 This lecture is part of an online algebraic geometry course about schemes and will be about quasi-coherent and sheaves. The choice of the word quasi-coherent is particularly unfortunate because these are the most basic sorts of modules over, sheaves of modules over a scheme and unfortunately landed with this five syllable monstrosity as a name for them. This is for complicated historical reasons. Coherent sheaves originally appeared in the study of complex analytic manifolds and quasi-coherent sheaves were a sort of variation of them which turns out to be the really basic sort of sheaf. But anyway, we're stuck with this word so we have to make the best of it. So I suppose X is a ringed space with a sheaf of rings OX. Now the idea is that sheaves over a space behave like sets. They're a sort of weak model of set theory and the sheaf of rings should be sort of as a ring in this funny model of set theory and over a ring you should consider the modules over it so we should consider modules over this ring on the ringed space and a sheaf of OX modules is defined in what is essentially obvious way. We have a sheaf F such that F of U is a module over O of U where U is open in X and these all have to be compatible with restriction homomorphisms. So if V is contained in U, we have matched from OU to OV and we have matched from F of U to F of V and let's call both of these restriction maps row and then if we've got an element R in here and an element F in here so F is a element of this module then row of R F should be equal to row of R times row of F. In other words the action of this ring on this module should be compatible with restriction homomorphisms in the obvious way and a morphism of modules is again defined in the obvious way. If we've got two sheaves of modules F and G then a morphism of sheaves of modules from F to G would just be a morphism of sheaves from F to G such that F of U mapping to G of U is a morphism of modules and modules over a sheaf of rings behave much like modules over a ring. You can do many of the standard operations. You can take kernels and images and direct sums and define exact sequences. The precise way of stating this is to say that the sheaves of modules over a sheaf of rings form an abelian category can be to find informally as a category that sort of looks very much like the category of abelian groups. So let's look at some examples. Here we're going to let's just take R to be a ring and let's take X to be the spectrum of R points prime ideals and what we're going to do is suppose we've got a module M over R. What we're going to do is to form a sheaf of modules over the spectrum of R and M twiddle is defined like this. So you remember spectrum of R has all these open affine sets D, F for F in R which can be sort of informally as the sort of points where F doesn't vanish or more precisely where F isn't contained in the prime ideal. So what we have to do is we just have to define what the sheaf is on each of these open sets and we just set M twiddle at this open set is just M F minus one. So this is a module over R F minus one. And as I said earlier, in order to define a sheaf you only need to define it on on this nice basis of open sets. So this is good enough to define it and we can check this forms a sheaf. Well, this is similar to the proof that similar to the proof that defining the value on D of F to be this forms a sheaf. So, for example, if we take R to be the integers, let's take the easiest non trivial example, and we might take M to be Z over two Z. Then we get a sheaf. So if you draw a picture of spectrum of Z, you remember it has a generic point zero and it's got all these primed two, three and so on. Then the stalk of the corresponding sheaf will be Z modulo two Z at this point here and be zero at all these other points and it's zero even at the generic point. So this is for Z modulo two Z. On the other hand, if you take and to be say the integers, you can check that the stalk at the point zero is now Q and the stalk at the local at the point two will be just equal to local ring Z two and so on. So you can turn any module into a sheaf. And next we can ask R or modules. The same as module sheaves of modules over O X where X is the spectrum of R. And it's easy to check that if M and N are modules then HOM of M to N is essentially the same as the homomorphism from M twiddle to N twiddle. So what we're really asking when we ask our modules the same as sheaves of modules, we're asking are they equivalent categories and the morphisms in both turn out to be the same. However, there's a bit of a problem. The problem is that the spectrum of R has modules that do not come from modules over the ring R. So there are too many sheaves of modules over the sheaf of rings of spectrum of R. So let's see an example of this. Let's take R to be just Z localized to two, which is the rational numbers M over N with N odd. So this is a discrete valuation ring and its spectrum looks like this. It's got a generic point corresponding to ideal naught and a close point corresponding to the ideal two. And let's think what about a module over a sheaf of module over spectrum of R is. Well, you have to provide a module over this open set and a module over the open set of that minus a point. So it consists of a pair of modules M up into N where this is a module over R. And this is a module over Q. So Q is the ring of functions associated to the open set where you remove this. So if you unwind the definition of a sheaf of modules, it turns out to be equivalent to this. You need to give a pair of modules and a homomorphism between them. Compatible with the embedding of R into Q. However, if M comes, if you take the sheaf of modules coming from an R module M for M hat, N would have to be M inverted at a half. So the sheaves of modules coming from modules over a ring are very special. Instead of being allowed to choose an arbitrary module over the rationals here, it would have to be of this special form. For example, we can now take M to be Z2 and N to be zero. And this would be a sheaf of modules not coming from a module over R. So this is a real problem because we want modules over the spectrum to be the same as modules over the rings. So how do we fix this? Well, we say a module, an OX module is called quasi-coherent. There's that long word again. If it is locally of the form M twiddle. Well, what does that mean? So that means it locally looks as if it's a module over a ring in some sense. Well, there are two possible ways we could define this. We could say this means X is covered by a fine opens UI. So that M on UI is of the form MI twiddle where MI is module over RI and UI is the spectrum of RI. So this is saying we can cover it by open affine such that M looks like a module over a ring. Alternatively, we could say for all open affines U in X, we have M on U is of the form. Sorry, I shouldn't be calling that M. I should be calling this F. So that F on U is of the form M twiddle where M is a module over R and U is the spectrum of R. So there are two different possibilities. So this one has the advantage that it's a local definition. So a module M satisfies this condition if it satisfies it locally everywhere. So this one has the advantage that it gives the right answer for the spectrum of a ring R. It means that if you've got a ring R, then the modules over the spectrum of R are essentially the same as the modules over R. Fortunately, these two conditions are equivalent. So we don't have to worry about which of them is the most important. So I'll sketch the proof that they are equivalent. I'm going to miss out a certain amount of bookkeeping and we reduce to the key case, which says suppose R is a ring. And suppose spec of R covered by some open affines of the form DFI for FI and R, whereas usual this means roughly the points where FI doesn't vanish. Since the spectrum of R is compact, we can assume there are a finite number of these. And saying it's covered by open affines means the ideal generated by F1, F2, and so on is just R. And now suppose F is an OX module where X is the spectrum of R. And suppose F restricts to quasi-coherent module on each OFI. Then F is quasi-coherent. So this is sort of saying that for affine opens if something is locally quasi-coherent, then it's coherent. So I guess I should decide which notion of quasi-coherence we're using. Here we're going to use the definition of quasi-coherence that every open affine subset has the nice property. So what we've got to do is show that the following map is an isomorphism. So let's just draw a picture to focus our attention. Let's take a set DF1 and DF2 and DF3. So I'm going to cover the spectrum of the ring by three open affine sets, which will be enough to illustrate all the complications. And we've got a module homomorphism from M to F of DF1. Here M is F of X. So it's a set of global sections of F and we're restricting it to DF1. So this is just the restriction homomorphism. Now this induces a map from MF1 minus 1 to FDF1. And what we want to do is we want to show that this is an isomorphism. And if we show that for all I then it will be easy to show that the module F is quasi-coherent from whichever sense you want. So proving that this is an isomorphism is the key technical step of the proof. All the rest is kind of routine bookkeeping. So we do this in first two steps. We want to show it as injective and we want to show it as surjective. So let's first show that MF1 minus 1 maps to FDF1 is injective. Well, suppose M in M has image nought. Here we're taking M to be in here and I should really take it to be something in M with divided by a power of F1 but you can easily eliminate that and it's enough to do it just for M. Then on let's look at DF2. Let's draw a picture just to focus our attention. So here's DF2. So on DF2 it has image nought on DF2 intersection DF1. So that's this bit here. Well, it's obviously zero here. So obviously it has image zero here. Well now DF1, sorry DF2 is quasi-coherent or rather F restricted to this is quasi-coherent. Which means that if something vanishes on this open set then this means that M times some power of F1 is nought on the whole of DF2. So for each i, M times F i to the n i is equal to nought on DFi for some n i. Now we choose n large enough. We find M times F i to the n, sorry times should be minus n, that should be F1 is nought on all the DFi's. And this just means that M is nought in M F1 minus 1 by definition. So that shows this map is injective. Next we want to show the map from M F1 to minus 1 to F DF1 is surjective. So let's look at DF1 again. And we've now got DF2. And again because F on DF2 is quasi-coherent. This means that anything. So if A is a section of F on DF1 intersection DF2 then A times F1 to the n can be extended to DF2 for some n. It's more or less the definition of because if the sheaf is quasi-coherent on this it means the values on here are just given by elements of here multiplied by some negative power of F. So if M is in M, which you remember was F of X, then M to the F1 to the n1 can be extended to DF2. So in other words by multiplying M by power of F we can extend it to here. And similarly by multiplying it by another power of F we can extend it to DF3. But now we run into a problem because we don't know whether these two extensions are the same on this area here. And the answer is in general no. There's no reason why the extension to DF2 should be the same as the extension to DF3. So we seem to have a problem but fortunately this is easy to fix. What we do is we notice that on this let's magnify this little bit where we've got a problem. So we've got this space DF2 intersection DF3. And we've also got the intersection of them with DF1. And the extensions are the same here on the intersection with DF1. Well by applying the uniqueness result that we proved earlier to this open set here we see that the extensions are the same on DF1 intersection, so DF2 intersection DF3 if they're multiplied by some power of F1. That's just saying that if two things have the same image here then they're the same on here if you multiply them by power of F1. So if we multiply M by a sufficiently large power of F it can be extended to all the DFIs and the extensions are compatible. So this just means if you unravel all these this means that MF1 to the minus one maps to DF1 is surjective and therefore an isomorphism. So this ends the somewhat technical proof that the two possible definitions of quasi coherence are actually equivalent. So what we'll be doing next lecture is giving some more examples and properties of quasi coherent sheaves.