 This algebraic geometry video will be a review of completions of a ring. So the most obvious example is suppose we take the ring of polynomials kx, then one completion of it is the ring of formal power series in x. So polynomials in kx just look like a naught plus a1x plus anx to the n. Formal power series looked like a naught plus a1x. And so they just keep going indefinitely and don't have a greatest term. So how do we construct the formal power series ring from the polynomial ring? Well, what we do is we take the ideal of all multiples of x. So you can think of this as being all functions vanishing at the origin, if you like. And if r is this ring here, we look at r over i, which is just k. And then we can look at r over i squared, which is the ring of all polynomials, a naught plus a1x with x squared equals zero. So this is k with elements a naught. Then r over i cubed just looks like all set of all things in the form a naught plus a1x plus a2x squared, where we assume x cubed is equal to zero and so on. So an element of the completion means we pick an element of each one of these rings and these elements have to be compatible. So if we pick this element a naught plus a1 and r over i squared, it has to image a naught and r over i. So what's going on is we've got this sequence of rings r over i and then r over i squared maps to that and r over i cubed maps to that. And what we do is we pick an element in each one of these rings that are all compatible, commute with these all automorphisms and that's an element of the power series ring, which we call the completion r hat. So it's completion with respect to i. And this works for any ring r and any ring i, we can define the completion of a ring r as an ideal i. Another well-known example that turns up in number theory is where you take r equals z and you take the ideal to be the multiples of prime p, then the completion r hat at this prime p is just the ring zp of p-addict integers. But that's more to do with number theory than algebraic geometry. So what we usually do in algebraic geometry is we take r to be a local ring with a maximal ideal m and then r hat is usually defined as the completion with respect to this local ring. So we take r over m, r over m squared, r over m cubed and so on. And an obvious question is we've got a natural map from r to its completion. We can ask is this injective? And the answer is yes, if r is notarian. This covers most of the cases in algebraic geometry. However, if r is not notarian, then you get some weird examples when this map isn't injective. So let's just see a couple. So this is no in general. So let's take r to be say all smooth functions on the reals. Okay, this isn't actually a local ring, but never mind, we can easily localize it if we want. And we take i to be functions vanishing at zero. Then r over i to the n is just, you can think of this as power series where we ignore all terms of order greater than n. So the completion is just power series. However, the map from r to r hat, which takes any smooth function to its power series function is not injective. The reason being that you can have smooth functions that vanish to all orders at zero. In other words, all their derivatives vanish at zero. Well known example is e to the minus one over x squared for x not equal zero and nought for x equal zero. So this function is in the kernel of the map from r to its completion. Another example is a more algebraic example. We can just take the ring k, we can take all power series in x. Well, except I'm not gonna take just power series in x, I'm going to take power series in x to the half, x to the quarter and so on. What I mean by this is I'm just taking k power series in x and this is contained in the power series in x to the half. And this is contained in the ring of power series in x to the quarter and so on. And you can just take the union of these and I'm going to take the ideal to be generated by x, x to the half, x to the quarter and so on. And now you can see that i is actually equal to i squared. So if r is this ring here, the completion of r is just the ring k and the map from r to its completion fails rather badly to be injective. So what can you do with local rings? Well, the key property of completions of local rings is Hensel's lemma. Hensel's lemma isn't one lemma, it's about 17 different lemmas because there are innumerable variations of Hensel's lemma. So what they say is a typical example that they say that solutions in r over m can be lifted to solutions in the completion under some conditions. And there are many variations of Hensel's lemma depending on what you're looking at solutions to and what exactly these conditions are. So a typical one, so a typical example, you might want to try factorizing a polynomial. So suppose we've got a factorization F naught z equals G naught z times H naught z where F naught, G naught, H naught polynomials in k of z. So suppose you can factorize a polynomial, modulo the maximum ideal m. Then this lifts to a factorization, sorry, that should be F, that should be, so I should say this is equal to that mod m. So I want F to be in r of z. So suppose you factorized a polynomial, modulo m, then this lifts to a factorization F of z equals G of z times H of z where these are now in r of z if G naught and H naught are co-prime in k of z. So in general, a factorization doesn't lift but you need some extra conditions and a fairly typical condition is that these things have to be co-prime. If you're looking for roots of polynomials, you sometimes have a condition saying the root must not be a multiple root and so on. Hensel's Lemma is quite easy to prove. The various versions of it are quite easy to prove. So I'll just quickly sketch the idea of the proof. So the idea of proof, suppose F is equal to G times H modulo m to the n. So suppose I found a factorization modulo m to the n and we try to find elements a and b so that F is equal to G a, so G plus a times H plus b modulo m to the n plus one. Modulo m to the n plus one. Here a and b are going to be elements of m to the n plus one modulo m to the n plus two. So in other words, we've got an approximate solution. We're trying to refine it by modifying G and H slightly in order to make it more accurate. And if you do this, you need to solve, well, we expand all this out and collect terms together and so on. We find we need to solve an equation and perform G naught b plus H naught a equals some rubbish, something complicated. And the idea is you can solve this equation as G naught and H naught a co-prime. So this is related to the fact that if two polynomials over a field are co-prime, then they generate the unit ideal. So we can always solve G naught b plus H naught a equals one with b and a being polynomials. And a slight variation of this shows that you can solve this rather complicated expression. So the advantage of complete local rings over local rings is it's very easy to solve equations and find roots of equations and so on because you've got these numerous variations of Hensel's lemma. It's kind of like the reason why the real numbers can be easier to work with than the rational numbers. If you're trying to solve a polynomial of the rationals, it can be really difficult. If you're trying to solve a polynomial over the reals, it can be completely trivial. All you have to do is to say the polynomial is positive somewhere, negative somewhere else and therefore has a real root. So it's easy to prove equations have solutions over the reals and then much the same way, it's easy to show equations have solutions over complete local rings. So let's have a look at an example. Suppose I look at the curve y squared equals x cubed plus x squared, which I've used about a gazillion times before and looks something like this. Now, suppose we were kind of really short-sighted and peered at this curve near the origin, what we'd see if we magnified up is that it sort of looks like two lines crossing. So we would like to say that locally this curve looks rather like this curve here, which should just be the graph of y squared equals x squared. Well, you can't really see this just by looking at the local ring because if the local ring thought it was a union of two components, then the local ring would have zero divisors and the local ring of this doesn't have zero divisors because it's a subring of a field. So the local ring isn't really aware that near the origin, this thing kind of splits into two components. The local ring insists on seeing this loop here. So I'm going to take R to be the local ring at nought nought. So in other words, we're just taking the coordinate ring of this and inverting everything that doesn't vanish at zero. However, the completion of the local ring does see that this thing splits. So the completion of the local ring, you can think of it as just being, you can just take power series in x and y modulo, this ideal y squared minus x cubed minus x squared. Sorry, R hat, sorry, R equals, R hat equals polynomials in x and y modulo, this ideal here. Well, the point is that y squared minus x squared minus x cubed factorizes modulo the cube of this ideal. So here the maximal ideal is going to be generated by x and y. So this is equal y minus x times y plus x modulo i cubed. And we can either use Hensel's lemma to extend this factorization or just take square roots directly and extend this factorization to y minus x plus some power series times y plus x plus some power series. So over the completion of the local ring, we really do see that there are two separate components here because y minus x plus some power series is just this piece here. And y plus x plus some power series corresponds to this piece here. So it'd be this green line here. So this shows that the completion of the local ring is kind of looking more closely at this point than the local ring is. So the local ring is still seeing most of this curve, but the completion of the local ring focuses in much more closely and only sees very, very close to the origin. If you look the same, and what we say is they're analytically isomorphic, analytically isomorphic just means that the completions of the local ring are isomorphic with each other. We also notice that the completion of this local ring has zero divisors. So the local ring R has no zero divisors because it's a subring of a field, but its completion has zero divisors. And this gives you one of the ways in which completions can behave in slightly disconcerting ways. So a ring, a local ring that's an integral domain with no zero divisors may have a completion with zero divisors. Even worse, there are examples where the local ring is reduced, meaning it has no nilpotent elements, but its completion has nilpotent elements, which tends to cause problems. There's actually a sort of modified variation of completion called Henselization, which is somewhat better properties, but we're not going to discuss that. So you can think of a picture like this. If you've got a local ring R, then it maps to its completion R hat, and this maps to various rings, and this maps to R over M cubed, which maps to R over M squared, which maps to R over M, which is equal to K. And you can think of this series of rings as corresponding to finer and finer neighborhoods of a point. For instance, if R is the local ring of a point on the affine line, say you can think of R over M as being the coordinate ring of a point. You can think of R over M squared as the coordinate ring of a point with a tiny little bit sticking out. So you're looking at the first order neighborhood of the point. R over M cubed kind of looks like a point with a slightly bigger infinitesimal bit sticking out. You can think of R over M cubed as kind of being, you know, the point together with two more points in the same place, if you see what I mean. And the completion looks like a point with a sort of short, smooth curve sticking out of it. And finally, the local ring R kind of looks like the line with a finite number of points missing. So this sequence of rings, a local ring goes to its completion and then goes to all these successive quotients kind of corresponds to focusing in more closely and closely to a point on a variety. Okay, the next lecture will be on something completely different which is elimination theory.