 This lecture is part of an online algebraic geometry course on schemes and will be a review of the operation of localization so We recall that we defined the spectrum of a ring and Stated it was a locally ring space, but we haven't yet got around to actually showing that it's a locally ring space Which mainly requires checking the sheaf axiom in order to do this. We need to know a fair amount about localization of rings So we're going to review that this lecture. So we recall that suppose R is a commutative ring and S is a subset and We want to invert all elements of s so for example When you're constructing the rationals you take the integers and you want to invert all non-zero integers Well, if you can invert Two numbers you can invert their product. So we're going to assume that s is closed under Multiplication and you can certainly insert one. So we may as well assume that s contains one So the subset with these properties is sometimes called a multiplicative subset We don't really have to assume that but it's convenient and if you've got any subset you can just make it multiplicative by adding one and Making it closed under product. So so these are harmless assumptions So we're going to construct a ring R s minus one With the following properties first of all, there's a homomorphism from R s minus one and the images of S should be invertible and it should be universal with these properties So if we've got any other map from R to a ring T with the image of S being invertible Then we want to have a unique map from the localization to T and You can check this property actually Characterizes this ring here. So how do we construct this ring? Well, it's very easy All we do is we forcibly add inverses. So we can write R s minus one Is equal to R and now we just add variables t1 t2 t3 and so on and Quotient out by the ideal S1 t1 minus one is equal to zero s2 t2 minus one is equal to zero and so on So this this just forces Ti si to be equal to one So so we've sort of forcibly added an Inversed each element si Well, that's fine But there's a bit of a problem that it's rather hard to see how big this ring is So we can ask how big is R s minus one Well, it's quite easy to check you can write every element of this in the form R Times R over S for some S in S So that sort of gives an offer bound on it. The problem is When is an element of this form zero? So we've got a homomorphism from R to R s minus one rather obviously We can ask what is the kernel so the kernel is going to be some ideal and sometimes I can be zero so if R s equals naught for some S in S then Then R must be in the kernel I Because if s is invertible and R times s is zero then R must be invertible So it's certainly sometimes possible for this to have a non-zero kernel And the obvious guess is that the kernel consists exactly of elements killed by some element of S So let's put I to be the set of R Such that R s was not for some S in S And you can check this I is an ideal so not is an I because Nought times one is equal to zero This is where we use the fact that S contains one and if R one R two are in I And R one S one equals zero R two S two equals zero so R one plus R two Times S one S two equals zero and this is in S because S is multiplicative So I is closed under addition So this is why it's convenient to assume S is multiplicative It makes it easier to define this ideal I Um So we want to show That I is the kernel of the map from R To the localization R s to the minus one in other words an element Maps to zero in here only if it's really obvious that it maps to zero Because it must be killed by some S And we're going to do this by constructing this Localization in a slightly different way and we're going to do this in two steps So the first step is first assume that S has no zero divisors In other words, I is equal to zero and then we can just copy the construction of Q from Z. So you remember That when you construct the rational numbers from the integers or from that matter any quotient fields of an integral domain you Do it as follows So the elements of R s to minus one are going to be given by ordered pairs R s And we're going to write the ordered pair R s as R over s because if we write it as R comma s it's impossible to remember what it means Um modulo the equivalence relation where we say R one over s one equals R two over s two Different only if R one s two equals R two s one So we take all ordered pairs quotient out by this equivalence relation And now we can define additional multiplication in the obvious way So R one over s one plus R one two over s two R one s two plus R two s one over s one s two and R one over s one plus times R two over s two is defined to be R one R two over s one s two remember this this is really a definition of the sum on the product and these things are abbreviations for the ordered pair Consisting of that element and that element or rather the equivalence class of that ordered pair And now to show this is a ring we have to check a large number of things. So we want to check first of all that this is an Equivalence relation Secondly, we want to check that addition and multiplication and subtraction are well defined I guess we should have defined zero and one but it's obvious how to define those and three we want to check the axioms Or a ring Okay, well, that's a dozen or so things to check and I'm not going to go through and check them all because Almost all of them are completely routine. There's one subtle point so You may say where in all this do we have to use the fact that s is no zero divisors well everything goes through No matter what s is except for the following question is the equivalence relation transitive So this turns out to be the one Routine check that isn't quite as routine as you might think So suppose R one over s one is equivalent to R two over s two and R two over s two is equivalent to R three over s three Well, this means R one s two equals R two s one R two s three equals R three s two And now if you can multiply this one by s three and this one by s one and subtract you get R one s two s three Is equal to R three? S two s one so s two but R one s three minus R three s one is equal to zero So R one s three minus R three s one equals zero if S two is not the zero divisor So this is the point where we need to use the fact that S is no zero divisors and of course this implies that R one over s one is equivalent to R three over s three and this proof really does break down if If s has zero divisors and if s has zero divisors this construction does not actually work the equivalence relation isn't transitive and We don't get a ring So how do we fix this? So let's now do the case suppose S has zero divisors well now we just replace R by R modulo i where i is as as above i is a set of R such that R s equals north for some s and s And now the image of S in R over i Has no zero divisors so we can form so we can form R over i And then invert all the elements of s or rather the images of the elements of s Well in the previous construction we noticed that if i is equal to naught the map from R to R over i Sorry to R s minus one is actually injective R is actually a subring of this And you can check this directly from the equivalence relation. So we see what is happening is we're getting a map from R to R over i And this is a subset of R over i s to minus one and We've also got the ring R s to minus one and you can check that these are actually the same In fact, you can define that ring to be that ring if you like so So I is the kernel of the map from R to R s to minus one So this just says that R over s is equal to naught if and only if Rt equals naught for some t in s So so this gives us control over the localization We can tell whether something is zero and of course if we can tell whether something is zero We can tell whether or not two things are equal just by subtracting them There is actually one step construction Which is the usual one you'll find in books This is where you define R1 over s1 to be equivalent to R2 over s2 if R1 s2 minus R2 s1 times s equals nought for some s Well, that sort of works, but it's a rather unintuitive definition of the equivalence relation It's not at all clear why you suddenly put s in there or why it works So I think it's clearer to construct localization in two steps You first kill off this ideal i and you then do the construction for when s has no zero divisors So an important special case is when s is the complement of a prime ideal P you remember a prime ideal has the property that if x y is in P Then x is in P or y is in P and if you think about it This just means the complement of P is multiplicative if x and y are not in P Then their product is not in P And in this case the localization r s minus one is often denoted by rp That's often a gothic letter, but I can't draw gothic with with these pens um so um, it's easy to confuse rp with the quotient r over p and in fact these are Completely different operations. They're sort of almost complimentary operations And to see this let's look at a few examples And so first of all, let's just look at the example when r is equal to c of x Then and then we have two different sorts of prime ideals p. We could take p to be zero Or p to be x minus alpha for some alpha We don't really care what it is And what I want to do is I want to draw the spectrum of the quotient And the spectrum of the localization so you can see what the difference is And so first of all the quotient by zero is just the ring r So it's spectrum sort of looks like this. It's got a generic point And it's got a lot of points on it and it's just the affine line um and If we're quotient out by this ideal, then we've just got the field c So we've just got a point here, which might be say this point So that's the quotient If we take a localization what we're doing If we localize at naught, then we've just got the field of rational functions and this only has one point So we just end up with a single point, which i'm drawing as a line because it's a very big point and If you localize at x minus alpha it means you're taking all rational functions Whose pole is not at this point here So this is the point alpha And as we saw earlier the spectrum is two points. It's got a generic point And uh one point there And if you look at the dimensions of these you find this is dimension one. This is dimension zero This is dimension zero and this is dimension one This point sort of looks one dimensional, but it's really zero dimensional because it's got nothing inside it um so um It's not quite obvious what's going on yet, but it may be clear if we look at a more complicated example So now let's look at r equals c x y We have three sorts of ideals. We can take p equals naught or p equals multiples of an irreducible polynomial or p equals So x minus alpha y minus beta Let's draw pictures of all these well, um, let's draw pictures of r over p And the localization r p in order to see what the difference is well as usual r over p It's just the spectrum of the whole space which you remember has um Some it has a generic point it has a lot of curves on it And it has some closed maximal points um, and as before if we localize At zero we just get the field of rational functions, which now only has one point And now we see this is dimension two and this is dimension zero If we quotient out by f then we're just getting the spectrum of the coordinate ring of this curve So the spectrum just looks like a curve look like whatever curve we The curve f equals zero and and and it still has these maximal points on it And finally if we Quotient out by this ideal then we're quotient out by some that then we're just Uh quotient is just a field so we just get a single point So so this point might be this point and it might be say this point So You can now see these have dimensions these quotients of dimension two one and zero Now let's take the localization. Well, we worked out this earlier and the localization here kind of Looks like this. It's got a point and it's got a lot of ghosts of one dimensional lines And it's got a generic point And it's sort of two dimensional And so remember dimension means it means the length of the biggest chain So we can have points in a line and uh and a plane. So we've got a chain of Of three prime ideals, which is length two um this one you remember we've worked out its spectrum and its spectrum has two points so it has the The maximal ideal p and it has a zero ideal Which sort of looks like this And now you may think this yellow bit is two dimensional the blue bit is one dimensional In fact the blue bit is zero dimensional and the yellow bit is one dimensional And it's got stretched out a bit so You see in each case the dimension of the quotient Plus the dimension of the localization is equal to two Which is the dimension of the thing you're started with and what is happening is If you've got some closed sub variety given by a prime ideal The quotient is kind of telling you what's going on inside this sub variety. For instance, if you take this curve Then taking the quotient of the coordinate ring is telling you what's going on inside this curve Or is taking the localization is telling you what's going on outside in the sort of neighborhood of the curve So and again if you pick a point here this localization is telling you what's going on outside this point nearby As I mentioned earlier, this is the reason for the name localize Localization you're sort of picking a point here and just looking locally at it So you might be looking locally at this line and if you look locally at this line You only see this bit here and if you look locally at this point, you only see this bit there. So this is a sort of Magnificent these are magnifications of neighborhoods of these points in some sense The other thing to be aware of is that when you go from ideals to Algebraic sets you invert the order so bigger ideals correspond to smaller algebraic sets. So here the um, what you're doing is when you Take a quotient You obtain you're left with all ideals that are bigger than this ideal But when you look at it geometrically You're looking at all sub varieties that are smaller than the sub variety you started with so so here for instance the localization will that's on the quote the prime ideals of the quotient the all prime ideals bigger than this whereas the spectrum contains all Irreducible sub varieties that are smaller than this blue one On the other hand when you localize It's kind of the other way around so here When we localize we only keep the ideals that are smaller than this And geometrically that this means we only keep the sub varieties containing the point where we localized So it's it's kind of all rather confusing because everything gets turned upside down when you switch between ideals and sub varieties um So you you can think of Taking a quotient as making P minimal In fact, you're not just making minimal you're making it the zero ideal whereas localizing is making P maximal You invert everything not in p which obviously forces p to be a maximal ideal and similarly When you localize at a sub variety you're kind of making that sub variety minimal So if we localize here, we're making it a minimal sub variety and somehow getting rid of all the points at it When you take a quotient you're making the sub variety maximal by throwing away everything outside it Okay, that's enough about localization and next lecture we will be Discussing why the spectrum of a ring is indeed a ring space