 This talk is the third of three lectures about the Bernstein-Sato polynomial. So in the first lecture we defined the Sato Bernstein polynomial, but we didn't prove it existed. In the second lecture we defined holonomic modules over the vial algebra and proved Bernstein's inequality and quickly defined holonomic modules. What we're going to do this lecture is to study holonomic modules and use their properties to prove the existence of the Bernstein-Sato polynomial. So let's just recall what a holonomic module is. We have the vial algebra A, which is noncommutative polynomials in variables x1 to xn and their partial derivatives, so this is short for differentiation with respect to x1 and so on. And if m is a module over the vial algebra, Bernstein inequality shows that the dimension of m is at least n if nm is not equal to zero. Here the dimension of m doesn't mean its dimension is a vector space, which is usually infinite, but it's dimension defined as the degree of a Hilbert polynomial associated to it. So in the commutative case, modules have dimension at least zero and the modules of dimension zero are all of finite length, at least if they're finitely generated. What we want to do is to show that something similar holds here, that the modules of finite length, which are the holonomic modules, so holonomic means finitely generated plus dimension equals n for zero. The zero module is also considered to be holonomic. So what we want to do is to prove the first theorem, holonomic modules have finite length. So what is going on here intuitively is that if a module has dimension k, then any filtration can have only a finite number of subquotions of dimension k bounded by some constant, but it can have infinitely many subquotions of dimension less than k in general. So if there are no modules of dimension less than k, it has to have finite length. Now over commutative rings the only interesting case is when k is equal to zero, in which case the module is zero dimensional and has finite length. In the non-commutative case we get this funny bound due to Bernstein showing that modules can't have dimension less than n unless they're zero and this will turn out to mean the modules of dimension equal to n have finite length. And its proof isn't very difficult. So first of all the Hilbert polynomial is additive on exact sequences. Which is very easy to show. So the leading coefficient is additive on exact sequences. For holonomic modules as the Hilbert polynomial has dimension exactly equal to n unless it's zero. If we weren't talking about holonomic modules what might go wrong is that in an exact sequence nought goes to a, goes to b, goes to c, goes to zero. One of these modules might actually have dimension less than the others and then we couldn't really say very much about the relation between the leading coefficients because there would be coefficients of different degrees of monomials. But as long as a, b and c all have the same dimension and the same leading coefficient, the same highest degree, the leading coefficients must be additive. Well, so the multiplicity is additive. So you remember the multiplicity is the dimension factorial times the leading coefficient. And again the multiplicity is certainly not additive if you don't work with modules that aren't of the smallest possible dimension, which is n in the case of modules over the vial algebra. And the point is this multiplicity is always an integer greater than or equal to zero and it's equal to zero if and only if the module is zero. So in particular the length of a module must in fact not only be finite but it's less than or equal to the multiplicity because if it has several composition factors, each of these composition factors is the same dimension so the multiplicity of the module is the sum of the multiplicities of the composition factors and the multiplicities of the composition factors must all be at least one. So the length is bounded by the multiplicity which is finite. So next we're going to have to deal with modules that we don't know are finitely generated. So suppose m is a module over a. So not known to be finitely generated. What we want to do is to prove that it's finitely generated. And suppose it has a filtration, m0 contains m1 and so on, with such that the dimension over the complex numbers of mi is bounded by some polynomial in i of degree n. Then m is finitely generated and so holonomic because if it's finitely generated and it's Hilbert polynomial's degree at most n then we have to find it to be holonomic. So the key point is to prove finite generation. And this is easy because first of all every finitely generated sub module is holonomic of multiplicity less than or equal to n factorial times leading coefficient of the polynomial we had up there. And this implies m is finitely generated. Otherwise we can find a chain of finitely generated sub modules. I can't call them m sub i so let's call them m superscript 0 contained in m superscript 1 and so on. These would be strictly increasing, they're all finitely generated. And this is impossible because the lengths of the mi must be bounded by n factorial times the leading coefficient of this polynomial. So in other words in particular the multiplicity of this module m is at most the dimension factorial times that this leading coefficient. So since we can't find an infinite chain of finitely generated sub modules this immediately implies that our module is finitely generated and therefore holonomic. Now we're going to have a corollary of this, let's call it corollary. Let's let the module m be the ring of polynomials over the complex numbers. Now this is a module over the vial algebra because we can multiply these by polynomials and differentiate them and I'm also going to add in p to the minus 1, where p is some polynomial and if m is this then m is holonomic. So if we didn't have the p to the minus 1 this would be kind of obvious because we could just work out the Hilbert polynomial of this but if we allow inverses of a fixed polynomial so we're localising at one polynomial then we've got to be a little bit more careful. So proof, we're going to define a filtration. We put mk is the set of all polynomials that form f over p to the k where the degree of f is at most m plus 1 times k where m is the degree of p. And then we see that ai mk is contained in mk plus i so we've got a perfectly good filtration. And we can also see that the dimension of mk is at most m plus 1 k plus n choose n. I've got that right. Which is a polynomial in k of degree less than or equal to n. So by the previous lemma m is finitely generated so holonomic. So all we're saying is that this module in some sense doesn't grow faster than a polynomial of degree n which we saw automatically implies it's holonomic. In fact we can also see its multiplicity is at most m plus 1. If you care about that. Now finally we can prove existence of the Bernstein polynomial. So let's just quickly recall what the Bernstein polynomial is. So we want the Bernstein polynomial of some polynomial p which is a polynomial in several variables x1 to xn and what we want to do is to find a polynomial b so this is the Bernstein's SATO polynomial and b of s times p to the s should be equal to some polynomial in the xi and the delta i and s times p to the s plus 1. Well what we do is we work over the field k which is the ring of rational functions over the complex numbers in s. So this is going to be rational functions and we're going to cheat a little bit because we were proving results about the vial algebra over the complex numbers but in fact that everything works for the vial algebra over any characteristic zero field such as this field here. And now what we're going to do is we're going to take the field c of s and then we're going to look at all polynomials x1 up to xn and we're going to put p to the minus 1 and then we're going to cheat again. We're going to multiply this by p to the minus s and we're going to say this is holonomic over the vial algebra a which is now we now work over the field of rational functions in s and have x1 up to xn delta 1 up to delta n. Now this more or less follows from the previous lemma except first of all we've put an extra factor of p to the minus s in here and this doesn't really affect the proof and secondly we're not quite working over complex numbers but over this other characteristic field and this doesn't really matter either. The previous proof just goes through for this and shows it's holonomic. Now we have a chain of sub-modules. We look at ap to the s which is contained in ap to the s plus 1 which is contained in ap to the s plus 2 and so on inside this module. You notice p is invertible so that's fine and this is a decreasing chain in a module of finite length and you may think being finite length is this sort of rather technical condition but in this case it's incredibly powerful because if you've got a decreasing sequence in a chain of modules of finite length so it must eventually be constant so it's eventually constant. In other words this says that ap to the s plus k is equal to ap to the s plus k plus 1 for some k. So p to the s plus k is equal to p to the s plus 1 plus k times some polynomial p in the xi and the delta i where this may be calling it p isn't a terribly good name let me call it q so you don't get it modelled up with this p where this is a polynomial with coefficients in and now the coefficients may actually be rational functions of s. So p to the s is also some polynomial in the xi and the delta i of p to the s plus 1 with coefficients in c to the s we can just change s to s plus k it's quite harmless. Now these are rational functions in s and we let bs be a common denominator so here we've got a polynomial it's got only a finite number of coefficients each coefficient is a rational function so we've got a finite number of denominators and we can multiply them all together and get this hugely complicated common denominator and we just multiply by the common denominator and we get bs p to the s is now a polynomial in xi delta i and s times p to the s plus 1. So this is now a Bernstein-Sato polynomial so we found a property a polynomial b of s with the property that we needed for the Bernstein-Sato polynomial if we apply it to p to the s we get some polynomial in these variables times p to the s plus 1 so that proves the existence of the Bernstein-Sato polynomial