 This algebraic geometry lecture will be about products of affine and projective varieties. So we will first do products of affine varieties, which is fairly easy. First of all, if you want to take the product of affine space a m times a to the n, this is pretty obviously just going to be affine space of dimension m plus n. We've got a pretty obvious map if you take, if you've got points x1 up to xm and points y1 up to yn. This just gives you a vector x1 up to xm, y1 up to yn, not very difficult. In terms of coordinate rings, you've got the coordinate ring here, which would be the ring of polynomials in m variables, and you've got a ring of polynomials in n variables. And the ring of polynomials of this affine space is just more or less the tensor products of these. So we've got kx1 up to xm tensored over k with ky1 up to yn, which is just more or less the ring of polynomials in all these variables. If you're looking at algebraic sets, it's not much more difficult. So I suppose we've got an algebraic set defined by an ideal i in this ring, and suppose we've got an algebraic set y defined by an ideal j, so let's call that algebraic set x, then x times y is just the set of points divided by the ideal generated by i and j in this product ring. So i and j are both ideals of this, and you can just take the ideal generated by them. So that's pretty straightforward. Now let's try the analog from projective space. So the first question is, what is the product of projective space of dimension m and n? If we can figure this out, then it's not too difficult to do the case of projective sets. Well, is it pm plus n? Let's try the first obvious guess. And the answer is no. The problem isn't pm plus n. In fact, this isn't true even topologically. For instance, p1 of r times p1 of r, the reals, is just a torus s1 times s1, so it looks something like this, so it's a surface. On the other hand, p2 of r is non-orientable, so I can't really draw it because non-orientable surfaces don't fit in three space. So topologically, these are quite different things. Similarly, over the complex nums p1 of c, p1 of c, is isomorphic to a product of s2 times s2. In particular, the second homology group is z plus c, whereas p2 of c can be written as a union of a point plus a copy of c, plus a copy of c squared in a nice way, and you can read off its homology from that, and its second homology group is just z. So p1, pm times pn and pm plus n are not the same. They're still closely related, in particular, they're birational. So this contains a copy of a m times a n as an open subset, and this contains a copy of a m times a n as an open subset, too. So these have dense open subsets that are isomorphic, which is more or less what is meant by birational. So they're similar but not the same. So let's try copying. So you remember we had a map here from a m times a n to a m plus, and what happens if we try and copy this for projective space? So we take x0 up to xm in projective space, pm, and we take the point y0 up to yn in pn, and let's try mapping this to x0, x1, xm, y0, up to yn. Well, if you think about it a few seconds, you'll realize this is a stupid thing to do because it's not well defined. The problem is that this has to be the same point if you multiply all the x's by constant lambda, but if you multiply all these x's by lambda, it's not the same point in p whatever this is, n plus n plus 1. So this construction just doesn't work at all. Cross it out. It's just a blunder. Well, the trouble was that these numbers were not homogeneous of the same degree in x. So these have degrees 0 in x and these have degree 1 in x. So in order to get a map from x from the x's times y0 up to yn to something, we've got to arrange for everything to be of the same degree in x and y. Well, there's an obvious way to do this. We can just take x0, y0, x1, y0, xm, y0, and we just keep on going x1, y0, x1, y1, and so on. All the way up to xmyn. And this now gives us a perfectly map from projective space pm times pn to, well, there's a bit of a problem here because this is projective space of dimension pmn plus m plus n, which has completely the wrong dimension. So this map is certainly not on 2. So if we call this, well, it's not on 2. We'll see in a moment that it's a closed subset. And to see it's a closed subset, we want to write down some relations between all these. So let's number these as what? w0, n0, w1, n0, and so on, wm, n0. All the way up to wm, nn. I guess that should be coons, not commas. And now there are some obvious relations between all wij's. So we've got wij is equal to xiyj. So we see that w... wij times wkl is equal to wi... Let me just get this right around. This is a bit confusing. wik times wjl. Wait a minute, I think I have got confused. Let me see, wij should be wil and that should be wkj. Incredibly easy to get muddled up about this. Okay, so we've got some relations between the wij's. So we would like to check this as on 2. Before checking it's on 2, let me just give a slightly more abstract way of defining this map. So if you've got a point in k to the m plus one, another point in k to the n plus one, then you can get a point. So let's call this point v and this point w. Then we get a point v, v tensor w in the space km plus one tensor kn plus one. So a point of projective space is defined by non-zero vector in m plus one-dimensional affine space. So v and w more or less represent points in projective space if you think of a line spanned by v rather than w. So we can find a point in this space here which represents a point in the corresponding projective space pm plus n plus pmn plus n plus n. And if you unravel everything, you find this map taking v and w to v tensor w is more or less just this map down here. So this is a sort of computational way of defining this map and this is a more abstract way of defining the map. It's just taking a tensor product of vectors. Anyway, so we've got a map from pm times pn to some subset and we would like to know is this subset defined by all these equations here? Well, we want to show the map is onto. So we want to show the map from pm times pn to the vectors w naught, nought, octon, wmn satisfying w i, j, w k, l equals w i, try and get it right around this time, w i, l, w, j, w k, j. Yeah, it's still getting confused by this. So we want to show this as onto. So let's take a point like this. We can assume w naught, nought equals one because one of these w i's must be non-zero and just by rearranging coordinates we may as well take it to be w nought, nought. Then we find w k, k, l is equal to w nought, l times w k nought, which is equal to y l times x k assuming which we'll write as y l times x k. So all these w k l's are determined by the numbers w nought, l and w k nought and we can fix these to be anything we like by choosing the y's and the x is correct. So any point satisfying these equations here is in the image of p m times p n. I should say this map is called the Segre embedding. So we've managed to show that the products of two projective spaces is a projective variety given by a surprisingly large number of equations in a space of quite high dimension but whatever it works. As I said, now that we've embedded product of projective space it's really completely straightforward to show that the product of any two projective varieties is also a projective variety just by more or less copying the argument we did for affine spaces. Anyway, let's see what this map looks like in the simplest non-trivial case. So let's try and embed p1 times p1, see what happens. Well, here we've got points x nought x1 and here we've got a point y nought y1. So we're going to map this to p3. So this is 1 times 1 plus 1 plus 1 and we map these two points to the point x nought y nought x nought y1 x1 y nought x1 y1 in p3. And if we call these numbers, let's call these numbers w nought w1 w2 w3. We find there's only one, we only get one relation between these, which is w nought w3 equals w1 w2. So this is a quadric in p3. Now if we're working over an algebraically closed field, any two non-singular isomorphic, it just means a quadratic homogeneous degree to polynomial and saying it's non-singular means it essentially means it must use all the variables and in fact any two non-singular quadrics by completing the square or something from linear algebra must be just x nought squared plus x1 squared plus plus x n squared equals zero. Possibly let's say algebraic closed field of characteristic not equal to two because things always go wrong in characteristic two and I'm too lazy to think about what happens in this case. So we see over again over an algebraically closed field of characteristic not two, any quadric in p3 is isomorphic to p1 times p1. In particular it's got two sets of lines on it because p1 times p1 obviously is two sets of lines you can fix a point in this and then you get a lot of lines there or you can fix a point there and then you get a lot of lines in that one. For example, suppose we take a sphere x squared plus y squared plus c squared equals one. So this is over the real numbers this is just a sphere and what I'm going to do is I need to claim that this sphere actually has two sets of straight lines on it. That seems a little bit odd because a sphere in r3 obviously doesn't have any straight lines whatsoever on it like I don't mean g or d6 as in differential geometry I mean actual straight lines. Well it has lines over the complex numbers it doesn't have lines over the reels of course for instance I can give you one example of a line there are a couple of there are a few lines that are easy to spot we can just set x equals one y equals i times z. This is obviously a straight line in the complex numbers because all the numbers are just linearly dependent on z and it obviously satisfies this equation here so I'm going to leave it as an exercise find all straight lines on the sphere x squared plus y squared plus z squared equals one over the complex numbers of course. Might want to notice that this equation can be put into more or less the same form as this equation if we write it as x plus iy times x minus iy equals one minus c one plus c. So if we homogenize we would change these ones to some variable say w and then this would be something times something equals something times something which is optimal linear transformation the same as this equation here.