 This algebraic geometry lecture will be about Newton's rotating ruler and its relation to resolution of singularities of an algebraic curve. So let's start by taking a polynomial f of x, y with complex coefficients. Then the result that Newton proved was that we can write y is a power series in x to the 1 over n for some n. So it's not quite a power series in x, it's a power series possibly in some rational power of x. These things are more or less called pre-se series. Actually pre-se series also allow negative powers of x. The pre-se series would be the union over all n of power series in x to the 1 over n, where you also allow x to the minus 1 over n. The pre-se series were introduced by pre-se in 1850, but they were actually used also by Newton. I think he introduced them in 1676 or so, so he was a couple of centuries before pre-se. They really ought to be called Newton series, but nothing is ever named after the person who first discovered them. So let's see a couple of simple examples. If f of x, y is y squared minus x squared minus x cubed, then we can just write y is equal to x times the square root of 1 plus x. And we can write this as a power series x times 1 plus x to the half of that and so on. So in this particular case, we can write y as a power series in x. In general, you do need fractional powers of x. For instance, if f of x, y is y squared minus x cubed, then you can't write y as a power series in x, but you can write y as x to the 3 over 2, which is a power series, and x to the half are rather trivial power series because it's only got one nonzero term. So in general, for more complicated f, you'll get some power series in some fractional power of x. So how do you prove this? Well, what you do is, first of all, we should assume f is not a polynomial in x because if it's a polynomial in x, then you can't write y as a function of x. So we can assume that f x, y has a term that perform y to the n, but none, but not y to the i for i less than n. So we take the smallest power of y that occurs in f. If everything in f is divisible by x, then we just divide by f and continue, so that's harmless. And then what we do is we plot all monomials of the form x to the i, y to the j that occur in f. So what we do is we sort of draw this big diagram. Here we'd have one x, x squared, x cubed. So we just draw the square lattice of all possible monomials. So here's y, y squared, x, y. And what we do is we draw, mark the ones that occur in f. So maybe this one occurs, and maybe this one occurs, and this one occurs, and this one occurs, and so on. So here this is going to be y. We're going to take this as our term y to the n, so in this case n equals 3 because this is the smallest power of y. And now we get Newton's rotating ruler, so here's Newton's ruler. And what you do is you put the ruler on this point here and rotate it until it hits the first point. So this is Newton's rotating ruler. I guess strictly speaking, this is my rotation ruler, but whatever. And you draw this line through these points. In general, you can sometimes continue this. We don't need to for this particular example, but you can continue by then going to this point and rotating your ruler again and getting another point. And then you could continue at this point and so on. And the polygon you get like this is called the Newton polygon. There are actually several slightly different variations of what the Newton polygon actually is. But for this lecture, we're only going to use the first edge of the Newton polygon. We don't really care about these, but the other ones are important in other applications. And now, well, there's a bit of a problem here because you notice this edge doesn't pass through a point on this line. And this is a bit inconvenient. So what we want to do is to add extra points so that it passes through a point for y to the n minus 1. So here's y to the n minus 1. And we want it to pass through this point. And that's easy to arrange. So what we have to do is to change x to x to the 1 over r for some suitable value of r. So r, in this case, will be 2 because we have to go down by 2 to get to that point there. And changing x to x to 1 over the r is harmless because you remember we're using Prisa series where you're allowed to use rational powers of x. So we can assume the Newton polygon looks like this. So here we've got y to the n, and we've got 1. And the Newton polygon, by filling around with x, we've arranged that the edge of the Newton polygon actually passes through this point y to the n minus 1 x to the something. So this will be y to the n minus 1 x to the some other number m. So this term will occur in the polynomial. This term may or may not occur. And there may be several other monomials that occur. And what we do is we think of these as the smallest degree terms. Well, here, by degree, we have to put degree of y is equal to m and degree of x equals 1. So we're not giving y and x both the same degree. But we think of the terms here as being the smallest degree terms. And the things on this line form a homogeneous polynomial in y and x to the m. And we can factorize it. So we get y minus alpha x to the m, y minus beta x to the m, and so on. And now we're going to make the change of variable. We just change y to y plus alpha x to the m in order to get rid of one of these factors. So our polynomial now becomes y times y minus something x to the m times y minus something x to the m and so on. And what this means is that if you take this line here and extend it down where it meets the x-axis, this coefficient vanishes. So there's no term in x to the mn. And so we now have y as a root. And it may be a root. So not y as a root. Zero is a root. And zero may be a root of high multiplicity. We're not quite sure. So now what we do is we make the following change of variable. We change y to x to the m times y. Now if we do this, all the monomials y, y to the n minus one. So y to the n will change to x to the m, y to the n, y to the n minus one, x to the m will change to, so x to the mn, x to the mn, y to the n minus one, and so on. And these are all divisible by m, x to the mn. So we can then divide by x to the mn. And let's look at what this does to the coefficients. So what happens is that the coefficients on this line that we were interested in all get shifted. So y to the m times x are just fixed. So these stay where they are. These coefficients get shifted like that. And this coefficient gets shifted there and so on. And these coefficients get shifted like this. This coefficient gets shifted like that. So all the coefficients on the Newton ruler line get shifted to here. And this coefficient will be zero because we made it zero by changing y to y plus something times x to the n. And some of the other coefficients might be zero. And this coefficient is definitely not zero. Now, if any of these coefficients are none zero, then we've changed our original problem to one with a smaller power of y. So if these are none zero, we've changed y to the n to y to the power of y to something smaller. This corresponds to the coefficient of y to the n. And you see if one of these is none zero, we've now got a smaller power of y. And so we can just go back and start again. So we start again with the smaller power of y. So that happens if this original root alpha was of multiplicity less than n. So we have to think further about what happens if this original root alpha happens to have multiplicity exactly equal to n. This can happen quite a lot, especially if n is say one. But it occasionally happens if n is bigger than one. So what happens in this case is, let's draw these points again. What has happened is that if it is equal to one, then actually this coefficient was one and actually all these coefficients were in fact zero. And so by making the change of variable, we either reduce the smallest power of y or we make all the coefficients on Newton's line zero. Well, if they're all zero, then Newton's ruler doesn't get stuck here. We can now rotate Newton's ruler a little bit further. And we now run into a different point here somewhere. So what happens is what we've done is we've made a change y goes to y plus something times x to the n. And either this makes n to something smaller or we can rotate Newton's ruler a bit further. If we rotate Newton's ruler, we can repeat this whole business again. So we can now change y to y plus something times x to the n plus one. And again, either we will obtain a power of y less than n or we will kill off all the coefficients on this line and then we can sort of continue like this. So we can keep changing y to y plus a power of x plus a higher power of x plus a higher power of x. And this either continues forever or we reduce the smallest power of y. Now we can only reduce the smallest power of y a finite number of times because it starts with n and can't get below zero. So eventually we must hit a point where this process here continues forever. Well, what happens if we do that? So we're sort of changing y to y plus something times x, then we're changing y to y plus something times x squared and so on. And the result of all this is we change y to y plus something times x plus something times x squared plus something times x cubed and so on. And the result of all this is that all powers of the form y to the ix to the j for i less than n and now zero after this change. So our series is divisible by y to the n, but by the time you've got to this point n is usually equal to one. So anyway, it's divisible by y. Well, this means the original series, so the original series f of x, y is divisible by some power series in x to the one over r for some r. So you remember, at various points in the calculation we had to change x to x to the one over r in order to ensure that this line went through a point with coefficient y to the n minus one. So we've managed to find a power series solution of the form y. So we've managed to find a power series solution y equals power series in x to the one over r which satisfies f of x, y equals zero. So that concludes the proof of Newton's theorem that you can find solutions to algebraic equations that are power series and some fractional power of x. So what can you do with this? Well, the first consequence is that the field of pre-sust series is algebraically closed. So you remember the pre-sust series are things of the form that can be written as c of x to the one over r, x minus one over r. So these are power series and x to the one over r which are allowed a finite number of negative exponents. And the reason for this is you see Newton's method works perfectly well if you've got something in x and y where this is a pre-sust series in x. You still find a solution y. And you can think of this as being a polynomial in the field k of y where k is the field of pre-sust series. So what this is saying is that if you've got a polynomial in y whose coefficients of pre-sust series, then you can find a root that is also pre-sust series. In other words, this is just saying the pre-sust series are an algebraically closed field. It's one of the very few examples of an algebraically closed field you can write down explicitly other than the complex numbers. In fact, it's an example of a quasi-finite field. Sorry, the pre-sust series are an example of a quasi-finite field, which means that the haze is very much like a finite field in the following sense. So suppose you take the pre-sust series, the field of pre-sust series and you look at all its algebraic extensions. Well, by Newton's theorem, the only algebraic extensions are ones you get by adding a root of x. So we get k and then we're going to join x to the half or we're going to join x to the quarter or we're going to join x to the third or x to the one over sixth and so on. So the collection of field extensions of k is very easy to describe. You just get one extension of degree n for every n and they're embedded in each other in this way. You remember this is exactly what happens for a finite field of order p. It's contained in the finite field of order p squared and the finite field of order p cubed and these are contained in the finite field of order p to the sixth. This is contained in the finite field of order p to the fourth and so on. So in both cases, you can write down the absolute Galois group. It's just a sort of inverse limit of cyclic groups and it's just the completion of the integers. Whoops, sorry, that should be z hat, which is a sort of almost a cyclic group except it's not quite cyclic. So the field of Prisa series behaves very much like a finite field in that they have the same absolute Galois group. And finally, I should explain what Newton's theorem has to do with resolution of singularity. Suppose you've got a plain curve singularity. So what's the curve going to do? It does something complicated at the origin. It's got a singularity here. And what Newton's theorem says is that if we change x to x to 1 over r, this kind of looks as follows. So we've got a singular point here and we've now got several curves. So here's y is equal to f1 of x to 1 over r. And we've got another curve going through it. y equals f2 of x to 1 over r. And so on, we might have y equals f3 of x to 1 over r and so on. So these are Prisa series. So Newton's results showed that we can find one curve like this, but obviously we can just sort of iterate this and decompose this as a product of n curves like this where the smallest power of y is y to the n. And this is very nearly a desingularization of the singularity here because you sort of split up the singularity into several nice non-singular curves which just happen to touch each other. So you can imagine we can just take hold of these curves and just pull them apart and that would be a resolution of singularities of our original singularity. And of course Newton didn't state a theorem about resolution of singularities. I don't think people even use the term resolution of singularities. However, his method of proof is more or less, more or less shows you can resolve the singularities of an algebraic plane curve and you can modify it very slightly so that it does in fact become a resolution of singularities. So I think it's fair to say that Newton was the first person to show that you could resolve the singularities of a plane curve.