 This lecture is part of an online algebraic geometry course on schemes, and we'll be covering the scheme associated to a graded ring S, which is sometimes denoted by Proj S. This stands for projective scheme of S. So to motivate this, we will first recall the projective variety of a graded algebra over K. So here K is going to be an algebraically closed field, and we're going to look at a finitely generated graded algebra over K. So this will be Kx0 up to xn, and then we're quotient out by some graded ideal i. And let's first start by looking at the case when we just take i to be zero. So we look at the case of n-dimensional projective space, which is associated to the graded ring Kx0 up to xn. And what we're going to do is recall how projective space corresponds to this graded ring. First of all, we look at the points of projective space. These just look like x0 up to xn in projective coordinates, where this is considerably the same as lambda x0 up to lambda xn. And this is not all coordinates are zero. So these are the points of projective space. And what do they correspond to in this ring here? Well, they correspond to homogenous maximal ideals among those in the set of ideals not containing the ideal x0 up to xn. So I don't mean maximal ideals of this ring. I mean maximal ideals in this set here. So maybe I should put maximal in inverted commas because we're not actually talking about maximal ideals of the ring. For example, the .100 corresponds to the ideal generated by x1, x2 up to xn. Because these are the elements all vanishing at this point here. And of course, this isn't a maximal ideal of this ring because it's contained in this ideal here. But it's maximal among the set of homogenous ideals not containing this. Next, we want to know what is the topology. So we can have a base of open sets. It's going to be given by the sets df, where f is homogenous and a degree greater than zero. And this consists of the maximal ideals, sorry, the ideals not containing f. Again, they're not quite the maximal ideals of the ring. They're these ideals here. For example, if f is x0, then dx0 is just the set of all points. With x0 not equal zero. So you may as well take x0 to be one. So it'll be these points here, which is isomorphic to the affine plane an. In general, df is just the complement of the hypersurface f not equal zero in n dimensional project space over k. And it's an affine open subset. So we've got the points and we've got a base of the open sets. And finally, we want to know what are the regular functions on the open sets, open sets u. So the regular functions on an open set u will denote it by o of u. And we're only going to define it for these special open sets. So o of d of f is going to be kx0 up to xn. And then we're going to invert f. And then we're going to take the degree zero elements. And you can see this works. For example, let's see what happens if we take f equals x0. Then we take kx0 up to xn, x0 minus one. And we want the degree zero elements of it. Well, every homogenous monomial here can be multiplied by a unique power of x0 in order to make it have degree zero. So this is just essentially the same as kx0 x1 up to xn. Well, I guess isomorphic to this. I guess strictly speaking, what we're doing is we're thinking of this as being kx1 over x0 up to xn over x0. You can sort of make x0 equal to one because you're just taking the degree zero elements. So this gives us a locally ringed space. We've defined a set of points, a base for the open sets on the points, and we've defined a ring on a base of open sets. And you can check without too much difficulty that this satisfies the axioms for a locally ringed space. And we can also work out the stalk of a point corresponding to the ideal m. It's not too difficult to figure out that this is just you take kx0 up to xm, xn. You localize it at m, and you take the degree zero point. So this means degree zero, and this means localize at m. In other words, invert everything not in this maximal ideal. And this gives a one, almost one to one correspondence between finite dimensional graded algebras over k with no nil potents with projective varieties over k. And this correspondence is given as follows. What you do is if you've got a finite dimensional graded algebra over k with no nil potents, it looks like kx0 up to xn modulo i. And this of course just corresponds to the subset of p to the n with i equals zero. Now what we're going to do is just copy this construction for more general graded rings s. So we're going to take a graded ring s, which is sum over n greater than or equal to zero of sn. And first of all, s nought need not be a field. Secondly, s need not be finitely generated. And thirdly, s might have nil potents. So just as we extended the definition, the correspondence between affine varieties and certain algebras over fields to arbitrary commutative rings, we're going to extend the construction of projective varieties from graded algebras to a construction for arbitrary graded rings. So we still keep the ring to be graded. And now we just copy this construction. So the points of our scheme, we're going to call our scheme roge s for the projective scheme of s in some sense. So the points of s, well, previously we took certain graded ideals that were maximal among the set of ideals not containing all positive things here. So we're going to take these to be prime ideals of s, not containing sum over n greater than zero of s n. So this is a prime ideal of s, but we kind of want to throw it out. It kind of corresponds to the point of projective space with all coden at zero, except this isn't actually a point of projective space. So we throw it out. So that's the point. Now we want a base of open sets. The base is going to consist of sets df, where f is homogeneous and degree greater than zero. df is going to be the set of primes as above, not containing f. So informally we think of this as being the points where f does not vanish. Only of course it's not really a set of points where f doesn't vanish because f isn't quite a function on the points that we sort of pretend it is. And finally, we want to define the regular functions and we just copy the definition that we had for projective varieties. We just define o of d of f, one of our special basis sets to be, you take the ring s, then you invert f, and then you take the degree zero element. So this says we invert f, and this says we just take degree zero elements of it. Then proge s is a scheme. And we're not going to verify this. The verification is straightforward but involves checking lots and lots and lots of little conditions which isn't terribly exciting. And it's rather similar to the check we did to show that if you take a commutative ring, then you get a scheme out of it. So I'll just leave this as an extended exercise. And we can cover it by open affine sets. In fact, each df is an open affine subset isomorphic to the spectrum of this ring here, s f minus one zero. We have very good control over this scheme. We know explicitly what a large collection of open affine subsets are. Well, there are rather a lot of homogenous polynomials, but we don't really need all of them. So if f one f two and son generate some of n greater than zero. Then as an algebra, sorry, as an ideal over. So as an ideal, then, then proge s is covered by the sets d fi. So normally we only need a finite number of these. Sets to cover it. So for example, if we take k x naught up to x n, then some over n greater than zero of s n is generated by x naught up to x n. So we only need d x naught up to d x n to cover proge of s. So although it looks as if we're building this out of a vast infinite collection of open affine subsets in practice, we often only need a fairly small finite number of them. So notice that the variety of this graded ring isn't quite the same as the scheme of this project of ring. So if we take s equals k x naught up to x n, we can look at the variety p n. And this is points. I'm just looking like that in, in n dimensional space. So it's points are just things that form x naught up to x n. So this is of course the same as lambda x naught up to lambda x n. However, if we draw the points of the scheme p n corresponding to s, I guess of course we call this proge s, then we not only get all these points here, but it also has various points corresponding to closed subsets and it has a sort of large generic point and so on. So the scheme contains all the points of the variety, but it also contains points corresponding to irreducible subsets of dimension greater than zero. So it's not exactly the same. So let's look at this construction explicitly in the case when s is just the ring of polynomials in two variables over a ring are then proge s will be covered by dx naught and dx one because x naught and x one generate this so we should work out what these are. Well, dx naught is equal to r x naught x one x naught minus one and then we take the degree zero elements of this which is just r x one over x naught. So it's just polynomials in this variable here and similarly dx one is just polynomials in x one over x zero and we want to know the intersection so dx naught intersection dx one is just dx naught x one which is we take polynomials in x naught x one x naught minus one x one minus one degree zero which is just all polynomials in x naught over x one and its inverse. So we've got maps between these two rings and this ring and this corresponds to having a map. I'm getting a bit tired of writing x one over x naught. So I just call this why so we've got a map from our spectrum of our y y to minus one to spectrum of our y and spectrum of our y and sorry our y to minus one. So these are two copies of the affine line and this is more or less the affine line minus the origin so we're gluing together two copies of the affine line over the affine line minus the origin in order to get p one in order to get prong of s. Well this is exactly the construction we had for p one in the previous lecture so prong of s is just p one of the previous lecture. So this construction generalizes the construction of the projective line we had to high dimensional projective space. So as another example let's just work out the points of p k n where p k n is just going to be prong of k x naught up to x n. So this is a scheme that corresponds to n dimensional projective space over k and we want to think what do we mean by the we're taking the not the points the k valued points. And the k valued points you might think of these as being the points from spec of k which is a point to p k to the n. But as we saw last lecture this isn't quite right because this allows all sorts of funny automorphisms of k. So what we should really do is consider these both as schemes over the spectrum of k. In other words we've got natural maps like this. And all we're looking at is is morphisms from this scheme to this scheme that make this diagram commute. So these will be the k valued points of p to the n k. And more generally we can take points with any algebra over k. So if we have an algebra r over k then we can look at map spectrum of r to spectrum of k to p k to the n. So we want to find all the maps here that make this commute. And this is why we write this apparently rather silly morphism from spec k to spec k which is just the identity map. If we were just interested in the k valued points this would be kind of pointless. But we're also interested in more general r valued points where r might be an algebra over k. And in this case we still want this map to commute. It's again eliminating silly automorphisms of k that we're not really interested in. Well as we saw last lecture even for the one-dimensional line these sorts of points can really be quite complicated for general r. So we'll just take r to be a local ring which makes the calculation much easier. And the point is if r is a local ring that m be a maximal ideal then m is in the closure of all other points. So you remember a point of the spectrum of r is just a prime ideal and its closure is all the prime ideals containing it and as m is maximal m certainly contains it. So I suppose we map spec of r to some scheme s. And suppose this scheme s is covered by open sets. What we do is we look at the image of m. So we look at this point f of m. And now all other points of r that image, the closure of the image, must contain f of m. So if f of m is in some open set u then f of spectrum of r is a subset of u. So this means we don't have this problem that the spectrum of r might not be contained in one of these open sets but might pass through several of them. Which makes it very complicated to work out what spec of r is. So anyway we know that pr of m is covered by these open sets dx0 up to dxn. And dxi is just the ordinary spectrum of kx0 over xi, which is just 1xn over xi, which is just the affine space over k. So the morphism from the spectrum of r to dxi are just given by points of r to the n, which we can write as r0, r1, 1. So this is in the i-th position up to rn. And we have to glue these all together for the various different, we have to glue the copies of dx0 up to dxn together. And if you do this, we see that the points from spectrum of r to pkn, that commute with the maps to spectrum of k, correspond to points r0 up to rn, where sum ri is invertible. And this point is equivalent to the point lambda r0 up to lambda rn, where lambda i is invertible. This is very similar to the usual definition of projective space over a field, except instead of saying all the ri's and non-zero, we just say that at least one of them has to be a unit. In other words, these generate the unit ideal. So that describes the k-valued points or projective space over k with values in any local ring. In general, projective varieties over k kind of correspond to certain schemes over k of this form, prunch of kx0 up to xn over i, where this ring has no null potents. In other words, k is a radical ideal. Again, I'm not going to write out the details of this because they're kind of routine and not terribly exciting. Morphisms of projective varieties, v to w, don't quite correspond to morphisms of schemes, s to t. Because morphisms of schemes, as I said, allow the extra funny things like automorphisms of k. So what you really want to do is look at morphisms of schemes that are over spectrum of k. So morphisms of varieties from v to w correspond exactly to morphisms of the corresponding schemes from s to t that make this diagram commute. Okay, next lecture will probably be a discussion of the functor of points associated to a scheme.