 So this lecture is part of an online algebraic geometry course on schemes and will be about morphisms of a scheme to projective space. So we'll start by just recalling morphisms of a scheme X to a space Y, which is affine. So I suppose Y is the spectrum of some ring B. Well, first of all, if X is equal to the spectrum of the ring A, then we saw that morphisms from X to A are just the same as ring homomorphisms from B to A. And remember the ring homomorphisms going the opposite direction to the scheme homomorphisms. I hope I've got it the right way around. So if X isn't affine, suppose X is covered by open affines and XI. So X1, X2 and X3. So what is a homomorphism? Sorry, what is a morphism from X to Y look like? Well, we have to give compatible morphisms from X1, X2, X3 and so on to Y, which means we need compatible maps from homomorphism of B to A1 and homomorphism from B to A2 and so on. So we have to give a section over X1 and a section over X2 and a section over X3 that are compatible on each of these intersections. And if you think about it, this is just the same as giving a homomorphism from B to the global sections of regular functions of the sheaf of regular functions on X. In other words, we just take ring of regular functions on X and we look at ring homomorphisms from B to that ring and that's the same as homomorphisms of schemes from X to Y. So the point is that if Y is an affine scheme, it's really easy to describe morphisms from the scheme X to Y. In fact, we can just as easy to describe the functor from X to sets represented by Y. For example, if Y is say Z of X1 up to Xn, then morphisms from X to Y are the same as n-tuples of regular functions on X. So morphisms to affine schemes are easy. And now we want to look at morphisms from X to a projective scheme, say n-dimensional projective space over Z, say. We work over Z for simplicity, although similar things apply to other base rings. And we discussed this earlier and couldn't really give a completely satisfactory answer at that time because we didn't have the enough technology about modules, but now we do. So we're going to fix this problem and give a proper description of these morphisms. And first of all, let's quickly recall what we said earlier. First of all, Pnz is a union of n copies of affine space. And if X is something like the spectrum of a field or the spectrum of a local ring, then morphisms from X to P to the nz is just the union of morphisms from X to these various copies of An. However, this fails in general. If you think of Pn as Pn, so this fails in general. If you think of Pn as being the union of various copies of affine space, then that's fine if the image of X just lies inside one of these open sets, which happens if X is the spectrum of a field or a local ring or something like that. However, if the image of X happens to lie something like that, so it's not in any one of these open sets, then obviously you can't describe morphisms like that. In particular, this fails if X is equal to the spectrum of z, just the integers, for example. We saw an example earlier where it wasn't like that. However, for the spectrum of z, we can still give a description in terms of projective coordinates. So we can think of a map from X. We can try describing a map from X to the P to the nz as a set of tuples, An0, A1, An. Where the ai are in z or whatever ring we're working over, and this is optoisomorphism. So we identify this with lambda A0, lambda A1, opto lambda Am, where lambda is a unit. So this could be z or whatever ring we're working with. Now this works for fields, local rings, works the integers. It works for pretty much any easy ring you can think of. So it'll work for polynomials in several variables over a field. However, it doesn't quite work for all rings. I mean, if you take something like this, it certainly defines the morphism from spectrum of a ring to projective space. The problem is there are some rings for which there are extra morphisms that can't be described quite like this. So here we want the ai's over a field. You can't have them all zero. So instead, we should say A0, the ideal generation by A0 up to An is equal to the whole ring R. So let's see the simplest example where this description doesn't quite work. Here we're going to take R to be z root minus five, which is a standard counter example to an awful lot of things for very obvious reasons. And we're going to take a morphism from R to P1 over z indicated by two one plus root minus five. So what does this mean? You notice first of all that two one plus root minus five, the ideal generated by this is not equal to R. So this isn't the obvious way of describing morphisms where we insist that these should be the ideal R. So what does this mean? Well, we have to cover R by two open sets. And we can take the open sets D2 where two is invertible and D3 where three is invertible and these two open sets cover R. And over D2, we can write this as one one plus root minus five over two and over D3 where three is invertible. We can write this as one minus root minus five over three from one. And this is in one of the affine space over z because of the one there. And this is in the other affine space over z because of the one there. And you can easily check these two morphisms are compatible in the intersection of D2 and D3. So they define a morphism from R to one dimensional projective space. But you can't do it by taking co-prime, by taking elements of R that generate unit ideal. What you notice is instead, these elements two one plus root minus five generate an invertible ideal. Or if you want an invertible sheath and this is the extra key idea we need to describe all morphisms to projective space. So what we're going to do is we're going to describe them as elements like this, optoisomorphism, where lambda is a unit except that the SI, the sections of an invertible sheath rather than just R or an invertible module if we're working over a ring. And the SI have to generate L where we'll explain what that means in a moment. Now the reason why the previous description worked for various rings like Z or fields or local rings, you remember for these rings, the Picard group was trivial. So any invertible sheath is just isomorphic to the ring we first thought of. And we may as well just take these elements SI to be in the ring. So this is an extra problem we have to deal with in cases of rings and for that matter, schemes that have a non-trivial Picard group. So let's describe exactly what happens. So the description is that morphisms of X to n-dimensional projective space, let's do it just over Z, correspond to isomorphism classes consisting of the following data, an invertible sheath L and sections S naught up to SN of L, this is over X of course, that generates L. And as promised, I need to say what this means. That the restrictions to each stalk generate the stalk. So the stalk of L is just a module of rank one over the local ring at that point. So, and we treat this as an example treat this all up to isomorphism. So in particular, if two line bundles L are isomorphic then this gives us the same morphisms to projective space and furthermore line bundles L have automorphisms. And in the special case we had earlier when we just took A naught, A1 and An to be elements of R, automorphisms of the invertible module R are just units of R. So saying we're taking these up to isomorphism is the same as saying we're considering this to be the same as this with all coefficients multiplied by lambda the unit. So this condition about multiplying by units essentially corresponds to just giving an automorphism of our line bundle L. Now I'm going to explain why morphisms of X to P to the N correspond to isomorphism classes with invertible sheaves and we should really check this because we've sort of almost overlooked one sort of morphism corresponding to invertible sheaves and we'd better make sure we haven't overlooked any others. First of all, suppose we're given morphism from X to projective space. We want to construct an invertible sheaf on X. Well, projective space has a special invertible sheaf O1. So you remember invertible sheaves over projective space can be constructed from certain graded modules and we get O1 just by taking the obvious graded module and shifting the grading by one in one direction or the other. And I'm not going to say which direction because I always get it wrong. And this invertible sheaf has M plus one special sections which are just the coordinate functions X naught, X1 up to XN. So X naught, X1 and some aren't functions on projective space what they are sections of this line bundle over projective space. And now all we do is we've got a map F from X to projective space. So we just pull back F. We take F up a star of O1 and this is invertible on X. And we can sort of pull back the sections X naught up to XN to X and these sections X naught up to X and generate O of one. And you can check their pullbacks generate the pullback of this line bundle on X. So it's easy to go from a morphism of X to projective space to an invertible sheaf plus a whole bunch of sections. Now we need to go the other way. Suppose we're given an invertible sheaf L and sections S naught up to SN. And we want to construct a morphism from X to projective space. So how do we do that? Well, we can pick some point X in X. Now some SI does not vanish at X. So when we say it does not vanish what we mean is we look at the stalk at X and the stalk will be a module over the local ring and saying SI doesn't vanish means it's not in the unique maximal sub module of the stalk. And then we can find so an open set open affine set with X in U such that SI does not vanish on the whole of U. And now we can define a morphism from U to projective space by just mapping it to one of the open affines in P to the N. So we map U to some N dimensional affine space over Z and we map it to N dimensional affine space. Well, that's easy because U is affine and this is also affine and we know how to describe morphisms of affine space and this can be described as by taking S naught over SI S one over SI and so on up to SI minus one over SI SI plus one over SI. So we missed out SI over SI up to SN over SI. And now you notice that these quotients are all in the global sections of regular functions on U. So S naught and SI are not functions on U that sections of a line bundle over U but their ratio is a function on U because SI is none zero. So this just gives a morphism from U to one of the affine spaces covering PN. So we can do that for every point X of S and now there's a certain amount of routine checking we have to do. We have to check that these morphisms from open neighborhoods of X are all compatible and the morphisms are the same when they're into sections and we have to check that this map is the inverse of the one we defined earlier from morphisms to line bundles with sections and there's a page or two for routine checking which I'm simply going to not bother with I'll leave this as an exercise or something. So should notice that this correspondence between line bundles and morphisms projective space also occurs in algebraic topology where you find for reasonable topological spaces complex line bundles on X correspond to homotopy classes of maps from X to infinite dimensional projective space over the complex numbers. So here we saw that maps from a scheme to finite dimensional projective space are closely related to line bundles together with a choice of sections. So somehow complex projective space has something to do with complex line bundles in both of them. So let's do an example. Let's find morphisms from P1 over Z to P1 over Z. Well, morphisms are given by a line bundle together with some sections and line bundles at O, N. And if N is less than naught there's a rather shortage of sections. If N is greater than or equal to naught then the sections are spanned by polynomials homogeneous polynomials in two variables which I'll just write as X and Y so we have X to the N, X to the N minus one Y and so on. So sections correspond to homogeneous polynomials in two variables of degree N. And we want two homogeneous polynomials so we have a homogeneous polynomial PXY and QXY. So our morphism is going to be indicated something like this once you've chosen N. And now we need P and Q should generate the line bundle and what does this mean? Well, it means they should have no common factors. And if we, well, they're homogeneous in two variables it's easier to forget about them being homogeneous in two variables and just write them as polynomials in one variable. So we've got two polynomials in one variable and what we want now is they have no common factors and at least one has degree N. So saying degree N means they also generate the line bundle over the point infinity. So we've really just rational functions almost except there's one slight extra problem. So for O, N morphisms, the morphisms we get from O, N morphisms to P1 and Z correspond to rational functions P over Q with PQ co-prime and one of degree less than or equal to N and one has degree N except this is not quite correct. This is for N not equal to zero for N equals zero we also have to include the map taking P1 to the point infinity in P1 roughly speaking and that's where you take, I guess, where you take P to be one and Q to be zero. So there's one extra slight morphism to remember about. In particular, we can also work out the automorphisms of projective space over say a field by copying this well for an automorphism, the line, the inverse image of the line bundle O1 obviously has to be O1. So we have to take N equals one. So the morphisms we get correspond to rational functions AX plus B over CX plus D that are co-prime. In other words, we want AD minus BC is non-zero. And furthermore, we should remember that this is going to be the same morphism if we multiply ABC and D by non-zero constant. So these give you the group GL2 over the field K but we should then quotient out by the diagonal matrices in this field. So this is the projective general linear group over K. So next we can ask, suppose we've got a line bundle L over X and some sections S naught up to SN. What if S naught up to SN do not generate L? Well, that doesn't really matter too much. We don't get a morphism from X to projective space. We get a morphism from U to projective space where U is the open subset where not all the SI vanish. That means as usual that they're not in the over each stalk, they're not in the maximal sub-module at that stalk. For example, let's just take X to be two-dimensional projective space over say a field. Now, if we take the line bundle L to be 01, then the sections over it are generated by the three coordinate functions X naught, X1, X2. And if we take all of these, we just get the identity mapped to P2, which isn't very interesting. So suppose you just take X naught equals S naught equals X naught and S1 equals X1 and don't take S2. Then the set U is the set of points where S naught and S1 don't both vanish. So it's really just P2, minus a point. So we get a morphism from projective space minus a point to one-dimensional projective space. So you can think of this as just being the map that takes X, Y, Z to X, colon Y. And it's obviously not defined at the point 001. So if you don't mind your morphism not being defined on the whole of X, then we can find quite a lot of morphisms from open subsets of extra projective space. Finally, what happens if we're just given a line bundle and somebody doesn't bother to give us any sections of L, what can we do? Well, how about we just take a basis of sections of global sections of L. So if X is projective, this is a finite dimensional space. And this gives us a perfectly good morphism to projective space. And then we've just commented that it might not be defined on the whole of X, but it will at least give us a morphism from some open subset of X to L. So we get a morphism from U contained in X to projective space, where N is the dimension of the global sections minus one. And it doesn't really matter which base we take of the space of global sections of L because taking a different basis just corresponds to doing an automorphism of projective space. So let's see what happens if we do this in the case where X is the projective line. And then we take L to be O of N and see what we get. Well, if N is less than naught, then the set where it's defined is just the empty set. We get a rather boring map from the empty set to projective space. If N is equal to zero, then we just get a map from P one to P naught, which is a point not very interesting. If N is one, the global sections of L just correspond to X and Y. So we map X colon Y to X colon Y, which is just a map from the identity map from P one to P one. Again, not terribly interesting. If N equals two, we map X, Y to X squared, X, Y, Y squared. Well, this is a map from P one to P two and the image is a conic. So you can see that this corner squared is that one times that one, which is the equation of a conic. If N equals three, we map X colon Y to X cubed, X squared, Y, X, Y squared, Y cubed. And this is the famous twisted cubic, an embedding of the projective line into three-dimensional projective space. And of course, you can go on for this for other values of N, but it's quite similar. This sort of idea is used to classify curves and surfaces and so on. So suppose you've got to curve X. You can pick a line bundle L, for instance, you might take L to be the tangent bundle or the cotangent bundle. And then you can do this trick and get a canonical map from X to projective space. And sometimes this map will embed X as a closed immersion in projective space. And if you do that, then you've sort of got good hold of what X is. Sometimes you don't get a closed immersion. For example, if X is an elliptic curve and you take L to be the tangent bundle, well, the tangent bundle is just trivial. So the space of global sections is just one-dimensional and you just map the elliptic curve to a point. So this certainly doesn't always work. But in many cases, it will give you a sort of canonical embedding of a variety X into projective space. Okay, the next lecture will say a bit more about relation between line bundles and embeddings into projective space.