 This algebraic geometry video will give a few examples of morphisms of varieties. So the first example will be the twisted cubic. And what we're going to do is to show that this is isomorphic to one-dimensional projective line. So let's just start by recalling what the twisted cubic is. This twisted cubic can be thought of as the set of all points s cubed, s squared t, s t squared, s cubed in three-dimensional projective space. So we think of a number of these coordinates as w, x, y, z. There's an alternative way of describing it. Instead of describing it as a collection of points, we can describe it as the set of zeros of some polynomials. And we recall that it's the set of zeros of the following polynomials, x, z equals y squared and w, z equals x, y. So in particular we get a graded ring coming from k, w, x, y, z, modulo, all these polynomials. So first of all there's an obvious isomorphism from p1 to the twisted cubic. And this just takes the point s t to s cubed, s squared t, s t squared, s cubed. It's very easy to check that this is a morphism. So this is an isomorphism of topological spaces. So we've got a regular map from p1 to the twisted cubic and it's an isomorphism of topological spaces. But as we saw in an earlier example, this isn't enough to prove that it's an isomorphism of algebraic sets. So we need to construct a regular map from the twisted cubic to p1. And the question is how do we do this? Well, the idea is we cover the cubic by open affine sets. Now functions on open affine sets are very easy to write down because they're just given by elements of the coordinate ring. And so we choose a function. Let's call these open affine sets ui on each ui where by function I mean a function to p1 and check they're the same. On each ui intersection uj. So this is a very general way of constructing maps from projected varieties. You can cover them by affine varieties, construct the function on each affine variety and check they're all compatible. So let's first cover the twisted cubic by affine varieties. We know p cubed is covered by three copies of a3 because we can have w not equal nought or x not equal nought or y not equal nought or z not equal nought. And in particular the twisted cubic is covered by the intersections with these three sets. However, we notice that we don't really need two of those. The twisted cubic is covered by two affine sets. So the two sets can be w not equal nought or z not equal nought because you remember if we've twisted cubic is s cubed s squared t st squared t cubed. So you can see if w and z are both equal to zero then x and y would also both have to be zero. So we don't need two of these open sets. So on w not equal zero, we can define a map w, x, y, z maps to w colon x. And this is well defined because w is zero. On z not equal zero, we map w, x, y, z to y, z. So these are both in p1. So here we've defined two maps from open subsets of the twisted cubic to p1. Now we want to check they're compatible. Suppose we look at the intersection w not equal zero and z not equal zero. Well, we notice that w, z equals x, y on the cubic. And so w colon x is equal to y colon z in p1. So we have a map. So we have a regular. So we have a morphism from this cubic to p1. And now it's completely straightforward to check that this is the inverse of the map from p1 to the cubic we defined earlier. So the twisted cubic really is isomorphic to projective space. Now we recall for affine varieties to affine varieties isomorphic if and only if they're corresponding coordinate rings are isomorphic. And we can ask if this is true for projective varieties. So any projective variety in projective space has a graded ring associated with it. And it's natural to conjecture that two projective varieties isomorphic if and only if they're corresponding graded rings are isomorphic by analogy. However, this is simply false. If we look at p1, the corresponding graded ring is just kx, sorry kx, y. So it's polynomials, two variables, degree x equals degree y equals one. On the other hand, the twisted cubic. We have the graded ring k, w, x, y, z. Sorry, they shouldn't be colonized. They should be commas modulo the ideal generated by w, y minus x squared x, z minus y squared w, z minus x, y. Again, the degree of w, x, y, z equal to one. So this is a graded ring. And p1 is isomorphic with a twisted cubic. However, the graded rings are definitely not isomorphic. For instance, in degree one, this ring is two dimensional, but in degree one, this ring here is four dimensional. So projective varieties don't determine the corresponding graded ring. Turns out the graded ring actually depends a bit more than on the projective variety. It depends not only on the projective variety, but also on a line bundle over the variety in some space of sections of the line bundle. So the next example, I mentioned a few times that although you remember a1 minus zero is affine. It's just isomorphic to the hyperbola x, y equals one. The space a2 minus zero zero is not affine. So what we're going to do now is to show that this space is not affine by calculating the ring of regular functions on it. So let's find the regular functions on a2 minus the origin, which are the same as morphisms to one-dimensional affine space. So how are we going to do this? Well, we know the regular functions on any affine variety. So what we're going to do is much the same as we did for the twisted cubic. We're going to cover a2 minus zero zero by two affine sets. Let's call them u and v. And any regular function is going to be determined by a regular function on u and a regular function on v. That are the same on the intersection u, intersection v. Well, it's pretty obvious what we can choose for u and v. So u is going to be the affine plane minus the x axis. And v will be the affine plane minus the y axis. And now we can find the coordinate rings of these two spaces. So here are the coordinate rings. And the coordinate rings of the affine plane minus the x axis is, of course, just kxy to the minus one, because if we remove the x axis, it's just the same as saying we can invert y. And similarly, the coordinate ring of this affine space is kxy to the minus one. So we want to look at the intersection. So the intersection u, intersection v, coordinate ring will be kxy x minus one, y to the minus one. So again, the intersection is also an affine space in the obvious way. So we've got maps from the intersection u, intersection v to u and v. And these induce maps going the opposite direction between the coordinate rings. And in this case, the maps of the coordinate rings are just the obvious one, x goes to x and so on. So what we want to do is to find all ways of choosing an element of this ring, an element of this ring, which have the same image in here. Well, this is particularly easy because these maps are just injective. So all we want is an element of this ring that can be written as a polynomial in xy to the minus one and also a polynomial in xy to the minus one. It's kind of really obvious what the answer is. We just get kxy. So this is the regular functions on a squared minus the origin. And now you notice this is exactly the same as the coordinate ring of the affine plane. Well, we have a map from, we have a regular, not a regular map, a morphism from a squared minus 00 to the affine plane. It's just the obvious inclusion. So this gives us a corresponding map from the coordinate ring of a squared minus the origin. Sorry, sorry, from the coordinate ring of a squared to a squared minus the origin. Now, since morphisms of affine varieties correspond exactly to morphisms, the homomorphisms of the corresponding coordinate rings, if a squared minus 00 were affine, this would be its coordinate ring. And since this map is an isomorphism, a squared minus 00 would have to be isomorphic to the affine plane, which is kind of ridiculous because it doesn't contain the origin. The only conclusion we get from this is that a squared minus 00 is not affine. It's just a quasi-affine variety that isn't affine. The difference between these two examples is that here we're removing a subset of co-dimension 1 and here we're removing a subset of co-dimension 2. And roughly speaking, if you remove a subset of co-dimension 1 from an affine variety, it generally remains affine. All that happens is you change the coordinate ring by adding the inverse of whatever vanishes on this co-dimension 1 subset. However, if you remove a subset of co-dimension 2 or more from an affine variety, it's quite often no longer affine. There's a sort of analog of this in high-dimensional complex analysis that functions can't have zeros that have co-dimension 2 or more. So if you take a closed subset of dimension 1 in a complex analytic variety, you can quite often find a function vanishing on it. But if you remove a subset of co-dimension 2, then there's no function with that as a zero. So there's a big difference between co-dimension 1 and co-dimension bigger than 1. So...