 This video is part of an online course on commutative algebra and will be about a gadget called the COSL complex. So I'm first going to recall a little bit about regular sequences, and then I'll explain what the COSL complex is, and then I'll explain how to use the COSL complex to solve a problem about regular sequences. So we recall that if we've got a ring R, a regular sequence is a sequence of elements x1, x2 and so on, such that x1 is not a zero divisor in R, and x2 is not a zero divisor in R over x1, and x3 is not a zero divisor in R over x1, x2 and so on. And there's one other condition that I forgot to mention last lecture, which you have to add the non-triviality condition that R over x1 up to xn is not zero, because if you allow this to be zero, then there are lots of rather silly examples of regular sequences. So, well, the definition of regular sequence depends on the order, and we can ask the following question, is a permutation of a regular sequence also regular? And if it is, that would make life much easier, because we wouldn't have to worry about the order of the regular sequence. And the answer is no in general. So here's an example. Let's just take a polynomial ring in three variables and quotient it out by xz. So this is just the coordinate ring of a union of two hyperplanes meeting along a line, sorry, two planes meeting along a line. And now we can look at the sequence of elements x-1 and xy, and this is regular, because x-1 is not a zero divisor here, and if we question out this ring by the ideal x-1, we're just setting x equals to one, so we set z equals to naught and just polynomials in y, and this just becomes y, and it's not a zero divisor. On the other hand, xy x-1 is not regular, and this is because xy is a zero divisor, because if you multiply it by z, it becomes zero. So it's a little bit funny that this element is a zero divisor in our original ring, but stops being a zero divisor if you quotient out by something. So you can see that if you permute a regular sequence, it need not be regular in general. However, it's true if r is a local ring. When I say true, I mean the permutation of a regular sequence is regular. So for a local ring, the condition that r divided by the regular sequence is none zero just says that all elements x-1, x-2 and so on x-n of the regular sequence in the maximal ideal m of r. Well, the proof that for r being a local ring can't be entirely trivial because it's not true for all rings, so we must somehow use the fact that r is local improving this. And to do this, we will define the causal complex of a sequence. You can define the causal complex for any sequence, but it will only really be interesting if x-1 up to x-n is a regular sequence. So let's first take n equals 1 and look at the causal complex of an element x-1. Well, the causal complex in this case is just nought goes to r, goes to r. Here we multiply by x-1, goes to r over x-1, goes to zero. So it's a rather obvious exact sequence. Notice that it is exact if x-1 is not a zero divisor, which is most of the condition that x-1 should be a regular sequence. For n equals 2, we take a sequence of two elements and we define the causal complex like this. Nought goes to r, goes to r squared, goes to r, goes to r over x-1, x-2, goes to zero. And in this case, the differential is defined as follows. It takes the element 1 to x-1, x-2 in r-2 and it takes an element a, b in r-2 to x-2, a minus x-1, b. And the map from r to r over, that is the obvious map. And note by the way there's a minus sign there which you need in order to make this differential followed by this, sorry the composition of these two differentials equal to zero. And in general, the causal complex is defined in a rather similar way. What it looks like is the following. You take nought goes to r, goes to r to the n, goes to r to the n, choose 2, so on all the way up to r to the n, choose n minus 2, goes to r to the n, choose n minus 1, goes to r to the n, choose n, goes to r over x-1 up to x-n, goes to zero. So another way of writing these is that these are just the exterior powers of r. So you can think of this as being r, the first exterior power of r, the second exterior power of r and so on. And what we'd like to do is define this and also check it's exact if the sequence is regular. Well the easy way to define it and prove it's exact is to get the causal sequence for n elements by splicing together two copies of the causal sequence for n minus 1 elements. So I suppose we've got the causal sequence for x-1 up to x-n minus 1. So this is going to map nought goes to r, goes to r to the n minus 1, goes to, and go to r to the n minus 1, goes to r, goes to r over x-1 up to x-n minus 1. So here's a copy of the causal sequence for n minus 1 elements. Now we're going to take a second copy of it. So it's exactly the same as the first row. So what's the point of writing this out twice? Well now what we're going to do is we're going to define a map between these two things and this map is going to be given by multiplication by x-n here. And we have to be a little bit careful because what I want to do is I want to put in some minus signs. So the signs here alternate. So you remember on the previous sheet we had a little sign here and the signs in general come by making them alternate. And the reason for this is we can now define the causal complex where we almost define it by taking the sum of these two bits here as our causal complex. So the terms of our nth causal complex are going to be the sums here. So we're going to get nought goes to r, goes to, well if we take r plus r to the n minus 1 we get r to the n and so on. And the differential is given by taking the obvious maps from this thing to this thing as a sum of that and that and that. And the reason we put minus signs in here is so the composition of two differentials is zero. Well there's one thing we have to be a little bit careful about. What we actually do is we cross out this bit. So the causal complex ends with r, goes to r over x1 up to x to the n. So we're not going to cross off this bit but we replace this, we're quoting out this bit by xn. So we sort of quote out this by the image of that and then delete that. So we do a little bit of twiddling right at the end. And otherwise we just get r to the exterior powers of r and a differential and so on. So that sort of more or less defines what the causal complex is. If you write out a formula for the differential explicitly we see that the causal complex does not depend on the order of x1 up to xn up to isomorphism. And to see that you have to sort of work out what the differential is explicitly and you see that it's really independent of the order. And I'm not going to do that because it's slightly tedious and I will get it wrong if I try and do it. So now we want to show that if x1 up to xn is regular it implies the causal complex is exact. Well, most of this is fairly easy. If you define the complex consisting exactly of these pink things then it's a very easy exercise to check it's exact. And the only problem is we're not quite defining the causal complex to be these pink things here because we sort of crossed off this term here and twiddled the last term very slightly. It turns out that this means the causal complex is obviously exact except right at the end. So we've got this slight problem that we're going from... If I write out the last bit we've got r here, r to the n-1 goes to xn goes to r and this goes to r over x1 up to xn-1. And the only problem is exactness at this term here that we need to know that if something here maps to zero in r then it's the image of something here. Well, if it maps to zero in this bit here I guess that yellow isn't quite visible so let me do it in green. If it maps to zero in here then it's certainly the image of something there that's easy to check but as I said we deleted this. So all we know is that its image is zero in here and we want to know it's in the image of something there. Well to do that we would like to know that the image in here is zero and we can argue as follows, suppose we've got an element of this represented by an element a here and an element b here whose image here is zero. Well then the image of b here must also be zero. Now if this map here is injective this implies the image of b in this group here is also zero which makes it easier to prove exactness. So xn injective on r over x1 up to xn minus 1 implies exactness. That's assuming we've shown exactness for the causal complex on n minus 1 elements as long as xn is injective on this then the causal complex on n elements is injective. Well this is exactly the condition that x1 up to xn is regular. Well at least it's most of the condition that x1 up to xn is regular. So we see that if x1 up to xn is regular this implies the causal complex is exact. And this is very nice because it gives us a way of producing exact resolutions of modules and because you see the causal complex is a resolution of this module here by free modules so we see that if x1 up to xn is regular we get a finite free resolution of the module r over x1 up to xn and this is very exciting if you do homological algebra because you can use this free resolution to calculate x groups and tor groups of this module here. Of course not all modules over a ring can be constructed even if they're generated by one element they aren't necessary of this form with x1 up to xn a regular sequence. For instance if it has a finite free resolution of length n this implies that a lot of all its tor groups for sufficiently lot for anything bigger than n vanish but there are plenty of examples of modules over rings with an infinite number of non-zero tor groups. So this is actually not all modules can be written like this. Well what we would like to do is ask if the causal complex is exact is x1 up to xn regular and the answer is no in general. The reason for this is that we actually gave a counter example to this at the beginning of the lecture. If we take a sequence such that x1 x2 is regular then the causal complex is exact but if x2 x1 is not regular it has an isomorphic causal complex so the causal complex of this is still exact but if x2 is a zero divisor as in the example at the beginning of the lecture then the causal complex will not be exact. However the answer is yes if r is a local ring and the consequence of this is that r local implies a permutation of a regular sequence is regular. So this will prove the result we state at the beginning of the lecture very useful it means you don't need to worry too much about the order of a regular sequence as long as you're just working with local rings in which case all the elements of the regular sequence are in the maximum ideal. So to finish off with I'm just going to explain roughly why when r is a local ring the exactness of the causal complex implies x1 up to xn is regular and what we do for this is in fact we don't need the full exactness of the full causal complex we only need exactness of the h1 term so let's write hi for the ith homology of the causal complex actually I should have put hi of x1 up to xn for the ith homology of the causal complex on x1 up to xn and then we find we get the following long exact sequence so I'll first write down the long exact sequence and then explain roughly where it comes from so the only bit of it we're interested is going to be h1 of kx1 up to xn-1 goes to h1 of x1 up to xn-1 goes to h1 of x1 up to xn goes to m over x1 up to xn-1 goes to m so that should be an r my notation muddled up over x1 up to xn-1 so this map here is multiplication by xn and this map here is also multiplication by xn and now this exact sequence holds whether or not the ring is local and where it comes from is if we go back and look at the construction of the causal complex on x1 up to xn we get it by splicing together two copies of the causal complex for x1 up to xn-1 and if you sort of do a bit of diagram chasing on that it's not too difficult to find out you get a sort of long exact sequence where in general this would be an hi and an hi and an hi and here would have an hi-1 and it would sort of go on like that I'm not going to give the proof of this because most proofs of diagram chasing are very easy to do and very boring to watch somebody else do them so I'll just leave it as an exercise anyway the key point is here we're assuming the causal complex on x1 up to xn is exact this means the homology vanishes so in other words this term here would be equal to zero and let's see what consequences that gives well first of all when it gives two consequences the first one is that this implies that this map here is injective so if x1 up to xn-1 is regular so is x1 up to xn so that side of it reduces us to showing that x1 up to xn-1 is regular well to show that x1 up to xn-1 is regular we would like to show this term here is zero so we can ask is this zero well what information do we have about this well if this vanishes we know this map here is on to now for general rings the fact that we've got the map multiplication by xn is a surjective map on modules doesn't really tell us very much about the modules there's lots of modules where multiplication by some ring element is surjective however in the special case when r is local we have the famous Nakayama's Lemma which says that if m is a finitely generated r module and m times m equals m where m is the maximal ideal then m equals zero well let's go back and apply this to this module here so let's apply it to m being h1 of x1 up to xn-1 then xn from m to m is on to by assumption and xn is contained in the maximal ideal because for a local ring elements of a regular sequence would be in the maximal ideal so m times m equals m so m equals zero and m was of course just this module here so h1 of x1 up to xn-1 equals zero so by induction so using some sort of inductive assumption x1 up to xn-1 is regular which is what we wanted to prove so this completes the sketch of the proof that for regular sequences over a local ring the order doesn't matter okay I think that's enough about regular sequences for the moment the next lecture will be about the next type of local ring which are Goromstein local rings which is a slightly stronger condition than being Cohen-McCawley