Bézout’s Theorem

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We introduce Bézout’s theorem, a classical result in algebraic geometry, in this post. Intuitively, suppose two curves \(C,D\) on the plane are given. Then the number of intersection points of \(C\) and \(D\) depends on their degrees: for example, a conic \(\y=x^2\) defined on the plane and a line generally meet at two points. Bézout’s theorem is a generalization of this observation.

Proposition 1 (Bézout) Over an algebraically closed field, inside \(\mathbb{P}^n\), if hypersurfaces \(H_1, \ldots, H_n\) of degrees \(d_1, \ldots, d_n\) have no common component, then

\[\deg(H_1 \cap \cdots \cap H_n) = d_1 \cdots d_n\]

holds. Here the intersection is taken with multiplicity into account.

In particular, inside \(\mathbb{P}^2\) two curves of degrees \(m\) and \(n\) meet at \(mn\) points. One must be somewhat careful that they must not have a common component; for instance, one cannot use this to compute the intersection of two identical curves.

Example 2 (Two conics) Consider two conics in \(\mathbb{P}^2\)

\[C_1 = Z(\x_0^2 + \x_1^2 - \x_2^2),\qquad C_2 = Z(\x_0\x_1)\]

\(C_1\) is the projectivization of a cone, and \(C_2\) is the union of the two lines \(Z(\x_0)\) and \(Z(\x_1)\). Since these two curves share no common component, Bézout’s theorem predicts \(2 \times 2 = 4\) intersection points. Computing the intersection explicitly, when \(\x_0 = 0\) we get \(\x_1^2 = \x_2^2\), yielding \([0:1:1]\) and \([0:1:-1]\); when \(\x_1 = 0\) we get \(\x_0^2 = \x_2^2\), yielding \([1:0:1]\) and \([1:0:-1]\). Thus we verify that they meet in exactly four points.

Proof

Rather than proving the general case, we prove Bézout’s theorem only in \(\mathbb{P}^2\). For this purpose we use the following lemma.

Proposition 3 (Hilbert polynomial) For the homogeneous coordinate ring \(S(X)\) of a projective variety \(X \subseteq \mathbb{P}^n\), we call the function \(H(t) = \dim_\mathbb{K} S(X)_t\) the Hilbert function of \(X\). By the Hilbert–Serre theorem, this function coincides with a polynomial \(P_X(t)\) for \(t \gg 0\), and we call this polynomial the Hilbert polynomial of \(X\).

In particular, for a curve \(C = Z(F) \subseteq \mathbb{P}^2\) of degree \(d\), the Hilbert polynomial of \(S(C) = \mathbb{K}[\x_0, \x_1, \x_2]/(F)\) is

\[P_C(t) = dt + \frac{d(3-d)}{2}\]

Here the degree of \(P_C\) is \(1\), the dimension of \(C\); the leading coefficient is \(\deg C = d\); and the constant term \(P_C(0) = \frac{d(3-d)}{2} = 1 - \frac{(d-1)(d-2)}{2}\) is the arithmetic genus of \(C\).

The Hilbert function \(H(t)\) counts the number of independent elements remaining after removing those homogeneous polynomials of degree \(t\) that vanish on \(C\)—in other words, the number of homogeneous polynomials that act as distinct functions on \(C\). As \(t\) grows, this number behaves like a polynomial; its degree equals the dimension of \(C\), namely \(1\); its leading coefficient is proportional to the degree \(d\); and its constant term equals the arithmetic genus \(1 - \frac{(d-1)(d-2)}{2}\).

Proof

Let \(S = \mathbb{K}[\x_0, \x_1, \x_2]\). The dimension of \((S/(F))_t\) equals the dimension of the space obtained from \(S_t\) by removing the multiples of \(F\). Since multiplication by \(\cdot F: S(-d) \to S\) is injective, we obtain the following short exact sequence:

\[0 \to S(-d) \xrightarrow{\cdot F} S \to S/(F) \to 0\]

Because \(\dim_\mathbb{K} S_t = \binom{t+2}{2}\), comparing dimensions in degree \(t\) gives

\[\dim_\mathbb{K} (S/(F))_t = \binom{t+2}{2} - \binom{t-d+2}{2}\]

Expanding this yields

\[\frac{(t+2)(t+1)}{2} - \frac{(t-d+2)(t-d+1)}{2} = dt + \frac{d(3-d)}{2}\]

as desired.

This result will be used crucially in the proof of the following Proposition 5.

Proposition 4 For a degree \(d\) curve \(C = Z(F)\) in \(\mathbb{P}^2\), the intersection \(C \cap L\) with any general line \(L\) consists of exactly \(d\) points (counted with multiplicity).

This proposition is the simplest special case of Bézout’s theorem. It provides the geometric intuition that a degree \(d\) curve meets a general line in \(d\) points.

Proof

Without loss of generality, let \(L = Z(\x_2)\). Since \(C\) is defined by a degree \(d\) homogeneous polynomial \(F(\x_0, \x_1, \x_2)\), substituting \(\x_2 = 0\) in \(L \cap C\) gives \(F(\x_0, \x_1, 0)\). This is a degree \(d\) homogeneous polynomial in \(\x_0, \x_1\), and over an algebraically closed field it has exactly \(d\) roots (counted with multiplicity). Since \(L\) is general, \(F(\x_0, \x_1, 0)\) is not the zero polynomial; otherwise \(\x_2\) would be a factor of \(F\), so \(C\) would have \(L\) as a component, contradicting the hypothesis.

We now prove Proposition 1. The key consists of two points. First, we show that the sum of intersection multiplicities agrees with the dimension of a certain global algebraic object, and second, we compute that dimension precisely as \(mn\) via the Hilbert polynomial.

Proposition 5 (Proof of Proposition 1) Two curves \(C = Z(F)\) and \(D = Z(G)\) of degrees \(m\) and \(n\) in \(\mathbb{P}^2\) with no common component satisfy \(\sum_p i_p(C, D) = mn\).

Proof

We divide the proof into two steps.

Step 1. We first show the following equality.

\[\sum_{p \in C \cap D} i_p(C, D) = \dim_\mathbb{K} (\mathbb{K}[\x_0, \x_1, \x_2]/(F, G))_t \qquad (t \gg 0)\]

That \(C \cap D\) is a finite set is known from the assumption that \(C\) and \(D\) have no common component. (§Dimension, ⁋Example 14) For a point \(p = [a:b:c] \in C \cap D\), assuming \(c \neq 0\), we have \(p = (a/c, b/c)\) in the coordinates on \(U_2\), and for the dehomogenizations \(f, g \in \mathbb{K}[\x, \y]\) of \(F, G\),

\[i_p(C, D) = \dim_\mathbb{K} \mathcal{O}_{\mathbb{A}^2, p}/(f, g)\]

Since \(V(F, G)\) is finite, \(f, g\) generate a zero-dimensional ideal \((f, g)\) in the affine ring \(\mathbb{K}[\x, \y]\), and by the Chinese remainder theorem,

\[\mathbb{K}[\x, \y]/(f, g) \cong \prod_{p \in V(f,g)} \mathcal{O}_{\mathbb{A}^2, p}/(f, g)\]

Thus \(\dim_\mathbb{K} \mathbb{K}[\x, \y]/(f, g) = \sum_p i_p(C, D)\).

On the other hand, the Hilbert function \(H(t) = \dim_\mathbb{K} R_t\) of the quotient \(R = S/(F, G)\) of \(S = \mathbb{K}[\x_0, \x_1, \x_2]\) becomes constant for \(t \gg 0\) (proved in Step 2), and this constant value equals \(\dim_\mathbb{K} \mathbb{K}[\x, \y]/(f, g)\). This is because when \(t \gg 0\), there exist polynomials of degree \(t\) that can independently adjust the values at each intersection point, so the evaluation map \(R_t \to \mathbb{K}^{\lvert V(F,G) \rvert}\) is surjective.

Step 2. We now show that \(\dim_\mathbb{K} (\mathbb{K}[\x_0, \x_1, \x_2]/(F, G))_t = mn\) for \(t \gg 0\). Write \(S = \mathbb{K}[\x_0, \x_1, \x_2]\). Since \(F, G\) have no common irreducible factor, the multiplication maps \(\cdot F: S(-m) \to S\) and \(\cdot G: S/(F)(-n) \to S/(F)\) are both injective, and we obtain the following two short exact sequences.

\(0 \to S(-m) \xrightarrow{\cdot F} S \to S/(F) \to 0\) \(0 \to S/(F)(-n) \xrightarrow{\cdot G} S/(F) \to S/(F, G) \to 0\)

Reading the degree as \(m\) from Proposition 3 (Hilbert polynomial), the Hilbert polynomial of \(S/(F)\) has the form \(P_F(t) = mt + c_1\). Applying the Hilbert polynomial to the second exact sequence, the Hilbert polynomial of \(S/(F, G)\) is

\[P_{F,G}(t) = P_F(t) - P_F(t - n) = \bigl(mt + c_1\bigr) - \bigl(m(t-n) + c_1\bigr) = mn\]

Thus for \(t \gg 0\), the dimension of \((S/(F, G))_t\) is the constant \(mn\), and by Step 1 we have \(\sum_p i_p(C, D) = mn\).

Generalization

So far we have proved Bézout’s theorem only for curves in \(\mathbb{P}^2\). To extend this to arbitrary projective spaces and general projective varieties, we need the Chow ring. The key fact is

\[\operatorname{CH}^\ast(\mathbb{P}^n) \cong \mathbb{Z}[H]/(H^{n+1})\]

Here \(H\) is the hyperplane class, and a variety of codimension \(k\) and degree \(d\) has class \(dH^k\). In particular, a hypersurface of degree \(d\) corresponds to \(dH\), so the intersection product of \(n\) hypersurfaces \(H_1, \ldots, H_n\) is

\[[H_1] \cdot [H_2] \cdots [H_n] = (d_1 H)(d_2 H) \cdots (d_n H) = d_1 d_2 \cdots d_n \cdot H^n\]

Since \(H^n\) is the class of a point in \(\mathbb{P}^n\) and its degree is 1, we obtain \(\deg(H_1 \cap \cdots \cap H_n) = d_1 \cdots d_n\). Under this intuition, the generalized Bézout theorem is stated as follows.

Proposition 6 (Generalized Bézout theorem) For two projective varieties \(V, W\) in \(\mathbb{P}^n\),

\[\deg(V \cap W) \leq \deg(V) \cdot \deg(W)\]

holds. Here \(\deg(V \cap W)\) is the sum of the degrees of the irreducible components of \(V \cap W\). Equality holds when \(V\) and \(W\) have a proper intersection (that is, when \(\operatorname{codim}(Z) = \operatorname{codim}(V) + \operatorname{codim}(W)\) for every irreducible component \(Z\) of \(V \cap W\)); in this case, assigning an intersection multiplicity \(m_Z\) to each component \(Z\) gives \(\sum_Z m_Z \deg(Z) = \deg(V) \cdot \deg(W)\).

Example 7 (\(\mathbb{P}^3\)) Consider two quadric surfaces \(Q_1, Q_2\) in \(\mathbb{P}^3\). Since each has degree 2, when they intersect properly the intersection \(Q_1 \cap Q_2\) is a curve of dimension 1 and degree 4. Concretely, taking \(Q_1 = Z(\x_0\x_3 - \x_1\x_2)\) and \(Q_2 = Z(\x_0\x_2 - \x_1\x_3)\), the intersection decomposes into four lines, and the sum of their degrees is still 4.

The proof of Proposition 6 relies on the general theory of intersection theory via the Chow ring. For details, see §Intersection Product. The inequality from §Dimension, ⁋Example 14 reappears as one concerning the codimension of the components.

Applications

Cayley-Bacharach Theorem

Proposition 8 (Special case of Cayley-Bacharach theorem) Let \(C_1 = Z(F_1)\) and \(C_2 = Z(F_2)\) be two cubic curves in \(\mathbb{P}^2\) with no common component, meeting in proper intersection at nine distinct points \(p_1, \ldots, p_9\). Then any cubic curve \(C_3 = Z(F_3)\) passing through \(p_1, \ldots, p_8\) also passes through \(p_9\).

Proof

Assume two cubic curves \(C_1, C_2\) meet in proper intersection at nine distinct points \(p_1, \ldots, p_9\). The space of degree-3 homogeneous polynomials \(\mathbb{K}[\x_0, \x_1, \x_2]_3\) on \(\mathbb{P}^2\) has dimension \(\binom{3+2}{2} = 10\), and the condition that a curve pass through each point \(p_i\) is a single linear condition, so \(V = \{F \in \mathbb{K}[\x_0, \x_1, \x_2]_3 : F(p_i) = 0 \text{ for } i = 1, \ldots, 8\}\) is a subspace of dimension \(\dim V \ge 10 - 8 = 2\). On the other hand, \(F_1, F_2 \in V\) and \(C_1 \neq C_2\), so \(F_1, F_2\) are linearly independent. If the eight points are in general position then \(\dim V = 2\), and \(F_1, F_2\) form a basis of \(V\). Hence for any \(F_3 \in V\) there exist constants \(\alpha, \beta\) such that \(F_3 = \alpha F_1 + \beta F_2\). Substituting \(p_9\) into both sides gives \(F_3(p_9) = \alpha F_1(p_9) + \beta F_2(p_9) = 0\), so \(C_3\) also passes through \(p_9\).

The intuition behind this result is as follows. The condition that a cubic pass through eight of the nine intersection points of two cubics imposes eight linear constraints on the space of cubics (which is 10-dimensional), leaving a 1-dimensional family whose members all pass through the ninth point as well. This shows that the \(3 \times 3 = 9\) of Bézout’s theorem is no accident.

Pascal’s theorem

Proposition 9 (Pascal’s theorem) For six points \(A, B, C, D, E, F\) on a conic, if the three intersection points

\[P = \overline{AB} \cap \overline{DE},\quad Q = \overline{BC} \cap \overline{EF},\quad R = \overline{CD} \cap \overline{FA}\]

all exist, then these three points are collinear.

Proof

Let us denote the conic by \(\Gamma\). Define two cubic curves

\[X = \overline{AB} \cup \overline{CD} \cup \overline{EF},\quad Y = \overline{BC} \cup \overline{DE} \cup \overline{FA}.\]

Each is a union of three lines, hence a degree-3 curve. Under a general position assumption, \(X\) and \(Y\) have no common component.

Since \(X \cap Y\) contains \(A, B, C, D, E, F\) and \(P, Q, R\), it contains at least nine distinct points. By Bézout’s theorem, \(\sum_{p \in X \cap Y} i_p(X, Y) = 3 \times 3 = 9\), so \(X \cap Y\) consists exactly of these nine points, and the intersection multiplicity at each point is 1.

Now define a new cubic curve \(Z = \Gamma \cup \overline{PQ}\). This is a degree-3 curve passing through eight of the nine points of \(X \cap Y\), namely \(A, B, C, D, E, F\) and \(P, Q\). By Proposition 8 (Special case of Cayley-Bacharach theorem), \(Z\) must also pass through the ninth point \(R\). Since \(R \in Z = \Gamma \cup \overline{PQ}\), we have either \(R \in \Gamma\) or \(R \in \overline{PQ}\).

If \(R \in \Gamma\), then we would have \(R = \overline{CD} \cap \overline{FA} \in \Gamma\). However, by Bézout’s theorem, \(\overline{CD}\) and \(\Gamma\) meet in at most two points, and since we already have \(C, D \in \Gamma\), we get \(\overline{CD} \cap \Gamma = \{C, D\}\). Similarly, \(\overline{FA} \cap \Gamma = \{F, A\}\), so \(R \in \Gamma\) is impossible. Therefore \(R \in \overline{PQ}\), i.e., \(P, Q, R\) are collinear.

Maximum Number of Double Points

Bézout’s theorem gives an upper bound on the number of singular points of a plane curve.

Proposition 10 The maximum number of ordinary double points (double points with two distinct tangent lines) that an irreducible plane curve of degree \(d\) can have is \(\binom{d-1}{2} = \frac{(d-1)(d-2)}{2}\).

Proof

Suppose an irreducible curve \(C\) of degree \(d\) has \(n\) ordinary double points \(p_1, \ldots, p_n\). The genus (geometric genus) \(g\) of a curve is the genus of the smooth curve obtained by normalization. The genus of a smooth projective plane curve is given by the genus-degree formula as \((d-1)(d-2)/2\), and in the presence of singularities the genus decreases by the \(\delta\)-invariant \(\delta_p\) of each singularity \(p\). The \(\delta\)-invariant of an ordinary double point is \(\delta_{p_i} = 1\). Therefore

\[g = \frac{(d-1)(d-2)}{2} - \sum_{i=1}^n \delta_{p_i} = \frac{(d-1)(d-2)}{2} - n\]

and since the geometric genus cannot be negative,

\[n \leq \frac{(d-1)(d-2)}{2}\]

This upper bound is achievable. For example, taking a smooth curve of degree \(d\) and projecting it generally onto \(\mathbb{P}^2\) yields an irreducible curve with exactly \(\frac{(d-1)(d-2)}{2}\) ordinary double points.

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References

[Hart] R. Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics, Springer, 1977.
[Ful] W. Fulton, Intersection Theory, Springer, 1984.
[Sha] I. R. Shafarevich, Basic Algebraic Geometry I: Varieties in Projective Space, Springer, 2013.