 So this talk is part of an online course on commutative algebra and will be about co-limits and exactness. So more specifically, it'll be about the following problems. So suppose we've got a lot of exact sequences of modules over a ring indexed by elements I inside some category J. And suppose we've got suitable morphisms between all these elements. So we've got a functor from J to modules. Then the question is, is the following sequence exact? So if we take zero, it goes to the co-limit over J of AI, goes to the co-limit of BIs, goes to the co-limit of all the CIs. And we can ask, is this exact? So this is part of a general problem but pretty much whenever we've got a functor involving modules, we want to know whether it preserves exactness. So first of all, it preserves right exactness in all cases. In other words, if we miss out this zero here, then if all these are exact, then this is exact. And there are two ways to see this. So method one, we could observe that taking co-limits is left adjoint to a diagonal functor. So what we've got here is a functor from our category of modules to the category of functors from our category J to modules. And if we've got any module M, then we can just have a functor from J to modules that just takes everything in J to M. So if J is say, a category with two objects and two morphisms, then the corresponding functor from J to M will just be to take everything here. And the corresponding left adjoint functor takes some sort of function from J to M to some sort of functor from J to modules to the co-limit, which is in this case, will just be the quotient of B by whatever you need to make these two morphisms the same. And left adjoints are always right exact. So in particular, co-limits are left adjoints so they preserve right exactness. So that's the first method. Method two is as follows, that the point is that co-limits commute with co-limits. So the point is if you've got an exact sequence, A goes to B, goes to C, goes to zero, then C is just the co-limit of these two maps. So we've got the map from A to B that we started with and we've also got the zero map from A to B. So what we want to say is that co-limit of a quotient is equal to a quotient of co-limits. And as a quotient is a special sort of co-limit, the fact that co-limits commute with quotients is a special case of the fact that co-limits commute with co-limits. So the next question is why do co-limits commute with co-limits? Well, the easiest way to see this is to just look at a sort of example. So suppose one of the categories just has two objects and two morphisms and the other has say three objects and a couple of morphisms that looks like this. Then you can consider the sort of product of these two categories which is going to look like this. And you can take co-limits of the first category and this will give you some objects here. So here we're taking the co-limits along each row, so to speak. And we can also take co-limits along the column. And if we do that, we will get some things there. And then we can take the co-limits of these things or the co-limits of these things and we want to know if they're the same. So we get some sort of thing here. And you can see that this, whatever object this is, is both the co-limits of this diagram and of this diagram because it's in fact the co-limits of the whole thing. So the co-limits of this will be this object here. And this is fairly easy to check. I'm not going to actually do it. And the only thing I'm going to observe is that this is rather similar to Fibini's theorem in integration theory. So in Fibini's theorem, if you've got a function of two variables, fx, y, dx, you can either integrate with respect to x and then integrate with respect to y or you can integrate with respect to y and then integrate with respect to x. And then integrate with respect to x or you can do a sort of double integral, two-dimensional, the big integral. And all these three integrals are the same under suitable conditions on F. And the same thing sort of happens with co-limits. You can either take co-limits horizontally, then take them vertically or you can take them vertically, then take them horizontally. Or you can take a sort of super co-limit where you do both at once and all three of these produce the same answer. So this is a sort of Fibini's theorem for co-limits. In particular, we see that the co-limit functor over J is right exact. And if you've got any right exact functor, it has a derived functor which could be written as co-lim1 over J. So I'm not gonna say very much about this derived functor because nobody seems to use it very much. So what we now want to ask is, what about left exactness? So we can ask, is a co-limit of injective maps injective? In other words, is this co-limit functor left exact and therefore exact? Because we've just shown it's right exact. And the answer is no. And it's quite easy to give examples of this. So suppose we take the co-limit of two maps, Z mapping to two or zero to Z. So the co-limit, and let me put the co-limit in green. So the co-limit is going to be Z modulo two Z. On the other hand, we can have a map from the rationals to the rationals which takes as multiplication by two. And the co-limit of these is zero. And now the integers are a subset of the rationals. So these maps here are injective. So here we're taking a co-limit of injective maps and this map here is not injective. Well, you might quibble that we're taking a co-limit over a funny category instead of taking a co-limit over a directed set. Even if you take a co-limit over a directed set, it doesn't preserve co-limits. So here's a second example. Suppose you take Z mapping to Z, mapping to Z where this is multiplication by two or you could take Q mapping to Q and Q with multiplication by two there. And here you find the co-limit is the rationals and here you find the co-limit is Z modulo two Z plus Z. And again, this group here is not a sub-module of this group here. So co-limits don't preserve exactness even if you just take co-limits over directed sets. However, a co-limit of, so a co-limit over a filtered category, let's call filtered filtered does preserve exactness. So this is a very important case in which taking co-limits over filtered categories differs from taking co-limits over possibly unfiltered categories. So in order to see this, we first need a special property of filtered co-limits or filtered co-limits. And the key point is here is that a filtered co-limit of modules MI is can be written as the following. You take the disjoint union of all the MI and you question that by the following equivalence relation. We say that MI is equal to MJ here. MI in MI and MJ in MJ. If we can find a map from I and J to K, so that MI and MJ have the same image in MK. So here's the element MI and here's the element MJ. And you can look at their images in here and they should both have the same image. So this gives a relation on the disjoint union. And the key point is that this relation is an equivalence relation. And the proof that it's an equivalence relation uses the fact that J is filtered. And it's easy to see this by drawing a little diagram. So suppose you've got these elements MI and MJ and suppose MI is equivalent to MJ. What this means is that from I and J you can find maps to a third element of the category such that MI and MJ become equal. And then MJ is equivalent to MK. And this means that if we take MK here, then MJ and MK become equal in this category here. Well, now what we can do is we can use the fact the category is filtered to find a new object here and arrows like this. And now is the image of MI in this equal to the image of MK in this? Well, not necessarily. But what we can do now is we've got two maps from MJ to this element here. And we can now find another object here in a map here making these two maps equal if you compose them with this map. And now you can see that MI and MK have the same image. So MI is equivalent to MK. And this proves the relation is transitive and proving it is symmetric and reflexive is very easy. So it is in fact an equivalence relation. So the fact that the category is filtered means you can define the co-limit in this way. Notice that if the category is not filtered then you can't generally do this. For example, if your category just looks like this then the co-limit of two modules is going to be MI direct sum MJ. And there is usually absolutely no reason why the direct sum of two modules should be their union quotient out by something. In fact, it never will be unless one of the two modules happens to be zero. So this result definitely fails for non-filtered categories. So now we can use it to show that a filtered co-limit of injective maps is injective. And this is quite easy. Suppose we've got elements MI contained in NI for I in your category J and suitable maps between them. And we want to show that the limit or the co-limit of the MI is contained in the co-limit of the NI. So let's put a question mark to minus that's what we want to prove. So suppose we've got some element MI in MI whose image in the co-limit of NI is zero. So we've got MI and it's going to map to some element NI in capital NI. And because this has image zero, it means there must be some element J in the category such that NI has image zero in NJ. Well, then we have the corresponding image of MI in MJ must map to zero. So MJ equals zero as the map from MJ to MJ is injective. So MI is zero in the co-limit because its image under some map of J is zero. So this shows that filtered co-limits are of injective or injective. So in terms of the derived front, we can say that the first derived front of a co-limit vanishes if the category J is filtered. So I should add a sort of awful warning here. So we can do the same for limits instead of co-limits. So co-limits are right exact and we find that limits over a category J is always left exact. However, it need not be right exact even if the category is filtered. So we'll see an example of this fairly soon in an upcoming lecture. So in other words, the derived functor of the, sorry, that should have been a inverse limit. So the inverse limit is always left exact. However, limits or inverse limits, their derived functor need not banish even if the category is filtered. Finally, we want to discuss co-limits of flat modules. So what we're going to show is that a filtered co-limit of flat modules is flat. Notice that an unfiltered co-limit of flat modules need not be flat, of course. A fairly simple example of this is the fact that any module is a co-limit of flat modules, not necessarily filtered because we can just take a resolution of it by free modules. So for any module, it's a quotient of a free module and the kernel is the image of another free module and these modules are both free and therefore flat. So M is the co-limit of R to the star goes to R to the star. It's a co-limit of two maps between free modules. So the fact that the limit is filtered here is absolutely essential and this is quite easy to prove. So suppose that M is the co-limit of some modules MI. So for MI, MI is flat and this is a filtered co-limit and suppose we want to show that M is flat. So let's take an exact sequence. North goes to A, goes to B, goes to C, goes to zero and then we notice that North goes to A, tends to MI, goes to B, tends to MI, goes to C, tends to MI, goes to North is exact. We're assuming this is exact, of course and this follows because MI is exact and now we notice that North goes to the co-limit of A, tends to MI, goes to co-limit of B, tends to MI, goes to co-limit. C, tends to MI, goes to North is exact and this follows because co-limit preserves exactness and here we are assuming that this is a filtered co-limit. So this is where we use the fact that we're taking filtered co-limits and finally we see that North goes to A, tends to co-limit of MI, goes to B, tends to co-limit MI, goes to C, tends to co-limit. MI goes to North is exact and this is because taking tends to product commutes with taking co-limits. So we've shown that if we start with an exact sequence then tensioning it with the co-limit of the MI is also exact, which means the co-limit of all the MI is not exact flat. Incidentally, there's a converse to this theorem which says that any flat module is a filtered co-limit of finitely generated free modules and what we've proved is that any co-limit of finitely generated free modules is flat. So Lazard's theorem says that a module is flat if and only if it's a co-limit of finitely generated flat modules. I should add a slight warning here that the proof of this theorem in Eisenberg's book appears to have a slight gap in it. So if you want to see a complete proof you can check Lazard's original paper. Okay, so that's enough about co-limits. Next lecture we will probably be discussing limits and flatness.