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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-28
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A quotient by a Weierstrass polynomial is a finite module over the smaller germ ring

Statement

Let W be a Weierstrass polynomial of degree d in zm. Then the quotient Om,0/(W) is a finitely generated Om1,0-module, where O0,0=C when m=1, generated by the residue classes of

1, zm, , zmd1.

Facts & Assumptions

Given: A degree-d Weierstrass polynomial W.

[L1]

A Weierstrass polynomial is the monic degree-d polynomial in the last variable from Weierstrass polynomials in the last variable.

[L2]

Weierstrass division gives unique quotient and remainder of degree <d upon division by W (Weierstrass division theorem).

[L3]

Noetherian-module language is that of Noetherian commutative rings and modules.

Proof

technique · direct
1.1

By [L1] and [L2], every germ fOm,0 can be written uniquely as f=qW+r0+r1zm++rd1zmd1 with rjOm1,0. Modulo (W) this becomes [f]=r0[1]+r1[zm]++rd1[zmd1], so the listed residue classes generate the quotient as an Om1,0-module.

L1L2L3
2.1

The same division theorem [L2] makes the remainder unique, so those generators give a canonical normal form for every class in the quotient. Since there are only d generators, the quotient is a finite Om1,0-module.

step 1.1L2

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources