 So this talk is part of an online course on commutative algebra and will be about Nakayama's lemma. So we will start with some examples just to motivate the problem we're going to solve with it. Suppose R is a local ring of analytic functions near a point, say sin near the point nought in the complex numbers, for example. So the maximal ideal, m, is just the functions vanishing at the point zero. And now we see the intersection of all powers of the maximal ideal is equal to zero because m to the n is equal to the functions vanishing to order n at the origin. And you know that if a power series vanishes to infinite order, then, so if an analytic function vanishes to infinite order, then it must actually be zero. Now let's look at the case where R is the local ring of smooth functions near point nought, let's say in the real board complications. And again, m is maximal ideal of functions vanishing at zero. And this time the intersection over m to the n is not equal to zero. And that's because we can have functions vanishing to infinite order at zero, which aren't, which don't vanish near zero. For instance, we can have e to the minus one over x squared for x not equal zero and nought of x equals zero. And you can see at this function here, all its derivatives at the origin vanish. So it lies inside this ideal. So we have the following problems, sometimes the intersection of all powers of a maximum ideal is zero and sometimes it isn't. So we'll just have a third example before discussing this. So here I'm going to take R to be the union of all formal power series in x to the one over n for all integers n greater than or equal to one. And again, we see the maximum ideal is now going to be the things with constant term one, sorry, constant term zero. And this time the intersection of m to the n is not zero. In fact, it fails in an even more drastic way because m is actually equal to m squared because you can, if you've got a power series in x to the one over n, then it's the square of some power series and x to the one over two n. So in fact, the intersection of all maximal ideals, all powers of the maximum ideal is just equal to m itself. And the intersection of this fails to be nought in an even more drastic way than before. In general, we have a map from ring R to its completion. Where the completion of R is the inverse limit of R over m, R over m squared, R over m cubed and so on. That means you take an element of each of these rings and make sure the elements are compatible. For example, if you've got the ring of polynomials over x and you take your ideal m to be the ideal generated by x, then this ring is k, this is kx over x squared, this is kx over x cubed and so on. So you can see the inverse limit is just going to be the ring of all formal power series in a polynomial x. So the question is, is this map injective? And if it is, then the formal power series ring is a reasonable approximation of ring R and if it isn't then you've lost a lot of information. And you can see the kernel of this map is just the intersection of all powers of m. So the question as to whether a map from a ring to its completion is injective is the same as this question we've been asking about whether the intersection of all powers is of a maximum ideal is zero. And our aim is to show that the intersection of m to the n is zero if m is a maximal and R is a notarian local ring. And to do this, we're going to need two lemmas, one of which is Nakayama's lemma, which we're going to do today. And the other is the Art and Rhys lemma, which I'm going to do next lecture. So for the rest of this lecture, I'm just going to be discussing Nakayama's lemma. So Nakayama's lemma says the following that says that if m is a finitely generated, that should be capital M, module over a local ring R, and with maximal ideal m, then m times m equals m implies m is equal to zero. And it's got a really simple proof and it's a very useful lemma. It turns up something like half a dozen times in Eisenbud's book in various places. It's traditional at this point to point out Nakayama wasn't actually the first person to prove Nakayama's lemma. It seems to originally do to a crow. And furthermore, Nakayama apparently really disliked having this lemma named after him, but the name seems to have stuck and you can't really do much about it. And the proof is really easy. Suppose m is generated by a one up to a n. And then we're going to take n minimal. And if n is greater than or equal to one, then we can put a one is equal to some of m i a i because m is equal to m times m. So here m i is in m, sorry, little m and a i is in m. Too many m's here. And this means that one minus m one times a one is equal to sum over i greater than one of m i times a i. And we notice this is a unit because in a local ring, one minus anything in the maximum ideal is automatically a unit. So a i, so a one is a combination of a two up to a n. Well, this means we can drop a one from the collection of generators, which means that a number of generators wasn't minimal, which is a contradiction. So the number of generators must actually be zero. So this is the end of the proof. That may be why Nakayama didn't like having a lemma named after him since the proof is kind of almost trivial. It's just a couple of lines long. And I suppose it's a subtle way of insulting someone to name something completely trivial after them. Anyway, we noticed by the way that all we need for this to work is that one minus anything in this ideal is a unit. So it also works for m being the Jacobson radical of the ring R. So the Jacobson radical is just the intersection of all maximal ideals. And the intersection of all maximal ideals is the property that if you add one to anything in it, the result is a unit. So the same proof goes through. And there's a useful corollary of Nakayama's lemma, which says that if m is a finitely generated module over the local ring R, and m1, mn generate m over m times m, then they generate the whole module m. And again, for the proof of this is almost trivial. What we do is we just put n to be m divided by the submodule generated by m1 up to mn. And then we notice that n is finitely generated because m is, and furthermore, n is equal to m times n, as you can easily check from this property here. So n is equal to zero. And n equals zero says that m is generated by m1 up to mn. So that's the end of the proof. By the way, there's a sort of traditional mistake people make here. We say that if m is finitely generated and m over the maximal ideal times m is generated by these, then it's generated by these elements here. It's not true that if m, a module over a local ring, has the property that m1 up to mn, generates m over m times n, then m1 up to mn generates m. So here we've dropped the condition that m is finitely generated. And if we do this, there are lots of counter examples. For instance, we could take the ring r to be z localized at two, and we could take m to be the rationals. And here the maximal ideal is just the ideal generated by two. And you can see that m over m times m is in fact zero. So this is certainly generated by the empty set of generators and this one obviously isn't. So we notice that this is of course, not finitely generated. So I think Lang's algebra book, I think the second edition actually claimed this was true as part of the theorem, but it just isn't. So we're now given application of Nakayama's lemma. So we want to say suppose s contains r and its integral over it. So suppose r contains s r rings and s is integral over r. Now, last time we talked about the geometric meaning of an extension being finitely or integrally closed, but we didn't quite say what the geometric meaning of being integral was. And well, here's one possible interpretation. The application we want to show is that the spectrum of s mapping to the spectrum of r is on to. So in order to do this, so let's prove this, what we do is we just pick some prime r, some prime p in the spectrum of r. And we want to find a prime of s whose intersection with r is equal to p. And first of all, we can localize at p. This means invert all elements of r, not in p. And we do the same in s, we invert all elements of s that are not in p. And by doing this, we can assume p is maximal. You remember localizing at a prime is just a way of kind of turning that prime into a maximal ideal. So the next step is we're going to check that p s is not equal to s. Well, if p s is equal to s, then we can write one is equal to sum of p i s i for some p i and p s i in s. And we're going to let m be the sub-algebra of s generated by the s i, sub-algebra over r. Then one is contained in p m. So p m is equal to m. We also notice that m is finitely generated as an r module. I mean, it's obviously finitely generated as an r algebra because we wrote down a finite set of generators. Now we want the stronger condition that's generated as an r module. And this is because s is integral over r. And you remember that if you've got an algebra generated by finite number of integral elements, then it's finitely generated as a module. So this is where we're using the fact that s is integral. So now we've found that m is finitely generated and it has this property here. And if we take these two properties, then Nakayama's lemma applies m, that m equals zero, which is a contradiction. So if we assume that if p s is equal to s, then by applying Nakayama's lemma, we get a contradiction. So we know p s is not equal to s. And now we can just pick the prime of s containing the image of p rather a maximum. I guess we could pick a maximum idea of s rather than the prime. We can do that because p doesn't generate the whole of s. And if you call this q, then q intersection r must be equal to p because it can't be the whole of r and p is a maximum ideal in r. So p in the spectrum of r is the image of q in the spectrum of s. So this shows that the spectrum of s mapping to the spectrum of r is on two. Okay, so next lecture, we're going to prove the art and Reese's lemma and then combine that with Nakayama's lemma to prove that the intersection of all powers of a maximum ideal and notary in local ring is zero.