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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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A nonconstant rational map has total fibre multiplicity equal to its degree

Statement

Let R:C^→C^ be a nonconstant rational map of degree d. Then for every value w∈C^ the total multiplicity of the fibre R−1(w) is exactly d.

Facts & Assumptions

Given: A nonconstant rational map R=P/Q with coprime polynomials and degree d=max⁡(deg⁡P,deg⁡Q).

[L1]

A complex polynomial of degree n≥1 has exactly n roots counted with multiplicity (A complex polynomial of degree n has exactly n roots counted with multiplicity).

Proof

technique · direct
1.1

For a finite value w, the finite preimages of w are exactly the roots of P−wQ. If deg⁡(P−wQ)=d, then [L1] gives exactly d such roots. If deg⁡(P−wQ)=m<d, then ∞ is also a preimage and its multiplicity is exactly d−m in the infinity chart, so the total multiplicity is still d.

L1givenalgebra
1.2

For w=∞, the finite preimages are the roots of Q with multiplicity. If deg⁡P≤deg⁡Q=d, then [L1] gives all d preimages in the finite chart; if deg⁡P>deg⁡Q, then ∞ contributes the remaining multiplicity deg⁡P−deg⁡Q. So the total multiplicity is again d.

L1givenalgebra
2.1

Every sphere value is either finite or ∞, and both cases give total fibre multiplicity d.

given∎

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