 This lecture is part of an online commutative algebra course and will be about the relationship between a ring R and its localization R S to the minus one for a multiplicative subset S. So the first thing we want to do is to look at the relation between ideals of R and ideals of R S to the minus one. Of R S to the minus one. So suppose we've write down the map from R to R S to the minus one as being F. And then if we've got an ideal I of R, we can go from ideals of R to ideals of R S to the minus one by taking I to F of I, except F of I is not an ideal in general. For example, we could take the map from Z to Q and then if you take the ideal two of Z, its image in Q is not an ideal. So instead of mapping this to F of I, we map it to the ideal generated by F of I. On the other hand, if we've got an ideal J of, and the localization, then we can take the inverse image of J in the ring R and this will indeed be an ideal of R. We don't have this problem that we need to fiddle around with it a bit. Incidentally, this operation is sometimes called the contraction of J and denoted by JC. And this ideal here is sometimes called the extension of I and denoted by I E. And I'm not going to use this notation because I always get E and C muddled up. And these operations aren't necessarily in versus of each other because there's not a bijection between ideals of these two rings. However, they're still quite closely related. What we're going to show is that if you start with an ideal here and then move it to an ideal of I and then move it back to here, you get back the ideal we first thought of. In other words, if J is an ideal of R estimizes minus one, then J is equal to the ideal generated by F of F to the minus one of J. And this is easy to check. It's obvious that F, F to the minus one of J is contained in J, this is trivial. So we want to show that the ideal generated by this contains J. So suppose X is in the ideal J. Then X is equal to R over S for some R and S. So R over one, which is equal to R over S times S is in J. So R is in F to the minus one of J. Notice R over one, we're writing for an element of R S to the minus one R for an element of R. So R over one is in F of F to the minus one of J. So R over S is in the ideal generated by F F to the minus one of J because S to the minus one is in R S to the minus one. So what this means is that the map taking J to F to the minus one J is an injective map from ideals of S, so ideals of R S to the minus one to ideals of R. So we can think of the ideals of R S to the minus one as being a subset of the ideals of R. So an immediate corollary of this says that if R is notarian so is R S to minus one. And this follows from almost any of the criteria for a ring R being notarian. For example, notarian corresponds to the ascending chain condition for ideals. And if the ascending chain condition holds for ideals for the ordered set of ideals of R, it holds the ordered set of any subset of it. So it's true for this. And by the way, you notice this ideal is generated over R by some elements. So if the set S is finite, then this fact follows from the Hilbert-Base theorem. But you notice that in general, R S to the minus one is not usually finitely generated as an R algebra. So the Hilbert-Base theorem saying that finitely generated algebras are notarian doesn't work or it does not help. The other thing to notice about this example is it's a special case of the fact that localization is really nice. Pretty much any nice property of a ring will also hold for any localization of the ring. So this is one of the reasons why commutative algebra, people doing commutative algebra like localization so much it's just this really well-behaved operation that doesn't cause any trouble. And there are other operations like completion which are continually causing odd technical problems but localization is just really well-behaved. So we can now see that the spectrum of R S to the minus one is homeomorphic to a subspace of the spectrum of R. Well, we've seen it's a subset. Well, it's not really a subset. It's really isomorphic to a subset but mathematicians are always a bit sloppy about identifying spaces. So it's isomorphic to a subset of the spectrum of R C above. All we've got to do is to check the topology and the topology of spectrum of R S to the minus one has a basis of open sets of the form U R over S. So you remember this is the primes not containing R over S and this is the same as the set U R over one because S is invertible. And the set U of R over one is obviously going to correspond to the set U of R in spectrum of R. So you can see that the topology that the spectrum of R over S just has the subspace topology of the spectrum of R. So the picture of the spectrum of R over S is as follows. So what you do is you sort of think of the spectrum of R as being some sort of space like this. And then we've got some elements S and you can think of the elements S as being sort of functions on the spectrum of R. They're not really so. So let's pretend elements of S of functions on the spectrum of R. So these functions will have zeros. So, and the zeros, you can think of as being some sort of space of co-dimension one. So this might be zeros of F in S. But by a zero of F in S, what we mean is that F is in the prime corresponding to some point here. So you remember the points of the spectrum of R are really primes. And we sort of pretend that elements of R are functions and pretend that if they're in the corresponding prime then they vanish there. So spectrum of R is some space and we've got all these lines here. But what's the spectrum of R S minus one? It's just equal to the spectrum of R minus all the zeros of elements of S So you can see this because when I said that spectrum of R is homeomorphic to a subset. So when I said the spectrum of R S minus one is homeomorphic to a subset of the spectrum of R you can ask which subset this is. This is just the subset of primes P which is a disjoint from S, which is, leave this as an easy exercise because I'm feeling too lazy to prove it. So that's what spectrum of R of S looks like. It's informally obtained from the spectrum of R by throwing away a number of co-dimension one pieces whatever that means. So if you're throwing away a finite number of these then this will be an open subset of the spectrum of R if S is finite or finitely generated. In general, it would be the intersection of an infinite number of open subsets of R. So it's not open in general but it sort of behaves a little bit like an open subset and it's not too far from being an open subset of the spectrum of R. Well, we should take a look at some examples and the main examples I'm going to look at are the following. So the most important special case is RP which is defined to be RS minus one when S is equal to the complement of a prime ideal P. So we briefly mentioned this last lecture and we recall the complement of a prime ideal is a multiplicative subset. In fact, that's more or less the definition of a prime ideal. So we remember this is called the localization of R at prime P. Notice here the localization of R at S means you invert all the elements of S. The localization of R at P means you invert all the elements not at S. So the terminology is a little bit mixed up as usual. So we've had several examples of this before so I'll just quickly recall the examples we've had before. So if we take Z localized at zero this was just the rationals. If we take Z localized at say two this is just the set of all rationals that form A over B for B odd. The other examples we had were if you take C of X localized at zero this is just the ring of all rational functions and C of X localized at the ideal X is just rational functions that are defined at zero where the point zero in the complex numbers corresponds to the ideal of all multiples of X. Well, what I really want to do is not just review this one-dimensional case but do a two-dimensional case. So we're going to take R to be C of X, Y. As usual, I'm going to pretend the complex numbers are just one-dimensional. So I can draw the complex plane. Well, yeah, the terminology is a bit modeled up. When I say the complex plane I mean something with two complex variables not complex plane with one complex variable. Yeah, terminology has really got messed up. Now let's look at some prime ideals of this. Well, it's got three different sorts of prime ideals. So here are some primes. We'll classify the primes of this ring a bit later but for a moment I'm just going to state what we get. First of all, there's the prime consisting of all multiples of zero. So this is the zero prime. Next, we've got the prime of all multiples of F where F is an irreducible polynomial. For instance, F might be say Y squared minus X cubed. Say what could be any other irreducible polynomial. And thirdly, we've got the primes of the form X minus alpha times Y minus beta. So where alpha and beta are complex numbers. So there's a three different sorts of prime ideal. And now let's try and draw a picture of the spectrum of CXY. Well, I drew a picture of this before. So what we can do is we can think of the points of this form as kind of correspond to points of the complex plane C2 which I'm going to draw as a real plane. So we've got a real plane with lots of these points in it. Then I'm going to draw the prime here will correspond to a point whose closure contains all the pairs of points where F vanishes. So you can think of this as being some sort of curve. And you can think of this as being the curve of all points where Y squared equals X cubed. Say whatever else the equation is. So might be another curve there and so on. And finally, there's the point corresponding to the point zero whose closure contains everything. So you can think of this as being a sort of huge two-dimensional point whose closure contains absolutely everything. So the spectrum of this string has zero dimensional points which I've indicated in red and one dimensional points in green and a huge two-dimensional points in yellow. Yes, I know points are zero dimensional, not two dimensional, but you just have to put up with this. Now what I'm going to do is I'm going to draw pictures of the localization of this at various points. So let's localize at point zero. Well, this means I'm just going to invert all functions that aren't zero. So I'm getting the field of rational functions and the spectrum is only one point and looks like this. Notice that you shouldn't confuse this with the quotient CXY over the ideal zero. So this is CXY localized zero. And this is the quotient of CXY at zero is quite different because this is just CXY. So its spectrum looks like this. It's got all the various points of the spectrum. So the localization has very little to do with the quotient. So now let's look at what happens if I localize at this green point. So let's take CXY and localize at ring F and see what we get. Well, this means I invert, I'm allowed to invert all functions except those that vanish along this curve. And what this means is the spectrum still contains this ideal corresponding to F and it still contains the generic point, but everything else has been killed off. Again, this is quite different from the quotient because if I take the quotient CXY over F, what does its spectrum look like? Well, its spectrum looks like this. We keep the green point corresponding to F and we keep all the red points on it because the prime ideals of this just correspond to the prime ideals of CXY containing F which correspond to the points contained in the closure of F. So again, you can see the localization has nothing to do with the quotient. Finally, let's localize at, oops, sorry, wrong color. Let's localize at C, let's localize the point X minus alpha Y minus beta. Well, this time I'm going to keep the point corresponding to alpha beta and we also keep a sort of ghost of all the one-dimensional points passing through it and we also keep the yellow generic point. So that's the spectrum of the localization of one of these red points. On the other hand, the quotient, if we take the quotient by this ideal, so if we take CXY modulo X minus alpha Y minus beta, this is just a field, in fact, it's just the field of complex numbers. So it's spectrum just looks like a point there. So these are what you get if you take the spectrum of a quotient and these are what you get if you get the spectrum of a localization. Let's see what the difference is. If you get the quotient of something, you're sort of taking all, if you take a quotient at some prime ideal, you're sort of taking all points inside that prime ideal in some sense. So here we're localizing this prime ideal. So we take all points inside it, which corresponds to all primes containing it. On the other hand, if you localize, what you're doing is you're taking all points outside but nearby. So here if we take this green line, the only point outside of nearby is the yellow point. Whereas if we take this red point, the points outside it are the red point itself and the yellow point and all green points passing through it. So we get something that looks a bit like that. So localization is the sort of opposite to taking a quotient. So what we have is a spectrum R over P. You can think of us being points in the closure of P. So here a point means a prime and saying it's in the closure of T just means it contains P. So the spectrum of R over P corresponds to primes containing P. On the other hand, the spectrum of R localized at P is the points whose closure contains P and these points of primes, these are primes contained in P. So what this is doing is it's turning P becomes a maximal ideal. On the other hand, if we take the quotient by P, P becomes minimal. In fact, it's definitely minimal because it just becomes a zero ideal in the quotient R over P. So quotients and localizations are sort of almost opposite operations. One of them is making P maximal and the other is making P minimal. So here P is the unique closed point of the spectrum. And here, on the other hand, P is in the closure of, no, it's not. The closure of P contains all points. So P is about as far as it can possibly be from being closed because if you take it closure you just get the entire spectrum. So let's just have a final example of looking at localizations. So let's take the localization. Let's take the ring C, X, Y, modulo the ideal generation by X, Y, and let's localize it at the ideal generation by X and Y. And we asked how many points are there in the spectrum? And what I want to do is to do this problem without any calculation whatsoever just by drawing pictures. So what we're going to do is just draw pictures of what's going on and this will hopefully make it obvious how many points that are in the spectrum and what they are. I mean, if you try and do it algebraically it's not at all clear what happens if you take this quotient and then localize and then try and figure out the prime ideals. But so let's draw pictures of the spectrum. So first of all the spectrum of C, X, Y over X, Y looks like this. Well, it's that the places where X, Y are zero is really just the X-axis and the Y-axis. So we draw the X-axis and the Y-axis and we draw all the points on them. So this is the spectrum of this ring here. It's got two one-dimensional points. So this is the ideal where either this is the ideal X and this is the ideal Y. So the spectrum is a unit of two irreducible subsets and then it's got all these various points. For instance, this is the point generation might be the ideal generator by X and Y minus two say this might be the ideal Y, X minus three and here we've got the ideal X, Y. So now we want to localize to the point that the ideal point nor which corresponds to the ideal X, Y and you remember localizing means we take the point that we look at the prime that we're localizing at and just keep the points outside that are nearby. So we end up like with this. Now you can see there are just three points in the spectrum. So the point is if you try and visualize these rings geometrically by thinking about what the spectrum is this question about what is the spectrum of this localized quotient becomes completely obvious. Okay, so next lecture we will be reviewing a bit more about the spectrum of a ring which is how to regard functions of the ring sorry how to regard elements of the ring as being functions on the spectrum of a ring.