lemmaroutineunresolved_ref

Private odd valuation for each radicand

Lemma·alg:private-valuations·§8.4 A separate odd valuation for each radicand

Statement (paper)

For every ℓ∈{1,…,j}\ell\in\{1,\ldots,j\} there is an integer-valued valuation vℓv_\ell of MM such that vℓ(Dℓ)=1,vℓ(Dm)=0(m≠ℓ). v_\ell(D_\ell)=1,\qquad v_\ell(D_m)=0\quad(m\ne\ell). Consequently the classes of D1,…,DjD_1,\ldots,D_j are independent in M×/(M×)2M^\times/(M^\times)^2.

In plain English

For each ℓ∈{1,…,j}\ell\in\{1,\ldots,j\} there is an integer-valued valuation vℓv_\ell of MM with vℓ(Dℓ)=1v_\ell(D_\ell)=1 and vℓ(Dm)=0v_\ell(D_m)=0 for m≠ℓm\ne\ell. Hence the classes of D1,…,DjD_1,\ldots,D_j are independent in M×/(M×)2M^\times/(M^\times)^2.

Why it's here

Each radicand DℓD_\ell gets its own valuation, which sees only DℓD_\ell to odd order and treats the other DmD_m as units. The construction splits by the shape of the limiting locus: the plane (orders along irreducible curves QℓQ_\ell), a general curve (power-series branch at a smooth point), a line, or a hyperbola (simple zeros of Lℓ−4L_\ell-4). Since a square has even valuation under every integer-valued valuation, any nonempty product of the DmD_m is a non-square, which gives independence modulo squares.
Classified routine:
A case-by-case construction of discrete valuations (orders of vanishing along curves or at smooth branch points) that separate the radicands. Independence modulo squares then follows by the standard parity argument: a square has even valuation, as in Kummer-theory independence criteria.

Uses

Nothing else in the paper

Used by

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Inside the proof

Sources

PaperOpen
Section
§8.4 A separate odd valuation for each radicand
Page
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Label
alg:private-valuations
Line
sections/07-algebra.tex:438
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

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    unresolved_ref
    Excluding b=1 uses the unit-circle exclusion equation alg:unit-circle and deletion hypothesis argued in Section 8 text, not a labeled statement.