z Um incidentally, it's not always true that the picard um group of a product is the product of the picard groups Of the variety so we can ask suppose we've got a picard group of x And a picard group of y We can map this to the picard group of x times y And ask is this an isomorphism? And the answer is sometimes Well, it need not be injective And it's not injective for really stupid reasons. We could just take y to be empty um, it need not be surjective And again, we could just take say x to be p1 with picard group z and y to be two points Then you can easily check that The Picard group of y is just trivial So this side is just c and this side is z plus c Well, you might sort of fuss about this I mean the first example is definitely cheating by choosing an empty variety in the second example Well, this isn't connected. So you might think maybe if it's connected then Maybe something would happen But there's another example with where it's not surjective You could just take x equals y to be say a curve of genus greater than zero And then what happens is if you look at x times y and draw it as x times y You can get several sorts of devices. First of all, you can get these vertical devices and you can get these obvious horizontal devices So that gives you a z for the vertical devices and a At least a z for the horizontal ones, but you can also get Diagonal devices So you can just have the point x equals y in x times y and that gives you more devices Which don't come from products of the Picard group of x and y Well, they do if x and y have genus zero, but they they don't in general Next we can look at um a degree three surface in p3 And in this case the Picard group Turns out to be z to the seven And I'll just explain very roughly Where this seven comes from Well, the degree three surface is the famous cubic surface Which is isomorphic to a projective plane blown up at six points and the projective plane has Picard group z and blowing up at a point means you replace the point by a copy Of p1 roughly speaking and this p1 gives you a new divisor And if you analyze what's going on it gives you a new factor of z In the Picard group. So every time you blow up a point of a surface You increase the Picard group by a factor of z So so a cubic surface is obtained by blowing up p2 and six points So so the Picard group is a free group on seven generators And it's kind of obvious you can carry on blowing up points as much as you like and obtain Surfaces whose Picard group is a free abelian group of rank as large as you like um Next we can find a surface with the Picard group sick click Of some finite order And this is easy We just take um The projected plane p2. Let's take a non-singular Curve of degree D let's call this curve c and take our surface s to be p2 with c removed Then um, we've got a map from the Picard group of p2 to the Picard group of s so If you take any divisor in p2 Well, it may have a component in s and if it's got a component in s you can just sort of shift it a bit to make it not so So it may have a component in c and if it's in c you can just shift it a bit to be not in c And check this is well defined So we get a map from Picard of p2 to Picard group of s and this is obviously onto because if you've got a divisor in s If you've got an irreducible divisor in s you can just Take its completion and get a divisor in p2. So this group is onto um the kernel is just generated By the curve c it's pretty obvious if you take the curve c in here then it's becomes trivial in in in s so um The image of the curve c Is just d times the divisor here because that's the degree Of of c. So we get a map z goes to z Goes to picard group of s Goes to zero. So the picard group of s Is just z over dz um I should say I've been kind of cheating with these examples of surfaces because Every single example of a surface I've calculated the picard group of is actually a rational surface That means it's birational to the projective plane and surfaces that are birational for projective plane It's particularly easy to calculate their picard groups You may have noticed that the picard group was always a discrete group and didn't have some sort of continuous component Similar to the Jacobian of a curve