 This lecture is part of the online algebraic geometry course on schemes and we'll cover products of schemes So let's first just recall what a product in any category is So if you've got two elements x and y in the category You can ask is there a product x times y of them? And what a product is is a sort of universal element with maps to both x and y So this means if we've got any element z and Maps to x and y Then there's a unique morphism from z to x times y and you can check that products exist for sets groups and so on and And you can also see that any such product is unique up to isomorphism and Has most of the properties you expect from products of sets and what we want to do is to show that Products of Schemes exist Well more generally we don't just want products of schemes. We want products of Schemes Over a base scheme Yes, so what this means is that x and y Should both be schemes over some base scheme s and we want a product of x times y over s which is usually denoted like this so It's you it's a universal element that if we take maps If we take an object z together with morphisms to x and y that commute with these morphisms here Then there's a unique element like that and and of course the these two maps should also commute with each other So so that's what we want to prove given schemes x y and s and morphisms like that We want to show the existence of a scheme with this sort of universal property And and we first do it for affine schemes Well affine schemes are more or less the same as rings Except that the direction of all morphisms gets reversed. So what we want to do is to Have Something like a product of rings with all arrows reversed. So given rings a and b And we want to find Some sort of object here, which is universal for maps of a and b to it So if we've got any ring c and Hormomorphisms from a to c then there should be a unique map like that. Well Just as before we really want to do a relative case. So we'd really like to work with rings that are algebras over some fixed ring are And this will be the tensor product a tensor over rb So just have a quick review of tensor products. So you remember that if m and n are modules Over are we have a tensor product m tensor n which is universal For bilinear maps, in other words, if we've got any bilinear map from m times n to some module X say Then this factors through A map from m times m to the tensor product m tensor n So this map some of these maps are linear and some are bilinear. So this one is bilinear And this one is also bilinear, but this map here is is linear So it's a sort of a tensor product is a sort of way of converting Bilinear maps into linear maps and its existence is kind of obvious. You can see there's a sort of universal Module generated by a bilinear image of m times n because you can just define the tensor product m tensor n is generated by elements m tensor n for m in m and n tensor n and then you Questioned out by the the relations To make the map from m times n to n tensor n bilinear. So we'd add relations m1 plus m2 tensor n equals m1 tensor n plus m2 tensor n and rm tensor n Is equal to rm tensor n are the same For n so this gives us a map from m times n to m tensor n which takes m times n to m tensor n and This map is These relations force this map to be bilinear and you can see it's sort of universal because if you've got any map from m times n to some map x then you can just Extend this to a linear map by taking tensor n to the image of m times n so we have a tensor product of modules and if a and b are algebras then a tensor over rb is also an r algebra Because you can define a product on a tensor b by a1 tensor b1 times a2 tensor b2 Equals a1 a2 tensor b1 b2 and you can easily Check that this extends to an algebra structure on a tensor b with By by using the universal property of the tensor product Furthermore, this is universal um given a and b and r Then If we have algebra maps from a and b to any r algebra x Then you can extend this to a map from a tensor rb to x Just by saying, you know, if this map is g f and this is g And then this map is just defined by taking the image of a tensor b to f a tensor g b so a tensor over rb is a co-product In the the category of commutative rings Co-products that just is it's just like a product except you reverse all the arrows. So it's sort of universal for um Maps like this. It's also sometimes this is also sometimes called a push out Because you think of a tensor rb is being pushed out from a and b Anyway, if we reverse this we see that for affine schemes There is a product um in fact a product over um Over any um affine scheme s because given schemes of the form Spec of a and spec of b mapping to spectrum of r then the um product over spectrum of r is just the spectrum of a tensor over rb So for affine schemes the existence of products follows from the um behavior of tensor products of commutative rings um a product of arbitrary schemes times over sy follows from First of all the product for affine schemes plus lots of bookkeeping So um, what I will do is I'll just sketch how you do this Um, because filling in all the details will merely cause everybody to fast forward to the end of the video so First of all, we've done the case when x y and s are affine Um by using a tensor over r of b Now we look at the case when x and s are affine But y might not be and all you do is you cover y by open affines y i and you glue together The the the schemes times y i so the idea is that y i might be covered by um say a couple of affine open sets Like this so so here's y And um, what you do is you take the product of these with x so this is y1 This is y1 times x This might be y2 and y2 Times x Might look like that and then you glue together y1 times x and y2 times x and um you get a scheme That's Has the property of x times y And we can do that because here we're we're we're just taking a product of two affine schemes Over s I guess I should have done this all over s um Then you can take s to be affine And x and y anything and again what you do is you cover x by schemes x i which are open affine And we glue all the schemes x i times over s y Which we can do because we know how to glue affine schemes times any scheme over s and finally As you can probably guess we cover s by open affines s i And we put x i to be the inverse image of s i in x And y i you can sort of guess what it is And we form the products x i times Sorry x i times over s y i And glue them together to get x times over s y so there's a lot of routine checks I've missed out but Nothing unexpected happens in this construction So now we should have a few examples to see what this looks like So suppose you want to take a product of two affine lines So let's just take the product of two affine lines You might expect that this is the affine plane, but it usually isn't As we will see later What you need to do is you need to take the affine line and you need to take the product over a point Of the affine line and this is then the um affine plane and we can see this because these these are all just affine schemes So we can just look at what's happening at the level of rings and here um We've got the coordinate ring k x and if we tensor this with the coordinate ring Of k y what we get is k tensor with k x y and that's what we'll see in a moment. This need not be k um, this is not usually K so so we can't just take products of schemes. We need to take products of schemes over another scheme so um This corresponds to taking k of x tensor over k with k of y and this is indeed isomorphic to k of x y which corresponds to the affine plane So this is why we We don't just want to take products of schemes We want to take products over a base scheme because if we don't take products over a base scheme We can't even get the product of two affine lines, right? um Next you notice that the product is a bit weird. So first of all um the product apology The product of a one times a one does not have The product apology so if we look at a one times a one Um, which is the affine plane I guess I should take a product over a naught then it has all sorts of closed sets Are given by curves Which are not closed sets in the product apology Um, this of course happened for variety. So shouldn't be a big surprise But the next thing you want to note is that the product Of a one times over a naught a one um does not have the same set as the product of the sets a one and a one. So not only does the topology not Not only is that do we not get the product topology? We don't even get the product set and this follows for the same reason this closed set is the closure Of its generic point So there's a generic point of the affine plane Corresponding irreducible closed set and this generic point doesn't come from a a product of A point in a one and a point in a one in any reasonable way I mean occasionally it will if if we take a vertical line or a horizontal line then that will come from A product of um a point in here and a point in here. So the the scheme theoretic product has It doesn't look very much like the ordinary set theoretic or even the topological product of two spaces And it gets even weirder because The product of two points Well, the product of two points surely that must be a point. What else could it be? Well, no It need not be a A point And by a point I mean a scheme whose underlying space is just a point And to see this, let's look at some examples. What we can do is let's just look at the spectrum of k um tensed over the spectrum of k of the spectrum Of l where here we've got three fields k um contained in Big k and contained in l and let's take k l to be finite Algebraic well So and therefore algebraic extensions Of little k and see what happens. So so um, you notice that um the spectrum of all these three spaces is just a point So these are just points with extra structure and um So that should be a Ordinary product not a tensor product sign. So what we're going to look at is k tensed over little k with l And we need to work up what this is. So so let's first suppose k is the separable extension Of little k and what happens if it isn't separable? Well, we'll see a little bit later But separable extensions are already weird enough This means that k is equal to k of x Modulo p of x where p of x is irreducible and has no multiple roots because um, it's separable And now we can work at what k tensor with l is so k TensorFlow little k with l is then just l of x modulo p of x The tensor products over fields are very easy to work out And now this factorizes as l of x Modulo p1 of x p2 of x and so on where p now factorizes into irreducible factors over l. So these are irreducible over l um And by the chinese remainder theorem um This factors as a product so all these because all these p i's are co-prime and have no multiple factors because the extension is separable so k times over k of l is isomorphic to the product um l of x over p1 of x times l of x Over p2 of x and so on and this is just a product of fields and this corresponds to a disjoint union of um spectra You remember the product of rings has nothing to do with the product of schemes The products of rings corresponds to taking disjoint union of schemes so so so speck of k times over little k of l is a finite discrete set with one point for each irreducible factor of p over the field l so in particular As we said the product of two points isn't necessarily a point. It might be a a finite collection of points Um, so let's have a couple of examples. For instance, if you take q of root two tense it over q with q root two Well, this is just a field. It's the field q root two root three So here the product of two points is indeed a point Fine no problem on the other hand if we take q of i Say tensor with q of i Well, if you take the polynomial x squared plus one and factor it over q of i we get two factors So this splits as a product of two fields. It's q of i Times q of i so on the left we've got a tensor product on the right We've got an ordinary product and it just happens that two times two is two plus two So so we get this Slightly odd thing. So the spectrum of this looks like two points um, well I assumed earlier that k was separable and I sort of promised I would show you what happens if k was inseparable so suppose k and l both inseparable over Little k well if you do this things can get even weirder. So for example Let's take k to be equal to l to be equal to fp of T And let's take little k to be fp of t to the p so um The degree of k over big k over little k is just equal to p And if we put big t equals t to the p just for simplicity then k is equal to k of x Quotient it out by x to the p minus t So we're taking so we form both k and l by taking a p-th root of a generator of little k And then we find that um k tensor over little k of l is Given as follows well, um what we can do is we can just take l of x Over x to the p minus t But in l t is a power big t is a power of little t. So this is just l of x Over x to the p minus t to the p And since we're in characteristic p. This is just l of x over x minus t to the p And now we see that x minus t is nil potent In k times over k l. So this is not a product of fields In fact, it's a non reduced scheme So k tensor over k l Is not reduced So this has given an example of two reduced schemes Over a reduced scheme whose whose fibered product is not a reduced scheme. Um, so this is One of many reasons why trying to work in characteristic p is a sometimes a little bit bizarre um So in particular we see from this that if you take a A field k then the spectrum of k Tents times the spectrum of k can be large in general And if we want to get the right product of two points If we want to say a point sensor with a point is indeed a point We need to take the spectrum of k times Over the spectrum of k With the spectrum of k and this will then be equal to the spectrum of k So the product of a point and a point is indeed a point provide you to remember To work over something Okay, next lecture we will give some more applications of products and fibered products