claimroutineinferred_dep
Fractional-linear case gives private simple zero
Claim·§8.4 A separate odd valuation for each radicand·inside lemPrivate odd valuation for each radicand
Statement (paper)
In the fractional-linear case, with one has , and has two distinct simple zeros at finite points with , the roots of . For a corresponding point , algebraic over , no other radicand , , vanishes or equals at , so this simple zero of is shared by no other radicand, and order at its abscissa gives the required valuation over .
In plain English
In the hyperbola (fractional-linear) case, with we have , and has two distinct finite simple zeros with , the roots of . At a corresponding point no other radicand () vanishes or equals , so this zero is private to and its order gives the valuation over .
Why it's here
This handles the hyperbola branch of the dichotomy from the same-curve claim, where and . It produces, for each , a simple zero of that no other radicand shares, so the order of vanishing there is an odd valuation supporting the square-class independence of the in the parent lemma.
Classified routine:
A direct calculation on the hyperbola branch from the preceding same-curve claim: it substitutes the fractional-linear form, checks that the quadratic has distinct simple roots, and verifies that the zero is not shared by other radicands, which is standard valuation bookkeeping.
Uses
Used by
Sources
PaperOpen
- Section
- §8.4 A separate odd valuation for each radicand
- Page
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- Label
—- Line
sections/07-algebra.tex:560
Leannot aligned
Phase 2 aligns this node to a Lean declaration.
Friction
- infoinferred_depoutline: "Fractional-linear case private simple zero" rests on claim-alg-private-valuations-5