Bézout's theorem

Bézout's theorem is one of the theorems I was scared of. Let alone its proof, I was always uncomfortable about the word "multiplicity" because I was not sure what it really meant. What I am writing here is more or less a poor regurgitation of a good exposition in Vakil's book, which made me realize that this theorem was not so bad after all.

Bézout's theorem. Over any field $k,$ any two curves $C_{1}$ and $C_{2}$ in $\mathbb{P}^{2}$ such that $\dim(C_{1} \cap C_{2}) = 0$ meet at precisely at $\deg(C_{1})\deg(C_{2})$ points counting with multiplicity.

Really, most of the effort in understanding this theorem goes into understanding what some of the words mean in the statement: "curve", "degree", and "multiplicity". For the rest, we will be working over a fixed field $k.$

Degree. Let $i : X \hookrightarrow \mathbb{P}^{n}$ be a closed subscheme. Then $p_{X}(m) := h^{0}(X, \mathcal{O}_{X}(m))$ is a polynomial in $m$ when $m$ is large enough. This is called the Hilbert polynomial of $X$ in $\mathbb{P}^{n}.$ The coefficients of this polynomial are in $\mathbb{Q}.$ It turns out that the leading coefficient of the Hilbert polynomial $p_{X}(m)$ of $X \hookrightarrow \mathbb{P}^{n}$ times $d!$ is always an integer (18.6.G in Vakil), where $d = \deg(p_{X}).$ The degree of the closed subscheme $X \hookrightarrow \mathbb{P}^{n}$ is the leading coefficient of its Hilbert polynomial $p_{X}(m)$ times $d!,$ which is a non-negative integer.

Dependence on the embedding. Note that the Hilbert polynomial and the degree of a closed subscheme $i : X \hookrightarrow \mathbb{P}^{n}$ depends on the embedding $i$, not just the scheme $X.$ For instance, the Hilbert polynomial of $\mathbb{P}^{1}$ in itself is equal to $m + 1$ with the degree $1,$ but when $\mathbb{P}^{1}$ is considered as a closed subscheme of $\mathbb{P}^{3}$ given by the Veronese embedding, its Hilbert polynomial is $3m + 1$ with degree $3.$ (See Example 1 on p.490 and 18.6.D in Vakil.) After all, the dependence on $i$ should not be so surprising because $$i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m) \simeq \mathscr{O}_{X}(m).$$ This isomorphism is more or less a definition.

Example. The notion of degree will be used to express what it means to the "multiplicity" of a point at the intersection of two curves. For instance, consider the intersection of the parabola $y = x^{2}$ and the $x$-axis $y = 0$ in the affine plane in $(x, y)$-coordinates. When we intersect these two curves, we have $(x^{2}, y) = (0,0),$ so $(x, y) = (0, 0)$ is the only point at the intersection. Well, this is correct, except the fact that when we work with schemes, we need to remember $(x^{2}, y) = (0, 0).$ That is, we are intersecting two affine schemes whose data are equivalent to the following two rings: $k[x, y]/(y-x^{2})$ and $k[x, y]/(y).$ Hence, their intersection corresponds to $k[x]/(x^{2})$ not $k[x]/(x) \simeq k,$ although as topological spaces they are equal $$\mathrm{Spec}(k[x]/(x^{2})) \simeq V(x^{2}) = V(x) = \{(x)\} \simeq \mathrm{Spec}(k[x]/(x)),$$ a single closed point of the affine line $\mathrm{Spec}(k[x]) = \mathbb{A}^{1}.$ Note that this issue comes in even if one only considers the most classical algebraic geometry, namely when $k = \mathbb{C}.$ In this case, the degree of the point at the intersection is equal to the dimension of the ring $k[x]/(x^{2})$ over $k,$ which is $2$.

To fit this example to our situation, we need to projectivize these curves. Namely, the parabola $y = x^{2}$ becomes $yz = x^{2}$ in $\mathbb{P}^{2}$ and the line $y = 0$ becomes $y = 0$ in $\mathbb{P}^{2}.$ More technically the parabola is given by $$\mathrm{Proj}(k[x,y,z]/(yz - x^{2})) \hookrightarrow \mathrm{Proj}(k[x,y,z]) = \mathbb{P}^{2},$$ and the line is given by $$\mathbb{P}^{1} = \mathrm{Proj}(k[x,y,z]/(y)) \hookrightarrow \mathrm{Proj}(k[x,y,z]) = \mathbb{P}^{2}.$$ The intersection of these two curves is given by $$\mathrm{Proj}(k[x,z]/(x^{2})) \hookrightarrow \mathrm{Proj}(k[x,z]) = \mathbb{P}^{2}.$$ If we just look at this as a topological space, then this is identical to $V_{\mathbb{P}^{2}}(x),$ which corresponds to the $k$-point $[x : z] = [0 : 1]$ in $\mathbb{P}^{2}.$ However, the point $[0 : 1]$ also corresponds to $$\mathrm{Proj}(k[x,z]/(x)) \hookrightarrow \mathrm{Proj}(k[x,z]) = \mathbb{P}^{2},$$ but the former point has degree $2,$ and the latter will have degree $1$ (in the sense of the degree defined with the Hilbert polynomial).

More generally, any degree $d$ hypersurface of $\mathbb{P}^{n}$ (i.e., the locus given by a single degree $d$ homogeneous polynomial) has degree $d$ in the sense of Hilbert polynomial, and this is great because the latter notion of degree is more intrinsic to the scheme That is, it does not require explicit description of polynomials. (See 18.6.H in Vakil's book.)

As this example illustrates, when the closed subscheme $X$ we are dealing with consists of finitely many closed points on a projective curve on $C$, then we may consider $X$ as a divisor on $C,$ namely $$X = n_{1}x_{1} + \cdots + n_{r}x_{r},$$ where $x_{i}$ are closed points of $C,$ and the degree of $X$ is precisely the degree of $X$ as a divisor: $$\deg(X) = \sum_{i=1}^{r}n_{i}[\kappa(x_{i}) : k].$$ Thus, in this case, the degree can be computed by local information even though we needed somewhat global setting (i.e., projective situation) to define it. This is because for any closed point $x \in C,$ if we consider $\{x\} \hookrightarrow C$ as a closed subscheme (and hence a closed subscheme of some $\mathbb{P}^{n}$), then its degree is precisely the degree $[\kappa(x) : k]$ of the residue field $\kappa(x)$ of $x$ in $\mathbb{P}^{n}$ (or in $C$).

Multiplicity at intersection. Due to the discussion that follows from the previous example, if we have two curves $C_{1}$ and $C_{2}$ in $\mathbb{P}^{2}$ and $C_{1} \cap C_{2}$ is a finite set of closed points, then $\deg(C_{1} \cap C_{2})$ is equal to the sum of degrees of the residue fields of the points in $C_{1} \cap C_{2}$ over $k,$ and this is really what we should mean when we say the "number of points with multiplicity". Sorting this out lets us formulate our statement more rigorously as follows.

Restating Bézout's theorem. Let $C_{1}$ and $C_{2}$ be curves in $\mathbb{P}^{2}.$ If $\dim(C_{1} \cap C_{2}) = 0,$ then $$\deg(C_{1} \cap C_{2}) = \deg(C_{1}) \deg(C_{2}),$$ where the degrees are given as the closed subschemes of $\mathbb{P}^{2}.$

Remark. By saying that $C$ is a curve in $\mathbb{P}^{2},$ we mean that $C \hookrightarrow \mathbb{P}^{2}$ is an $1$-dimensional closed subscheme. Requiring that $\dim(C_{1} \cap C_{2}) = 0$ means that $C_{1}$ does not pass through none of the generic points of irreducible components of $C_{2},$ and vice versa. This is equivalent to saying that these two curves do not share any irreducible components or that they intersect at finitely many points. Under any of these equivalent hypotheses, it is necessarily true that $C_{1} \cap C_{2}$ consists of finitely many closed points. If $C_{1}$ is irreducible, then these conditions are equivalent to saying that $C_{1}$ is not contained in $C_{2}.$

Reduction. Say we have an $1$-dimensional irreducible closed subscheme $C \hookrightarrow \mathbb{P}^{2}.$ Considering the standard affine open cover of $\mathbb{P}^{2},$ we have a standard affine open subset $U \subset \mathbb{P}^{2}$ such that $U \simeq \mathbb{A}^{2}$ and $C \cap U \neq \emptyset.$ Note that $C \cap U$ has codimension $1$ in $U \simeq \mathbb{A}^{2}.$ Since every height $1$ prime in a UFD is principal, there is a single polynomial in two variables that defines $C \cap U.$ The vanishing locus of the homogenization of this polynomial is precisely $C.$

Using this argument, we see that $C_{1}$ and $C_{2}$ are hypersurfaces of $\mathbb{P}^{2}.$ What's great about hypersurfaces of $\mathbb{P}^{2}$ is that they don't have embedded points. Thus, our hypothesis says that $C_{1}$ and $C_{2}$ do not intersect at any associated points. (The language of associated points may be found Section 5.5 in Vakil's book.) This lets us see that the following is a more general statement of Bézout's theorem.

More general version of Bézout's theorem. Let $X \hookrightarrow \mathbb{P}^{n}$ be a projective scheme of dimension $\geq 1$ and $H$ a hypersurface of $\mathbb{P}^{n}$ that do not pass through any associated points of $X.$ Then $$\deg(H \cap X) = \deg(H)\deg(X),$$ where degrees are given by the relevant closed subscheme structures in $\mathbb{P}^{n}.$

Proof. Let $d := \deg(H)$ and $f \in \Gamma(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(d))$ the defining homogeneous polynomial of $H.$ Denoting $i : X \hookrightarrow \mathbb{P}^{n}$ for the given closed embedding, we see that $H \cap X$ is given by the vanishing locus of the section $i^{*}f \in \Gamma(X, \mathscr{O}_{X}(d)).$ The hypothesis that $H$ avoids all the associated points of $X$ implies that affine locally the section $i^{*}f$ is a non-zerodivisor. Hence, we see that $H \cap X$ defines an effective Cartier divisor of $X,$ which gives the following exact sequence: $$0 \rightarrow \mathscr{O}_{X}(-d) \rightarrow \mathscr{O}_{X} \rightarrow i_{*}\mathscr{O}_{H} \rightarrow 0.$$ Applying $(-) \otimes_{\mathscr{O}_{X}} \mathscr{O}_{X}(m),$ we get $$0 \rightarrow \mathscr{O}_{X}(m-d) \rightarrow \mathscr{O}_{X}(m) \rightarrow i_{*}\mathscr{O}_{H}(m) \rightarrow 0,$$ so we have $$p_{H}(m) = p_{X}(m) - p_{X}(m-d).$$ Computing the leading coefficient of the right-hand side gives the result. $\Box$