 This lecture is part of an online commutative algebra course and will be about the relationship between tensor products and localization. So, first of all, we define a module M is called flat. If taking tensor products with M preserves exactness, this means that if zero goes to A, goes to B, goes to C, goes to zero, is exact. This implies zero goes to M tensor A, goes to M tensor B, goes to M tensor C, goes to zero, is exact. As we saw, this bit is always exact, whatever M is. So, the key point is that if A is submodule of B, then M tensor A is a submodule of M tensor B. For example, over the integers, we saw that Z modulo two Z is not exact, because as we saw about three or four times, there's an example where this doesn't preserve exactness. Over any ring R, R itself is exact. Because R tensed over R with the module A, it's just isomorphic to A. So, tension with R doesn't actually do anything to this. Flatness as a concept was introduced by Seher in the 50s, and growth and declatives showed that it was absolutely fundamental in algebraic geometry. It's a bit of a mystery why it is so important. It's not at all an intuitive concept. You can see it's not obvious, because people were studying commutative algebra for about a century before Seher introduced this concept. Informally, what flatness means, so M being flat means the modules Mp very nicely. So, what is Mp? Well, this is the localization of M at a prime ideal P, and we're going to explain what this is fairly shortly. So, we can sort of think of M as being something like a family of modules Mp over local rings, and flatness kind of says this family varies nicely without jumping suddenly or anything like that. So, flatness turns up a lot when you are studying families of modules or families of schemes. You tend to want to put a flatness condition on this family in order to force it to be well behaved, where well behaved is defined to be flat. So, the main result we're going to prove, this lecture, is that the localization R s to the minus one is a flat R module. This is a special case of the fact that localization, such as this, is always nice. It has pretty much any good property you can think of, and flatness is one good property, so it turns out that localizing R produces a flat R module. So, what we're going to do is define a module M s to the minus one. So, we recall we defined R s to the minus one for R a ring and s a multiplicative subset of R. And the elements are of the form R over s, where this is really a sort of equivalence class of ordered pairs, and R over s is equal to zero, if and only if R times s one equals naught for some s one in s. And addition is defined using the usual rules of high school arithmetic. And we're going to define M s to the minus one in the same way. So, what we do is we're just going to copy the construction of R s to the minus one, replacing R by s, except we're not actually going to do this because we've already done it for R and doing it again for M is almost exactly the same and not very interesting to watch. So, I'll just summarize what happens. The elements are all of the form M over s for M in M and s in s, of course. And M over s is zero, if and only if M times s one equals naught for some s one in s. And from this, you see that M one over s one is equal to M two over s two, provided that M one s two minus M two s one times s equals naught for some s in s. So, this is just like the definition of the localization of R, except you're replacing R by any module. So, the key point is that M mapping M to M s to minus one preserves exactness. In other words, if we've got a sequence, naught goes to A, goes to B, goes to C, goes to naught, which is exact. This implies naught goes to A s minus one, goes to B s minus one, goes to C s minus one, goes to naught is exact. And the main problem is to show that if we've got an element in here, suppose some element B s minus one in B s minus one has image zero in C s minus one. Then B has image C in C for some element C and B s minus one maps to C s minus one, which is equal to zero. So, C s one equals zero for some s one in s, by definition of when something is zero in this localization. So, B s one has image zero in C. So, B s one is in the image of A. So, it's the image of say some element A in A. But then B s minus one equals B s one, s one minus one s minus one equals A s one s minus one is in the image of A s minus one. So, this proves that localization of modules preserves exactness, which is one of the reasons why localization is such an easy thing to deal with. We define the localization of M at a prime P to be M s minus one, where s is equal to the complement of P. So, M P is a module over R P, and can be thought of as a sort of stalk of M at prime P. So, just as R becomes a sheaf of rings over the spectrum of R, M is a sort of sheaf of modules over the sheaf of rings. So, this is very heavily used in algebraic geometry. You turn modules over rings into sheaves over the spectrum. We're not going to use this very much, but I'll just sort of sketch what you do. So, just as for rings, for each open set U of F, you assign it to ring R F minus one. For modules, for each open set U of F, you assign it to module M F minus one, which is, of course, a module over this ring. And for each prime, we have a stalk, which is the localization of R at P. What this does is construct something called a quasi-coherent sheaf. And you don't need to remember what that is or how to spell it because we're not going to use it anymore this course. So, informally, as I mentioned earlier, you can think of M as being a sort of, when you think of it as being the sheaf of modules, you think of it as being the sheaf of modules. You think of it as being a family of these modules over local rings, parameterized by points of the spectrum. Anyway, let's get back to the proof that R S is flat. So, sorry, we want to show that R S minus one is flat. Well, we want to show that, we want to show that mapping a module A to A tensed over R, R S minus one preserves exactness. So, let's put a question mark because that's what we're trying to prove. We know that A goes to A S minus one preserves exactness. And now you can guess what we have to do. All we have to do is to show that A S minus one is, when I say equal to, I mean chronically isomorphic to, A tensed over R S minus one. And this is easy to do because we can construct inverse maps in each direction. So, first of all, we can take A S minus one to A tensed with S minus one. And on the other hand, we can take A tensed R S minus one to A R S minus one. And it's easy to check that these two maps are inverses of each other. So, these two things are really the same. So, this shows that R S minus one is flat as an R module. And we've also seen that tension with this is the same as localizing with respect to S, which is also kind of useful. So, localizing is flat. I better just have a warning that taking quotients doesn't preserve flatness. So, we've seen that R over I is not usually flat as an R module. So, we've seen an example of this. We've seen Z over two Z is not flat. Similarly, taking quotients of modules by ideals does not preserve exactness. So, if zero goes to A goes to B goes to C goes to zero is a sequence of R modules. You can try looking at the modules, Nought goes to A over I A goes to B over I B goes to C over I C goes to zero and ask whether this is exact. And as you can probably guess by now, it isn't exact because it sort of fails to be exact here. In fact, A over I A is really just the same as A tensored over R with R over I. And we've just seen that R over I isn't flat in general. So, taking quotients is sort of a bit of a difficult property, but localizing is a much easier and better behaved property. So, what do flat modules actually look like? Well, we can answer that in some simple cases. So, suppose R is an integral domain. Actually, what I'm about to say works for all rings, but you have to be a little bit more careful about definitions. So, I'll just take it to be an integral domain for simplicity. Let me notice that Nought goes to R goes to R goes to R over A R goes to Nought is exact where this is multiplication by A and A is non-zero. Now, let's tensor this with M and suppose M is flat. Then we get Nought goes to M goes to A M is exact. We can map it on to M over AM, but I don't really care about this. And what this means is that multiplication by A is injective on M. So, flat modules are torsion free. So, torsion means MA is equal to Nought for some A not equal Nought in R and M not equal Nought. So, if this happens, we say the module has torsion with M. So, torsion free means that no element of M is killed by anything other than zero. The converse of this is sometimes true. So, for Z, torsion free modules are flat. And I may or may not prove this later. This is actually true for all principle ideal domains and also for some slightly more general rings. So, for Z, torsion free is just the same as flatness. In particular, for finitely generated modules over Z, torsion free, sorry, flat modules are just the same as free modules because those are the same as torsion free modules. So, why don't we just use torsion freeness instead of flatness? Well, it turns out that for most rings, torsion free doesn't mean flat. We'll see lots of examples later on. And torsion free modules don't really behave all that well, but flat modules do behave very nicely. So, we should really use torsion. We should use flatness, not torsion freeness in general. So, I'm going to finish with three useful properties relating flatness with localization. So, the first says that vanishing is local. So, there's a sort of catchphrase. What does that mean? It means the module m is equal to zero, if and only if m localized at p is zero for all primes or for that matter for all maximal ideals. So, we can test whether a module is zero by looking whether it's zero locally at each prime. And this is easy to show. Suppose m, m is equal to naught for all maximal ideals and pick a little x in m, then x is equal to naught in the localization mm. This means the annihilator of x is not contained in the maximal ideal m because x is naught in this if and only if x is killed by something not in m. So, the annihilator of x is not in any maximal ideal. Here we're assuming that mp is naught for all maximal ideals. So, the annihilator of x is equal to the whole ring r, so x is equal to zero. So, we've shown that if all localizations are zero and x is an element of m, then x must be zero. So, if all these vanish, then m vanishes. The next useful property, useful property number two, this exactness is local. What this means is that naught goes to a, goes to b, goes to c, goes to naught is exact. If and only if naught goes to ap, goes to bp, goes to cp, goes to naught is exact for all prime p. And as before, we could just take maximal p if we wanted. And the first implication follows because localization preserves exactness as we showed in the first part of the lecture. For the other implication, what we do is we just observe the kernel of b goes to c, modulo the image of a goes to b, vanishes if and only if this sequence is exact in the middle by definition. So, if we localize this at m, then we see this is equal to the kernel of b goes to c, localize to m, modulo the image of a goes to b, localize to m as rm is flat. And now, we see that this is zero, if and only if this is zero. So, if we miss out the m here, we see that this is zero, if and only if this is zero for all m, using the fact that this vanishes if and only for all its localizations vanished. So, that shows that this sequence is exact if and only if this is exact for all maximal ideals p. So, that's very nice because we can test whether a sequence is exact just by testing it locally at all primes. The final result says that flatness is local. In other words, m is flat if and only if mp is flat for all primes p, and by now you can guess we can replace this by all maximal elements p. So, again, this is very useful because we can test whether a module is flat just by looking locally at each prime and seeing whether it's flat. Well, we're just going to prove one of the implications because the other is pretty easy. In fact, both are pretty easy. So, suppose nought goes to a, goes to b, goes to c, goes to nought is exact. We want to show that this tensor with m is exact, assuming that mp is flat for all primes p. And all you do is we know that nought, all we have to do is to show that nought goes to a, goes to b, tends over, tends with m, localized m goes to b, tends with m, localized at m, goes to c, tends with m, localized at m is exact. So, we just need to show this is exact for all maximal ideals by part two that we showed on the previous piece of paper. And then we just observe that this is equal to m tensed over rm with mm. This should be a tensed over r. And this is equal to something similar and this is equal to something similar. So, if m localized at little m is flat, then this sequence is exact for all little m. So, this sequence tends with m is exact. So, we've shown that m is flat, if and only if it's localization at all maximal ideals is flat. Okay, so next lecture, we're going to be proving another useful property of flatness and then we'll be, our next lecture after that, moving on to artinium rings.