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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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Intersection with a smooth curve is a vanishing order

Statement

Assume the Axiom of Choice. Let C be a plane projective curve smooth at p, and let D be a plane projective curve whose local equation g at p does not vanish identically on C (no common local component through p). Then

Ip(C,D)=ord⁡p(g∣C),

the valuation of the image of g in the discrete valuation ring OC,p.

Facts & Assumptions

Given: AC, a plane projective curve C=V(F) smooth at p, a plane projective curve D=V(G) with local equation g at p whose restriction g∣C is nonzero, and a local equation f of C at p.

[F1]

The local ring OC,p=OP2,p/(f) is a discrete valuation ring with valuation ord⁡p, and the class of g in it is the restriction g∣C Uniformising parameters at smooth points of a plane curve, Discrete valuation rings, Uniformising parameters.

[F2]

Length is unchanged on passing to the quotient by the equation of the curve, because an O/(f)-module has exactly the same submodules over O and over O/(f): OP2,p/(f,g)≅OC,p/(g∣C) by localisation commuting with quotients Localisation commutes with quotient rings: S−1R/S−1I≅Sˉ−1(R/I), so Ip(C,D)=ℓOC,p(OC,p/(g∣C)) Local intersection multiplicity of two plane curves.

[F3]

In a discrete valuation ring V with uniformiser π, a nonzero element x=uπn with u a unit has ℓV(V/(x))=n; a nonzero element of the DVR is a nonzerodivisor, so the quotient is a finite-length module exactly of this length Every nonzero fraction is a unit times a power of a uniformiser, Length and valuation in a DVR, Composition series and length of a module.

[F4]

Since g does not vanish identically on C, its class in the DVR is nonzero, so the finiteness hypothesis of the definition is satisfied and [F3] applies Finite local length exactly when no common local branch.

Proof

1.1F1F2F4given

The ring OC,p is a discrete valuation ring by [F1], and the restriction g∣C is its nonzero element. The quotient identification of [F2] gives Ip(C,D)=ℓV(V/(g∣C)) with V=OC,p.

1.2F1F3algebra

In the discrete valuation ring V with uniformiser π and valuation ord⁡p, the element g∣C has the normal form g∣C=uπn with u a unit, and the length of V/(g∣C) equals n=ord⁡p(g∣C).

2.1step 1.1step 1.2F3given∎

Combining steps 1.1 and 1.2, Ip(C,D)=ℓV(V/(g∣C))=ord⁡p(g∣C); and by the convention of the definition the value is 0 exactly when g∣C is a unit, i.e. when p∉D. This proves the displayed equality.

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Sources