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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-27
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Algebraic Bezout formula as a sum of local scheme lengths

Statement

Assume the Axiom of Choice. Let k be a field and let F,G∈k[x0,x1,x2] be nonzero homogeneous forms of positive degrees d and e with no common nonconstant factor, and put X=Proj⁡(k[x0,x1,x2]/(F,G)). Then ∑x∈XℓOX,x(OX,x) [κ(x):k]=de, a finite sum of local lengths weighted by residue degrees. If moreover k is algebraically closed, then [κ(x):k]=1 for every x∈X, and therefore ∑x∈XℓOX,x(OX,x)=de.

This is the algebraic length statement supplied to the later plane-curve page. It asserts nothing about local equations other than the dehomogenised F,G, nothing about invariance under other choices of equations for the same local curve, and no geometric intersection formulation.

Facts & Assumptions

Given: The Axiom of Choice, a field k, nonzero homogeneous forms F,G∈k[x0,x1,x2] of positive degrees d,e with no common nonconstant factor, the standard graded quotient S=k[x0,x1,x2]/(F,G), and X=Proj⁡S.

[L1]

Assume AC. X is nonempty and finite, every chart ring is zero or of Krull dimension 0, and the total length satisfies len⁡k(X)=de (Two coprime projective plane forms meet in total length equal to their degree product).

[L2]

Assume AC. For a zero-dimensional X the total length is the finite sum len⁡k(X)=∑x∈XℓOX,x(OX,x)[κ(x):k] over the finitely many points, each local ring OX,x being a finite-dimensional local k-algebra of finite length and each residue field κ(x) being finite over k (Total length of a zero-dimensional projective scheme, A zero-dimensional projective scheme has finitely many closed points with finite-dimensional local rings, The residue field at a point of an affine scheme, The degree [K:F]=dim⁡FK of a finite field extension).

[L3]

A field F is algebraically closed exactly when it has no nontrivial finite extension; equivalently F is algebraically closed if and only if every finite extension F⊆K satisfies K=F (An algebraically closed field: every nonconstant polynomial has a root in the field, A field is algebraically closed exactly when every nonconstant polynomial splits, equivalently when it has no nontrivial finite extension).

[L4]

Assume AC (declared for consumers of this corollary). The Axiom of Choice

Proof

technique · direct
1.1

By [L1] the scheme X is finite, nonempty and has total length len⁡k(X)=de.

L1
1.2

By [L2] the total length of X is the finite weighted sum len⁡k(X)=∑x∈XℓOX,x(OX,x)[κ(x):k] over the finitely many points of X, with every residue degree [κ(x):k] finite over k.

L2
2.1

Combining steps 1.1 and 1.2 gives ∑x∈XℓOX,x(OX,x)[κ(x):k]=len⁡k(X)=de, which is the displayed formula.

step 1.1step 1.2
3.1

Now assume that k is algebraically closed; by step 1.2 each κ(x) is a finite extension field of k, so [L3] gives κ(x)=k and hence [κ(x):k]=1 for every x∈X; substituting into step 2.1 gives ∑x∈XℓOX,x(OX,x)=de.

L3step 1.2step 2.1
4.1

The weighted sum equals de over an arbitrary field by step 2.1, and over an algebraically closed field it collapses to the unweighted sum of local lengths by step 3.1; the Axiom of Choice is inherited from [L1] and [L2] and is declared here for users of the corollary as the standing assumption [L4].

L1L2L4step 2.1step 3.1given∎

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