 So this lecture is part of an online course on commutative algebra and what I would be doing today is trying to explain what an affine scheme of a ring is. So informally, this is a way of pretending that R is a ring of functions on the spectrum of R. So this will be a continuation of the previous lecture, so I'll just very quickly recall what we did there. So the spectrum of R has open sets U of F for F in the ring, which you can think of informally as being the points where F is non-zero. And for each of these open sets we assigned a ring O of U of F, which was just R localised at F, which we think of as being functions on the open set U of F. At least it's going to behave as if it were a space of functions on U of F. And in order for this to behave like functions on the open set U of F, you remember last lecture there were three properties it had to have a restriction property and a pre-sheaf property and a sheaf property. So we're going to recall these properties and just check that these things do have the properties. So first of all, let's just check restriction. So restriction just means if we've got a map, say U of F1 F2, which is contained in U of F1, then we should have a restriction map from R of F1 F2 to the minus one. So the restriction map goes in the opposite direction restriction map from R of F1 to minus one to R of F1 F2 to the minus one. So this just corresponds to restricting a function on this open set to this smaller open set. So since there's an obvious map from this localization to that localization, we do indeed have restriction maps. You can check that they satisfy some obvious conditions for restriction maps like composition if you've got the chain of three open sets. The next was the pre-sheaf property, which says that if U of F is covered by sets U of Fi. So here's U of F and it might be covered by three sets U of F1, U of F2, U of F3. And suppose G in the ring of functions on U of F is zero in the ring of functions on U of Fi, or I, then G should be zero. So that's just saying that if a function is zero on all sets of an open cover then it should be zero. So we better check that it satisfies this condition. So here this is just R of F the minus one. And now we should think about what it means for the U of Fi to cover U of F. Well, first of all, we can replace R by R F minus one. And doing this we can assume that F equals one. So we're talking about covering the whole spectrum of the ring by these sets U of Fi. And then the sets U of Fi cover the spectrum of R. Well, this means that no maximal ideal or no prime ideal contains all the Fi. Otherwise it would be a point of spectrum of all not contained in these sets. So this implies that the ideal generated by the Fi is the whole of R because it's not contained in a maximal ideal. So this implies sum of Ai Fi equals one to sum Ai. So this is the condition that says these open sets cover the spectrum of the ring. Now suppose that let's take G to be an element of R and suppose that G equals naught on all these rings O of U of Fi. Well, what this means is that G times Fi to the Ni equals naught for some Ni because this is the condition for G to be naught in the localization R Fi minus one. And now we know that F1 to Fn generate the unit ideal. And we'll also F1 to the N1 up to Fn to the N, that's a rather bad notation, Nn also generates the unit ideal of R because we can just take sum of Ai Fi equals one and raise this to some high power and we'll discover that one can also be written in terms of powers of the Fi. So sum of Bi Fi to the Ni equals one for some Bi. Well now let's just multiply this by G. So we find G is equal to sum of Bi Fi to the NiG. Well this bit is just naught by assumption up there, so G equals zero. So in other words, our definition of the ring of functions on an open set does behave reasonably in that if a function is zero on all open sets of a cover then it's zero. Next we have the sheaf condition which is a bit trickier. So this says that suppose the U of the Fi cover U of F, so here we might have U of F1, U of F2, U of F3 and U of F might be the union of these. So suppose we're given Ri over Fi to the Ni on the regular as a regular function of Fi and suppose these are the same on the intersection U of Fi, intersection U of Fj. In other words Ri over Fi to the Ni is equal to Rj over Fj to the Nj on this set. Then we can find sum R over F in U of F equal to Ri over Fi Ni on U of Fi. So what this is saying is if we're given functions on each of these open sets of a cover and they're equal on the intersections of these open sets then we can glue them together and get a function on the big set. And we'd better check that our definition of functions on open set satisfies this condition. Well as we're going to prove this in several steps. So step one, it's the most important one is to cheat. We're going to assume R is an integral domain for simplicity. The result is still true even if we don't assume this but it's a little bit trickier. In fact it's so tricky that I always get it muddled up whenever I try and do it for R not an integral domain. So we're only going to do a simple special case. So step two, as before we can assume that F equals one by replacing R by the localization R F to the minus one. So in other words, we're assuming that the UFIs cover the spectrum of R. Again, as before we have sum of Ai Fi equals one for some Ai as the UFI cover the spectrum of the ring R. And step four, we can assume that all the Ni equals one just replace Fi by Fi to the Ni. So we're given elements Ri over Fi and the exponent up here we're just taking to be one. And the problem, the goal is to find some element R in R with Fi R is equal to Ri. So in other words, we're just saying R is equal to Ri over Fi on the localization. And we know that some of the Ai Fi equals one. So if we had an R with this property, what we're going to do now is just sort of do figure out what R, suppose we have found such an R, let's try and figure out what it is. So that sum of Ai Fi is one. So sum of Ai Fi R would have to be equal to R. And if Fi Ri is equal to Ri, this would have to be equal to sum of Ai Ri. So this suggests, so we define R to be sum of Ai Ri because it has to be that if it has this property. Now we have to check that Fi R equals Ri. This calculation doesn't prove this yet. I mean we assumed Fi R equals Ri in order to reduce that R was equal to that. But the implication the other way is something a little bit tricky that we still need to prove. So we can prove it as follows. So we've got to show Fi R equals Ri. So we calculate Fi R is equal to sum of Aj Rj times Fi. And this is because, just like this in here, here we're assuming the fact, here we're using the fact that R is equal to sum of Rj Aj by definition of R. And this is equal to sum of Aj Fj Ri, which looks like a misprint, but we'll come to that later. And this is equal to Ri. And this is because we're using the fact that sum of Aj Fj is equal to one. And you remember this followed because our open sets covered the spectrum of R. Now we come to this middle inequality. So what's going on here? Well, here we seem to be saying that Rj Fi is equal to Ri Fj, which looks like a misprint. Well, this is true because Ri over Fi is equal to Rj over Fj on O of Fi intersection O of Fj. And here we're also using the fact that R is an integral domain. If R is not an integral domain, then these two are equal on this intersection. It's a little bit more complicated than this because you've got to put factors of Fi and Fj on both sides and things get into a real mess. So anyway, we've proved that R has restriction Ri over Fi on O in the ring O of U of Fi, which shows that our definition really does behave as if it is functions on open sets of Ri. So open sets of the spectrum. What we have essentially done is construct something called the affine scheme of the ring. So the affine scheme just means you take the space spectrum of R and each open set of the form O of U of Fi. You assign the ring R of Fi to minus one, and then you check these behave as if they were functions on the open set U of Fi by doing the check we've done above. So the spectrum of the ring R is used very heavily in algebraic geometry. So what I'm going to do is for the rest of this lecture is give you a sort of dictionary relating the ring R to the affine scheme spec of R together with the rings I associated before. So this side is going to be algebra and this side is going to be geometry. And what we're going to do is have a sort of way of converting between algebraic facts and geometric facts. So first of all, a ring is going to correspond to this topological space spectrum of R together with this collection of rings on open sets. And what does a prime ideal correspond to? Well, a prime ideal, that's easy. It just corresponds to a point because we define the points of spectrum of R to be prime ideals. So that's not a big deal. Maximal ideals, well, we've done that they correspond to closed points because we've got this rather weird topology on the spectrum where points need not be closed. What about an element of R? Well, this corresponds to a sort of function. You remember whether or not it was a function was a little bit hazy because the space it took values in sort of varied. So for each point P, we have to take values in the local ring Rp and this local ring can vary. So it's a rather funny sort of function. If the ring has no null potents, then instead of this local ring, as we saw, we can take values in the field. A function can also correspond to a sort of hypersurface of zeros, which is the set of P such that F is not in P. However, this is not one to one because if you multiply the function by some say scalar in a field, you'll get the same hypersurface. So that's not an exact correspondence. What about an ideal of the ring? Well, ideals correspond to closed sets. As we saw earlier, because if you've got an ideal, we can just take a closed set sort of corresponding informally to the points where all elements of the ideal vanish. And again, this is not a one to one correspondence because different ideals can actually give rise to the same closed set. What about a local ring Rp where we localize it at a prime? Well, this corresponds to local rings of the spectrum of R, which is informally functions defined near the point P. So what we do is we take a small neighborhood of P and look at the functions defined on that neighborhood and then kind of take a direct limit as that neighborhood gets smaller and smaller. What about a localization Rf to the minus one? Well, this corresponds to a special open set of the form u of f. So that would just be u of f. So these form a basis of the topology. So not all open sets are necessarily of this form. And I guess different elements f can give the same localization. And again, this isn't quite a one to one correspondence, but we can think of localizations as something to do with open sets. Or generally, we can have a localization R of s to the minus one for some set s. And this sort of vaguely corresponds to an intersection of open sets. So we might have an infinite intersection of open sets and this is something to do with a more general localization of R. So you can take an idempotent a with a squared equals a and these correspond to closed and open sets. So this corresponds to the set where informally where a is equal to zero. So a squared equals one, you can check that the prime ideals containing a form a set that is both open and closed in the spectrum of R. So this kind of corresponds to something called cloak and sets. So finally, one thing that I'm not going to say so much about because you haven't quite covered it yet is that modules over the ring will turn out to correspond to certain sheaves over the spectrum. But since we haven't really discussed sheaves, I'm going to just leave that to an algebraic geometry course. Let's have an example and see what this looks like for two rings as usually I take the two basic rings spectrum of C of X and spectrum of Z and show that these are as usual rather closely analogous. So the spectrum of C of X looks a bit like the complex plane so it's got points nought in C corresponding to ideal X and the point three corresponding to ideal X minus three and so on. And it's got a generic point corresponding to ideal zero. And we could look at an open set we might look at the open set. So we're going to be responding to say the function f. Let's take X to be X minus one X minus four and look at you have F will you have F will just be the open set where we take the complement of the points. That should be the point one and change that to X minus one. We take the complement of the points one and four in the complex numbers. So, you have F is this blue set here, and then the, the range of view of F will just be C of X, where we localize at X minus one minus one and X minus four. So we allow, you can think of this as being functions where we allow poles at one and four. For instance, a typical element might be X squared over X minus one squared X minus four. And this will be in the ring of regular functions on the set you have F. So you see it's regular except at one and four. So you have F doesn't have the points one and four and that's okay. We can, we can think of it as having a pole of order two at this point and a pole of order one at that point if we like. And then we can do something very similar for spectrum of Z. So we might, for example, take various points on spectrum of Z corresponding to prime ideals as usual. And we might take F to be say the function 14, and then U of F will be the spectrum of Z with the numbers two and seven emitted. And what will O of U of F be? Well, it will be Z where we invert two and we invert seven. So you should think of this operation here as correspond to this operation here. And a typical element of this ring might be say nine over 28. So this will be in the ring O of U of F. So we can think of as being three squared over say two squared times seven. And you can think of this rational number as being a sort of function which is defined on spec of Z except at the point two and the point seven. And you see it's not defined at two because it's got a pole of order two at the prime two in the same way that this has a pole of order two at the prime X minus one. And it's got a pole of order one at the prime seven. So this is the very geometric way of thinking of rings. You think of elements of rings as being functions on the spectrum of Z. And you think of elements of certain localizations of the ring as being functions on open sets of the spectrum of the ring. Okay, and next lecture we will be reviewing tensor products in preparation for discussing the relation between tensor products and localization.