 This lecture is part of an online course on commutative or homological algebra and We will be talking about the groups tau i r of a b Where r is a ring and a and b are r modules If r happens to be non commutative then a should be a right r module and b should be a left r module But we usually do the commutative case. So we're not worried about that too much so in the first couple of lectures we calculated this group when r was the ring of integers and In the case of integers these are none zero only for i equals zero or one and you've generally missed out the subscript i So first we give the definition and this is similar to the definition for the integers only more so So for the integers we took a resolution of a by R to the n naught and R to the n1 now over the integers if we take a free module then in the sub module of a free module is also free So if we want we can stop the resolution here for more general rings. We have to keep going If we want the sequence to be exact And so on so that we have a possibly infinite sequence here where where this is exact So the image of each map is the kernel of the next map So that's step one step two is we tensor this with b So we get b to the n3 goes to b to the n2 goes to b to the n1 goes to b to the n0 and we drop out the a and In fact we put a zero there That's not a misprint. I really am missing out the a and the third step is we take the homology groups of this sequence So so if we call this map d and we're taking the homology of the operator d which means the We're looking at the groups where we take the the kernel of b to the n i to b to the n i minus one and Quoting it out by the image of b to the n i plus one goes to b to the n i So this is the ith homology group of This and by definition it's equal to tour i are of a and b So just as the integers we have several questions first of all, is it well-defined In other words, is it independent of the choice of resolution? Secondly, we want to show that it's a functor in a and b Thirdly, we want to show it's symmetric in other words Tour i a b equals tour i b a I'm missing out the superscript r out of laziness So The fourth condition the very persistent ant So the fourth condition is the long exact sequence So if we've got a short exact sequence of modules nought goes to a goes to b goes to c goes to zero The ant is back again Either unusually Stupid ant were unusually interested in homological algebra. Anyway, um, so from this we get the exact sequence of modules and As for the integers, it's not necessarily exact on the left We can fix this with the first tour group and then we get a tour b with m and Tour 1a with m, but then this map here Um, need not be exact and we can fix that by putting a tour to see m and We just sort of keep going infinitely long So I'm not going to prove any of these this lecture although I probably do so next lecture What I'm going to do instead is give some examples of tour So we will first give some examples of tour groups turning up in other areas of mathematics and Then give some examples of calculations So example one There p and says definition of Intersection multiplicities So if you've got two Sub varieties of an algebraic variety they might intersect for instance You might have a curve here and a curve here intersecting in this point and you can see in some sense You should really count. This is having intersection multiplicity two that are sort of two Points of intersection of the same place and it was quite a problem for a long time to define it an analogous Intersection multiplicity in higher dimensions and say I came up with this rather nice definition It's just sum over I have minus one to the I of tour I R over I R over J except I guess we should really take the length of this So here R is a local ring and I and J are ideals Corresponding so R would be the local ring of this point here and I and J would be ideals corresponding to some varieties meeting at that point So a second example Would be homology of a group So if G is a group then Island Bergen McLean to find its homology groups HIG with coefficients in a in a in a representation M of G and their original definition was rather somewhat round about but it turned out to be the same as tour over I of Z and M Where the ring here is the group ring So homology of groups is a special case of Of the tour factor a similar example is homology of Lee algebras So again, we can define the homology group of a Lee algebra a with coefficients and some module and this turns out to be Tour over I of UG and a And Talk live UG. So I'm able to R and where R is R is the reals So here UG is the universal enveloping algebra of The Lee algebra a so we can turn the homology of Lee algebras into a problem about tour of for modules over a certain ring example four is Hosh child Homology of an algebra so the hosh child co-homology of an algebra a With coefficients in the by module just turns out to be tour over I of a tensed with the opposite algebra Of a with coefficients in the module M so So these three examples homology of groups homology of Lee algebras and Hosh child homology were originally defined independently in different rather ad hoc ways and it was afterwards noticed that they were all special cases of the tour of modules over over a suitable algebra So tour the tour factor unifies three apparently different theories so now we'll come on to some calculation of Examples of tour groups if you look at the definition of tour it looks kind of really complicated and roundabout and Difficult to use but in fact in practice. It's usually rather easy to calculate tour groups So the first example we're going to take our to be The polynomial algebra over a Field in one variable only to make it easier. We're just going to quotient out by X squared So R is a two-dimensional vector space which has a base one and X and We're going to take the module M to be K Which we think of as an R module because it's our modulo the ideal X and What we're going to do is we're going to compute tour I over R of K with itself So how do we do this? Well, we look at this module here and we have to take a resolution of it so so let's take a resolution of K and Then we must map R to K and We can take the element one to one obviously and that maps on to the kernel of this is Obviously just the one dimensional vector space spanned by X. So we can make this exact by mapping one to the element X Now the kernel of this map is Again the element X which maps to X squared which is north so we can make the sequence exact by mapping one to X and We're obviously in a sort of loop here. We just keep going I'm just always mapping one to X. So so we get a resolution of K by Free our modules all of which have to be one dimensional So that's the first step. We take a resolution of K. The second step is we tensor with this module here that we're we're Calculating tour of well if we tensor R with K We just get K. So we get our exact sequence looks like our sequence looks like this And now we have to figure out what the maps are well this maps one to X which is zero in case of all these maps are just zero and Thirdly we take the homology of this complex. Well, since all the maps are zero the homology is obvious. It's just K So this is tour zero K of K. This is tour one K and K This is tour two and this is tour Three and so on so so we see that tour I Okay, and K is equal to K for all I greater than zero In particular this shows for example that there's no Finite resolution of K by free modules because if we had a finite resolution then all these groups would be zero Once we got past the length of that resolution, of course It's very easy to show there's no finite resolution without using tour in this particularly simple example So the next example Let's look at The ring R Which is now going to be polynomials in two variables And I'm going to take the module K zero zero to be R over XY That's a little bit careful here because K can be an R module in many different ways We could have K a B to be our modulo X minus a Y minus B So we can make change from module in For each point of the two-dimensional affine space over I anyway, let's start with this K first now We have to first have to find a resolution of K zero zero and the easiest way to do this is to draw a picture So let's draw a picture of R Right then one X X squared X cubed and so on then we have Y Y squared Y cubed XY X squared Y XY squared and so on and now We can map R onto K and the kernel Will be all this stuff here And the kernel is generated by X and Y so we can find Two free R modules a green one which is one dimensional another one dimensional one spanned by X and then these two modules Mapping on to this orange module will have a common kernel which is this module here so The picture of the resolution looks like this and now let me write it out more algebraic terms We've got this orange module mapping on to K and Then we've got this green free module R plus R where This element here one zero Maps to say X and zero one maps to say why and Then we've got this pink R module where we're mapping the element one to Say Y minus X and Finally It's easy to see that this sequence is exact. So here. We've got a resolution of K of length three And now we can tensor this With K zero zero so we get naught goes to K goes to K squared goes to K goes to naught And as usual we we sort of cross off this term here And we have to figure out what these maps are here. Well since X and Y are zero in K these maps are all zero So if we take the homology We just get K K squared K So this is tour zero This is tour one And this is tour two And so the fact that tour two is none zero shows that we couldn't have found a resolution of length less than this and The other thing we can do is Compute tour of K zero zero with one of these other modules here So let's try that. So we've got this resolution of Of R sorry of K which takes R goes to R squared goes to R goes to K goes to zero So you remember this took one to Y minus X and it took one zero to X and zero one to Y and we're now going to tensor this with say um Let's just do um K zero one just to be explicit and then when we tensor it with K we get K Goes to K squared goes to K However, these maps here are no longer zero because This is taking one to one zero And this is taking one zero to zero and zero one to one So in fact we see that this sequence is exact So if we take homology We get zero zero zero and here we see that um tour I of K zero zero with K zero one Is equal to naught for all values I And in fact you can see that if we take two of these modules corresponding to points in the plane Then these tour eyes will have the dimension one to one if these are the same points And will all be zero if the points are different um as an exercise you can do the same thing for a polynomial algebra in three variables not two variables and then you find you should find there's a tour three That is none zero So for the last example I'm going to calculate the homology Of a cyclic group of order n with coefficients in Say z So this is going to be group homology And we remember that this is defined to be tour um I of the group ring of z over n z of z over n z And z so I've got to calculate this well first thing is to figure out what the group ring is well suppose z over suppose um That that's right. We want to write a cyclic group of order n multiplicatively So let's take the group ring for a basis of elements one g G squared up to g to the n minus one with g to the n equals one So that gives us the group ring Now we have to find a resolution of um z over this group ring So we have to find something mapping on to z Let's take group ring g and we can map this on to z just by mapping the element one to one And the element g to the i will then also map to one because In case I didn't say so I'm taking the group g to be actually trivially on z Now we need to figure out what the kernel of this is Well, the kernel is generated by one minus g That's generated as a as a as a module over group g because one minus g obviously maps to one and it's Sorry to zero and it's easy to check that conversely anything mapping to zero must be Generated by this so we can make this sequence exact by Um mapping something onto one minus g. So let's just take the element one of this and map it to one minus g Now what's the kernel of this? Well the kernel It's not difficult to see is generated by one plus g plus g squared plus plus g to the n minus one so We can make this exact by taking another copy of z g here and mapping one to one plus g plus g to the n minus one And What's the kernel of this? um Well the kernel Is now generated by one minus g as you can easily check So let's continue We now map one To one minus g and now you see we're going around in circles because here We've got one mapping to one minus g and here we've gone one one to one mapping g to one minus g So the sequence just extends indefinitely to the left with with period two um now The next step is to tensor this sequence with z So we get z goes to z goes to z goes to z Etc goes to zero now We've got to work out what these maps are well This map here You remember it takes one to one minus g which is actually equal to zero in g. So this map is zero This map on the other hand takes one to one plus g plus Plus g to the n minus one and as g is equal to one this map here is multiplication by n Similarly, this map takes one to one minus g. So it's just the zero map so our map Is this is zero. This is multiplication by n. This is zero and so on and and since it's periodic This is also multiplication by n Now we take the homology groups Well, that's easy to work out. So the homology here is just z um the homology here is Z modulo n Z because the kernel is the whole of z in the image of the multiples of n Um here on the other hand the homology is zero because the kernel of this is zero And here the homology is z over n z And so on so so this is The zeroth homology of z over n z with coefficients in z This one here is the first homology Of z over n z with coefficients in z. This is the second homology group This is the third homology group and so on So the homology of a sick clip group at least with coefficients in z Is periodic apart from the first term, which is a little bit odd In fact, it's pretty obvious the homology with coefficients in any module Will have period two apart from the first term because the the resolution of z was was periodic Um, okay, that's enough examples for the moment So next lecture will probably be on checking the basic properties of Tor over a ring such as the long exact sequence and the symmetry and so on