# Algebraic Geometry Jottings 7

Fulton characterizes the intersection number, I(P, EF), with seven properties. Let me just repeat the last one:

1.  I(E∩F) = I(E∩(F+AE)) for any A.

As I mentioned in post 5, we can justify this using the “level curve” idea from post 3. The values of F on a branch of E, parametrized by t, are given by a power series F(t). At t=0 the branch passes through the level curve F=0. The order of F(t) gives the multiplicity of the branch-curve intersection. Since the branch is part of the level curve E=0, we have

(F+AE)(t) = F(t)+A(t)E(t) = F(t)

Since all branch-curve intersections have the same multiplicity, so does the curve-curve multiplicity.

Theorem 3.3 of Fulton (p.37) says that there is a unique function I(P,EF) of P, E, and F satisfying Fulton’s seven properties. Uniqueness follows from an algorithm implied by the properties. For existence Fulton provides an explicit formula. Let’s have a look before we discuss it: $I(P,E\cap F) = \dim_k(\mathcal{O}_P(\mathbb{A}^2)/(E,F)_P)$

Here k is the ground field we’re working over, and $\mathbb{A}^2$ is the affine plane, i.e., k×k. $\mathcal{O}_P(\mathbb{A}^2)$ is the local ring at P; I’ll say more about that in a bit. (E,F)P is the ideal generated by E and F in $\mathcal{O}_P(\mathbb{A}^2)$, i.e., all elements of the form AE+BF where A and B are arbitrary elements of $\mathcal{O}_P(\mathbb{A}^2)$. Finally, $\mathcal{O}_P(\mathbb{A}^2)/(E,F)_P$ is the quotient ring, which turns out to be a vector space over k, and dimk is its dimension.

(Actually, Fulton doesn’t distinguish notationally between the ideals (E,F) in k[x,y] and (E,F)P in $\mathcal{O}_P(\mathbb{A}^2)$. I think this is unfortunate. He does write $I\mathcal{O}_P$ for the smallest ideal in $\mathcal{O}_P$ containing an ideal I in k[x,y]. I’ll let $(\mathbb{A}^2)$ be implicit; I don’t think any confusion will ensue.)

Property (7) offers an entry point to understanding the formula. It won’t come fully into focus until we start dissecting proofs of Bézout’s theorem, but let’s go down this road a bit.

Our justification for (7) involved considering the values of F along a branch of E. Let’s look more generally at polynomial functions restricted to the curve E. The restriction map is clearly a homomorphism, and the kernel clearly includes all multiples of the polynomial E (i.e., the principal ideal (E)). The quotient ring, k[x,y]/(E), is called the coordinate ring of E. It’s denoted Γ(E).

It’s tempting to think of Γ(E) as being the restricted polynomial functions, but that’s not usually correct. For a simple counter-example, let’s work with the affine line, and let E(x)=xn. Then the “curve” xn=0 is just the origin, and (E) consists of polynomials whose lowest (nonzero) term has degree at least n. Then Γ(E) = k[x]/(xn) amounts to the polynomials of degree less than n: each element of Γ(E) is an equivalence class of polynomials, with a unique representative of degree less than n. The homomorphism π from k[x] to Γ(E) keeps all terms of degree less than n and throws away the others. Multiplication is done “mod xn“. Congruence notation helps: xn≡0, so we multiply the usual way, and the product is congruent to a unique polynomial of degree less than n. Γ(E) is a ring, but it’s also a vector space of dimension n over the ground field, with basis {1,…, π(x)n–1}.

Compare with the restriction homomorphism. The “curve” E is just a point (namely 0), and restricting a polynomial p(x) to E gives you the value p(0), aka the constant term. The restriction kernel consists of all polynomials with constant term 0. So the restriction homomorphism keeps the 0-th coefficient, while the homomorphism from k[x] to Γ(E) keeps the first n coefficients. (Or equivalently, enough info to compute the value and the first n–1 derivatives of p(x) at 0.)

Now let’s enlarge our scope to the affine plane. Let’s take the case E(x,y) = yxn. So yxn mod (E), and any polynomial p(x,y) is congruent to a polynomial without any y‘s. Congruences mod (E) turn into equalities in Γ(E), so Γ(E) is isomorphic to k[x].

How about the intersection of two curves E and F? This time the ideal (E,F) is clearly zero on the intersection. So the homomorphism restricting functions in k[x,y] to EF has a kernel that includes (E,F). But this restriction kernel could be larger. Let’s define Γ(EF) to be k[x,y]/(E,F).

Take the case where E(x,y) = yxn and F(x,y) = y–c, where c is a constant belonging to the ground field. We have yxn and yc, thus xnc. It follows that any polynomial in x and y is congruent to a polynomial in x alone of degree <n. So {1, π(x), …, π(x)n–1} is a basis for Γ(EF) over k. (Well, a rigorous proof calls for some futzing around. But let’s go with it.) The dimension is n, and Bézout says that E and F should have n intersections!

For example, let’s say n=2 and c=1. Then EF is the intersection of the parabola y = x2 with the horizontal line y = 1, two distinct points. Naturally an element of Γ(EF) is determined by its values at these points. But if c=0, there’s only one point in the intersection, and an element of Γ(EF) stores info about both the value and the derivative. Finally, if c is negative and our ground field is ℝ, then the intersection is empty, but an element of Γ(EF) holds info about what would happen if we extended the ground field to ℂ. Intriguing!

In post 4, I observed that branch-branch multiplicities are as granular as it gets. Bézout on the other hand surveys the entirety of EF. The coordinate ring likewise treats all intersection points equally. When we start digging into Fulton’s proof of Bézout’s theorem, we’ll see that dimkΓ(EF) is the number of points in E∩F, provided that (a) k is algebraically closed; (b) none of the intersections lie at infinity; and (c) intersections are counted with multiplicity.

Let’s see now how the local ring confines attention to just one point. Start with the nice easy case of E(x,y) = yx2 and F(x,y) = y–1. Then yx2 and y ≡ 1, so x2 ≡ 1, so (x+1)(x—1) ≡ 0. If we could divide by (x+1) we would have x ≡ 1. Combined with y ≡ 1, this would mean that {1} was a basis for—something. (More on that in a moment.) The intersection number at (1,1) would be 1, as we want. Likewise at (–1,1).

You allow division by passing from the ring k[x,y] of polynomials to the field k(x,y) of rational expressions. Now, k(x,y) is infinite dimensional over k, just like k[x,y]. Γ(EF) has the right dimension because we divided out the ideal (E,F). But the field k(x,y), like all fields, has only two ideals: itself and the zero ideal. Dividing out, we get either the zero ring or the field k(x,y).

The local ring to the rescue! Let P be the point (1,1). At P, we’d like to allow division by (x+1), but we have no desire to give (x—1) the same privilege. The local ring at P (denoted $\mathcal{O}_P$) consists of all rational expressions p(x,y)/q(x,y) such that q(P)≠0 (i.e., q(1,1)≠0). These are precisely the expressions we can evaluate at P without things blowing up. An analyst would say that the elements of $\mathcal{O}_P$ don’t have a pole at P.

We have $k[x,y]\subset\mathcal{O}_P\subset k(x,y)$. Also (E,F) is an ideal in k[x,y]. But it’s not an ideal of $\mathcal{O}_P$ because it’s not closed under multiplication by arbitrary elements of $\mathcal{O}_P$. Clearly the right thing to do is to enlarge (E,F) to an ideal in $\mathcal{O}_P$. Easily accomplished: $(E,F)_P = \{aE+bF : a,b\in\mathcal{O}_P\}$ $= \{\frac{h}{r} : h\in(E,F), r\in k[x,y], r(P)\neq0\}$

is the smallest ideal of $\mathcal{O}_P$ containing (E,F).

OK, some congruence calculations. Starting with yx2 and y ≡ 1 we get x2 ≡ 1 and then (x+1)(x—1) ≡ 0 and finally x—1 ≡ 0. That last step is kosher only because we are using (E,F)P instead of (E,F) for our congruences. Passing to the quotient ring $\mathcal{O}_P/(E,F)_P$, we have a vector space of dimension 1 over k, with basis {1}.

We’ll be keeping company with $\mathcal{O}_P/(E,F)_P$ quite a bit. I’m dubbing it the modded local ring.

I’m hungry! Time for a break.