 So this lecture is part of an online commutative algebra course and will be about completions of rings. So we will first define what a completion of a ring is. So we've got a ring r and it has an ideal i and its completion is just the limit of the rings r over i to the n. This is sometimes called a projective limit or inverse limit or something. So what this means is we're taking the ring r over i and there's a map from r over i squared to this and the map from r over i cubed to that and so on and we just take the limit of these which means we must take an element a0 of r over i and an element a1 of r over i squared and element a2 of r over i cubed such that a1 is the image of a2 and a0 is the image of a1 and so on. And it's easier to see what's going on if we look at an example. So let's choose the basic example. Let's take r to be the ring of polynomials over a field and we take i to be the ideal generation by x and then we see r over i is just k and r over i squared is just kx over x squared. So the elements of this look like an element a0 and the elements of this can be all represented as a0 plus a1x. We're sort of taking x squared to be zero and similarly r over i cubed looks like kx over x cubed and its elements can be represented as well. They're not quite degree to polynomials because you're really saying x cubed is equal to zero. So what we have to do is we take a constant and then a polynomial of degree one and then a polynomial of degree two and so on and these all have to be compatible. So the constant must be the constant term of this polynomial and this polynomial must be this polynomial modulo x squared and so on. So if we put these all together we're obviously just getting a formal power series so this sort of goes on forever. So the completion of a ring which is normally denoted by r with a little hat on it so we see that the completion of k of x with respect to the ideal i is just the formal power series ring which is usually indicated by putting two square brackets around it. Similarly if we take the ring of polynomials in two variables and complete it with respect to the ideal generated by x and y this just turns out to be formal polynomials in two variables and so on. Now let's look at a different example where we take the ring to be the integers z and we take the ideal to be the ideal of all multiples of 10 and then let's figure out what the completion is. Well the completion is we have to take z modulo 10 and z modulo 100 and z modulo 1000 and we take an element in each of these and we kind of stick them all together. Well an element of z modulo 10 can be represented as some sort of we can just represent the last digit of an integer so it's an integer mod 10. Something in z modulo 100 might be represented by a number up to 100 or integer modulo 100 z modulo 1000 we might have an integer modulo 1000 and if we go on like this we see what we get is an sort of infinitely long integer so it's a sort of infinitely long decimal integer. So you can think of this as being sort of the opposite of the construction of real numbers so for real numbers if we take a real number we have a finite number of digits before the decimal point and an infinite number after. For the completion of z we have no digits after the decimal point there's an infinite number before and you can you can check the usual rules of additional multiplication actually work perfectly well for these except of course additional multiplication now infinite operations. Well the ring this is called the ring of 10 added integers and people don't actually use it much for the following reason and by the Chinese remainder theme z modulo 10 to the n is isomorphic to z modulo 2 to the n times z modulo 5 to the n so if we take the inverse limits of all these it's basically so the inverse limit of z modulo 10 to the n is just the inverse limit of z modulo 2 to the n times z modulo 5 to the n which is the inverse limit of z modulo 2 to the n times the inverse limit of z Z modulo five to the N. So the 10-addict integers, so the 10-addict integers might be denoted by Z, 10. So the 10-addict integers are just isomorphed to the two-addict integers times the five-addict integers. And you can do this sort of reduction for any composite number. So you may as well just focus on the N-addict integers so these are the two-addict integers and these are the five-addict integers. And more generally, the P-addict integers for P prime are very widely used in number theory. And you should think of the P-addict integers as being a sort of analog of the ring of power series. So you remember that the ring of polynomials is very similar in some ways to the ring of integers. So these are both principal ideal domains, for example. And a maximal ideal Z kind of corresponds to a maximal ideal two in the integers. And if we take the completion of polynomials with respect to this ideal, this is just the ring of formal power series. And this is analogous to the completion of the integers at two, which is the two-addict integers. So rings of formal power series and rings of P-addict integers are in many ways rather similar. By the way, it's common in algebraic topology to write Z modulo two, Z two for the ring of integers mod two. In number theory and algebraic geometry, you never do this because you use, tend to use Z two for the ring of two-addict integers. And the problem with this notation is it's rather ambiguous. This notation for the integers mod two has the advantage that it's completely unambiguous. No one could mistake this for anything else. Anyway, notice by the way that Z 10 has zero divisors because it's equal to Z two times Z five. So you can take an element that's one in here and zero in here and multiply it by an element that's zero in here and one in here and their product is just going to be zero. So even though Z does not have zero divisors, its completion at this ideal does have zero divisors. Instead, it's not terribly easy to write down the zero divisors directly. You have to find a series of integers that are one modulo two to the N but divisible by five to the N. So it's, well, it's not exactly difficult but it's not completely trivially. So there's an alternative construction of the completion. So you know in analysis, you define the reals to be the completion of the rational numbers and that's certainly not a completion in the sense that I've talked about here. However, they are quite closely related. So you remember when you construct the ring of reals from the ring of rational numbers, what we do is we have a metric on the rational numbers where we define the distance from X to Y to be the absolute value of X minus Y. And then we can define the completion of any metric space. So R is defined to be the set of Cauchy sequences. Modulo, the set of Cauchy sequences that 10 to zero. And I'm not going to remind you of the definition of Cauchy sequences because it's got far too many quantifiers in and I always get muddled up by it. And now we can do the same for any other distance. So we can put a distance on the ring R by putting the distance from X to Y to be the absolute value of X minus Y. Well, what's the absolute value of X minus Y? Or rather, what is the absolute value of X in the ring? Well, we define the absolute value of X to be C to the minus N if X is in I to the N and X is not in I to the N plus one. And it's equal to naught if X is in all I to the N for instance, if X is equal to zero. And now it's not too difficult to check that the completion of the ring, of the ring R in this metric, this is just another way of writing down the inverse limit of R over I, R over I to the N. So that gives you an alternative construction. And there are some differences. For example, for the real numbers, we know that the absolute value of X minus X plus Y is less than or equal to the absolute value of X plus the absolute value of Y. So the distance from X to Y is less than or equal to the distance from X to Z plus the distance from Z to Y. For the, so this is for the absolute value of the reals. For the absolute value of R with respect to an ideal, we actually have a stronger thing. We know we find the absolute value of X plus Y is less than or equal to the maximum of the absolute value of X and the absolute value of Y. And the distance from X to Y is less than or equal to the maximum of the distance from X to Z and the distance from Z to Y. So this is called the ultrametric inequality. So the distance function that we've defined for rings has rather stronger properties than the usual distance function for the reals. Anyway, there are sort of three themes that appear for completions. So first of all, completions of rings are similar to the reals in many ways. For example, over the reals, we can define Bessel functions and Gamma functions and powers and we can sometimes define powers X to Y, Gamma functions, even Bessel functions for some rings that are completions. I mean, you can't always do this for all rings but you can do a certain amount of analysis on completions of rings. This is particularly popular when you take R to be, when you take the completion to be the ring of piadic integers and there's a whole branch of piadic mathematics where you go through all real analysis and generalize it to piadic numbers. So the second major principle about completions is that it is easy to solve equations, solve equations in completions of rings. We will be talking about this later where we use something called Hensel's lemma which is the specific thing that makes it easier to solve equations. The third principle is that completion is a sort of stronger version of localization and this is what I want to talk about. That's the third one that I want to talk about a bit more in the rest of this. First of all, we notice that if R is a unit, let's put that this is R in the ring, R is a unit in R over P to the N, P to the N, it is a unit in R over P to the 2N and this is quite easy to see because if we have one plus R S is in P to the N so this means R is a unit in R over P to the N then we just square if we find that one plus R S squared is in P to the 2N, well, this is equal to one plus R times something or other, don't really care what the something or other is, is 2S plus R S squared or something like this. So this indicates that R is a unit in R over P to the 2N and the inverse is unique so the inverses of R in R over P to the N for N greater than or equal to one are compatible and give an inverse to R in the completion. So this gives a way of finding units in the completion. All we have to do is to show that something is a unit in R modulo sum power of P and now suppose M is a maximal ideal and we're going to look at the completion of R at the maximal ideal M. Well, in this case, we find that in this case, R hat is now a local ring. This is because all elements of R not in M are units in R over M, which is a field. So any element of R over I to the N modulo sum R over M to the N, not in M over M to the N has an inverse. Also, we notice we get a map from R M to the completion where this is just the localization because this means you just invert all elements not in M and we've just said that if something in R is not in M, then it has an inverse in the completion R hat. If R is notarian and an integral domain, then the intersection of I to the N is equal to zero so R is actually contained in the completion. These can fail if M is not a maximal ideal. So all these fail if P is a prime ideal. For example, if we take P equals naught in Z, then the completion Z hat is just Z, which is not a local ring. So the completion of a ring as a prime ideal need not be a local ring in general. This doesn't matter because if you've got a prime ideal, you can first localize it to make it maximal and then take the completion if you want. Another example is if you take the ring Z and again, if you now localize it at zero, this is equal to Q and now this is not the subring of the completion. So the localization is a maps to the completion for maximal ideals, but not necessarily for prime ideals. We also get weird things happening if the ring isn't notarian. For example, suppose we take the ring R to be, say the ring of formal power series in X to the one over N for N greater than zero in Z. So it's the sort of pre-certed expansions. Then this is local and if we take the maximal ideal to be generated by all powers of X, all positive powers of X, then the completion is just C because M to the N is equal to M for all N. So the localization is not containing the completion even though we're completing the maximal ideal. But the problem here is the ring isn't notarian so all sorts of weird things go wrong. So generally the picture you get is that for M maximal, we get this series of ring homomorphisms R maps to the localization, which maps to the completion, which maps to R modulo M to the three, which maps to R modulo M squared, which maps to R modulo M. And you can think of all these as being a series of rings getting closer and closer to this field. And if you try and draw the spectrum of these rings, well, the spectrum of R might be something like that if R is the spectrum of say a nodal curve. The spectrum of its localization is sort of looks like this except we miss out the finite number of points. You remember the spectrum of the localization kind of looks like this except we delete all the points. So it's kind of, we've removed some points. The spectrum of the completion at say this maximal ideal looks much finer. It kind of looks like this bit here. So the spectrum of the completion in some sense just looks like this bit. So you can think of informing that we've got maps between these. Now here the spectrum just is a point and here the spectrum is a sort of infinitesimal neighborhood of the points. So in this case it might have sort of two infinitely close points, whatever that means. And this will be a slightly bigger version of this and so on. So you can imagine we're kind of getting, in the spectrum of all these rings is really a point but you can think of it as being a sort of multiple point in some sense. So what's happening is that we're getting sort of bigger and bigger spectra as we go in this direction as we go from the ring to its localization to its completion to various quotients by powers of M. If we look at a slightly more and a slightly more explicit example, suppose we take R to be say the ring K X Y modulo Y squared minus X cubed minus X squared. So in this case its spectrum really does look like that. And let's work over a field of characteristic zero. And now we can ask what does the completion of R look like? Here we're taking the completion at the ideal M, which is just generated by X and Y. Well, this is going to be the formal power series ring in two variables, quotient it out by Y squared minus X cubed minus X squared. Well, over formal power series, this actually factorizes. So this is now equal to Y minus X times the square root of one plus X times Y plus X times the square root of one plus X. So you notice this is a formal power series, which is a perfectly good element of this formal power series ring. So we see that this ring here has zero divisors because in this ring Y minus X times the square root of one plus X times Y plus X times the square root of one plus X is just equal to zero. And as I said, the spectrum of this kind of looks a bit like this, where we've got two copies of the spectrum of a formal power series ring. You remember the spectrum of this as just two points, a generic point and a closed point. And we take two copies of this and identify the closed point. And you see what is happening is that we're just picking out a sort of infinitesimally small piece of this spectrum and that's the spectrum of this completion. And you can see that this is reducible. It's the union of two closed subsets, pretty obviously. This corresponds to the fact that this completion has zero divisors. So in particular, we notice that the if a ring R if a ring R is an integral domain, then a localization R S minus one is also an integral domain. But the completion need not be, even if you're taking a completion as a maximal ideal. And geometrically, this corresponds to the fact that if you're localizing at a point, you're sort of throwing away some points in some sense, but you can still somehow see most of the ring and you can see that it's still irreducible. But if you complete, you're focusing in much more closely to this point. And now you can't see this bit of the spectrum and you think that the spectrum is now reducible. So I'll finish by trying to draw a picture of a completion. And the completion I'm going to draw is just the two adding integers. So this is the inverse limit of Z modulo two to the N for all N. And you can draw a picture of it a bit like this. First of all, the last digit can be zero or one. And then if the last digit is zero, the next last digits can be zero, zero or zero, one. And here last digits could be one, zero or one, one. And then inside here, we have each of these, there are two possibilities of what the three last digits are. And then there are more possibilities for what the four last digits are. So you can think of the two added integers as being the sort of intersection of all these closed sets. So the first closed set is just going to be this disk and then we take these two disks inside it and then four disks inside that and take the intersection of all these. And you can see that this is now essentially what we've got is homeomorphic to a contour set, which is also homeomorphic to two point set to the power of infinity. That's a countable infinity if you care about the difference. There's an alternative picture of the two added integers or rather the three added integers, which is drawn by Fomenko, which I've got a picture of here. This comes from Koblitz's book on periodic numbers, which I strongly recommend it if you want to find out more about them. And this is Fomenko's picture of the three added numbers. I'll focus in on it a bit more so you can see. And it looks really impressive, although frankly I have no idea why this is a picture of the three added numbers, but that'll, I guess it's an exercise for you to puzzle out why this is the three added numbers. Okay, so that's the introduction to completions and next lecture, we'll probably be looking at Hensel's lemma.