 This lecture is part of an online commutative algebra course and will be about Hensel's Lemma. It's actually a continuation of the previous lecture about Hensel's Lemma. So we just quickly recall that Hensel's Lemma says that if you've got a local ring r and a maximal ideal i, so r over i is a field, and if you've got a monic polynomial f of x, then we can lift a root a of f of x in r over i can be lifted up to a root in the completion r hat. And what we're going to do now is to just give an application of this, which is what is the structure of the group of units of the piadic integers. So you remember the piadic integers is the completion of the integers at the prime ideal p, which is a sort of limit of z over p to the n over all n. So we can think of elements of the piadic integers as being numbers written in base p, except they go on to the left and infinite number of spaces. So typical 7-adic integer might look like that. So how do we find the structure of the group of units in this? Well, the first thing we do is we can identify some roots of unity in zp star. Because we notice that z over pz has p minus 1 roots of unity. In fact, every non-zero element of this satisfies the equation x to the p minus 1 minus 1 equals 0, because the non-zero elements of this just form a group of order p minus 1. And now we notice that all the roots of this are distinct mod p. Well, obviously because they're all the different numbers mod p. So we can apply Hensel's lemma or roots lift to zp. So in the ring of piadic integers, or rather the group of units, we have p minus 1 roots of x to the p minus 1. And they are congruent to 1 to up to p minus 1 mod p. So what this means is the group zp star is isomorphic to z over p minus 1z. So this is the p minus 1 roots of 1 times. And now we have the elements that are congruent to 1 mod p. Because if a number is congruent to a mod p, we can find a root of unity congruent to a mod p and the quotient will be congruent to 1 mod p. So we're reduced to trying to figure out what is the structure of the group of piadic integers that are 1 mod p. And we need to distinguish two cases because p equals 2 is slightly different from p odd. So let's first look at the case p odd. In this case, the elements of zp star that are congruent to 1 mod p is isomorphic to the group of piadic integers that are congruent to 0 mod p under addition, where this is under multiplication. And in fact, this is isomorphic to the group of piadic integers because you can just multiply by p, but I'm taking the numbers congruent to 0 mod p for a reason we'll see in a moment. And the reason for this isomorphism is that we have an exponential map from this group to this group and its inverse is a logarithm map. So how do we define the exponential logarithm for piadic numbers? We just define x of x to be 1 plus x plus x squared over 2 factorial plus x cubed over 3 factorial and so on. We have to be a little bit careful here about convergence. So convergence of this for the real numbers is really easy because these terms go to 0. However, in the piadic numbers, these terms don't always go to 0 because you remember piadic numbers are large if they have large powers of p in their denominator like this. So we need to actually check when this converges. Well, it turns out that it converges if p divides x. Here we're taking p not equal to 2 because it actually, for p equals 2 we'll see in a moment the convergence is actually a bit trickier. Now, to figure out the convergence we need to know how many powers of p divide n factorial. Well, that's quite easy to figure out how many powers of p are in n factorial. Well, if we look we've got 1, 2, 3, opt to n and we multiply them all together. And there are n over p, this means the integer part of n over p of these divisible by p. So we get at least n over p or the integer part of this powers of p in this. But then we also have to take n over p squared of these divisible by p squared. So that gives us an extra n over p squared powers of p and we get n over p cubed of n divisible by p cubed and so on. So altogether the number of powers of p in n factorial is n over p plus n over p squared plus n over p cubed and so on. Which is less than or equal to n over p plus n over p squared plus n over p cubed and so on. Which is equal to n over p minus 1. And now we see that if p divides x and p is greater than 2 then the 1 plus x plus x squared over 2 factorial and so on converges because the number of times p occurs in x to the n increases more rapidly than the number of times p occurs in n factorial. If p is equal to 2 then you see this sort of grows like n which is the same rate of growth as x to the n if x is divisible by 2. So we see if p equals 2 and p squared divides x then x of x converges. So the exponential series for p adding integers converges provided p or possibly p squared divides x. It's considerably easier to check that if p divides x then the series for log of 1 plus x converges. So if we go back here we see that as long as p divides x and x is odd then these two maps here are inverses of each other and it's easier to check these actually given isomorphism between these two groups. So we have the following result about the group of units of the p adding integers. This is isomorphic to z over pz, sorry z over p minus 1, z times zp. And we notice that this is torsion free. Of course this is p naught equal to 2. And this is the set of elements of finite order. We notice this is very similar to the result we get for real numbers. The real numbers is isomorphic to z over 2z where these are roots of unity times r under addition. And again the exponential map is isomorphism from r under addition to the positive numbers here. You may wonder what happens for p equals 2. Well here it's slightly more complicated. So zp star is isomorphic to z over 2z. These are the numbers plus or minus 1 times the elements of zp star that are congruent to 1 mod 4. So now we have a 4 rather than a 2. And this group here is isomorphic to z. I guess that should be a 2. This is isomorphic to z2 under addition. So the prime 2 really does behave a little bit differently. We would get slightly more roots of unity than you might expect from the general case. So that's an example of using Hensel's lemma to identify the structure of the units of the periodic numbers. Now what I'm going to do is discuss some minor variations of Hensel's lemma. One of the problems with Hensel's lemma is that there are about 17 different theorems all called Hensel's lemma. All of which look rather similar and it's kind of difficult trying to remember which is which. So the one we've been working with says that if A is a simple root of fx equals 0 in r over i where f is going to be monic, then A lifts to r. Here r is going to be a complete local ring under the maximal ideal i. This condition is a little bit stronger than you really need. I'm just making the condition fairly strong for simplicity. Well, there's a generalization of this which is often also called Hensel's lemma, which says that suppose f of x equals g nought of x times h nought of x in r over i of x. So suppose f actually factorizes modulo i. Here as before we're taking f to be monic, which means leading coefficient 1. And suppose g nought and h nought are co-prime in r over i of x. Then the factorization lifts so we get f of x equals gx times h of x in r of x. So what this says is that factorization's modulo, your maximal ideal i, can be lifted to factorizations in the complete local ring provided g nought and h nought are co-prime. If we take g nought to be of degree 1, then this is essentially the previous case of Hensel's lemma we had because saying f has a root just says that f factorizes as a linear factor times something else. And the condition that g nought and h nought are co-prime just corresponds to saying this root is simple because saying the root is simple just means that the root isn't a root of h nought, so g nought and h nought would be co-prime. The proof of this is quite easy and there are a couple of ways of proving it which are rather like the proof of the previous version of Hensel's lemma. So let me just sketch how you prove it. The idea of this proof is that suppose f of x equals g n of x times h n of x modulo i to the n. Then we want to lift this to, so we want to lift g n and h n to g n plus 1 which might be g n plus a and h n plus 1 equals h n plus b so that f of x equals g n plus 1 of x times h n plus 1 of x modulo i to the n plus 1. And if you keep going like this an infinite number of times, then you'll get a factorization modulo i to the n for all n and you can glue these together and get a factorization in the complete local ring. Well, in order to solve this equation, what we find is we have to solve g n plus a of x times h n of x plus b of x is equal to f of x modulo i to the n plus 1 and we need to solve for a of x and b of x. And if we expand this out, we find that what we have to solve is g n of x times b of x plus h n of x times a of x is equal to some incredibly complicated mess. So this mess is going to be some element in i to the n modulo i to the n plus 1 because we're just working modulo i to the n plus 1 here. So you get this just by expanding this out and seeing what you get. And we don't need to work out what this mess is because what we can do is we can solve this equation for a x and b x no matter what this mess is. And the reason for that is that we have g nought of x times c of x plus h nought of x times d of x is equal to 1 modulo i for some c d as g nought and h nought of co prime modulo i. So if we've got two polynomials over a field that are co prime, then there's some linear combination of them that's equal to the identity polynomial. That's because polynomials over a field form a principal ideal domain. And now, using this equation, we can just solve this because we just multiply both sides of this by this complicated mess. So we get this mess here modulo i to the n plus 1 and then we get g nought times something plus h nought times something is equal to this mess, which is what we wanted to solve. So you can find the polynomials g n and h n just by repeating this construction over and over again. Now I want to say a little bit about Henselian local rings. So a Henselian local ring is a local ring where Hensel's lemma holds. Well, as I said, the problem is there are dozens of versions of Hensel's lemma in the literature, so I better say which version of Hensel's lemma. Well, it's the version I just stated. It says that if f monic equals g nought times h nought modulo i, where i is the maximal ideal in a local ring, then the factorization lifts to r of x if g nought and h nought are co-prime. And one example of a Henselian ring is just any complete local ring. So here Hensel's lemma or one version of it just says that complete local rings are Henselian. Other examples are things like you can get some rings of convergent formal power series and things like that. Now, Nagata showed that any ring, so any local ring has a Henselization, which is roughly a smallest Henselian ring containing it. So if you've got a local ring here, you can think of it as being contained in the Henselization, which in turn is contained in the completion. And the Henselization can be thought of as more or less the algebraic part of the completion of the local ring. The completion of a local ring is usually absolutely huge. I mean, if r is countable, the completion will usually be uncountable, and it contains enormous amounts of stuff that we're not really that interested in, at least if we're algebraists. And the Henselization sort of just keeps the algebraic part of it that we're interested in. For example, if r is, say, a ring of polynomials over a field, then the completion will be the ring of formal power series over the field, but the Henselization would be the algebraic power series. So this is sort of very analogous to the algebraic closure of the rationales. We can take the rationals, and it's contained in the complex numbers, which is algebraically closed, but there's a much smaller algebraic closure. So we can just take the algebraic numbers, which will be also an algebraically closed field, but it's only countable, whereas the complex numbers are uncountable. So the Henselization of a local ring is kind of related to its completion in much the same way that algebraic numbers are related to the complex numbers. So there's also a concept you sometimes find called a strictly Henselian ring, which I'll just mention. This is just a Henselian local ring where the field r over i is separately closed, and the reason why Henselian local rings turn up is these are the analogs of local rings for the etal topology in algebraic geometry. So if you want to define a local ring of a point of the spectrum of a ring, what you do is you essentially take the direct limit over all open neighborhoods of the point of the coordinate ring of that open neighborhood. And this gives you the local ring of a point in the Zariski topology. In algebraic geometry, the Zariski topology is sometimes not really good enough, and there are more complicated topologies or rather growth and dip topologies, such as the etal topology and the strictly Henselian local rings are the analogs of local rings for the etal topology. So that's probably the main way in which they turn up. Okay, so that's enough about Hensel's lemma. The next lecture will be about flatness of completions.