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 V\subset\mathbb{P}^n be a projective variety. Then for any hypersurface H not containing V, we can write V\cap H=Z_1\cup\ldots\cup Z_\ell for some minimal set of varieties Z_j. We’ll define the intersection multiplicity of V along H to be i(V,H;Z_i)=\mu_{I(Z_i)}(S/(I(V\cap H))), 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:

Theorem: \sum_{j=1}^\ell i(V,H;Z_j)\cdot \deg Z_j=(\deg V)(\deg H)

Proof: Take f to be the defining polynomial of H. It is of degree d. We get a short exact sequence of modules 0\to S/I(V) (-d)\to S/I(V)\to M\to 0, where the first is just the second with the grading shifted down by d and M=S/I(V\cap H).

Taking Hilbert polynomials, we get P_M(n)=P_V(n)-P_V(n-d). And now we’ll compare the lead coefficients of each side. Let’s denote the degree of V by e. 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 \frac{de}{(r-1)!}n^{r-1}, and so the intersection has degree equal to de=(\deg V)(\deg H).

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 M and pick a filtration like the nice ones we know exist. The quotients are of the form S/\mathfrak{q_i}(\ell_i), and so we can write P_M=\sum P_i where the P_i 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 I(Z_j). Each of these occurs \mu_{I(Z_j)}(M) times, that is, i(V, H;Z_j) 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 \sum_{j=1}^\ell i(V,H;Z_j)\deg(Z_j). And so the result follows. QED

Now, if we specify that we’re living in the projective plane, then we take Y,Z be distinct curves of degrees d and e, that is, irreducible one dimensional projective varieties who intersect in a finite collection of points \{P_1,\ldots,P_s\}, we have that \sum i(Y,Z;P_i)=de. 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 C\subset \mathbb{P}^2 be a plane curve of genus g and degree d. Then the Hilbert Polynomial of C is dn+(1-g). 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 P_C(0), and g=1-P(0).

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 0\to (f)\to k[x,y,z]\to k[x,y,z]/(f)\to 0. 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. \binom{n}{d} is the number of ways to choose d elements from a set of n elements. It is equal to \frac{n!}{d!(n-d)!}. If we fix d, then this will just be a polynomial in n. In fact, we can prove that Hilbert Polynomials all must be rational coefficient sums of these.

Now, as graded modules, (f) and k[x,y,z] are isomorphic, though with a degree shift of d. So H_{(f)}(n)=H_{k[x,y,z]}(n-d). 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 H_{k[x,y,z]}(n)=\binom{2+n}{2}. And so the Hilbert Polynomial of the curve will be \binom{2+n}{2}-\binom{2+n-d}{2}. Plugging in n=0, we find that P(0)=1-\binom{d-1}{2}.

Thus, the arithmetic genus is \binom{d-1}{2}=g. So given a curve of degree d, it has Hilbert polynomial dn+1-\binom{d-1}{2}.

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.

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