 This algebraic geometry video will give some more examples of resolutions of singularities. So the first example we're going to look at is the equation x to the four plus y to the four equals z squared. So this is a historically rather famous equation that was used by Pierre de Fermat in his proof that x to the four plus y to the four equals z to the four has no integer solutions apart from the trivial one. He in fact showed this equation here has no integer solutions. Anyway, we're not going to discuss Fermat's work on this equation. Instead, we're going to look at the singularities of this equation. So we can find the singularities by differentiating with respect to x, y and z. So we find four x cubed equals naught four y cubed equals naught and two z equals naught for a singular which obviously shows the only singular point is the origin. So now we're going to resolve the singularity of the origin by blowing up at the origin. So we're going to blow up the point zero zero zero. So this is a subset of a cubed. So we're going to work inside a cubed times p squared. And we're going to introduce new variable big x b y and big z for p squared. And we get the equations x times big y equals y times big x x times big z equals z times big x and y times big z equals z times big y. And now as usual this copy of projective space is covered by three open affine sets because we can take big z equals one or big y equals one or big x equals one. If we take big z equals one then we write x equals z times x y equals z times y and we find z to the four x to the four plus z to the four y to the four equals z squared or z squared x to the four plus c squared y to the four equals one. And you can easily check this is non-singular. So let's look instead at y equals one. So this time we put x equals y times big x and z equals y times big z. Now the question becomes y to the four x to the four plus y to the four equals y squared z squared. Now we can divide this out by y squared and we get y squared x to the four plus y squared equals z squared. Now if we check for singularities of this equation we see we check for singularities by differentiating with respect to x, y and z. So we have y squared times four x cubed equals naught. We differentiate with respect to x. If we differentiate with respect to y we find two y times x to the four plus one equals naught and if we differentiate with respect to z we find two z equals naught. And now you notice that we get singular points whenever y equals z equals zero. So these are all singular. So we have an entire line of singular points and if we put y equals one we get something very similar. In fact we get more or less the same line. What is in fact going on is we've got an entire projective line of singular points and this is one of the affine pieces covering it. So this is different from the duval singularities we had in the previous lecture where we blew up a point of the singularity and we kept on getting singular points. So here we've blown up a point and now we've got a singular line so you might think this has made the singularity worse because the dimension of the singularity has got bigger but that's misleading. The nastiness of a singularity is really more to do with its co-dimension than with its dimension. So singularities tend to be easier to deal with than singularities of co-dimension too. So turning a point into a line is actually good. Anyway if we want we can continue and finish resolving this so we've got the equation y squared x to the four plus y squared equals z squared and now we can resolve this singularity by blowing up along this line y equals z equals zero. So let's introduce two new variables s colon t for a copy of p1 and we introduce the equations s times z equals t times y and now we can for example cover this affine line by two sorry this projective line by two copies of an affine line we either set s equals one or t equals one if we set s equals one we find z equals t y and our equation becomes x to the four plus y squared equals t squared y squared we can now divide by y and we get x to the four plus one is equal to t squared which you can easily check is non-singular and similarly if we look at t equals one we get a non-singular line. So we've resolved the singularity of our original equation in two steps first we blow up a singular point and then we blow up a singular line and when we've done this we end up with a non-singular variety. The next example well I've been giving a lot of examples where blowing up a singular point or line makes the singularity better if you choose your blow up badly you can actually make the singularity worse so let's have a look at x squared minus yz equals nought so this is just a cone it's got a conical singularity of the origin which is easy enough to resolve by blowing up the origin as we did in an earlier example well suppose instead of doing that let's blow up along the line y equals z equals zero so what we do is we introduce two new coordinates y comma z in the projective line so this is in a cubed and we're now going to be working in a cubed times p1 and we introduce the usual extra equations for the blow up y times big z equals z times big y and now as usual we take one of the affine lines that forming a cover of p1 so we can take either big z equals one or big y equals one so if we take big y equals one we make the substitution z equals y times big z and our equation becomes x squared minus y squared z equals zero so we started off with a fairly simple conical singularity and now this more complex singularity so this is now the Whitney umbrella which has definitely got rather more complicated you see this equation only has degree two and by blowing up we've actually turned it into a degree three equation which is clearly more complicated than this one so the point of this is you need to be a little bit careful about where you blow things up it doesn't automatically make things better it can make them worse next we have an example from number theory so in number theory we can look at the ring z and join the square root of minus three and this ring has the problem it doesn't have unique factorization um for instance we have two times two equals one plus root minus three times one minus root minus three and unique factorization doesn't hold and so on um well what does this have to do with singularities and algebraic geometry well what we're going to do is pretend that this is a coordinate ring of a variety now actually it isn't a coordinate ring of a variety it's actually a coordinate ring of a scheme that we will be studying later but we haven't actually covered schemes so we're going to pretend it's the coordinate ring of a variety and um what are the points of this variety well the points are going to go to respond to maximal ideals so although it's not really a variety with the points we can still look at the maximal ideals and it's got a maximal ideal generated by two and root minus three um this isn't a principal ideal domain so the maximal ideals aren't necessarily um um generated by one element and what we're going to do is try and show that in some sense the variety corresponding to this or scheme corresponding to this has a singular point at this point here where we're pretending that maximal ideals are points so what we should do is we should look at the local ring near this point and we can um take the local ring to be z2 root minus three so z2 is the ring you get by inverting or um or uh primes not equal to two um so this is now a local ring and the maximal ideal of this local ring is now generated by two and root minus three minus one uh sorry this maximal ideal should have been two root minus three minus one this thing isn't a maximal ideal it's the whole ring so we have a maximal ideal that I'm going to call m and if let's call this ring let's call uh this local ring r then r over m is a field it's just the field with two elements as you can use um and r over m squared has order eight again it's not too difficult to check so m over m squared has order four and is a two-dimensional vector space over um r over m and this space here you remember is really the cotangent space at um this point corresponding to the maximal ideal so we can work out the dimension of this ring by which we mean the crawl dimension which is the greatest length of chains of prime ideals and um the only prime ideals are the zero or maximal ideal so this ring r is dimension one and now we've got a point where the cotangent space at this point m has dimension two which is bigger than the dimension of the ring so it corresponds to a singular point or at least it would if this was a variety which it isn't but never mind and the fact that this is not a unique factorization domain turns out to be related to the fact that it has a singular point um roughly speaking the fact that it's got a singular point implies that this can't have unique factorization it almost doesn't actually hold there are some rings of algebraic integers that have no singular points but still don't have unique factorization but anyway we can now resolve the singularity and there are several ways of doing this we can resolve this singularity by blowing up um we won't do this um in number theory you resolve the singularity by taking something called the normalization of the ring or the interval closure and taking the interval closure of z root minus three gives you the ring z root minus three sorry root minus three plus one over two see if we call this element here omega then omega squared plus omega plus one is equal to zero so omega is an algebraic integer lying in the field of this and that's what you do in algebraic number theory to take the um integral closure anyway this ring does have unique factorization which you can go to any algebraic number theory course and its points are non-singular so we can say this is non-singular so taking the interval closure of this ring is sort of the same as taking a desingularization of a variety if we pretend this is the coordinate ring of a variety then this would be the coordinate ring of a desingularization of that variety except of course as I said they're not really varieties they're really schemes um so um so the the process of taking the integral closure of an order of an algebraic number field in number theory is really almost the same as um resolving the singularities of an algebraic curve um next I want to give an example of application of resolution of singularities so here here a great theorem that you could resolve the singularities of any variety in characteristic zero and it's supposed to have large numbers of important applications so I thought I would give a rather sketchy description of one particular application of it um so you all know the gamma function gives the interval zero which is minus t t to the s minus one dt and this converges for the real part of s greater than zero and you can extend it to an anamorphic function of s by um integrating by parts in the usual way and this question posed we've got a polynomial in several variables f of x1 up to xn and we multiply it by some smooth compactly supported function phi and we're going to take this to the power of s so this is going to be some polynomial and integrate with respect to dx1 up to dxn so this is smooth compact support and um this um is going to be a holomorphic function of s if the real part of s is reasonably large and the problem is can we analytically continue to all s so you can see this is more or less the same as the gamma function so f would correspond to the very simplest polynomial um just t and phi would correspond to e to the minus t and yes i know e to the minus t doesn't have compact support but frankly i don't really care um and the problem arises because f might be zero somewhere you see if f is never zero there's absolutely no problem um analytically continuing this because this is just a perfectly well-defined finite number so the problem occurs at the zeros of f and there's zeros of f will be some sort of hypersurface with singularities and the more complicated the singularities are the more difficult it is to figure out what on earth is going on in this integral and a tear pointed out there was a very easy way to settle this question what you do is you just resolve the singularities of the equation f equals zero so you use Hieronarchist theorem to resolve the singularities of the variety f equals zero now what you end up with when you do that isn't actually a non-singular space because when you blow up singularities according to Hieronarchist theorem you introduce all these exceptional curves now the exceptional curves don't really matter because they all cross transversely so when you resolve singularities by blowing up repeatedly strictly speaking you don't quite resolve them what you end up with is a lot of singularities that look things like x k equals zero so we just have a lot of hyperplanes all intersecting at right angles to each other so we can doing this we reduce the case of integrals like x one x two x n to the s times phi of x one x n and so on dx one to x n equals zero and this is really easy to do because this just splits as a lot of integrals of the form like x one to the s times f of x one dx one equals zero and you can do this one in just the same way that you do the gamma function by integrating by part so resolution of singularities allows you to prove meromorphic continuation of integrals like this actually shortly after or maybe at the same time that a tier found this proof using resolution of singularities Bernstein found a much more elementary and very ingenious way of proving analytic continuation of these functions using polynomials he introduced called Bernstein polynomials also called Bernstein's Sato polynomials because Sato discovered them at a similar time so this particular example is mainly maybe kind of cheating slightly because it turned out you don't really need the full power of Hieronarchis theorem to prove this analytic continuation but at least shows a sort of typical way in which you can use resolution of singularities to prove things okay the next lecture will be about completions of local rings and what they have to do with singularities