 This video is part of an online Algebraic Geometry course on Scheme Theory and This lecture will be about showing that the spectrum of a ring really is a locally ring space So we'll just recall the definition the spectrum of a ring so the points of The spectrum are just prime ideals The topology is given as follows The sets D F which the set of prime ideals P with F Not in P former base for the topology You can think of these as being Informally something like the points where the function f doesn't vanish except that f isn't a function It's just an element of a ring but if you if you pretend it's a function on speck of R then this would be the set of points where the function doesn't vanish and The sheaf structure is defined by saying that the Sheaf at of the open set OD of F is just our Sort of localized at F. So you forcibly invert the element F And the question is is this a sheaf? Well, the answer is it isn't a sheaf yet because we've only defined it on some of the open sets Not all of the open sets. However, the first point is that we can define a sheaf On a set X By defining it on elements of a base Because then it's automatically defined on any other open set because you can cover the open set by open sets in the base And we would like the base to have the property that It's closed under Intersection this makes things convenient and you notice these open sets form a base that is closed under intersection because Df Intersection DG is the prime ideals Not containing F or G which is just the same as the prime ideals not containing FG and to do this we just need to check the sheaf condition for Covers of a base element you buy base elements Ui So in other words if we've got an element you that's a union of your elements you I then given Elements of the sheaf for each of these open sets you which you can think of informally as being functions on these open sets If they're compatible on the intersections, then there's a unique element of the sheaf of you So a key point in handling sheaves Of schemes is to restrict attention to these rather special open sets because they're much easier to handle And this property means that you don't really need to pay too much attention to the other open sets I'm not going to prove this. It's just a fairly routine piece of bookkeeping Anyway before going on this leads to the question What are what do you do if you for some reason you need open sets not of this form? It's not very common for this In fact anybody using open sets not of the form DF is probably up to no good But if you do need to handle them you can handle them as in the following example So let's take R to be C XY and Which you can think of as being an affine plane and I'm going to take two sets first of all, I'm going to take you one You can set you one to be the set of points where X is Not zero at least the set of closed points where X is not zero and you too Will be the set of points where Y is not zero Well, more precisely you one is just the set DX and You two is a set dy The prime ideal is not containing X or not containing Y. So the sheaf condition says that Supposedly want to know what oh of you one union you two is So this is going to be Roughly speaking, it's the complex plane minus a point if you would it ignore the none closed points and The sheaf condition says that an element of this is given by an element of oh you one and an element of oh you two Which have the same image when you restrict them to oh you one intersection you two So we need to work up what these various rings are Well, oh you one is just C of X Y X the minus one polynomials in X Y next the minus one and oh you two It's just C X Y Y to the minus one and it's obvious what oh you one intersect you two is This is just C X Y X the minus one Y to minus one so An element of this ring here consists of a polynomial F in the first ring and a polynomial G in the second ring Which have the same image in here and in this particular case It's very easy because these two rings are both sub rings of that one So the polynomials must actually be equal. So you must have some rational function Which only is poles where X is norton only as poles where Y equals norton. It's kind of obvious that this implies F is polynomial in X and Y so the For the sheaf Spec of R for this ring if we want to know what oh of this set is it turns out to be just the ring of all polynomials again, so it's in fact the same as the Ring you get associated to the whole of the spectrum of R anyway, let's continue with the Proof that the spectrum of R really is a sheaf so we need to show Suppose we're given a Basis of an open set that's even the union of some other open sets So here's here's DF and it's going to be the union of several other things So this might be DFI and so on And we need to check the sheaf condition in other words given Elements of the sheaf of all these open sets that are compatible. We have to show there's unique element of the sheaf of this open set and The proof of this isn't really terribly difficult, but it involves a lot of rather confusing bookkeeping So before giving the heart of the proof We're first thing to simplify the problem by by making several reductions. So the first reduction is we can replace R by R F to the minus one and the point is the spectrum Of R F minus one has underlying set essentially DF So instead of Proving it for this set for the ring R. We can Just just prove it for the whole of the spectrum of a ring if we replace R by that in other words We can assume that F equals one and DF is the whole of the spectrum of R So We suppose The sets DFI cover the spectrum of the ring R And what does this actually mean? Well, it means There's no prime outside all the sets DFI So look at the ideal Generated by these Fi F1 F2 and so on and What we've shown is there is No prime ideal Containing this ideal in particular. There's no maximal ideal containing this ideal. So this ideal must be the whole of R so a1 F1 plus a2 F2 and so on plus a n Fn equals 1 for some AI in R Incidentally this argument shows that the spectrum of R is quasi compact Which is means the same as compact unless you're French So the point is that if you've got a cover of the spectrum of R by open sets of this base it means that you can write one as a linear combination of all these elements Fi and That automatically means you can write one as a linear combination of a finite number of them So the spectrum of R is actually covered by a finite number of these basis open sets And by the way, this is an example of a partition of unity And so a partition of unity in analysis means you've got some open sets covering a set And you try and write the identity function as a combination of Functions each of which has support in one of these open sets And you notice that this function has support in the open set Df1 in some sense and this is supporting the open set DF2 in some sense So so this is an analog of a partition of unity in analysis and Okay Next we're going to check separability And this is fairly straightforward. So we need to show that given R in the ring R, which is just O of the spectrum of R, of course Show that if if the R is not in all the sets Dfi so these should cover the spectrum of R Then R is equal to zero So let's think about what this means Well R being zero in the sets D of I means R the image of R is not in the ring R fi to minus one and this means This is the same as saying that fi to the ni R equals naught for some Ni so this is an equality in R Because you remember that something is zero in R fi to minus one if and only if it's killed by something in the multiplicative set generated by fi On the other hand, we are given that a1 f1 plus a2 f2 and so on plus a and fn equals naught Because these open sets Dfi cover the spectrum of R and now we use a very useful trick. We can replace Fi to the ni by fi and The point is that the prime ideals Not containing fi are the same as the prime ideals not containing fi to the ni so So if the Dfi cover R then the Dfi to the ni cover R so We can Just assume that all these exponents ni are equal to one an alternative way of seeing that is you notice if you raise this to a High power then in every term at least one of the fi's occurs to some high power So we can assume that Fi to the power of one times i equals naught for all i And now we can just write R is equal to R times sum of a i fi Which is equal to sum over a i fi R Which is equal to naught because We said fi R was equal to naught and We're also using the fact that The sum of the a i fi is equal to sorry. There's a type of that should be a one not a zero Next we want to check the sheaf condition so So I suppose speck of R is The union of various open sets Df1 union Df2 and so on so picture it like this is Df1 and here's Df2 and this whole thing is the spectrum of R and We recall that the ring of functions on Dfi It's just R fi minus one and now Suppose given our i over fi to the ni in In this thing here and Suppose they're compatible Well, what does this mean? Well, it means they have to have the same restriction on Dfi intersection Dfj. So this means that our i over fi The ni is equal to rj over fj to nj in the ring R fi to minus one fj to minus one And what does this mean? Well, this means by definition of equality in this ring that fj to the nj times our i minus our J fi to the ni Well, we can't deduce that this is not what we can deduce is that Fi fj To the nij times this is equal to zero because that's the condition for something to be zero in this ring and so This is the data we were given and we want to show that there is Is some R in the ring R? so that R is equal to r i over fi to the ni in R fi to minus one and What does this mean? Well, it means that fi to the mi Times fi to the ni R minus our i Is equal to naught in R So for some mi So that's what we've got to show we suppose that we're given elements R i and fi and They satisfy this condition and they also satisfy the condition that Some of a i fi equals one for some A i so this is the covering condition then we want to find an R satisfying that condition So we've got a rather messy looking algebraic problem And the first thing we need to do is to simplify to bits that we can see what earth is going on So let's simplify before solving first of all we can replace High powers Fi to the to the sum thing by fi Using exactly the same argument we used before if we do this And we find the conditions become sum of a i fi equals one with some different collection of numbers a i And we just have fi fj times R i fj minus r j fi Equals zero. So we're given numbers a i fi and r i Satisfying these conditions. Well, this is still too complicated We've got rid of all these exponents, but there are still too many f's in this equation But we can get rid of them as follows. We put s i equals r i f i and then we find s i fj squared minus s j fi squared is equal to naught well Now we've again got powers of the fi so we can do this trick again replace fi squared by fi That's before so it will still satisfy this condition for some other values of a i So we find that s i fj minus s j fi is equal to zero Now we've simplified our conditions to something where we have some hope of actually solving them So let's summarize what we've got to do so summary given a i fi s i with some of a i f i equals one and F i s j equals f j s i So these are our conditions and we want to find R so that f j R Equals s j for all j Now you may wonder why I'm not putting in a power of f j on both sides of this Well, it turns out we don't need it We've raised fi to such a high power in our simplifications that it turns out We've now got enough information to solve this without any extra f j's on the left or the right So Question is what is R going to be well, we can sort of guess what R is so Um From this Let's suppose that we found an R then figure out what R is well from this We find that R is equal to R times sum of a i f i and now um if we Write this as sum of a i F i R We can now By this condition here Notice we've got that condition here and so we find this is equal to sum of a i s i So this suggests what R has to be So now what we do is we define R to be sum of a i s i and Now in order to finish the proof we just have to check That f j R is equal to s j for all j Which is what we have to prove there So let's just calculate we notice that f j R Well, this is equal to f j times sum of A i s i and this follows from the definition of R So here we've got a definition of R and this gives us this quality here Um next we see this is equal to sum of a i f i s j and this follows from This condition here So you may think I've actually muddled up all the subscripts i and j But it turns out you can swap the subscripts i and j by using this condition here and finally We notice this is now equal to Equal to s j and for this just follows from this condition here So it's a fairly straightforward calculation although I It's actually a really confusing one. I'm too embarrassed to tell you how many times I had to re-record this lecture before getting it right So this verifies that the Spectrum of a ring really does satisfy the scheme property There's one final thing to check we just want to know what the stock Of the sheaf o of What what the stalk of the sheaf A spectrum of R at a point p is Well, this is the direct limit of the O of r of f i Minus one Sorry, it's it's the direct it's the direct limit of O of d F i which is r of f i minus one as Where where the sets d fi form a base For the open sets containing the point p And you can see this is just the localization r p Of r at the prime P so this this localization just means you invert or f i that Are not contained in the prime p And this is a local ring So the spectrum of r is locally ringed space And there's an alternative construction for the spectrum of r in hot show and where he more or less does this whole construction backwards instead of starting with the sets d fi And constructing the scheme and then deducing that you have these local rings at all the points what he does is he starts with a Sheaf Defined by these local rings at every point and deduces that the Value of the sheaf on these sets d of f i is r fi to the minus one so you can do it either way round Okay, the next lecture we will be discussing morphisms of schemes