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CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Every meromorphic function on a plane domain is a quotient of holomorphic functions

Statement

Every meromorphic function on a plane domain is a quotient of holomorphic functions.

Facts & Assumptions

Given: A meromorphic function f on a plane domain Ω.

[L1]

A meromorphic function is holomorphic away from a discrete pole set (Meromorphic functions on a plane domain).

[L2]

Every discrete effective divisor on a plane domain is the zero divisor of a holomorphic function (Every discrete effective divisor on a plane domain is the zero divisor of a holomorphic function).

[L3]

A locally bounded punctured singularity is removable (Characterizations of removable singularities).

Proof

technique · direct
1.1

Let P be the pole set of f, with multiplicities equal to pole orders. By [L2], choose a holomorphic function h on Ω whose zero divisor is exactly P.

L1L2givenconstruct
2.1

On ΩP, define g:=fh. Near a pole aP, the zero of h has exactly the same order as the pole of f, so g is locally bounded on a punctured neighbourhood of a. By [L3], g extends holomorphically across every point of P.

L1L3step 1.1algebra
3.1

Away from P, one has f=g/h. Since both sides are meromorphic and agree on the dense open set ΩP, this quotient represents f on all of Ω.

step 1.1step 2.1algebra

Depends on

Used by

Dependency tree · two levels

13 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