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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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Coprime tangent cones force a power of the maximal ideal into the local ideal

Statement

Assume the Axiom of Choice. Let O be the local ring of A2 at the origin p over an algebraically closed field k, with maximal ideal m, and let f,g∈m have orders m=ordp(f), n=ordp(g). If the lowest-degree forms f∗ and g∗ have no common factor in k[x,y] (equivalently their zero sets meet only at the origin), then mt⊆(f,g)O for every t≥m+n−1. In particular O/(f,g)O is a quotient of O/mm+n−1, and the quotient map O/(f,g)↠O/((f,g)+mt) is an isomorphism for t≥m+n−1.

Facts & Assumptions

Given: AC, an algebraically closed field k, the local ring O of A2 at the origin, its maximal ideal m=(x,y), elements f,g∈m with orders m,n≥1, and their lowest-degree forms f∗,g∗ (the initial forms in the m-adic filtration).

[F1]

O is a two-dimensional regular local ring; its associated graded ring is gr⁡mO≅k[X,Y] with standard grading, in particular a domain, so initial forms multiply: (h1h2)∗=h1∗h2∗ associated graded ring of a regular local ring, embedding dimension and regular local ring, A local ring is a nonzero commutative ring with a unique maximal ideal, The Axiom of Choice.

[F3]

For coprime forms f∗,g∗ of degrees m,n the graded multiplication map k[x,y]t−m⊕k[x,y]t−n→k[x,y]t, (A,B)↦Af∗+Bg∗, is surjective whenever t≥m+n−1. Indeed, if Af∗+Bg∗=0, then f∗∣B and g∗∣A by coprimality in the UFD, so (A,B)=(g∗h,−f∗h) for h∈k[x,y]t−m−n when t≥m+n, and there is no nonzero syzygy when t=m+n−1; by [F2] and rank-nullity the kernel has dimension max⁡(0,t−m−n+1), while the domain has dimension 2t−m−n+2 and the target has dimension t+1, so surjectivity follows for t≥m+n−1 Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T, Module homomorphism and isomorphism, kernel, image and cokernel, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis.

[F4]

Coprime lowest-degree forms force f,g to be coprime in O: if a nonunit irreducible h∈O divided both, then h has order ≥1 and h∗ is a nonconstant form, and by [F1] h∗∣f∗ and h∗∣g∗, contradicting coprimality Irreducible and prime elements of an integral domain, Prime ideals and maximal ideals in a commutative ring. Clear the unit denominators of f,g to apply Finite local length exactly when no common local branch to polynomial numerators. Their ideal is m-primary; Proof 1.3 of that lemma gives mN⊆(f,g)O for some N.

Proof

1.1F3F4algebragivenF1F2

By [F4] there is N with mN⊆(f,g)O. By [F3], for every t≥m+n−1 and every form h of degree t there are forms A,B of degrees t−m,t−n with h=Af∗+Bg∗; lifting A,B, and using f=f∗+f>m, g=g∗+g>n with f>m,g>n of order at least m+1,n+1, we find Af+Bg=h+(order≥t+1). Hence every element of mt is congruent modulo mt+1 to an element of (f,g)O, i.e. mt⊆(f,g)O+mt+1 for all t≥m+n−1.

2.1step 1.1algebraF5

Fix s=m+n−1 and iterate the inclusion of step 1.1: ms⊆(f,g)O+ms+j for every j≥0, so choosing j=max⁡(0,N−s) gives ms⊆(f,g)O+mN⊆(f,g)O (if N≤s then ms⊆mN⊆(f,g)O directly). More generally the same argument with any t≥s in place of s gives mt⊆(f,g)O for every t≥m+n−1.

3.1step 2.1given∎

Consequently the quotient map O↠O/(f,g)O factors through O/mm+n−1, exhibiting O/(f,g)O as a quotient of O/mm+n−1, and for t≥m+n−1 one has (f,g)+mt=(f,g), so the quotient map O/(f,g)↠O/((f,g)+mt) is an isomorphism. This is the asserted containment and its two consequences.

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