 This algebraic geometry lecture will be about rational functions and rational maps. So let's first define rational functions. Suppose Y is an affine variety. Then we have the coordinate ring O of Y, which you can think of as being all polynomials on Y. And this is an integral domain because Y is a variety. So we can form the quotient field, or field of quotients, which is sometimes denoted K of Y. And we call these rational functions on Y. For example, if Y is the two-dimensional affine space, its coordinate ring is just the field of polynomials in two variables. So its elements are just polynomials in X and Y. And the ring field of rational functions on Y is the field of all rational functions in X and Y, which is usually denoted by round parentheses rather than square brackets. And its elements look like rational functions, so it'd be some polynomial divided by some polynomial, where of course this one is not identically zero. So the analog of rational functions for Riemann surfaces are just meromorphic functions. So if we have affine varieties and Riemann surfaces, then regular functions on affine varieties correspond to holomorphic functions on Riemann surfaces, whereas rational functions correspond to meromorphic functions. So meromorphic functions are the field of quotients of holomorphic functions, and rational functions are just the field of quotients of regular functions. Now let's look at what happens if Y is a projective variety. So now we're going to take Y a projective variety, and this time we can't do the previous construction because the ring of functions, the ring of regular functions on Y is just the field we're working over. So we've seen this for the projective line, and it's not too difficult to check for other projective varieties. So the quotient field, the field of quotients is again just K, which is far too small. This is really uninteresting. So we have to define rational functions on a projective variety in a different way. So instead what we do is we say a rational function is given by, first of all, a dense, well, let's take it to be affine open set, just simplicity. So it's going to be a dense open set, and secondly, a rational function on U. Except we need to say when two rational functions are the same. So if you've got two rational functions, let's call this rational function F. So suppose you've got a rational function given by a function F1 on a set U1 and a function F2 on a set U2. These are identified if F1 equals F2 on a dense open set contained in U1 and U2. So if you know about direct limits, you can say the rational functions is, got by taking the direct limit over dense open affine sets, U of the ring of rational functions on U, which I will just subtly call K of U. And any two dense open subsets, have formed a dense open set. So it's not too difficult to check that the rational functions on a variety X, which we know by K of Y, again, form a field. Again, we're taking Y to be an irreducible variety. If Y is not irreducible, people don't generally define rational functions. You can define rational functions on algebraic sets that aren't irreducible, but then the rational functions don't form a field. So it's a bit more complicated. Now we can define rational maps between varieties. So suppose X and Y are varieties as before they can be algebraic sets, but people don't usually define rational maps for algebraic sets. So a rational map from X to Y is a regular map from U to Y, where U is dense and open in X. And we identify two of these maps in the obvious way. So if we've got a regular map F1 defined on a set U1 and a regular map F1 defined on a set U2, we say these are equal if F1 equals F2 on an open set U contained in U1 in the section U2. So when I said open, these should be dense and open. So for example, rational maps from X to the affine line are the same as rational functions on X. And now obviously the next thing to do is to form a category whose objects are varieties and whose morphisms are rational maps. Unfortunately, this doesn't work. Rational maps do not form a category, category. And the reason for this is that composition need not be defined. And it's quite easy to find examples where the composition isn't defined. We can take a rational map from A1 to A1, just taking everything to zero. And then we can find another rational map from A1 to A1, which just takes X to one over X. So the image of this map is the point zero, but the domain of this map doesn't contain zero. So the composition is just not defined. So rational maps and on varieties don't form a category. However, you can modify this. We define a rational map to be dominant. So a dominant rational map means the image is dense or the image contains a dense open set more precisely. And then if you stick to rational maps whose image is a dense open set, then the composition is now well defined. So for example, this map from here to here, the image is zero, which is not a dense open set. So we're not going to count this as a real rational map when we form a category. Now that we've got a sort of category, we define two varieties to be birational if they're equivalent in this category of rational maps or rather dominant rational maps. So we say X and Y are birational. Means we've got rational maps from X to Y and Y to X. So the composition of this and this is, well, it might not be the identity map of X because it might only be defined on a dense open subset. But for rational maps, that counts as being the same as dense open subset. So this is a cruder relation than saying they're isomorphic. So if they're isomorphic, meaning there are regular maps between them who's better inverses of each other, then they're obviously birational because regular maps are rational maps. But there are plenty of examples of varieties that are birational but not isomorphic. So being birational is a sort of cruder equivalence relation than being isomorphic. So let's see some examples. So suppose we take the varieties, the affine line, the projective line, the hyperbola X, Y equals one in the affine plane and the curve X cubed equals Y squared. So we've seen earlier that these four varieties are all birational to each other. So A1 is P1 minus a point and this hyperbola we've seen can be identified as A1 minus a point. And we've seen there's a map from A1 onto this variety which is an isomorphism on A1 minus a point. However, no two of these varieties are equivalent. So this one has a sort of singularity and this one is A1 with a point removed and this one is P1 with a point removed. So they're not the same as each other. Similarly, we've seen some two-dimensional varieties that are birational. We can take P1 times P1, we can take projects of space P2, we can take A2 and we can take A2 minus the origin. And these are all birational to the plane A2. So these both of A2 is a dense subset and this is just A2 minus a point. But we've seen that no two of these varieties are isomorphic. So these are not isomorphic, they're birational. Well, the next question is, are there any varieties of the same dimension that are not birational? Since almost all the examples of varieties we've seen so far are birational to the affine plane. Well, that's because we've just been looking at the easiest examples. It's not too difficult to find examples of varieties that are not birational. Varieties that are not birational to an affine line. Varieties that are birational to an affine space are called rational varieties, by the way. So we want to find a non-rational variety. And in honor of Fermat, we will look at the elliptic curve X cubed plus Y cubed equals one. So it kind of looks like this. And we're going to show this is not birational to the affine line. More generally, there is no dominant map. There's no dominant regular map, sorry, no dominant rational map from the affine line onto this curve. So if we had a map from, suppose we had a map from A1 to this curve, which is dominant means it's not constant. What would this mean? We had rational functions X, T and Y, T with X, T of cubed plus Y, T of cubed equals one. Here we're using T as the variable in A1 and X and Y for the coordinates of this curve. So X and Y are rational functions in T. And we want to show X and Y have to be constant. Well, clearing denominators, so X and Y are going to be polynomials divided by polynomials. So by clearing denominators, we can find F of T cubed plus G of T cubed equals H of T cubed, where now F, G and H are polynomials. So you see this is very much like Fermat's last theorem for exponent three, except instead of working with integers, we're working with polynomials and we're trying to show there are no non-trivial solutions where the trivial solutions are where F, G and H are all constants. Well, we can assume these are co-prime and now let's work over the complex numbers or some algebraic closed field. And now we can factorize the left side as saying it's F T plus G T times F T plus omega G T times F T plus omega squared G T equals H of T or cubed. So again, this is what you do when you're trying to solve Fermat's last theorem over the integers. And now the polynomials over a field. So the polynomials over this field form a unique factorization domain. So these three are all co-prime and their product is a cube. So this must be a cube and this must be a cube and this must be a cube. Strictly speaking, we should say these are a cube up to units but the only units are constants. And since we're working over an algebraic closed field any constant is also a cube. So we don't really have this problem. So we can call this H one cubed, H two cubed. Sorry, H two cubed and H three cubed. So we get three equations F plus G equals H one cubed F plus omega G equals H two cubed and F plus omega squared G equals H three cubed. Now these are linear equations between F G and the cubes of the H. So we can eliminate F and G from these three equations and what this is going to do is it's going to give us a linear relation between H one cubed, H two cubed and H three cubed. And by multiplying by constants we get another relation between. So if we multiply H one, H two and H three by some constants, we get another relation between polynomials which says the cube of one plus the cube of a second is equal to the cube of a third. Well, now we found another equation of this form and you notice that if F G and H are not constants then H one, H two and H three have smaller degree than F G and H. So this gives us a new equation H one prime cubed plus H two prime cubed plus H three prime cubed. Sorry, equals that where these, this is H one times a constant and so on. So this gives a contradiction because for any solution of this equation we get a solution of smaller degree. You notice by the way that this works for any exponent that is at least three. It fails for exponent two and in fact the result is false for exponent two. For instance, we have the equation one minus T squared squared plus two T squared or squared equals one plus T squared squared. And this corresponds to the fact that X squared plus Y squared equals one is rational as we saw in lecture one of the course. And in fact, these polynomials here of course just come from the rational map from this to an affine line. And the reason the proof breaks down here is that H one, H two and H three no longer have smaller degree than F and G. So the induction fails. So thermal's last theorem for polynomials over an algebraically closed field is completely trivial. You can give a proof in a few lines whereas for the integers it's incredibly difficult and you know, while this proof is about 100 pages long. And the reason for the difference is that polynomials over a field are very much nicer than the cyclotomic field that you have to use for thermal's last theorem over the integers. In particular, polynomials over a field form a unique factorization domain and all units are nth powers for any n. That's why the proof is so much easier than for the integers. By the way, there's a much shorter proof using a bit more algebraic geometry that there's no rational map from A one to this curve X cubed plus Y cubed equals one. What you do is you notice this curve has genus zero and this curve has genus one. We'll explain what these mean later on. And it's a general result that there's no map, non-constant map from a curve to a curve of higher genus. So if you know about the genus of a curve there's a sort of one line proof that there's no rational map from A one onto this curve here. Okay, the next lecture will have a different proof that X cubed plus Y cubed equals one is not rational. So what we've done now is given an algebraic proof. The next lecture will give a sort of topological or analytic proof using elliptic functions.