 This lecture is part of an online course on commutative algebra, and will be about projective modules. Rather, it will be about the relation between projective modules and locally free modules. So let's just recall what both of these are. So locally free modules we defined last lecture, and these sort of correspond to vector bundles. So a vector bundle is a map from something to a space whose fibers are roughly speaking vector spaces. In the case of rings, we saw that a locally free module correspond to this is a module such that if you take a ring R and you cover it with open sets, cover its spectrum with open sets, spectrum of R i to the minus one, then M fi to the minus one has to be free over R fi to the minus one. And these open sets have to cover R, which means the ideal generated by all these elements is equal to the whole of R. So a projective module, module P is one with the following property that whenever you have a map from a onto a module B and a map from P to the module B, then it lifts to a map from P to a making this diagram commute. So some basic properties of projective modules, all of which are easy to prove is first of all free modules are projective. Very easy to check that free modules have this property because you just take a basis for the free module look at its images in B and lift those to elements of a that gives you a map from the free module to a. Secondly, it's also very easy to check that if P is projective, if P equals X plus Y is projective, then X and Y are also projective. So any sum and other projective module is projective, a sub module of a projective module is not projective in general. Similarly, any direct sum of projective modules is projective. Again, this is very easy to check and I'm not going to bother doing it. So we now have the following problem. Projective modules the same as locally free modules or more generally are projective things the same as locally free things so let's ask the question does is projective the same as locally free. The answer is yes and the other answer is no and which answer you get depends who you ask. So if you ask an algebraic geometry, they will tell you that there are lots of examples of locally free things. Projective if you are someone doing commutative algebra, they will tell you that these that locally free things are indeed equivalent if you are someone doing differential geometry. They again think that locally free things are projective. If you are someone doing complex geometry, say complex analytic geometry. But locally free things need not be projective. This is very confusing because quite a few people have kind of started off in one area and switched to some other area and got very confused about whether or not locally free things are projective. I want to try and explain why is there a difference and the problem is I'm not really going to be able to give a terribly satisfactory explanation because it really uses a certain amount of co homology so I'm just going to have to quote some facts from co homology So the basic thing is, let's talk about whether locally free things are projective in algebraic geometry. What you do is you get an exact sequence which looks like this. I'll explain this in a moment. So here A and B might be modules or they might be sheaves over a scheme or something if you're an algebraic geometry. So the basic thing you get here looks like this. So this X is something in homological algebra that I'll explain later and I sometimes write in capital letters and sometimes small letters because this X tier is supposed to be a shorter sheaf X and this X tier is just an Abelian group X. And this thing is a piece of the growth index spectral sequence of a composed funk to which you probably don't want me to explain in detail. Anyway, the key point is this X tier vanishes if a is locally free as a module or a sheaf or for that matter also vanishes if this is even locally projective. So this thing here. It vanishes for all. Or B if and only if a is projective. So this bit vanishes if a is locally free. And this bit. You want to vanish if a is projective and this is an exact sequence. So locally free things are projective. If this thing vanishes. Then locally free implies projective. And home of a to B is just going to be a module or a sheaf or whatever so I've got the question. Does this first co homology group, whatever that means vanish for all modules and the answer is yes. If you're doing commutative rings, or if you're doing smooth manifold. Why does it vanish if you have a commutative ring or a smooth manifold. Well it turns out that this vanishes. If you can define partitions of unity. So partition of unity means if you've got a smooth manifold covered by open sets. Can you write one as a sum of functions, each which is supported one of these open sets. If you're doing commutative rings and smooth manifolds. There is a sort of partition of unity. And this implies that this group here always vanishes and this implies that locally free things are projective. On the other hand, if you do algebraic geometry or manifolds, then in general you don't have partitions of unity. And this first co homology group that I was talking about doesn't necessarily vanish. And there are plenty of locally free objects that aren't projective. So this is a sort of explanation of why there seems to be so much confusion in the literature about whether triangles or locally free objects are projective. Sometimes they are and sometimes they're not. In fact, if you've done algebraic geometry, you may have noticed that even free things need not be projected which is a bit disconcerting, if you come from commutative algebra. It's not logical at all. It occurs for quite natural examples, even if you're working over one dimensional projective space, you take the free one dimensional module, so free one dimensional sheaf, that's not projective. Well, this is really a course in commutative algebra so we ought to be talking about the ring theory case. Locally free does indeed imply that a module is projective. And we can ask, does the converse hold? So does projective imply locally free? And it does quite often. So the answer is yes for finitely generated modules. And the answer is no in general. So what I'll do is I'll first quickly sketch why it's true for finitely generated modules and then give an example of a non finitely generated module where it's not true. Let's just quickly sketch the reason with a number of details missed out because I'm feeling kind of lazy. So if we've got a finitely generated projective module, it's quite easy to see that it's finitely presented. And with rather more effort, you can see that the stalks are free. So what does the stalks mean? Well, if you've got a module m, the stalks mean the localization of m as a module over r of p, here p is a prime of r and this is the local ring at p. The reason why the stalks are free is that projective modules over local rings are always free. That's actually a theorem of Kaplan ski, and I might talk about this later then again I might not. If you've got a finitely projected module then it's finite presented and the stalks are free. On the other hand, if you've got a finitely presented module and the stalks are free. This implies that it's locally free. And again, this is something I might prove later when we talk about modules with free stalks if I feel in the mood for doing this. The reason I'm not worrying too much about whether locally free implies projective is that in practice locally free modules that we come across are always obviously projective for instance they might be stably free and stably free modules are obviously projected because they're summands of projective modules so it's kind of nice to know this result but we never really need it very much. So, now I want to give an example of a projective module that is not locally free. And here we're going to take the following ring R, which is all functions from an infinite set X to a field with two elements. This is an example of a complete Boolean algebra. Complete Boolean algebras aren't actually used that much in commutative algebra. They tend to have some rather weird properties. They used very heavily in set theory and that they used a lot in Cohen's notion of forcing, which is used to construct lots of models of similar set theory with various weird properties. Anyway, so you can think of this as being a function from an infinite set X to Z over to Z, which is the same as a product of an infinite number of copies of Z over to Z and what this infinity is I don't really care can be countable uncountable anything. So if we take our module to be the following ideal, it's just going to be the functions or finite support. So you can think of an element of R as being an infinite set of zeros and ones with usually infinite number of zeros and ones whereas the elements I will be connected with all zeros after a certain point. So, first of all, let's show that I is projective. So for this. Notice that I is a direct sum of lots of modules I X where I X is a module of order two. And it is functions with support. The point X in our space X and this is a sum over all X. And now we notice that I X is projective. So R is equal to I X plus the module of of functions with support this joint from X. So this is the module R is projective. So I X is projected because it's a direct sum and of a free therefore projective module. The direct sum of projective modules is projective. And we've just seen up here that I is a direct sum of projective modules so I is indeed projective. And now we show that I is not locally free. And for this. What we have to do is to work out what does I F minus one look like over R F minus one. Well, after the minus one here F is any element of art is still a product of copies of Z over to Z. So unless so it's a product of a number of copies given by the support of of the element F so usually it would be just isomorphic to R. So what we find is that if F has finite support, then I F minus one is indeed free. In fact, it's just isomorphic to R F minus one. On the other hand, if F is infinite support. We can easily check that I F the minus one is over R F the minus one looks the same as I over on we find I is not so I F minus one is not free. So, we want to cover the spectrum of R by a finite number of forms spectrum of R F I minus one such that we want I I to be free over this so F I has finite support. But if F one up to F and have finite support, then F one up to F and obviously can't be the ideal are because these only generate functions with support the union of the supports of all the F I which is finite. By the way, you've got to be a bit careful here because you might think that the spectrum of this ring is is something to do with the set X. And if the set X is finite than the spectrum of R is indeed the set X. However, if the set X is infinite its spectrum is all sorts of weird extra elements not in X called ultra filters and if you want to know what an ultra filter is you can go and look at a book on set theory and forcing which goes on about them. So, the ideal I is a module that's projected but not locally free and of course it's not finitely generated. There's a final slightly funny example we can give using this ring, which is that for a vector bundles over a smooth manifold. So the property that the sum of finite dimensional vector bundles is a vector bundle where this sum is possibly infinite. That's sort of quite easy to check because all you need to do is notice that for each point on a smooth manifold you can find a contractible neighborhood and all vector bundles are trivial on that neighborhood and so on. And the analog for rings is false. The direct sum of locally free modules, even a finite rank need not be locally free. And we've actually seen an example of this. Which is the ideal I in the previous two slides contained in the ideal are because here the ideal I was a direct sum of the ideals I X and the ideals I X here we're all locally free. But if you take a direct sum of an infinite number of them, you get an I an ideal that's not locally free. So for modules that are not finitely generated have to be a little bit careful because your intuition about analog between vector bundles and locally free modules kind of sometimes breaks down a bit. Then, next lecture we might be looking at modules that are stalk wise locally free which is a slightly more relaxed condition of local freeness.