 So this lecture is part of an online course on commutative algebra and will be about limits and exactness of modules. So the basic problem we want to discuss is suppose we're given exact sequences of modules over a ring where the elements i are in some category j and we want to ask whether the limit of the ai, so the limit of these sequences is exact. So is this exact? And so last lecture we discussed the same problem for co-limits and much of the discussion of limits is very similar to the discussion of co-limits. So I'll first summarize the things that are similar to the discussion for co-limits. First of all, the limb function is left exact. This means that at least this part of the sequence is automatically exact and it's only the question of whether this limit maps on to this limit that we need to discuss. So the proof of this is the same as the proof that the co-limit functor is right exact except that you change the direction of all arrows and I gave two proofs of that last time and I'm not going to go through them again. In particular, limb has a derived functor, usually noted by limb 1, and if limb 1 of the ai is zero then the limit of the bi maps on to the limit of the ci. So the question we really want to ask is when is this derived function of the inverse limit zero on the sequence ai? Incidentally, in case you're wondering you can also get higher derived functions of the limit for various abelian categories, but they can be really weird. You start getting running into set theoretic problems when you try calculating them and as far as I know they're not used very much in ordinary commutative algebra, so I won't say much more about them. Now, we also saw that the co-limits need not be left exact and we gave some examples of this. Similarly, limit need not be right exact and let's start by having an example of this. A simple example is to take the sequence nought goes to z, goes to z, goes to z over 2z, which you may recognize as being the standard counter examples of absolutely everything. So this is an exact sequence and we're going to take another copy of it and then we're going to take another copy of this up there and we're going to just keep going. So we're taking infinite number of copies of this sequence. And now let's work out what the co-limits of this are. So the co-limit, I guess I forgot to say these maps here are going to be multiplication by 3. So the co-limit of this means you need to take an element of each of these copies of the integers such that this map to this map is multiplication by 3. So this element needs to be 3 times something, but it also needs to be 9 times something because it's in the image of something here and so on. So it has to be a 3 to the n times something for every n, which is only possible if it's 0. So we get nought goes to 0 and we get 0 again for the same reason. But here multiplication by 3 is an isomorphism, so we get z modulo 2z goes to 0 and it's certainly not exact here. So slightly worrying thing about this is that this is a limit over, well it's sort of filtered except it's the arrows are going in the wrong direction. So I guess you might call this co-filtered, if I can invent a term. So this is a limit over a co-filtered sequence of exact sequences and the result isn't exact. So a co-limit over a filtered collection is exact as we showed last time. However, the analog for limits isn't true that limits over co-filtered need not be exact. So the problem is that we want to find some condition that makes limits exact. And just before doing this, let's just give an example of why you might want limits to be exact. So we will soon be discussing the completion of a module M with respect to an ideal here. So this is just the limit of the maps M over I and this is mapped to by M over I squared, which is mapped to by M over I cubed and so on. So a particular special case of a limit is the completion and we want to know is if nought goes to A, goes to B, goes to C, goes to nought, is the completion nought goes to A hat. So should say the completion is M hat goes to B hat goes to C hat goes to nought exact. And the answer is it's exact except possibly here the map from the completion of B to the completion of C may not be surjective and we want to find conditions under which it is surjective. So we want to know when does the limit preserve exactness. In other words, well this is exact if the first derived functor of the limit applied to the A is zero. Well the answer is, I should say it's answer not A. This vanishes if A, number A I satisfy the Mitag Leffler condition. So who was Mitag Leffler and what was his condition. Mitag Leffler was first better point out that it was just sort of one name except that Leffler was his father's name and Mitag was his mother's name and to kind of join them together for some reason I don't know. So what was the Mitag Leffler condition. Well, Mitag Leffler was working well before anybody messed around with derived sequences or limits and he was actually working complex analysis. So it's the way his name got attached to this condition is rather round about it seems to originated with Borbaki's book on general topology. And if you look at one of his theorems here let me just try and focus in on it a bit so it becomes legible. So here we have a theorem credited to Mitag Leffler and you see he's got an inverse system of household uniform spaces and there's this funny condition here marked ML and if you look at it you will see that this is actually rather similar to the condition we're going to write a bit later. Well, it's pretty certain that Mitag Leffler didn't actually write down this theorem in any of his published papers because uniform spaces were only invented quite a long time after he died. But he had some argument in one of his papers about complex analysis which is sort of vaguely similar to the proof of this theorem so the connection with Mitag Leffler is rather indirect anyway it's now called the Mitag Leffler condition so what is the Mitag Leffler condition? Well you can state a Mitag Leffler condition for arbitrary filtered or co-filtered sets but we're just going to do the Mitag Leffler condition for taking limit over this category where you've just got elements one, two, three, four and so on. And this is by far the most important case it's what you need to take a limit over to construct completions. Sorry I forgot to reduce the magnification so that you can see it. So we're just going to do the Mitag Leffler condition for this case. And the Mitag Leffler condition is as follows it says that nought goes to the limit of the A i goes to limit B i goes to limit C i goes to nought is exact if A satisfies the following condition the image of A j in A i stabilizes for j large. So what's going on here is is we've got the space A i and this contains the image of A i plus one sorry contains which contains the image of A i plus two and so on. And one is that these maps, these inclusions here are eventually all equal. So we have the image of A n plus i equals the image of A n plus i plus one equals the image of A n plus i plus two and so on for some n, depending on i. An example of this or an example that doesn't satisfy this is if we take the A nought to be Z and A one to be Z and A two to be Z and so on. And we take all these to be multiplication by three, then you can see the image of this doesn't stabilize because the image of this in A zero is three Z and the image of this is nine Z and the image of this is 27 Z and so on. So this is the example. This appears in the example we had earlier where limits are not exact and you can see it doesn't satisfy Mitag Leffler's condition. Well, proving that if Mitag Leffler's condition holds then the limits are exact is a bit tricky to do directly. What we're going to do it is do it in three stages. So, so let's do first we'll do case one case one is when A i plus one goes to A i is onto for all i. So if this condition is satisfied then the Mitag Leffler condition is trivially satisfied because the image of any A j in A i is just A i. So let's look at what's happening. We've got nought goes to A i plus one goes to B i plus one goes to C i plus one goes to zero and we've got nought goes to A i goes to B i goes to C i goes to zero. And we're trying to show that the limit of the B i maps onto the limit of the C i. Well, what we have is we've got an element C i plus one and an element C i and various other elements for all the other subscripts. So if we pick an element in the inverse limit of the C i in particular, we've got two elements like that. And suppose we found an element B i mapping onto C i what we want to do is to lift it to an element B i plus one here. So let's put a question mark because we don't know whether that exists yet, which maps to B i and also maps to C i plus one and if we do that we can just sort of keep going. Well, so, so since this map is onto, we can pick an element x. Let me find it. So we can pick an element x mapping onto C i plus one. The problem is, it will map onto some element y which might not be equal to B i. However, since this map is injected, there's an element in A i mapping to B i minus y. So let's call this element Z. So Z maps to B i minus y. And this map here is onto. So we can lift Z to an element w mapping onto Z. And we can then take out element B i plus one to be x minus, sorry, minus x plus the image of W because that will then make this element equal to the element B i. So using the fact that this map is onto, we can repeatedly lift B i to an element B i plus one and by doing that an infinite number of times and muttering something about the action of choice, we can show that the limit of the B i to the limit of the limit of the C i is onto. So that does one case of the Mittag-Leffler result. Let's do case two, which is a sort of opposite result. Here, we're going to do the case when A i plus J mapping to A i is zero for J large. So this is the almost the opposite of these maps being onto. We're saying that these maps are eventually zero. Well, we can assume that A i plus one mapping onto A i is onto. And to do this, we just replace A zero, A one, A two and so on by some subsequence. So we can pick A zero and then pick a different A one to be the first element such that the map to A zero is zero and so on. So we can assume this condition here. And now, we need to use a different argument to show the limit of the B i mapping to the limit of the C i is onto. So here as before, we've got these elements. It was that sequence. And we suppose that we are given elements C i and C i plus one. We've got an element C i plus one mapping to C i. And previously, we started with an element in B i and trying to lift it to B i plus one. We can't actually do that here. What we do is we first lift C i plus one to some element X in B i plus one. And then we take the image of X here. And we're going to call this image B i and you can see that B i then maps onto C i. Well, the problem is, is this B i unique? And the key point is that it is. So let's just point out that B i is unique and does not depend on the choice of X. And the reason for this is suppose we've got another element X prime here. Well, then X minus X prime is the image of some element Y here. So B i goes to X minus X prime. And the image of Y here will be, so if X prime maps to B i prime, then the image of Y here will be B i minus B i prime. However, this map here is just the zero map. So B i is equal to B i prime. And what this means is that this choice of B i doesn't depend on the choice of lifting here. And now we choose B i for all i like this. And it's easy to check that B i is the image of B i plus one. Notice that B i plus one depends on the space B i plus two up here and so on. And since the element B i maps to C i, that means we've constructed a canonical element mapping to the inverse limit of all these C i's. So in this case, the map from the limit of the B i to the limit of the C i is also on two. Now we need to do the third case, which is the general case. And to do this, we need to sort of combine the arguments for case one and case two. Fortunately, this is quite easy to do. So, what we've got is we've got an exact sequence, something rather A3 goes to A2 goes to A1 goes to A0. And we assume that this satisfies the Mitzag-Leffler condition. And then we put A i prime is the stable limit of A i plus j. You remember the image of A i plus j eventually stabilizes. So we now have these maps A3 prime goes to A2 prime goes to A1 prime goes to A0 prime. And notice these maps are all on two because A i prime is the stable limit and you can easily check that maps between the stable limits are on two. So we have an exact sequence nought goes to A i prime goes to A i goes to A i over A i prime. And these maps here satisfy the condition that A i plus one prime goes to A i prime is on two. So it satisfies case one that we covered earlier. On the other hand, you can check that these things here satisfy case two. So the Mitzag-Leffler condition applies to both these modules and to these modules. So we find that the derived functor of the limit applied to the A i prime is equal to zero and the derived first derived functor of the limit applied to A i over A i prime is also equal to zero. But now you remember from homological algebra that we have an exact sequence going from limb one of A i prime goes to limb one of A i goes to limb one of A i over A i prime. So if this is zero and this is zero, it follows that this is also equal to zero, which is what we wanted to prove. This means that any exact sequence with the A i's on the left, then if you take the limits, it remains exact. So this proves the Mitzag-Leffler condition that if the A i satisfy the Mitzag-Leffler condition, then the limit B i mapping to the limit C i is on two. We can have one example of this. Suppose we have maps nought goes to A i goes to B i goes to C i goes to nought with all the A i finite. By this I don't mean they're finitely generated or anything like that, I mean they're finite as sets. Then the limit of the B i mapping to the limit of the C i is on two. And the reason for this is that if all the A i are finite, this implies any decreasing sequence of modules stabilizes. So for a billion groups, provided the kernels of all these maps are finite, the limit of the B i maps onto the limit of the C i. Again, you remember we had an example earlier where the A i's were all Z, so if the A i's are infinite, the limit of the B i's doesn't necessarily map onto the limit of the C i's. Of course, this doesn't need to be finite. You can see all we need is a decreasing sequence of submodules has to stabilize, so this also applies for artinian modules. Okay, next lecture we will be moving on to study completions and during our study of completions we will need this result about when a limit of modules maps onto a limit of modules.