 This talk is part of an online commutative algebra course and will be about Hensel's Lemma. So, Hensel's Lemma is about how to find solutions of equations over completions of rings. So we remember last lecture we defined the completion of a ring with respect to an ideal i to be the inverse limit of all the quotients r over i to the n. So a fairly typical example is the periodic numbers, which is the inverse limit of z modulo p to the nz. Now, a common problem is to try and find roots of equations. So let's look at the following two problems. Let's try and find a... Does x cubed minus 5x minus 2 equals 0 have a root in the rationals? And the second question is does x cubed minus 5x minus 2 equals 0 have a root in the reals? Now, the second question, the answer is incredibly obvious because it's positive. This is positive for large positive x and negative for negative x. And so by the intermediate value theorem this must have a root somewhere in the reals. This one on the other hand you probably have to think about a bit to figure out whether it has a root or not. So the point is it's sometimes much easier to show that polynomials have roots over the reals than to show they have roots over the rationals. And part of the reason for this is that the real numbers form a complete field. Now the completions of rings are also complete in some sense and they have a sort of analog of this property that it is sometimes easy to show that they have roots. So Hensel's lemma says that if we can solve an equation modulo i to the n for suitable n we can solve it in the completion of a ring. So obvious question is what exactly is this number n? Well, number n sort of depends on which equation we're trying to solve as we will see in a moment. So for the real numbers the existence of roots tends to follow from the intermediate value theorem. So Hensel's lemma is a sort of analog for periodic numbers of the intermediate value theorem. It makes it easy to show that equations have roots. Anyway, we're now going to look at Hensel's lemma a bit more precisely. So here's the first version. So version one, suppose i is an ideal of a ring r with completion r hat. And suppose you've got a polynomial with coefficients in r hat. And suppose you've got a root a in r modulo i. So this means that f of a is congruent to zero mod i. Then a can be lifted to a root in r hat provided the derivative of f at a is invertible in r modulo i. So in applications r modulo i will often be a field because i might be maximal. So this just says that f prime of a is not equal to zero in r over i where i is maximal. So in this particular case it says roughly that a is a simple root of f not a double root or a root of higher order. We'll see a little bit later what happens if a is a double root. Well, this is quite easy to prove. It's enough to show that any root in r over i to the n can be lifted to a root in r over i to the n plus one. Because then we can stop with our root in r over i and then we get roots in r over i squared r over i cubed and so on. And we join these all together and get a root in the completion. So here we're just sort of doing it one step at a time. So suppose we've got a root modulo i to the n. So if f a is an i to the n, so a is a root modulo i to the n. And now we want to lift a to a root modulo i to the n plus one. So let's try and find a root f of a plus epsilon that is in i to the n plus one. So we want to find some epsilon with this problem. So we're starting with our root modulo i to the n and we're just trying to perturb it a little bit to make a root modulo i to the n plus one. So how do we do this? Well, we can expand this as it's f of a plus epsilon f prime of a and so on. So this suggests we should try epsilon is minus f of a over f prime of a. Well, we have a little bit of a problem, you know, we want to know what does this mean? Because f prime of a, well, it's not really a unit in not necessarily a unit in R or anything like that. Well, what we know is that f prime of a is a unit in our modulo i. So f prime of a times f prime of a minus one is equal to one in R over i for some f prime of a to the minus one. So this is really in our modulo i rather than in R, but we can sort of pretend it's in R. But now we can just define, we can then see that f of a plus epsilon with this, this value of epsilon is still going to be well-defined modulo i to the n plus one because this is the inverse here is defined modulo i and this is an element of i to the n. So we find that f a plus i is a multiple of, it's something in i times f of i, which is an i to the n plus one. So that's a subset of i to the n plus one. So whenever we've got a root modulo i to the n, we can just lift it to a root modulo i to the n plus one by doing this construction. So let's see an example of this. Well, let's just take square root of seven in z modulo three. So this is the completion of the integers at the ideal of all multiples of three. So this is the usual ring of three adic integers. Well, here the polynomial f of x is equal to x squared minus seven and we want to solve f of x equals zero. And we look at f prime of x. Well, this is just two x. Now it has a solution a equals one in z modulo three z. And we notice that f prime of a is equal to two, which is not congruent to zero in z modulo three z. So the condition is satisfied. So the root a mod three lifts to a root in z mod in the three adic numbers. And it's quite easy to work it out. You remember this this proof we sort of find roughly speaking we find the digits of a one at a time. So each time we found n digits we can do this construction and find n plus one digits. And if you do this we find the lift is you know the first few digits are two zero one one one. In case you're wondering this is this is in base three. So so if we look at that the root is going to be four modulo three squared and here it's going to be nine plus three plus one which is 13 modulo three cubed and so on. And we can do something similar for finding square roots in zp for p odd. So what about the ring of units in zp star here I'm going to take p odd for a reason that will appear very shortly and we want to know what are the squares. Well there are two obvious properties a square has first of all the number of zeros at the end of the p adic expansion is even. In other words if we've got a number two three one seven zero zero zero zero zero as its p adic expansion the number of zeros at the end must obviously be even. Because if you square something the number of zeros at the end is going to be even. So we may as well remove an even number of zeros and we're left with the problem of numbers whose last digit is odd. So obviously the first none zero digit is square mod p. So if we were looking at this mod p we'd have to have seven is a square modulo p. So conversely if these two conditions are satisfied then Henssel's lemma implies that the number is a square. And we may as well assume the number of digits at the end is zero and then all we're trying to do is to solve x squared is congruent to b modulo p where b is a square. Well we can solve this by assumption mod p because b is a square and the derivative is 2x which is none zero mod p. And the reason it's none zero mod p is because two is not equal to p. If p was two then this condition would break down. So this gives us the structure of the group of units in the periodic numbers modulo the squares and it's just z over 2z times z over 2z. Where this first z over 2z comes from the fact the number of digits at the end, number of zeros at the end is even. And the second z over 2z is the other obstruction coming from the fact that the last none zero digit or first none zero digit or whatever must be a square. Well we've had to use the fact that p is odd so obviously we should ask what happens if p is even. So we can ask what about p equals two. So let's try and solve x squared minus a equals naught in two add it integers. Well the problem a root modulo two to the n cannot always be lifted to a root modulo two to the n plus one. And we can see this easily because we have one squared is congruent to five modulo two squared. But a squared is congruent to five modulo two cubed as no solutions. So there's no way to lift this solution modulo two squared to a solution modulo two cubed. And our previous argument breaks down and it breaks down because if f of x is this thing here then f prime of x is just two x. And we saw earlier that this is no longer a unit and our entire argument breaks down. So we need to find a refinement of Hensel's lemma that will work for this case and tell us what squares in the two addict numbers are. So suppose that here I'm going to work in the ring zp of periodic numbers. There are versions of this for other completions but I'll just simplify notation a bit if I fix this example. So suppose f of a is congruent to zero mod p to the two d plus one for some number d. And suppose that f prime of a is not congruent to zero mod p to the d plus one. Then a can be lifted to a root in zp. So you notice that if d equals naught this is just the previous case. Then we would just say that f is zero mod p and its derivative is not zero mod p. So you can prove this in the same way as the case d equals zero but I'm going to give a slightly different proof. Well it's actually really almost the same proof but whatever. What we're going to do is to use Newton's method for finding roots. So you remember Newton's method for finding a root of a polynomial over the reals works like this. What we do is we stop with an approximate root x zero and then what we do is we draw the tangent line here and find a new root x one. And then we keep doing that we draw the tangent line and get x two and so on. And if you do a little bit of algebra you find that xn plus one is equal to xn minus f of xn over f prime of xn. So we can try and find a root of a polynomial over the reals by starting with an approximate value of the root and iterating this. And if you remember from the numerical analysis Newton's method sometimes behaves really well and sometimes behaves dreadfully. I mean it sometimes goes into wild oscillations if you don't start sufficiently close to a root or if the root is a double root or something. Anyway it turns out that for periodic numbers Newton's method behaves much better. So let's just have a look at Newton's method over the periodic numbers. So we write f of xn plus one is equal to f of xn plus minus f prime of xn times f of xn over f prime of xn. Plus f double prime of xn over two factorial times f of xn over f prime of xn squared and so on. So this is just the usual Taylor series expansion of f of xn plus one which is just f of xn plus minus f of xn over f prime of xn. Now you notice immediately that the first two terms cancel out and the third term is pretty small. So remember f of xn is divisible by p to the 2d plus one times something and f prime of xn is divisible by at most p to the d. So this stuff here is all divisible by p to the 2d plus two and there are further terms here which I haven't bothered to write out but they're divisible by even higher powers of p so you don't need to worry about them. And you might be feeling a little bit nervous about this two factorial because you seem to be dividing by something but we notice that f double prime xn over two factorial has coefficients in zp. You notice that if you take the second derivative of x to the n and divide it by two factorial this is just equal to nn minus one over two times x to the n minus two and this thing here is actually an integer. So these denominators don't really cause problems. Anyway what we see from this is that if fxn is congruent to zero mod p to the d plus one then, so 2d plus one then you iterate Newton's method once and it's now divisible by p to the 2d plus two. In fact more generally we see that if f of xn is divisible by p to the 2d plus k this implies f of x to the n plus one is zero mod p to the 2d plus two k. So in fact it's really great because we're nearly doubling the number of correct digits we have each step. You remember a similar thing happens for the real numbers when Newton's method is working really well and the number of correct decimal digits doubles at each step of Newton's method. So the same thing works periodically except it's much easier to prove. So using this we can now answer the question when does b in z2 have a square root? So first of all it must have an even number of zeros at the end and if it's got an even number of zeros at the end we can just cancel these out and assume the last digit is one. Well so suppose the last digit is one then it has a square root if and only if it's congruent to one modulo two cubed rather than two. And you can see this because we're trying to solve x squared minus b equals zero. So we need f prime of x which is 2x so this is fx should not be zero modulo two to the d plus one. Well so we can take d equals one because this will be two times something odd which is not zero modulo two squared. So a root exists if x squared minus b equals naught has a root modulo two to the 2d plus one which is two cubed. So an odd integer has a root in the two attic numbers if and only if it's congruent to one modulo eight. From this you can easily check that if you take the two attic units modulo the squares of the two attic units then this is z modulo two z times z modulo two z rather than being just z modulo two z. You remember zp star modulo zp star squared was just z modulo two z. If you don't insist that this number should be a unit then you get z modulo two z times z modulo two z times z modulo two z because you could also have some zeros at the end. I'll just finish off by giving a slightly more complicated example. Let's try and find out which numbers are fourth roots in the two attic numbers. So we're trying to solve x to the four minus b equals naught and let's assume b is odd because if it isn't you can just take out a number of factors of two if the number of factors of two take out isn't zero by four and it's not a fourth power. So here's our polynomial f of x and we know f prime of x is equal to four x cubed. And now we want f prime of x should not be zero modulo two to the d plus one. Well it's going to be four times something odd so we can take d equals two. So the fourth root exists if f of x equals x to the four minus b equals naught has a root modulo two to the two d plus one which is two to the five. So a two attic integer, an odd two attic integer has a root provided it's got a root modulo two to the five. So let's just check when this happens. We need to find the fourth powers in z modulo two to the five z. They turn out to be one and seventeen as you can easily check. And in fact you notice that these are just the numbers that are one modulo two to the four. So b has a fourth root if b is congruent to one modulo two to the four. So by sort of accident we can just reduce from two to the five to two to the four in this case. Okay that's enough about using Henssel's Lemma to find roots of a polynomial. Next lecture will be about using Henssel's Lemma in a slightly more complicated way to factorize a polynomial into possibly non-linear factors.