 So this lecture is part of an online course on homological or commutative algebra and will be about injective modules. So we just recall that a module is called injective. If whenever you have A as a sub module of B and there's a map from A to I, then you can extend this to a map from B to I here. I is the injective module. So in the last few lectures, we calculated derived functors in several steps. So first, we take an injective or projective resolution. Secondly, we apply some functor. This is for derived functors of some functor F. And thirdly, we take homology. And there was a sort of gap in our description of this that we sort of skipped over a bit. And this was the assumption that you could find an injective or projective resolution. So you can find an injective or projective resolution fairly easily if you can solve the following problem. So given a module M, can we find, first of all, a map from a projective module onto M or can we find a map from M into an injective module? Well, the first of these is really easy. That's because free modules are projective and it's really easy to find a map from a free module onto M. This one is a bit tricky and it's what we're going to talk about. So the first problem we're going to talk about is given a module M over a ring, can we find an injective module containing M as a submodule? If you do this, you can easily find an injective resolution of M because you map M into an injective submodule I0 and you then look at I0 over the image of M and you map this into an injective module I1 and then you take I1 over the image of I0 and you map this into an injective by two and you just keep on going like that and that gives you your injective resolution. So given a module, can we find an injective module? Well, the first problem is it's not very easy finding injective modules at all. So let's look at the ring R equals Z. So what are the injective modules? And it's not completely trivial to find any injective modules at all over the integers. Well, for the integers, the answer turns out to be that injective is the same as divisible. This definitely isn't true over more general rings. It just happens to be true over the integers and for that matter, principle ideal rings. So divisible means that if A is in the module and N is in the integers N not equal zero, then A equals BN for some B in the module. So you can divide any element of the module by any non-zero integer, although possibly not uniquely. And it's quite easy to show that injective is the same as divisible. First of all, let's show injective implies divisible. Well, for this, suppose I is injective, all you do is you look at nought goes to N, Z goes to Z, and so N, if you've got an element A in here, you can take a homomorphism from NZ to A just taking N to A. And this extends to a homomorphism from Z to I and the element Z will have image some element B and we can then see that NB is equal to A. So injective implies divisible by looking at this injective sequence and mapping it to I. If we want to show divisible implies injective, which is the more useful criterion. Suppose A is a sub-module of B and we've got a map from A to I. What you do is we pick some element B in B, little B in big B and we let N be the ideal of integers X with XB in A. So here we have NB is in A and now NB has some image in here. So say it as image I and then we can find some element J with NJ equals I. So NB goes to I and now we can extend this to a map by mapping B to the element J. So this extends to a homomorphism from the sub-module generated by big A into B to I. And now what we just do is keep repeating and you should mention something about Zorn's lemma in order to show that we can keep doing this until we run out of elements of B to apply this to. So we found a map from B to I. So divisible implies injective at least for the integers. And now it's easy to see that Z has enough injectives. What this means, but by saying enough injectives it means for any module we can embed it into an injective module. And all we need to do for this is to notice that Q over Z is divisible, so is injective. And now given M, if we pick some element M in M we can map M to some element not equal to zero of Q over Z. So let's say M is not zero. And that's because Q over Z has elements of every positive finite order. So if M has some finite order you can map it to some non-zero element of that order. And if M has infinite order you can map it to anything non-zero in here. So for any module we can find a map from any given element to Q over Z and we can then extend to a map M goes to Q over Z, where M goes to something non-zero. Well, obviously this map need not be injective yet but what we can do is we can map M to a product of copies of Q over Z for all M in M and a product of injectives is injective, that's very easy to check. So this is injective and we can map M to this by for each element of M we map it to something non-zero in one of these many products of Q over Z. So that gives you an injective map from M to some injective module. So summarized this shows the integers have lots of injectives and you may get the idea from this proof that the injectives over Z are quite complicated. I mean, we had to take these infinite products of things. Well, near the end of this lecture I'll explain that actually the injectives over notarian rings at least are reasonably easy to classify. But anyway, let's discuss injectives over a more general ring R. Well, fortunately we can reduce this to the case of injectives over Z by the following trick. This trick is in some ways very easy and otherwise it's very confusing because we're going to take HOM over the ring R here and this on the other hand is homomorphisms over the ring Z. And what I want to say is that this is more or less the same as HOM over Z from M to I. Here M is an R module and I is Z module. Now, this is actually rather trivial to check. As I said, it's just rather confusing because you've got to remember sometimes you're taking HOMs over R and sometimes you're taking HOMs over Z. And from this, we can now show that HOM over Z from R to I is an injective R module if I is an injective R module. Z module. Of course, we make this into an R module by using the R module structure of this bit here. And the reason for this is that if we've got nought goes to A goes to B and we've got some sort of and we've got the R module HOM over R from R to I, we're trying to find an extension from B to this in order to show that this is injective. Here A and B are R modules. Well, what we do is we look at this. So here A and B are considered as R modules. And here we just forget about the R module structure and think of A and B as Z modules. And then R module maps from A to this the same as Z module maps from A to I which as I is injective we can extend to a map from B to this. And that corresponds to an R module map from B to this R module. So this shows that this thing here is injective. So now we have lots of injective R modules because we can just take any Z module and this is, and this gives us an injective R module. So this gives us lots of injective R modules and using these R modules, injective R modules it's very easy to show that any R module is a sub module of an injective R module. And I'm going to skip this bit of the proof because it's very similar to what we did for the integers. So this more or less shows that there are lots of injective modules over rings. Well, so what do injective modules look like? Well, it turns out there's a really nice property of injective modules that any module is contained in a smallest injective module and which is in turn contained in any other injective module. Notice there's a note that this doesn't work for projective or free modules. So this fails for free or projective modules. For instance, if we've got the module Z modulo 5z over the integers, we can map Z to Z over 5z in at least two different ways. Here we can map one to the element one or we can map one to the element two and this gives two maps from a free module onto Z modulo 5z, but neither of them are smaller or bigger than the other in any sense or way. So there's no minimal projective module mapping onto a module but there is a minimal injective module that it maps into. It can be defined a little bit more precisely. So if we've got modules A contained in B, this is called an essential extension of A. If any non-zero submodule is contained of B has non-zero intersection with A. So if we had a submodule of B with zero intersection with A, we've sort of quotient out by that and get a smaller module containing A. So an essential extension is in some sense an extension you can't make any smaller. And we say, North goes to A, goes to I is called an injective envelope of A if first of all, I is injective. Secondly, A contained I is an essential extension. And what we'll show is that any module has an injective envelope and it's this injective envelope which is a sort of minimal injective module containing A that we can map it into any other injective module containing A. So how do we define, how do we construct this? Well, let's first of all do it over the integers. This is really easy. What we do is we have North goes to M. We pick an injective module containing M with I injective and then we pick E maximal contained in I with E intersection M not equal to zero. And we should invoke Zorn at this point because we're probably using Zorn's lemma somewhere. And now we just form North goes to M goes to I over E. So this is now an essential extension of M. And it is injective, it, when I say it, I mean I over E is injective as it is divisible because it's a quotient of a divisible module and over Z divisible implies injective. So over the integers it's easy to show that every module has an injective envelope. So what do these look like? Well, here's a couple of videos. So let's look at some of them. So what do these look like? Well, here's a couple of examples. So if we've got the integers, this is contained in the rational numbers and this extension is essential and this is divisible. So it's the essential extension of Z. What about Z modulo two Z? Well, this is a little bit trickier to find. It turns out the essential, the injective envelope of Z modulo two Z is the integers localized at two modulo Z. So this is isomorphic to the union of group Z modulo two Z which is contained in Z modulo four Z which is contained in Z modulo eight Z and so on. You notice that neither of these groups are finitely generated and in general, injective modules tend not to be finitely generated. I mean, they can be but they're usually not which makes them a little bit tricky to work with at times. So now let's see how to construct an injective envelope for a general ring. Well, first of all, we're going to need a lemma which says that if a module I has the following property, every essential extension I, contained in J is trivial by which I mean I is equal to J, then I is injective. And let's see why. Well, suppose we've got an extension nought goes to I goes to M. What we do is we pick maximal sub module E, so of M with E intersection I equals nought and again, we sort of have this ritual invoking of Zorn. And now we look at nought goes to I goes to M over E. If this is proper, I not equal to M over E, then E was not maximal. So I must be isomorphic to M over E, so M splits as I plus E. So any extension of I splits and this easily implies that I is injective. For instance, you can embed I into an injective module and that splits so I is a summand of an injective module and is therefore injective. So that's the lemma. If every essential extension is trivial, then the module is injective and this is what we're going to use to construct an injective envelope. So now pick a module M and let's choose nought goes to M, goes to I with I injective. And what we do is we pick E, contain I to be a maximal essential extension of M and let's write down Zorn as usual. So what we have is nought goes to M, goes to E, goes to I. And we want to show that E is injective. And for this, all we need to do is to show that E has no proper essential extension. So suppose we choose an essential extension E prime of E. So E contain E prime is essential. Well, now since I is injective, we can define a map F here. So F exists as I is injective. And now first of all, we notice that the kernel of F is not just zero as E is a maximal essential extension. And if the kernel was zero, then this would be a bigger essential extension of M. On the other hand, the kernel of F intersection with E is equal to nought as E is contained in I. So this means that E contained in E prime is not essential. So any extension of E is not essential. So E is injective. Well, as E was chosen to be an essential extension of M, this means that E is an injective envelope. So we've shown that an injective envelope exists. Now let's show that injective envelopes are unique up to isomorphism. So suppose we've got two injective envelopes. So we've got M and it maps to I and it maps to J and I and J are injective and M contained I and M contained J are both essential. And we want to show from this that there's a map from I to J that's isomorphism. And we do this in four steps. First of all, we can find F mapping from I to J. So here's a map F. And why can we do that? Well, that's because J is injective. So the green is the condition that we're using for this. Secondly, we show that F is injective as the kernel of F intersection M is equal to nought. And now we use the fact that I is essential. Rather M contained in I is essential. So if the kernel of F intersection M is zero, that means the kernel of F must be zero. So thirdly, we know I is contained in J. So J is equal to I plus some complement E. And this follows because I is injective. So if an injective module is a sub-module of another module, then this splits. Finally, we notice that E equals nought because J is essential. Or rather M contained in J is essential. So we have four conditions, I and J are injective and I and J are both essential extensions of M. And we've used these four conditions in our proof which is just as well because if we hadn't used one of these conditions, something would be wrong. So F is an isomorphism. So any two injective envelopes are unique. I guess I haven't quite proved that any injective envelope is, so that any injective sub-module containing M contains the injective envelope but I'll leave that as an exercise. I should have a warning, any two injective envelopes, M goes to I and M goes to J isomorphic but the isomorphism is not unique in general. It's quite easy to come up with examples of this. For example, if we take Z modulo two Z, this is contained in say Z localized at two modulo the integers and this has an automorphism just multiplying everything by minus one which is the identity on this map M here. So the injective envelope kind of non-trivial automorphisms which all attribute on M and this makes the concept of the injective envelope a bit iffy because this isomorphism isn't unique which makes it a bit difficult to pin down. It's kind of like the algebraic closure of a field. Every field has a algebraic closure that's unique up to isomorphism but the isomorphism isn't unique which makes it a bit difficult pinpointing elements of the algebraic closure of a field. I mean, the injective envelope has non-trivial automorphisms in general. So in particular any module M now has a canonical minimal injective resolution because we can take the injective envelope of M and then take the injective envelope of I mort over M and so on. And if we've got any other resolution of M there's a map from the minimal injective resolution to any other injective resolution. Of course the map from the minimal injective resolution to this other resolution need not be unique there may be two different maps but as we saw last time if we have two different maps they are at least homotopic. So there's a unique homotopic class of maps from a minimal resolution to any other resolution. Okay, I think that's quite enough about injective modules for the moment. Next lecture will probably be something about derived functions of direct and inverse limits.