 So in this algebraic geometry lecture, we will be discussing the relation between morphisms of varieties and homomorphisms of the corresponding coordinate ring. So we recall in the last lecture, we defined the notion of a morphism of varieties or algebraic sets or whatever. And in the lecture before that, we defined the ring of regular functions on a variety or algebraic set. So we have the following theorem. So suppose Y is affine, meaning it's an affine variety. Then morphisms from X to Y are same as homomorphisms of rings from O of Y to O of X. So here X can be any quasi-projective variety or algebraic set or whatever. Y has to be an affine variety. O of Y is the ring of regular functions and Y and the same for O of X. Another point to notice that morphisms are going here from X to Y, but the homomorphisms are going in the opposite direction from Y to X. So there's this very irritating reversal of the direction of arrows, which causes endless confusion. So the proof of this is one of these things that's not really difficult, but it's a bit confusing because the notation is something we're not terribly useful. So first of all, suppose phi is a morphism from X to Y. So we'll write this for the morphisms from X to Y. So let's see what that means. Well, that means we've got a map X goes to Y phi. Now this gives us an element, a homomorphism of rings from O of Y to O of X. And we do it like this. Well, an element of O of Y is just a regular function on Y. This is just a morphism from Y to A1. So if you call this morphism psi or something, it's there. And now if you've got a morphism from X to Y and a morphism from Y to A1, we can obviously just compose these. So this morphism here will be the composition of phi and psi. Notice this works for all X, sorry, for all Y. We haven't used the fact that Y is affine yet. So for any reasonable quasi-projective varieties X and Y, we get a map from this set to this set. That's very natural. And the problem, the hard part is to construct a map from hom O Y to O of X to morphisms from X to Y. I hope I've got all these things the right way around. It's very easy to get confused. So we do this as follows. Suppose that H is a homomorphism of rings from O Y to O X. Just stop and check, I've written everything down. And we can think of O of Y as being K X1 to Xn, modulo i for some ideal i, because Y is affine and contained in a n for some n. So it's automatically of this form. Now we're going to construct a map psi from X to Y as follows. So we should first draw a picture of everything to see what's going on. So we've got a map O of Y, mapping to O of X given by H and O of Y has functions X1 to Xn. Inside it. And in the other direction, we've got function Y here and a function X here. And all these coordinates X1 up to X and a regular functions on Y. So they're just maps from Y to A1 to the n. So this map is given by X1, X up to Xn. And there are lots of, there are lots of coordinate functions from A n to A1 given by taking X1 or X2 or whatever. And what we want to do is to construct a map from X to Y here. So this is the map we're trying to construct. I guess we should just recall this as K X1 up to Xn over the ideal i. So what we do is we look at H of X i. So here, all these X1 up to Xn elements of the regular map. So we can apply H them and get an element H of X i here. So if P is in X, so let's choose a point P in X, then H X1 of P H X2 of P and so on are all elements of K. So this is an element of K to the n. So this defines a map from X to A1, sorry, A n, which is just K to the n. And now this is actually the map psi. And now we have various things to check. First of all, we should check that the image is in Y because here we've just got a map from A to the n and we've actually got to make its image in the subset Y of A to the n. And this follows from the fact that H of i equals nought. And then we need to check that psi is a morphism. And the point here is that if we compose X i with psi, this is regular on X for each coordinate function X i. And from that you can check that psi is a morphism. You can also check that this map taking H to psi is the inverse of the map we defined on the previous piece of paper. I'm not going to give details of this. The details are kind of easy, but they're kind of really boring to listen to someone else go through. What you should do is you should work out all the details for yourself. They're sort of straightforward, but just involve a bit of bookkeeping. So let's see some applications of this. Well, the main application is that affine varieties over K are the same as finitely generated algebras over K with no nil potents. Well, what this means is you can map an affine variety to its coordinate algebra and go from the coordinate algebra to an affine variety. So there's one-to-one correspondence between the objects, but you also get a one-to-one correspondence between the morphisms between objects. So the same as means that the first category is equivalent to the opposite of the second. So you remember we have this problem that maps between varieties are in the opposite direction to maps between the corresponding algebras, but apart from this fact that all the arrows go in the wrong direction, affine varieties are really the same as these. I've used the concept of equivalence of categories. Equivalence of categories is actually a bit of a mess to define precisely. You can check a book on category theory like McLean's book on categories to the working mathematician if you're interested. Informally, it means you can just treat varieties as being more or less the same as these finitely generated algebras if you remember to reverse all the morphisms. So this is various consequences. For example, products of varieties or algebraic sets correspond to co-products of these algebras as above. Well, what do I mean by a co-product of an algebra? Well, a product in a category means for any elements A and B, you have an element C mapping to both of them, which is sorry, product A times B mapping to both of them, which is universal in the sense that if you've got maps from C to A and B, there's a unique map like that. A co-product is defined in exactly the same way except you reverse all the arrows. So it means if you've got maps R and S, then the co-product R of R and S is a universal thing with maps from both R and S mapping to it. So in other words, if you've got any element T and maps from R and S to T, then there's a unique map like this. Now in commutative rings, you should remember from commutative algebra that the co-product is just the tensor product. So we've got maps from R to R tensor S and S to this with all the arrows reversed. And since commutative rings or these sorts of algebras are really the same as affine varieties except that you've reversed the arrows, this means the definition of the product of affine varieties corresponds to the tensor product of commutative rings. By the way, this only works over algebraically closed fields. If you're working over non-perfect fields, this fails over non-perfect fields. If you've ever done a course in Galois theory, remember that everything goes wrong for non-perfect fields. And the problem is if K is an inseparable extension of little K, then K, big K tensor over little K, big K may have nilpotent elements. This is a weird thing that doesn't occur for separable extensions. And this means that the tensor product of two finitely generated algebras, even if they both have no non-nilpotents, may suddenly have nilpotent. So there's something really weird going on. So as another application of this, we can look at algebraic groups. So an algebraic group is defined in pretty much the obvious way. It's in some sort of algebraic variety or algebraic set with a morphism from G times G to G and a morphism from G to G, which is the called the inverse, and a map from a point, can't spell point to G, which is the identity. So this is the product. And these should satisfy the axioms for a group in a way that should be fairly, meaning should be fairly obvious. They have to satisfy associativity and the inverse should actually be an inverse and so on. So typical example is the algebraic group G of A. This is given by the field, sorry, by the affine line A1, which can be thought of as points correspond to the field K, and the map from A1 times A1 to A1 is just given by taking xy to x plus y, which is a perfectly good regular map because it's a polynomial. And similarly, the inverse is given by x goes to minus x and the identity is just zero. And it's completely obvious that these form a group, they're just, it's more or less just the additive group of the field K where you also consider it as an algebraic variety. Well, now let's look at this from the points of rings. Well, the coordinate ring of A1 is just K of Z. And the coordinate ring of A1 times A1 is K of x tensored with K of y. So the group operation from this variety to this variety corresponds to a homomorphism of rings going back in the other direction. Well, what is this homomorphism of rings? Well, it's the homomorphism taking z to x plus y, which defines a map from K of z to K of x plus K of y. You can see this if you unravel the definitions. You notice this homomorphism of rings is kind of encoding the addition of the algebraic group. And K of x, K of y and K of z are really, of course, all the same ring as just writing this because this is K of x, y, which is a polynomial ring in two variables. So what's going on is we've got two maps from K involving K x. We've got a map from K x tensored with K x to K x, which is the ring multiplication. And we've got another map from K x to K x tensored with K x, which corresponds to the group operation. And you can see that these are in some sense look dual to each other. And in fact, there's something called Cartier duality for certain algebraic groups that are finite dimensional over the field K. And you can take a dual group by switching the multiplication of the coordinate ring and this funny map that comes from the group product. Well, let's look at a slightly more complicated example of an algebraic group. Let's look at the multiplicative group GM. So here the underlying variety is going to be the non-zero elements of K. Well, that's really a quasi-projective variety or quasi-affluent variety. But you remember, this is actually isomorphic to an affine variety. This time the group ring is K x x minus one. Now the group product just takes x and y to the product x and y. And from this, it's not very difficult to guess what the co-product is. So we want to map from K x y, so K x x minus one to K x x minus one, tensor with K x x minus one. And it's not terribly difficult what this does. It just takes x to x tensor with x. So the third example, let's do a non-commutative group which is a little bit more interesting. Let's just do GL two of K and work out what the corresponding map like this is. So here the product is given by taking a one, b one, c one, d one and a two, b two, c two, d two, two, a one, a two plus b one, c two, a one, b two plus b one, d two, c one, a two plus d one, c two, c one, b two plus d one, d two. Okay, so and the inverse is given by a, b, c, d equals, what is it? It's one over a d minus b c times d minus c minus b a or something like that. So now we want to work out what is the coordinate ring? Well, the coordinate ring is K of a, b, c, d. Well, this would be the coordinate ring of all matrices. We don't want all matrices. We want the ones with non-zero determinants. So we should also add an a, d minus b, c inverse. So this is like taking an open subset of four-dimensional affine space. And you remember that this is really an affine variety because we can think of this as being K a, b, c, d, e, modulo the ideal a, d minus b, c times e minus one. So this ring here is a sort of informal way of talking about this ring here. So this is the coordinate ring of the general linear group of two by two matrices. And now we should have a map from this ring R to R tensor with R, corresponding to the group multiplication. And this map is going to take a to this element here, a one, a two plus b one, c two. So this element goes here, and b is obviously going to map to this element here, and c maps to this element here, and d maps to this element here. So we have a homomorphism of rings, which corresponds to the group operation on this general linear group. We can also write down the map for the inverse. So the inverse from gl two to gl two will correspond to a map from R to R. That's a homomorphism of rings, and it's obvious what this will do. It will take a to this element here, and d over ad minus bc, b goes to minus b over ad minus bc, and so on. So the technical name for this is that the coordinate rings of affine algebra groups like these are in fact things called hopf algebras, or possibly commutative hopf algebras. So hopf algebra is something that not only has a multiplication going from R tensor R to R, but also has a co multiplication going back the other way. And it also has things corresponding to the inverse, and the thing corresponding to the inverse is called an antipode of a hopf algebra. And hopf algebras satisfy some axioms that correspond to the axioms for a group. Okay, next example in the next lecture, we will show that the twisted cubic is isomorphic to the projective line.