 So, this algebraic geometry video will be about the degree of a projective variety. So suppose we've got a projective variety, v, contained in n-dimensional projective space. First of all, if v is a hypersurface, it will be given by vanishing of an equation f0 up to xn equals 0, where this is a polynomial. And we can just set the degree of v equal to the degree of f as a homogeneous polynomial in the xi's. Well, that's easy enough. We would also like to define the degree of v if v is of co-dimension bigger than 1. Well, you notice in this case the degree of v is equal to the number of intersection points of v with a hyperplane section h, except it sometimes isn't. So we should say this happens most of the time. And the reason why it sometimes doesn't happen is, well, first of all, suppose v is just a line. Then if you take another hyperplane section, it will have one intersection point with v. Well, except it sometimes doesn't, because if you take v to be a line and the hyperplane section is the same line, then it has an infinite number of points of intersection, so that doesn't work. Similarly, if v has degree 2, say it's some sort of conic, it might look like a parabola, then a generic line will have two intersection points with it. But what about this line? So instead of doing that, we could take this line here. And it's sort of got one intersection point of multiplicity 2, whatever multiplicity means. So defining degree to be number of intersection points is a bit iffy, because you have to say it's usually the number of intersection points, but it sometimes isn't. The way people got round this was by saying it's the number of points of intersection with a generic hyperplane. And then you run into the problem of what does generic mean, and you find that generic isn't really precisely defined. What it means is a hyperplane is called generic if it makes everything work properly. Well, that's not really satisfactory as a mathematical definition. But anyway, you can use this informal and non-rigorous definition to define the degree of v for any other variety. You say the degree of v, you want it to be the number of intersection points with a linear variety. That means a line or a plane or a hyperplane or something of co-dimension given by the dimension of v. So for instance, if v has dimension 2, then you'd take a linear variety of co-dimension 2 and count the number of intersection points with v. And again, you'd have this problem. You have to take the linear variety to be in sufficiently general position, whatever that means. So that was roughly the old classical definition of the degree of a variety, which is fine for informal arguments, but if you actually want to prove things, it becomes really rather tedious because there are all these exceptions and you're not quite sure when one of these exceptions takes place. And it turns out there's a much cleaner definition using the Hilbert polynomial. So the variety v corresponds to an ideal contained in the graded ring of projected space. So v is just a set of zeros of i. Let's call this ring r. And then we can look at r over i. So this is going to be a graded ring. And since it's a graded ring, it has a Hilbert polynomial. So we define the Hilbert polynomial from module over a graded ring, but we consider this ring as a module over itself, so we get a Hilbert polynomial. And then the degree of the Hilbert polynomial turns out to be the dimension of v. So this is a theorem from commutative algebra that says it's very close, well, it's very closely related to the theorem that says the dimension of a local ring is equal to the one plus the degree of the Hilbert polynomial of a graded ring m to the n over m to the n plus one. I'm not going to prove this theorem in this course because it's a somewhat technical theorem, which is probably best left to textbook some commutative algebra rather than lectures. But anyway, so commutative algebra shows that the degree of the Hilbert polynomial is just the dimension of the variety. Incidentally, this gives a fairly efficient way of calculating dimensions of varieties because the Hilbert polynomials are very computable objects. So if the dimension is equal to d, then the leading coefficient of the Hilbert polynomial is going to be a times x to the d over d factorial for some integer a, which we saw last lecture, and this integer a is called the degree of v. So property of integer-valued polynomials is that d factorial times the leading coefficient is always an integer, so a is an integer and it's also fairly obvious that it's a positive integer because we said the degree of the polynomial was d and we know that its value is a positive for large values of x. So this is a positive integer a and it gives a well-defined invariance of any projective variety. Notice this depends, when I say it's an invariant of a projective variety, it means it's an invariant of the projective variety together with its embedding into projective space. So this isn't an invariant of the abstract variety v. So let's see some examples. Let's take v to be projective space. Well, here we just take the graded ring in n plus 1 variables and the ideal is just nought. So we're looking at the graded pieces of this ring here and you see the graded pieces of dimension 1. Well, there are n plus 1 things of degree 1. There are n plus 1, n plus 2 over 2 things of degree 2, and n plus 1, n plus 2, n plus 3 over 3 factorial things of degree 3 and so on. So there's just a number of monomials of given degree. And in general, we're getting n plus k choose n, which is a polynomial in k of degree n and it's equal to k to the n over n factorial plus smaller terms. So we see the dimension is equal to n, which is this coefficient here, and the degree is equal to 1, which is this coefficient here. We've got to remember we multiply the coefficient of k to the n by n factorial. So n dimensional projected space does indeed have dimension n and degree 1, which is what we rather hope would be the right answer. So now let's look at a hypersurface of a degree D polynomial f x naught up to x k, sorry, up to x n. Well, here, we're now looking at the ring k x naught up to x n modulo all multiples of f. So the Hilbert polynomial is going to be n plus k choose n. So this is the number of monomials of degree k, and then we have to subtract n plus k minus d choose n. So these are monomials of degree k minus d and we have to subtract those because any of these multiplied by f will be of degree n. So that's degree, sorry, degree k not degree n. And now we can work out leading coefficient of this quite easy. It turns out to be d times k to the n minus 1 over n minus 1 factorial plus smaller terms. So we find the dimension is the degree of k, which is just as well a hypersurface really ought to have dimension n minus 1. And the degree is equal to d, which is again rather fortunate. So this coefficient here does indeed turn out to be the degree of the polynomial f. So the degree defined by Hilbert functions is the same as the naive degree defined by the degree of this polynomial f. So this confirms that defining the degree by Hilbert functions does seem to give the answer we expect. Now let's look at a slightly more complicated example. Let's look at the twisted cubic in p3. So you remember if we take the ring of polynomials to be W x, y, z for p3, then twisted cubic is defined by the equations W x, so W z equals x, y, x squared equals W y, y squared equals x, z. So we want to look at the polynomial ring quotient out by the ideal generation by these. So in general, what we can do is we can kill off any terms containing x squared, y squared or x, y, because we can turn them into smaller powers of x and y. So a basis for the quotient ring is just the set of monomials W i, z to the k minus i, W i, x, z to the k minus i minus 1 and W to the i, y, z to the k minus i minus 1. So these are degree k. This is for k reasonably large, greater than 3 or something. I don't quite know how long you have to go. So you see there are k plus 1 plus k plus k possibilities, which all together gives us k plus 1 plus k plus k equals 3k plus 1. That's the Hilbert polynomial. I guess we could write a divide by one factorial there. And now as we see the degree is equal to 3. And while we're about it, there's an exponent one here. So the dimension is equal to 1, which we already knew. So the twisted cubic curve in P3 has degree 3. Incidentally, you notice this shows the degree is not an invariant of the underlying projective variety because the twisted cubic is isomorphic as a projective variety to P1. And P1 only has degree 1. So the degree does actually depend on the embedding of the abstract variety into projective space. There's an invariant of projective varieties, sometimes called the Euler polynomial of V. And this can be defined as the constant term of the Hilbert polynomial. And this is a remarkable property that Xi of V depends only on the abstract variety, not on the embedding into projective space. So notice in the previous example that although the twisted cubic and P1 have different Hilbert polynomials, they both have the same constant terms. They both have the same Euler characteristic of 1. For historical reasons, people didn't used to quite use the Euler characteristic. They used minus 1 to the dimension of the variety times the Euler characteristic of V minus 1. And this was called the arithmetic genus. The reason for that is just people hadn't quite figured out what was going on in high dimensions where the arithmetic genus was defined. So they kind of got the definition a little bit muddled up. So the Euler characteristic is better than the arithmetic genus because it's the nice property that if you take the Euler characteristic of product of two polynomials, it's equal to the Euler product of two varieties. It's equal to the product of the Euler characteristics. So that shows it behaves a lot better than the arithmetic genus. The Euler characteristic is called the Euler characteristic because it's actually the Euler characteristic of coherent sheaf cohomology of the sheaf of regular functions. So you remember Euler characteristic and cohomology as a sort of alternating sum of cohomology groups. So when we do sheaf cohomology, we will find that there's another natural definition of this using sheaf cohomology. For example, we can work out the arithmetic genus of a plane curve of degree D. Well, the Hilbert polynomial of a plane curve of degree D is 2 plus K choose 2 minus 2 plus K minus D choose 2. And you can work out this as equal to 1 minus 2 minus D 1 minus D over 2. So the arithmetic genus is this number here. And this is where the term arithmetic genus comes from because in the special case of plane curves, the arithmetic genus happens to be the same as the topological genus, by the number of handles of the underlying topological space. Incidentally, the Hilbert polynomial is more or less the only discreet invariant of closed subvarieties of projective space. This is a theorem due to Robin Hart showing the author of the book with sort of vaguely following who proved that something called the Hilbert scheme is always connected and the Hilbert scheme is roughly a scheme parameterizing sub schemes of projective space with the given Hilbert polynomial. And the fact that this is connected means that the Hilbert polynomial is essentially the only continuous invariant of things in projective space. In the sense that if the invariant is a continuous function of the Hilbert scheme to a discrete set, then it's possible to write it in terms of the Hilbert polynomial.