 This algebraic geometry video will give some more examples involving morphisms of algebraic sets. So what we want to do is to show that you can find a product of any two projective varieties and that this is also a projective variety of isomorphism. So if we've got two varieties A in P to the M and B in P to the N, if we can find a product of P to the M and P to the N, then it's not very difficult to find a product of A and B inside this, essentially just by taking the equations of A and B and combining them. So the main problem is to show the product of two projective spaces is a projective variety. Well, as we saw earlier, there's something called the segregate embedding, which maps this into P, M, N plus M plus N. And what we want to do is to show the image of this really is a product in the category of varieties. So you recall that a product of two sets A and B is, sorry, two objects A and B. It's an object which is maps to A and maps to B and is universal for this property. So if we've got another object with maps to A and B, then it factors uniquely through the product. So we just have to check that segregate embedding has these two properties. So you better quickly recall what the segregate embedding is. So if P to the M has coordinates X naught up to X M and P to the N has coordinates Y naught up to Y N, then the image of the segregate embedding is given by X naught Y naught X naught Y one and so on up to X M Y N. So we will call these Z naught naught, Z naught one and so on. And we also remember that there are a whole lot of quadratic equations satisfied by these Z I's which say Z I J Z K L is equal to something. I always get the wrong way round. I think it's Z I L Z K J, but there's a pretty good chance I've got that wrong. So don't rely on it. So first of all, we have to check there was a morphism from the, from this variety here defined by all these quadratic equations, from this variety to projective space. So we want to define a map from, from this variety to projective space. So by symmetry, we may as well assume that one of the Z I's is none zero. So suppose Z naught naught is none zero. Then we can map the, that is segregate the segue variety to P to the M just by taking Z naught naught, Z naught one and so on to Z naught one. Sorry, the naught naught Z one naught Z two naught and so on. So this will be a point of P to the M, which is well defined because Z naught naught is none zero. On the other hand, instead of taking the set Z naught one, Z naught naught none zero, we might take a different open set C say Z naught one, not equal to zero and here the map is given by mapping it to Z naught one, Z one one, Z two one and so on. And now the point is we need to check these two maps. So we say do these coincide on the intersection of the two open subsets. So we want to, we want to know that these two points are actually the same. And this follows from all these relations z i j z k l equals z i l z k j. So these two points are actually the same in P and similarly, we have several other open subsets covering as part of the cover of the segue variety. And again, you can check that we get a well defined map to P M on each of these open subsets and these quadratic relations imply that all these maps are the same on the intersection. So we get a well defined map from the segue variety to the two copies of projective space. As you see, this proof is almost trivial. It's just consists of some slightly confusing bookkeeping. And the other thing we need to check is that we have the universal property. So here we've got the segue variety mapping to P M and to P N. And suppose we've got any other variety C with maps to P M and P N. Then we've got to show that there's a unique map from C to this segue variety. Well, defining maps from arbitrary projects of spaces to other projects of spaces is a little bit tricky. So what we do is we cover C by affine varieties. So we can think of C as being covered by several affine varieties. Now for each of these affine varieties, if we've proved this behaves like a product, this gives us a map from each of these open. So I should have said these open affine varieties. From each of these open subsets, we get a map to the segue embedding. And by the uniqueness of the map for the product, these must coincide in all the intersections. So we get a map from the whole of C to the segue embedding. So we can assume that C is affine, which makes life a bit easier. And of course, defining maps from affine varieties to P to the M is a little bit complicated. So we should also cover P to the M and P to the N by all these open affine varieties. So for example, a morphism from C to the open subset Z not equal 0 of P to the M. So Z naught not equal 0 is given by a set of functions. We take F naught equals 1, F1, Fm, where all these FIs are just regular on C. Similarly, a morphism to this subset. So it should be X naught not equal naught. Similarly, a morphism to the subset Y naught not equal naught on P to the N given by G naught equals 1 G1 up to Gn. So, well, in general, an affine variety C won't map to the open subset X naught not equal naught of P to the M. But what we can do is we can cover C by a lot of smaller open subsets where each of these open subsets maps to one of the standard open subsets of P to the M and one of the open subsets of P to the N and we can then just glue these all together. So we may as well assume that not only C is affine, but that its image is in the open subset X naught not equal 0 of P to the M and similarly for P to the M. Well, now we can define a morphism of C to the P to the segrate embedding just by mapping C to F naught G naught, F naught G1, F naught G2 and so on up to F M Gn. So this gives a well-defined map from C to projective space in fact to an open subset of projective space. And now we just have to do a lot of small checks checking that this is all well-defined and that all these maps are the same of the intersections of open subsets and so on and all of these checks are rather trivial and not terribly exciting to watch so I'm just going to omit them. Notice by the way that the segrate embedding is really a combination of two almost completely different constructions. First of all, we have a construction of a product P to the M times P to the N where we're sort of implicitly doing this by taking an open affine cover of P to the M and an open affine cover of P to the N taking products of these affine sets of the affine covers and then gluing them together. That gives us an abstract variety P to the M times P to the N. Secondly, we're constructing a map from P to the M times P to the N to a projective space P M N plus M plus N. And the construction of the product doesn't really have anything to do with the embedding into projective space. In fact, later on when we construct products of schemes the construction of products of schemes will be very similar to the first part of this that if you've got two schemes we're going to cover them by affine schemes take products of affine schemes and then glue everything together and then we don't bother with the second part of the construction because in general if we've got a scheme there's no reason why it should be embeddable in projective space.