 So this lecture is part of an online series of talks on homological algebra, and probably won't make a lot of sense unless you've seen most of the earlier talks. In fact, frankly, it may well not make much sense even if you have seen the other talks. So what we're going to do this lecture is look at the functor Tor A B, where now we're going to work over a ring I. So R is some ring. And what we want to do is to check basic properties. So the four properties we want to check are first of all, that it's well-defined. So I'll recall a definition in a moment. Secondly, we want to check its functorial in A and B. Thirdly, we want to check symmetry. It says Tor A B is Tor I B A, at least they're naturally isomorphic. And fourthly, there's a long exact sequence. So I'm not going to write out because I'm getting fed up with writing it out. So we'll start by recalling the definition of Tor. So you remember that over R equals Z, it's very easy. You take a resolution of or rather a presentation of A and then you tensor it with B and we get naught goes to B to the M, goes to B to the N, goes to zero. And this sequence here is exact, but this sequence here is not exact, not least because we've changed the A to a zero. And then we take the homology groups of this and the homology groups of this are Tor A B and A tensor B. Now, over arbitrary rings, we do much the same thing only more so. So we now take a resolution of A, only now we can't assume that this map is injected because a sub module of a free module is not injective in general. So we just sort of continue it like this and go on forever. And then as before we tense this with B, so we get B to the N naught goes to naught, B to the N one goes to that and so on. And then we take homology and these the homology groups of the groups Tor I of A and B. So as we did in the previous lecture. So the main difference between working over an arbitrary ring and working over Z is that instead of just having a presentation of A by two modules, you now have an infinitely long resolution and all the theorems we proved in the second lecture by working over the integers, we're now going to prove over arbitrary rings just by doing what we did in lecture two only more so. So first of all, let's check that this is actually well defined. You see the problem is that we've chosen a resolution of A, but there are lots and lots of different resolutions. And at first sight, it looks as if the results we get here is going to depend on which resolution we choose. So suppose we've got two resolutions of A. So here's one resolution of A, let's put that M naught, M one, M two and so on. And let's pick another resolution of A. So we have R to the N naught goes to R to the N one, R to the N two and so on. And there's obviously a map from A, which is just the identity map. And step one is we can find a morphism of these complexes. So complex is just a sequence of modules with a derivation D such that D squared equals zero. And we do this as follows. What we do is we first pick a map from R to the N naught to R to the N naught, making this commute. And it's easy to check you can do this because this is a free module. You can just have to define it on a basis in a suitable way. So that's step one. Then step two, we can find a map from this map group to this group making this square commute. And again, we just have to define it on a basis. And then we just continue like this. So we at least get a map from tour I of A and B to tour I of A and B, where this is computed using the first sequence and this is computed using the second sequence. So this map here is just going to be induced by these orange things here. So I guess I should have drawn this arrow in orange. So we can at least compare the two different definitions of tour. Well, now we've got another problem because this orange map is far from being unique. There are many different ways of defining this map here to make this commute in general. And there are many different ways of extending it. So suppose we take two resolutions of A and I'm getting rather bored of drawing these free modules. So I'm just going to draw them as points. So here we've got two resolutions of A and suppose we've got two maps between these resolutions. So the map from A to it is just the identity map which is fixed. And we've got an orange map between these diagrams and we've got a green map between these diagrams. And we're going to, we can compare the orange map and the green map by constructing a homotopy. So the homotopy is going to be this pink map going like this and it's traditionally called S. And if we call the this map F and if we call the green map G then the homotopy has the following key property that I'm getting my colors mixed up, sorry. F minus G is equal to SD plus DS where D is this map here. So what's going on here? I'm getting a bit short of colors but SD plus DS means you either go around like this or you go around like this. And F minus G is these two maps here. So it says the difference between these maps is got by going like this and then going like this and adding these. And what we do is we construct S step by step in much the same way that we construct this orange map. So we first define S on this group here in order to make this identity hold at this point. And then you can check, you can then define S on this map to make the identity hold here and you just sort of keep going like this. And this is one of the several things you should do yourself as an exercise to see what's going on. So this map S is called a homotopy from F to G. And it's a sort of algebraic analog of a homotopy between two maps in algebraic topology. Incidentally, you may have noticed the map S isn't unique either because you've got a lot of choice. And if you have two different maps, S say S1 and S2 they're related by a higher homotopy where this time we take a new sort of map going from that group to that group and from that group to that group and so on. And apologists all go kind of crazy thinking about these higher homotopies but fortunately we don't need anything beyond S this lecture so we will ignore those. And the nice thing about maps is if F and G are homotopic they induce the same map on homology. So here we've got two exact sequences and we've got two maps between them, the F map and the G map. And we can take homology groups of the top row and homology groups of the bottom row. So you remember the homology, the nth homology is equal to the kernel of D divided by the image of D and I'm being rather careless using D for two different maps because we take the kernel of this map divided by the image of this map. And this is easy to check because if F minus G equals SD plus DS of something means F of X minus G of X is equal to D of SX if DX equals naught. So if X is in the homology group then the difference between F and G is the image of D which means that zero in the homology group. Now you may be thinking to yourself that this is kind of stupid because if we look at this diagram that we wrote down it was by definition exact. So the homology of this row and this row is zero. So it's trivially true that F and G are the same map on homology because the homology groups are all zero. So what on earth is the point of this? Well, although this map, although the free resolution may be exact so is zero homology when we tense it with B so we're looking at the resolutions F naught, tense of B goes to F1, tense of B and so on. When we tense it with B the exact resolutions are no longer exact in general and the homology groups are now the torsion groups of A and B. So what we've shown is that the two homotopic maps, F and G induce the same map from the homology of the first complex which is the first core of A and B to the second core of A and B. So this is the tor computed using the resolution F and this is the tor computed using the resolution G. So what we've come up with is we have the first tor computed with the F's and we've got the second tor of A and B computed with the maps G and what we've done is we've constructed a map from this tor to that tor and then we can construct a map in exactly the same way from that tor to this tor and then you can check that the composite of these maps is actually the identity map on A and B because it's homotopic to the identity map of complexes. So tor A and B is isomorphic to tor A and B where this is computed using F and this is computed using G. So it doesn't matter which resolution you use to choose to compute tor the two tors you get with different resolutions are canonically isomorphic. So it's fairly safe just calling it tor I of A and B and not worrying about what the resolution is. Next we want to look at symmetry. So we can compute tor A and B in many ways. We can not only choose different resolutions of A but we can either choose a resolution of A and tense it with B. Well, we could also try choosing a resolution of B and then tensioning with A and then taking homology. And what we want to do is to show that we get the same results, whichever resolution we take. And so what we do is we take a resolution of A let's call it F naught goes to A, F1, F2 and so on. And we take a resolution of B. So here the FI and the GI are both free and these rows are both exact. And now how do we compare tensioning with B with tensioning this with A? Well, what we do is we take a tensor product, the two resolutions. And I'm going to need a big piece of paper for this. So this gets quite, well it's not exactly complicated but it takes up a lot of room. So I just did F naught, tensor G naught here and F1 tensor G naught and F2 tensor G naught. And here I put, let's try and get this the right way around F naught, tensor G1, F naught, tensor G2 and F1, tensor G1. And you can now see what's going on. We're getting this huge double diagram. And I'm getting a lot of board of writing out all these tensor products. I'm just going to put a lot of dots in to indicate what's going on. So we just get a sort of chessboard array of all these groups. Where all these maps are D. Incidentally, this is called a double complex and if I was doing double complexes properly, what you should really do is to put, is to sort of spray minus signs all over the place. So you might sort of put minus signs on all these ones and on all these ones or you might put minus signs on some of the horizontal arrows and no one can agree on whether you should put minus signs on the horizontal arrows or on the vertical arrows. Fortunately, for proving symmetry of tall, we don't actually need to put minus signs on any of these arrows that we can forget about this. I'm just mentioning this because in more complicated applications you usually do need to remember to put minus signs otherwise things don't work very well. Anyway, so we know the rows of this are exact because we're just taking exact sequence and tensioning with the free module and we know the columns are exact for exactly the same reason. And now these maps on onto because, so what we should do is we should just add a little bit at the end. So this maps on to A tense and G naught and this maps on to A tense and G one. This maps on to A tense and G two and so on. And similarly this maps on to F naught tense of B. I guess that's on two. And this goes to F one, tense of B and so on. And you've got to be a little bit careful because this row here is not exact. Let's point out that this is not exact by putting a pink circle around it. And similarly this column of course is not exact. So everything is exact except for the final column and the bottom row. And what we're trying to do is compare the torsion group using the resolution of A with torsion computed using the resolution of B. And what we notice is that if we look at the homology of this complex at this point, this is just Tor, this is just Tor two of A and B computed using a resolution of A. So this group here isn't Tor two. It's the homology of this sequence at this point which is Tor two. And similarly if we go up, where's it gone? Here I hope. So the homology of this is Tor two computed using a resolution of B instead of a resolution of A. And what we want to do is to get from this group to this group. And we do that by doing a sort of zigzag. So we go up to here and if we pick an element here we can lift it to here because this map is surjective and then we can take its image in here and the image in here is zero because the image of this element in here is zero because it's in the homology groups. So we can lift it to an element of this group and then we can take the image here and then the image of this is zero. So we can lift it to an element here and then finally we can take the image there. So we're going to zigzag pattern and you notice this zigzag is doesn't seem to be well defined because there was an ambiguity about this element because there's maybe more than one way to lift it. So it's ambiguous up to the image of something in here. And similarly, that's going to mean this element is ambiguous up to the image of something in here, but it's also ambiguous up to the image of something in here. So, and I guess this element is ambiguous up to the image of something in here, which is also the image of something in here. So all these orange circles indicate ambiguities that there are ambiguities in all these groups, which are the images of elements in these orange circles. For instance, the element here is ambiguous up to the image of something in this orange circle here. Anyway, this doesn't really matter because all these ambiguities just mean that this element here is ambiguous up to the image of something here, which is okay because this is a homology group which is only defined modulo the image of this. So the result of all this is that we get a perfectly well defined map from this tour group computer using a resolution of A to this tour group. And of course we can go back the other way and the composition of these is kind of obviously the identity. So these two tour groups are in fact canonically isomorphic and it doesn't matter whether you take resolutions of A or B. So the final thing we want to check is the long exact sequence. And as usual, I'm just going to give an overview of this and leave numerous details to anyone with a lot of spare time on their hands. So what we're given is we're given a sequence of modules A goes to B goes to C goes to zero. And we're going to be tension with M or rather taking tour of whatever it is with M. And we want to get some sort of long exact sequence out of this. So how do we do this? Well, the first thing we need to do is to choose compatible resolutions of A, B and C. So what we really want is we should have not goes to A goes to B goes to C goes to naught. And as usual, I'm getting rather fed up of drawing modules. I just draw them as points. What we want is exact sequences of free modules like this, such that all these vertical things are also exact. So we want the dots all free and the rows and columns exact. And it's not entirely trivial choosing compatible resolutions, but it's not that difficult either. So I'm going to sort of leave it as an exercise. So the idea is suppose you've got three resolutions off as far as that, you then check that the kernels of these three maps here form an exact sequence. And then you sort of splice together three free modules acting onto those kernels. And that's going to be these three modules. Then you sort of keep going. So anyway, I assume we've done that. And so we get this exact sequence. And now we're going to tense everything with M. And we're going to tense with M and we're going to throw away the bottom row. So what we end up with is something that looks like this. So we've got naught, naught, naught, naught, naught. And the rows are still exact. The reason for this is that all these rows were exact sequences of free modules. So they split. And if you've got a split exact sequence and you tensor it with M, then the result is still exact. However, the columns are very definitely not exact in general. So columns are not exact. And the homology groups of the columns are just tour of A, B or C with M by definition. So we've got the following problem. We've got this sort of three by N array of groups with the rows exact. And what we want to do is to show that we get a big exact sequence of homology. So the key step is we need to go from the homology of something here to something here. So what we want is we want to be able to go like that on homology. And to do that, what you do is suppose you've got some element in the homology of this column. Well, we can lift it to something here because this is onto. And then we can take the image there. And the image of that is zero because by definition, the image of the element here is zero. So it's in the image of there. So this orange arrow is really going like this. So we have a map from the homology of this element to the homology of that element. And it's not too difficult to check that it's well defined. So what's going on is if we draw the homology groups, there are natural maps like this, which are induced by the maps of these groups here. But there are also these snake maps. And it's pretty obvious why it's called a snake map. We've got a sort of long winding snake going like that. And what we want to do is to show that this sequence here is exact. So all these groups here are the homology of these homology groups of these columns here. And in order to show that as exact, you just do a lot of diagram chasing. And most of this I'm going to leave for anyone to do. I just sort of go through one of them as an example. So suppose you've got some element of this homology group, which maps to zero here. You want to show it in the image of something in this homology group. So how do we do this? Well, let's go up to this diagram. What we've got is an element of the homology here. So the image there is zero. And the image there is, well, it's not quite zero. It's only zero as an element of the homology group, which means it must be the image of something here. And so you take this element and take his image in here. And it's this element. You check this is in the homology, whose image is the element here that you first started with. So there are half a dozen little arguments like that. So what we've done is we've sort of checked it's exact at this point. You should also check it's exact at this point and exact at that point. And the argument in all three cases is kind of similar. So if we write out what these groups are, we see that this group would be tau i, a with m. This is tau i, b, m. This is tau i, c, m. And this one here is tau i minus 1, a, m, and so on. So this sort of long snake gives us the long exact sequence of tau groups. So that's a summary of how you prove the basic properties of tau groups. Next lecture we'll be looking at some generalizations of this. In particular, we will be looking at the so-called X groups, which are sort of related to homomorphisms of modules in the same way that tau groups are related to tensor products of modules.