This lecture is part of an online course on commutative algebra and will be about the Hilbert polynomial. So we will use the Hilbert polynomial in order to define the dimension of a local notarian ring. For this lecture we won't really worry too much about dimension or local rings or whatever. We will just define the Hilbert polynomial as some sort of abstract object. So suppose we've got a graded ring r. So it's going to be graded as a sum over integers n of components rn. And we're going to assume that r0 is notarian and r is finitely generated as an algebra over r0. So this implies r is notarian as well by Hilbert's theorem. And we're going to suppose that m is going to be a graded module over r. So it's going to be a sum of graded pieces mn. So this is a module. And we're going to assume that m is finitely generated as a module. Remember we have to distinguish between being finitely generated as a module or as an algebra. And the problem we want to discuss is how fast does mn grow as n tends to infinity. Well in order to discuss how fast it grows we need to know what is the size of this. So we need to have some way to measure the size of a module. And we're going to denote its size of mn to be as lambda mn which will be some sort of integer. And what is lambda of mn? Well we've got to decide that. So we can say what is lambda. Well here's an easy case. Suppose we take r0 to be a field. Then mn is always a finite dimensional vector space over r0. And we can take lambda mn just to be the dimension over r0 of mn. A slightly more complicated case. Let's take r0 to be an artinian ring. And then over artinian rings all finite e-generated modules have finite lengths. So we can put lambda mn is the length of mn. And lambda needs to have the following basic property. So if we've got an exact sequence of modules over r0 then we want that lambda of a plus lambda of c is equal to lambda of b. So this sort of says lambda is additive or additive on short exact sequences or whatever. And this is basically the only property of lambda we need, apart from the fact that lambda should be integer valued. If you know anything about k theory you can state this in a rather more exotic way saying that lambda must be a homomorphism from a k group of your ring r0 to the integers. But that's just a fancy way of saying it's additive. So now we can put together all these numbers lambda into something called a Poincare series. So we're going to define a formal power series in a variable t to be sum over n greater than or equal to 0 of lambda mn times t to the n. So this is just a formal power series with integer coefficients. And we've got a basic theorem that says that pt is a rational function. More precisely pt is equal to f of t divided by 1 minus t to the k1 times 1 minus t to the k2 up to 1 minus t to the ks. Well I need to explain what all these are. Well first of all this bit here is just a polynomial. And next we need to know what all these numbers here are. Well k1, k2, up to ks are the degrees of the generators of r. So we remember r was a finite generation algebra over r0. So we can pick generators of various degrees. So how do we prove this? Well it's fairly easy to prove. All we have to do is suppose the generators of the r0 algebra are rx1 up to xs. Where the degree of xi is going to be this funny number ki. We're going to use induction on s. So first of all for s equals nought this is trivial because then r is equal to r0 and mn is equal to 0 for n sufficiently large. So the Poincare series is just a polynomial.  Suppose s is greater than 0. What we do is we look at the following sequence. We take the exact sequence nought goes to kn goes to mn goes to mn plus ks goes to ln plus ks goes to 0. Well this is just multiplication by xs. And if we look at k which is the sum of all the kn's and l which is the sum of the ln's then these are modules over r in which xs acts as 0. So l is just the co-kernel of multiplication by x and k is just the kernel of it. So if we look at this exact sequence here then using the additivity of lambda we get the following exact sequence.  So lambda is additive on exact sequences with three terms and you can break any exact sequence into exact sequences of three terms. So when if you've got an exact sequence of modules like this then the alternating sum of the lambdas is 0. So now we convert this formula into a formula for the Poincare polynomials of k, l and n. So if we do this we find pk of t minus pm of t plus t to the minus ks pm of t minus t to the minus ks pl of t is equal to 0. So this formula is just the same as this formula except we've written it in terms of Poincare series. Well now we do a little bit of algebra and we find pm of t can be written in terms of these two terms as pl of t minus t to the ks pk of t divided by 1 minus t to the ks.  So this expression here is also a rational function except you've got to add an extra factor of 1 minus t to the ks in the denominator. So that proves the formula for the Poincare series of our module. Now we're going to look at the following special case. In this special case we're going to take all the xi have degree 1 so all the ki are equal to 1 in other words. So this is the most common case but it's sometimes usual to allow the xi's to have higher degree. So anyway this says that pt is equal to some polynomial in t divided by 1 minus t to the s where s is the number of generators. Well 1 over 1 minus t to the s of course. 1 over 1 minus t to the s can be expanded using the binomial theorem as sum over n greater than or equal to 0 of t to the n times n plus s minus 1 choose s minus 1. And we notice this is a polynomial in n if n is greater than or equal to 0. If n is less than 0 this becomes 0 and so this isn't a polynomial in n for all integers n but it is for n greater than or equal to 0. So the coefficients of pt are polynomials in n so let's say the coefficient of t to the n in pt polynomials in n for n sufficiently large. So you have to take n so that it's bigger than than the degrees of any of the terms in this polynomial here. But once you've done that all the coefficients of pt are going to be polynomials and these polynomials are integer valued. What this means is that p of an integer is an integer. So let's discuss integer-valued polynomials for a bit. So we can ask what are the possible integer-valued polynomials. This means they should take integers on integers. Well there are some obvious examples. We can just take all coefficients to be integers. So we take integer linear combinations of 1, t, t squared, t cubed and so on or so sum of n i, t to the i for n i and z. However these are not the only examples. For instance we could take n times n minus 1 over 2 and this is always an integer whenever n is an integer. Even though it's n squared over 2 minus n over 2 it doesn't have integer coefficients.  This is always an integer if n is an integer and i is greater than or equal to 0. And that's because this is just the binomial coefficient n choose i which is an integer because it's the number of ways of choosing an i from n n elements. So we can ask are there any other integer-valued polynomials and the answer is no this more or less gives all of them. So we have the following theorem. Any integer-valued polynomial f is an integer linear combination of these polynomials n choose 0, n choose 1, n choose 2 and so on. So this is 1, n, n minus 1 over 2, n, n minus 1, n minus 2 over 6 and so on. And let's look for proof of this. This is quite easy. Let's look at the values for various values of n. So let's take n to be 0, 1, 2, 3, 4 and look at the values of these polynomials. Well, this is going to be 1, 1, 1, 1, 1. This is going to be 1, 0, 1, 2, 3, 4. This is going to be 0, 0, 1, 3, 6. This is going to be 0, 0, 0, 1, whatever it is. Anyway, all we're really interested in are these values here and we can forget about all this rubbish down here. Now what we see from this is that given any integer polynomial, integer-valued polynomial f of degree d, we can find an integer-linear combination of n choose 0 up to n choose d with the same values n equals 0, 1, up to d. And the reason for this is we can fix the value at 0 by choosing a suitable linear combination of this and then we can fix the value at 1 by choosing a suitable linear combination of this and that doesn't affect the value at 0. Then we can add an integer multiple of this and fix the value at 2 without affecting the values at 0 and 1 and so on. So we've got this integer-valued polynomial f of degree d and this is equal to a 0 times n choose 0 plus a 1, n choose 1 and so on, up to plus a d, n choose d for some a 0, a 1, up to a d. And these are the same for n equals 0 up to d. And now we've got two polynomials of degree d, which are the same for d plus 1 values of the argument. So these must be the same for all n. So our integer-valued polynomial is an integer-linear combination of these binomial coefficients. Now there's a simple consequence of this. The leading coefficient of f, n is of the form d factorial, sorry, some integer divided by d factorial times x to the d. And that's because the leading coefficient of this polynomial here is 1 over d factorial. Of course, the leading coefficient need not be an integer as we've seen in the example n, n minus 1 over 2, for example. However, it isn't an integer if you multiply it by d factorial. So let's see an example of this. Well, suppose we have a projective variety given by some homogeneous ideal, i in k x 0 up to x n. So the variety is just going to be the set of points where all elements of this ideal vanish. Then we can take the ring to be k x 0 up to x n modulo i. So this is the homogeneous coordinate ring of our variety. And this is graded. So the dimension of Rn, so I guess I shouldn't call that n, the dimension of Rk is a polynomial in k for k large. And the degree d is the dimension of the variety. And d factorial times the leading coefficient is called the degree of the variety, which is an integer because we just said d factorial times the leading coefficient is an integer. So for an explicit example of this, let's just take a hyper surface in n dimensional projective space. So suppose i is just generated by f, where this is homogeneous and of degree m in n plus 1 variables.  Because these are the number of monomials in the polynomial ring, k x 0 up to x n.  And if you work out the leading coefficient of this, this turns out to be m k to the n minus 1 over n minus 1 factorial plus something or other. So the degree of the variety is equal to m, which fortunately is the same as the degree of this polynomial here. And the dimension is n minus 1, so the n minus 1 comes from this leading coefficient. So next lecture we'll be showing how to use the Hilbert polynomial to define the dimension of a local ring.