 This lecture is part of an online course on commutative algebra and will be about tensor products and exactness. So last lecture, we remember we had this following problem. If we've got a sequence such as nought goes to z goes to z goes to z modulo 2z goes to zero. And we tensor it for z modulo 2z. We get the sequence nought goes to z tensor z over 2z goes to z tensor z over 2z goes to z over 2z tensor z over 2z goes to zero. And this is z over 2z. This is z over 2z. This is z over 2z. And the problem is here that we can't really put a zero here in because this map here is multiplication by two and is not injective. So there's something wrong here. And there are two similar problems, because as well as tensioning with z modulo 2z, we can try taking say the homomorphisms from z modulo 2z to something. So what we're going to do is to take homes from z over 2z to this to this to this and see if we get an exact sequence like that. So let's try doing that. We'll get nought goes to hom z over 2z to z goes to hom z over 2z to z goes to hom z over 2z to z over 2z goes to zero. And let's examine this. Well, this is zero and this is zero. So that's fine. This map is injective. The problem is, this is z modulo 2z. So this map here is not onto and we'd better cross off that zero because it's not exact on the right. Now, as well as taking homes from z over 2z to something as a font, we'll also take homes from something to z over 2z. So we can take home of z to z over 2z and the same there. So let's write that out so we get home z to z over 2z and home z to z over 2z and home z over 2z to z over 2z and then we get zero. Now we have to be a little bit careful about which direction the arrows go in because the arrows actually go in this direction. So you mark them in reds. You notice there's something a bit unusual. And the reason for that is that if we've got a homomorphism from A to B and we look at the homomorphisms of B to X, then we can get a homomorphism of A to X by composition. So we get a map from home B to X going to home A to X and notice this is in the opposite direction. We've got a map from A to B here, but the map from homes from B to homes from A goes in the other direction. So all the arrows get reversed and this isn't incredibly confusing and I keep getting it wrong. Anyway, let's see what's going on here. So this is just z modulo 2z. This is z modulo 2z and this is z modulo 2z. So it can't be exact everywhere. And if you look again, this map here is multiplication by 2. So again, this is not onto and we'd better cross off this thing here. So in each case, if we start with an exact sequence and apply one of these three operations, we get something that's almost but not quite exact. It sort of fails at one end or the other. So operations like this that preserve exactness on the right are called right exact and operations like this that preserve exactness on the left are called left exact and operations like this I have no idea where they're called right exact or left exact because the trouble is the problem here is it's on the left of this sequence. But on the right of this sequence, well, it looks as if it's on the left of this sequence, but you remember I reversed all the arrows. So if I had the arrows going conventional direction be on the right. So I don't know whether this is called left exact or right exact. So what we're going to do now is just quickly check the sort of exactness properties for these two operations and then we'll use that to get exactness for tensor products, which is a little bit trickier. So suppose not goes to a goes to be goes to see is exact. And you notice that I'm not putting a note here on the right. Then we want to show that the following sequences also exact not goes to home of m to a is to home of m to be goes to home of m to see and again there's no zero on the right. And this is quite easy. Suppose we take f in home m to be what we're going to do is to show if something here has image zero there, and it's in the image of something there, and all the other bits of exactness even easier. So the image in home m to see is zero. This means the image of f is zero in C. So the image of f is in the image of a in B so f is really a map from m to a if we sort of pretend that a is a sub module of be a isn't quite a sub module of be but it's isomorphic to a sub module of be and usually we're a bit sloppy and pretend that it is a sub module. So f is in the image home from m to a. So there's nothing difficult about that proof. Now we're going to do the same thing for the other sort of home. So this time we take a to be to see goes to naught is exact and we want to show that home from a to m home be to m home seat m zero is exact and of course the arrows are going in the wrong direction. So now we take f in home be to m and suppose the image is not in home a to m well this means F vanishes on a where we pretend that a is a sub module of be so f is really a home from the modulo the image of a I guess the image of a is a sub module of be which is equal to C so so f is really in home from C to m so it's the image more precisely the image of something in home of C to m so those were both completely routine the proof of exactness for tensor products is a little bit trickier so suppose that a goes to be goes to see goes to naught is exact so we want to show m tensor a goes to m tensor b goes to m tensor C goes to zero is exact and what we're going to do is we're going to use the two previous results so first we pick any module x and we notice that home from a to x home be to x home C to x naught is exact and again the arrows are going the wrong way and now we're taking homomorphism from m to all of these so we get home m to home a to x and we get home be to x and home m to home C to x and naught there and now this is equal to home m tensor a to x and this is equal to home m tensor b to x and this is equal to home m tensor c to x so this sequence is exact now why do we get these identities here well the reason is that both of these are really just bilinear maps rather than the spaces of bilinear maps from m times a to x so the linear map from m tensor a to x are more or less by definition of the tensor product these these bilinear maps and for this one you notice that a bilinear map from m times a to x and we thought of the linear map taking element of m to a linear map from a to x so these space two spaces are canonically isomorphic by the way this is an example of something called adjointness so in category theory what you would say is the functor taking tensor products with m is left adjoint to the functor sorry tensor point with a is left adjoint to the functor from a to something or other which which essentially just means that homomorphism from m to this are the same as homomorphisms from this to x now in category theory functors that are left adjoint preserve things called co-limits and the quotient is a special case of a co-limit so the fact that this tensor product is left adjoint can be used to show that it's automatically right exact using a bit of category theory and what the argument here is really is really doing is just sort of giving the category theoretic arguments in this particular case so anyway we can now finish it off so we've got this map here and now we can observe that this says hom's from m tensor c to something are the same as hom's from m tensor b to something rather vanishing the image of m tensor a and what this is saying if you think about it a bit you'll say this is really more or less just saying that m tensor c is really the quotient of m tensor b by the image of m tensor a so saying that the homomorphism is something of the same as the homomorphism from thing 2 to thing 3 is just a way of saying this is the quotient of thing 2 by thing 3 so this gives us the right exactness of the tensor product just says that we have this exact sequence m tensor a just m tensor b just m tensor c just 0 so tensor products are taking tensor products has the right exact although it's not quite fully exact now we can use this to calculate tensor products by the way everything I've done here I'm really working over an arbitrary ring r so I'm taking tensor products over r and the homomorphisms or homomorphisms over the ring r but as you see these formulas are quite tedious enough to write out without putting all these subscripts as well so how do we calculate tensor products I suppose I want to calculate a tensor m well what I can do is I can take a map from a free module onto a and I can take a map from some other free module onto this and in general there's no reason why a submodular for free module should be free so this might not be injective well now I can just tensor this with m so I get r to the m tensor m goes to r to the n tensor m goes to a tensor m goes to 0 well a free module tense with m is just m to the m this maps to m to the n just a tensor m goes to 0 so there we've got a tensor m written as an explicit quotient of two modules and if you know enough about a and m you can use this to work out what the tensor product is so tensor products are quite highly computable and there's another useful property of tensor products for computing them which is that tensor products commute with direct limits this is actually another special case of the fact that tensor product is left adjoint to another function but we'll just show it by hand so what is a direct limit well I just give a special case of a direct limit suppose you've got modules a1 mapping to a2 mapping to a3 and so on then we can form a direct limit which is denoted by this and it's given by the union of all the ai except you have to quotient out by the following relation we say ai so this is in ai is defined to be the same as aj in aj if aj is the image of ai under one of these maps so if all these maps are injective we can pretend that each ai is a sub module of ai plus one and if we do that the limit really is the union but in general these maps might not be injective so we have to be a little bit more careful about it than we would expect well it's quite easy to prove that tensor products preserve exactness because you can see that bilinear maps from the direct limit of the ai times m to x are the same as a direct limit of bilinear maps from ai times m to x and by converting this into tensor products this shows that the direct limit of the ai tense with m is the same as direct limit of ai tensor m brackets around to make it clear what you're doing so let's have an example of this let's work out the tensor products of q times q and I'm going to work over the integers here so we can write q is a direct limit of z mapping to z mapping to z mapping to z and here these maps are going to be times 2 times 3 times 4 and so on so we can think of this as being z this as being all things of the form n over 2 this is the form all things over n over 24 and so on so you can see that q is just the union of all these modules here so now let's just tensor it with q well we tensor z with q we get q maps to q as multiplication by 2 and then we get multiplication by 3 and so on and these are all isomorphisms it's kind of obvious the direct limit is just q so q tensor over z with q is just the rationals well that's fine but there's a nasty trap that one can easily fall into if you're not paying attention let's try and work out the tensor product q tensor with z modulo 2z so and let's copy the previous argument so we write q is the direct limit of all these maps z maps to z maps to z maps to z and so on and now let's tensor with z modulo 2z so we find q tensor with z modulo 2z is the direct limit of z modulo 2z maps to z modulo 2z and so on and this since they're all the same this direct limit is obviously z modulo 2z right well no this is completely wrong to cross it out the problem is that we were being a little bit sloppy here so all these maps are injective so there's no harm in no great harm in missing them out however these maps here are not injective in general well they sometimes are so they're not always injective so you remember the problem with taking a tensor product is taking tensor products doesn't necessarily preserve injectivity so this is multiplication by 2 by 3 by 4 and so on so this is actually multiplication by 0 by 1 by 0 and so on so every second map here is actually the 0 map and this means the direct limit is actually 0 because if we take any element here for example well it's nothing much happens to it if we pass to here but if we then go to here it becomes a 0 element so every element in one of these z modulo 2z eventually becomes 0 so the direct limit of all these modules is 0 not z modulo 2z as you might think if you were being biased okay so next lecture we will be discussing the relation between tensor products and localization and we'll be introducing flatness