 This lecture is part of an online algebraic geometry course on schemes and will be an application of the Fibre product we studied last lecture to group schemes So we're just going to use Fibre products to define what a group scheme is So let's just recall the definition of a group So a group consists of a set G and it has a product Which we'll denote as a multiplication from G times G to G And it has an identity Which will be a map E from 1 to G where 1 is the set with one element as you can probably guess and It has an inverse map minus 1 from G to G and And these satisfy various axioms, and I'm going to write down the axioms in a slightly funny way so the axioms There's an associativity axiom which says that if you take a product of three copies of G You can map it to two copies of G in two different ways You can either take the identity map of G Times the product in other words you multiply these two and then take it to G or you can multiply the first two and To get that one and and just take the identity map on the third one And then these two you just apply the product and the associativity axiom just says this diagram commutes and The identity axiom is similar. So we have a map from G times 1 to G times G which is just The identity map on G times the So that the identity map from G to itself times the that the map from 1 to G and G times 1 can be identified with G and then we've got a multiplication map here So this diagram has to commute and there should be a diagram for the inverse which looks takes G to G times G to 1 to G and If I try and fill in these I'll probably get them wrong. So I won't and there should be two similar diagrams because We want the identity to be a left identity and a right identity and what the inverse to be left inverses and right inverses So this is a rather slightly strange way of defining a group the point of it. This makes sense for any category with products and A final element one so a final element just means an element such that any Thing in the category has a unique morphism to it so If a category has finite as a product of two elements then it has a product of any Positive finite number of elements and you can think of the final element as being the product of the empty set of elements So so the category has a product of any finite Possibly empty set of elements and then you can define groups for example You can just take the category of topological spaces that has products and a group in the category of topological spaces It's just a topological group Anyway, it should be pretty obvious what we're going to do. We're going to define groups in the category of schemes So we can define Groups In the category of schemes Because as we showed last lecture we have a concept of product of schemes Well, actually this is not a terribly useful construction better Is we can define groups in the category of schemes Over some fixed scheme s So this is this is a much more useful Now if G Is a group scheme over s By which we mean a group in the category of schemes over s then we can look at morphisms from any Element of the schemes over s to g so the morphisms from t to g over s former group Which we denote by g of t and we think of it as being the points of the group with values in the scheme s if T is the spectrum of a ring We put g of r To be the same as g of t by abuse of notation So g is a functor from schemes or possibly rings over s to groups Actually a covariant functor of rings and a contra variant functor of schemes. Okay, this seems to be a rather abstract way of looking at Groups, but in fact, it's very natural for example, we could just put Let g to be the Group scheme such that g of a ring is just the additive group of a ring r Let's work over let's take s to be the spectrum of z we'll just work over z's To make things a little bit simpler so we can sort of forget about s so So so here we've defined g as a functor from rings To groups and in order to show that it's a scheme We've got to show that this functor is actually represented by something and it's pretty obvious what it's represented by So the underlying scheme of s is just the spectrum of z of x because you can see that the homomorphisms of rings from z of x To r It's just the same as r So so taking taking any ring to its additive group is indeed a representable functor. So So the underlight so so we get a group whose underlying scheme is speck of z of r furthermore, we've got a maps from g times g to g and from g to g to g and from One to g where this means s which happens to be the spectrum of z in this particular case So these all correspond to Homomorphisms between the corresponding rings. So here we've got a map from z of x to z of x Tensored z of x and You can figure out that it's the map that just takes x to x tensor 1 plus 1 tensor x Similarly here we have a map from z of x to z of x which just takes x To minus x and here we've got a map from z of x to z which you can work out just takes x to 0 So these are the three Maps that give the group structure on this scheme So What can we do with this well one thing we can do with it is is change the base so g is a scheme over the spectrum of z and We can take another scheme mapping to the spectrum of z for instance we could take speck of fp Because there's a map from z onto a finite field of order p and we can now take the pullback of this So it can take g tensor over speck of z times spectrum of fp That's the product of course There's no real point in putting speck z in in this particular example because any scheme has unique map to spectrum of z however, the idea is that We're just working over speck z for simplicity and you should think that we can put in any scheme here and any other scheme here and sort of pull back a group over the first scheme to a group scheme over the second scheme so We've started with a group scheme over speck z and by using the pullback We've now got a group scheme over speck of fp Well, it's it's pretty easy to work out what it is in this particular case. So here. This is going to have underlying Spectrum just the spectrum of fp of x because this just corresponds for reducing things more p in this particularly particularly easy case and the Group multiplication maps will be given by essentially the same formulas as as we have here Well, another thing we can do with it is Take this group scheme here and take a subgroup of it So let's put Alpha p of r to be the elements of r elements x of r With x the p equals zero Under addition. So this is this is a group Well, it's a group provided. We remember to take r to be an algebra over fp. So we're really taking So this is going to be a group scheme over fp and So this is a functor so far and we want to know if it's actually represented by a scheme Well, it is because we can take the scheme spec of fp of x over x to the p and you can see that the morphisms from from the spectrum of r To the spectrum of fp x over x to p can be identified with the elements in r with x to p equals naught And There's a map from This to the spectrum of fp of x So this is a closed subscheme of this scheme here and we can think of this as being a closed subgroup Or a closed subgroup scheme of of the additive group we defined earlier Um, it's actually finite because we see that Alpha p maps the spectrum of fp is a finite morphism And it's a finite morphism because fp of x over x to the p is a finite module over Fp so we think of this as being a finite group scheme Um, it's sort of somewhat analogous to a finite group, except it's not quite a group because it's a scheme and We say its order is equal to the dimension of Fp of x over x to the p as an fp vector space Of course, it's easy to define the order because we're working over over a field if we're working over a more general scheme We'd have to think more carefully about what the order is to find us So this dimension is just p so we can think of this group scheme as being a group scheme of order p It sort of behaves in some ways as if it had p elements Um, and it's a little bit weird because if you look at fp of x We notice that this has nilpotent elements So this is the the spectrum of this Is a non reduced scheme And if you look at um how we defined it We were defining it just as a fiber product of Two things over spectrum of z and you can see this is non reduced. This isn't sorry. This is reduced This is reduced and this is reduced but this fiber product is not reduced So this gives another example of how taking products of reduced schemes without nilpotence may still give you nilpotence We had an example of this last lecture when we were fiddling around with tents of products of fields um so So we've got a funny group scheme of order two over the finite field of order p In fact, there are two other ways to produce group schemes of order p One is a discrete group Let's just define the group z over pz So z over pz is an ordinary honest group and we could just take A scheme consisting of p um Discrete points and just make these into a group. So this gives us a perfect good group scheme um and um It's um It's coordinate ring is going to be well you you just take the ring a and then you take elements a1 a2 up to a p Such that a i a j is equal to nought a i squared equals a i And the sum of the a i is equal to one so we can think of a i As being the function that's nought on the on most of the points except it's one on the ith points nought on all the others um and um You can do this you can make this into a group scheme over z In fact, you can do the same thing for any finite group whatsoever. You can turn a finite group into a group scheme Notice by the way that z over pz of a ring r need not have um have um p points um So for example if r is connected So sorry if r has no um idempotence other than one or zero So if r has idempotence not equal to zero or one then this May have more than p elements in it So although the group scheme has ordered p its values in a ring r might have more than p elements Um, and there's a third group you can form This time we're going to take the multiplicative group So we're going to take the multiplicative group is going to be a functor taking any ring r To the group of non-zero elements as before this is represented by the spectrum of z of x x the minus 1 because homomorphisms of this to any ring just correspond to the invertible elements of the ring and The map from g m times g m to g m corresponds to a map from this to the tensor product with itself and you can see that just takes x To x tensor with x and um the inverse map from g m to g m just corresponds to the map taking x to H the minus 1 and The identity map just corresponds to the map taking x to 1 Um, and this also has a closed subgroup called mu p to be which is the p th roots Of unity And this is represented by the spectrum of z of x over x to the p minus 1 Which you can see is a sort of subscheme of the spectrum of z of x h the minus 1 Because this is a is a quotient of of of this ring um, and again, we can reduce mod p so we look at Mu p over the field f p in other words We're just we're just going to reduce everything mod p and by being very sloppy We're still going to call this mu p. So it's its spectrum is f p of x Over x to the p minus 1 and if you look at this, this is again a non reduced scheme It's f of p x Over x minus 1 to the p and here we see there are some nil potent elements So, um, so mu of p is just the p th roots of 1. This means that mu p of r Is the p th roots of 1 In r So we found three different group schemes alpha p Mu p and z over pz So this takes any ring r to um Well any ring r without um idempotence to the cyclic group of order p This takes a ring r over the field with p elements the p th roots of 1 And this takes a ring r over the finite field p elements to the Elements whose p th power is zero um, and I'm going to leave a sort of exercise about these three groups that cartia duality now if we've got a um If we've got a ring suppose spectrum of r is a group scheme Well, the ring comes of maps r tensor r to r which is its multiplication um, and it comes with a map from One to r which is just the identity element of r and the group structure gives us a map from r to r tensor r And a map from r to 1 which corresponds to the multiplication and the identity of the group and finally we have this inverse map from r to r Now suppose r is a finite dimensional vector space over the field k What we can do is we look at look at its dual r star which are just the linear maps from r to k And now you notice that um, if we dualize all these if we dualize this map we get a map of this form involving r star instead so by dualizing r we find we still Sorry, this is a this means the dual not the non-zero elements of it so So r star is still spectrum So so so so spectrum of r star is also an algebraic Group scheme because you can check that all the The sort of actions for a ring are kind of dual to the actions for a group. So if we've got any group any group scheme Coming from a finite dimension algebra over a field you can just just take the dual of this And get another group scheme. It's a finite dimensional algebra over the scheme And the exercise that i'm going to end with is to figure out What's the dual of these three groups? It turns out the dual of each of these groups is is another one of these three groups I'm sorry leave it to you to work at which is which So next lecture we're going to move on and discuss the property of A scheme or a morphism of schemes being separable Which is an analog of a topological space being house dwarf