Now we’re ready for our first big geometric theorem. We’ll prove Bezout’s Theorem as a special case of another, more general theorem.
So now we let be a projective variety. Then for any hypersurface not containing , we can write for some minimal set of varieties . We’ll define the intersection multiplicity of along to be , where this is the multiplicity of the prime ideal on the module that we have previously defined.
So now, under these circumstances, we can prove the following:
Proof: Take to be the defining polynomial of . It is of degree . We get a short exact sequence of modules , where the first is just the second with the grading shifted down by and .
Taking Hilbert polynomials, we get . And now we’ll compare the lead coefficients of each side. Let’s denote the degree of by . Doing the calculation out (which is best done in private, as are almost all computations) we see that the right hand side has lead term , and so the intersection has degree equal to .
So now we’ll independently calculate the lead term on the left. Then the two must be equal and so we’ll have our equation. Look at and pick a filtration like the nice ones we know exist. The quotients are of the form , and so we can write where the are the Hilbert polynomials of the quotients. The shifts don’t affect the lead coefficients, and we can throw away any terms of nonmaximal degree because they won’t contribute, and so we get a sum over the ideals . Each of these occurs times, that is, times. As we can factor out the factorial from the start, we can treat the lead terms as just being the degrees. And so the lead coefficient on the left is . And so the result follows. QED
Now, if we specify that we’re living in the projective plane, then we take be distinct curves of degrees and , that is, irreducible one dimensional projective varieties who intersect in a finite collection of points , we have that . This is because points have Hilbert polynomial 1. (If we don’t require that they only intersect in a finite collection of points, things get trickier. However, a subject called Derived Algebraic Geometry appears to help understand this case, and you can read a bit about that here.)
In particular, this justifies the geometric notion that degree is the number of times that a variety intersects most hyperplanes, because hyperplanes have degree 1, and so the sum of the intersection multiplicities is the degree.
We’ll finish up by playing a bit with plane curves.
Let be a plane curve of genus and degree . Then the Hilbert Polynomial of is . This is because it must be linear, because curves are one dimensional, and the lead term must then be the degree. The constant term is , and .
Now, it’d be wonderful if there were some way of relating the genus and the degree of a plane curve. Then, we’d only need to know a single number related to the curve to know the Hilbert Polynomial. In fact, there IS a relation, but rather than just stating it, we’ll derive it.
We have a short exact sequence of graded modules given by . So to determine the Hilbert Polynomial of the last term, the one we care about, we must only determine the Hilbert Polynomial of the second and subtract that of the first.
First, we establish a bit of notation. is the number of ways to choose elements from a set of elements. It is equal to . If we fix , then this will just be a polynomial in . In fact, we can prove that Hilbert Polynomials all must be rational coefficient sums of these.
Now, as graded modules, and are isomorphic, though with a degree shift of . So . By a counting argument that I won’t make now (though it’s fairly nifty, so I might make it in the future) we have . And so the Hilbert Polynomial of the curve will be . Plugging in , we find that .
Thus, the arithmetic genus is . So given a curve of degree , it has Hilbert polynomial .
We’ll stop there for today, and next time we’ll visit a topic that is close to the heart of almost every algebraic geometer or number theorist: Elliptic Curves.