 So this lecture will be is part of a course on commutative algebra and will be about locally free modules. So last lecture, we talked about stably free modules. So what we want to do is to explain what is the difference between a locally free module and a stably free module. Well, first of all, locally free modules are sort of analogs of vector bundles. So I will stop by just reminding everybody what a vector bundle is. So a vector bundle is a vector bundle over a space map, space X is a map from a space V to X, such that the fibers are vector spaces. There are also some conditions saying that this vector space structure would behave nicely, but I'm not going to worry about that too much. And it should look locally like a map from R to the n times ui to ui where the ui open subsets covering X and R to the n is just an n dimensional vector space. For example, if we take X to be a smooth manifold, then we can just take the tangent space of X mapping to X. So this is just the tangent space whose elements are points of X together with a tangent vector at that point. Now this corresponds to a module over a ring R because we take R to be continuous functions on X. So we can take our module M to be sections of the vector bundle V. So in the case of the tangent space, the elements of the module M would just be tangent vector fields on X and you can, M is a module over R because you can just do point wise multiplication. So the analog of this for rings is where the space X will correspond to the spectrum of the ring R. So this corresponds to the space X we had before. So we want the spectrum of R should be covered by open sets ui and we may as well take this to be a basis of open sets. So we can take ui to be the spectrum of R fi to the minus one where fi is an element of R. And in order for these to cover R, you remember that the ideal generation by the Fi's should be the whole of R. And we want the module M to be locally free, meaning that it's free on each of the open sets ui. And what this means is that M of fi to minus one should be free over R fi to minus one. So this is an algebraic definition of a locally free module. You just specify a number of elements generation unit ideal of R and these modules should be free over these rings. So next we should discuss the relation between stably free and free module. So first of all, stably free implies free, sorry implies locally free. Well, if a stably free module is not finitely generated, then we know that it's free. So it's certainly locally free. If it's finitely generated, well, a stably free module that's finitely generated is a finitely generated projective module, because stably free modules are direct summands of free modules and therefore they're projective. And we will see later, or maybe we won't see later, depending whether I bothered to prove this, that finitely generated projective modules are locally free. So stably free implies locally free. We can ask the converse, does locally free imply stably free? And the answer is generally no. And there are two reasons for this. There's a boring reason, and there's an interesting reason. So what I'm first going to do is to give the boring reason why stably free, why locally free modules may not be free. And for this you think of it in terms of vector bundles. So we might have a space x, which is a union of two spaces x1 and x2. And we might have a vector bundle over this space x, which looks like r times x1 over x1 and looks like r2 times x1, x2 over x2. In other words, the dimension of the fibers of this chain. So here the fibers are one dimensional, then they're two dimensional. And this is certainly not of the form r to the n times x goes to x because that would mean the fibers all would have to have the same dimension n. So for vector bundles, they aren't necessarily stably free. And now what we do is convert this into an example for rings. So we're going to take, we want spectrum of r to be disconnected. So the big point of this example is the space x is two different components, so the dimension of the fibers can vary. And for this is quite easy, we could just take r to be z modulo 6z, which is z modulo 2z times z modulo 3z. So the spectrum of r just consists of two points. So we can take a module, which is going to be not dimensional over this point and one dimensional over this point and M might be say z over 3z. And then you can see that M is obviously not stably free, and it's locally free because it's free over this open set, and it's free over over this open set. Now we come onto the more interesting reason why locally free modules need not be stably free. The interesting reason is that the module M may be twisted, even if the spectrum of r is connected. And the basic example of this is the Mobius band. Okay, so the Mobius band sort of looks something like this. You see it's a, I'm thinking of this as being a two sided surface who probably all made copies of this in kindergarten or something where you take a strip of paper and glue it by itself with a twist in so it's so it's two sided and the Mobius band maps to a circle. S1. And if you take an open Mobius band you can see the fibers are just copies of one dimensional vector space so it's a vector bundle. And you can see it's not stably free. In other words, if you add a finite dimensional free vector bundle to this you're not going to get a vector bundle free vector bundle. And the reason is if you go around S1, then this you reverse the orientation of the fiber. That's sort of saying the Mobius band is a one sided surface means if you start here and go all the way around you end up with the one dimensional vector space now going in the opposite direction. And the same applies if you add a finite sum of free modules to this M1 it's still, it's still, you get this orientation reverses so it's not free. And you can see this is locally free, because if we just cover S1 by this open set, and by this open set say, you can see that over each of these two open sets, the Mobius band is just the product of R with the open set so so it's a low cost so it's locally free in this vector bundle. If you've got a good imagination you can see that if you take the sum of, let's call this Mobius band M, if you take the direct sum of two copies of this vector bundle that means that each point, you take a direct sum of two copies of the fiber. And you can see, this is isomorphic to the trivial or free vector bundle are two times S1. In order to see this what you do is you, you sort of think about the line, the sort of zero section and if you embed the Mobius band and R3, and kind of take normal bundle to the zero section you can see that gives you another copy of the Mobius band. If you add those two together and think about it a bit, you can see that the fibers of those can actually be trivialized, I'll leave that as a sort of exercise and visual imagination. So we can convert this as usual into a module over a ring by taking the ring to be continuous functions on the circle and the module to be sections of the Mobius band. So if I'm more algebraic example of a locally free bundle, so my second example, I'm going to take M to be the ideal generated by two and one plus root minus five in the ring R, which is Z root minus five. This example keeps turning up because it's the simplest example of a non principal ideal in a data kind domain so it's an example of an awful lot of things. So we recall that this is not a unique factorization domain as usual six is equal to two times three which is one plus root minus five times one minus root minus five and we're going to use this decomposition several times. First of all, we as we saw earlier M is non principal. So it's not free as a module. Being non principal doesn't always imply non free for arbitrary rings but it does for this particular ring. Now we want to show that M is locally free. So how do we do this well first of all, we have to cover our cover the spectrum of our bite open sets. And that's quite easy because we can just take the spectrum of our with the half inverted and the spectrum of our with a third inverted. So two and three generate the unit ideal, these open sets cover the spectrum of our so the union of these is the spectrum of our. And now we want to show that M, we want to show that M with two inverted is free over. The half inverted. And this is obvious because, because two is contained in M. So if two is invertible then M, a half is just equal to our our half. So this is just equal to M, the half included. And we also want to show that M, the third is free over the other open set our third. And to do this, we need to calculate very slightly more and we notice that here M one third is generated by one plus root minus five. So, and the reason for this is M is generated by two and one plus root minus five but to the other generator is one plus root minus five times. One minus root minus five divided by three, you remember this is two times three is the product of these two things, and three is now a unit. So, so the ideal is generated by one plus root minus five. This is a free module of rank one over R a third. So we've shown that M is locally free, we've covered the spectrum of all by two open sets, and check that M becomes free over each of these open sets. As a last example. We saw that from the Mobius band if you took a sum of two copies of the Mobius band, it became free well, we didn't actually see that because I left this as an exercise but whatever. So we can see for our example that M plus M is isomorphic to R plus R. And to see this, let's just you just sort of calculate a bit like this so you put a to be the element to plus one minus one plus root minus five in M plus M. So we let the element be be one minus root minus five plus to gain in M plus M. And now you just do a little bit of calculation you find that three a minus one plus minus five B is equal to zero plus one plus root minus five, and one minus five minus two B is equal to zero plus two. And now you see these two elements generate M. So the module generated by a and B contains not plus M. This easily follows that module generated by a and B in fact contains the whole of M plus M. So this easily implies that a and B generate M plus M. And now we can get an isomorphism from R plus R to M plus M just by mapping one plus zero to a and zero plus B and you can check that this is an isomorphism between these two modules. A similar thing happens for the ring of integers of any algebraic number fields, the non zero ideals in the number field are always locally free modules, and they're free if not if they're principle ideals. So if you take the sum of a finite number of that you can find a, if you take any such ideal then there's a, you can find a finite number, some of a finite number of these ideals which is this isomorphic to a sum of a finite number of copies of the ring of integers of the algebraic number field. In fact, you remember an algebraic number theory, you can form the isomorphism classes of ideals into something called the ideal class group where two ideals are considered the same if, if you can get what from one to another by multiplying by an element of R. So these, the ideals correspond to locally free modules of rank one and two metrically this corresponds to one dimensional vector bundles which are called line bundles. So the ideal class group of the ring of integers of an algebraic number field is a sort of analog of the group of line bundles and geometry which is called the Picard group. So next lecture, we will be looking at projective modules and how they're related to locally free modules.