claimroutine

Plane case: order along irreducible Q_ell

Claim·§8.4 A separate odd valuation for each radicand·inside lemPrivate odd valuation for each radicand

Statement (paper)

In the case where the left locus is the plane, the polynomial

Q_\ell=(z-z_{r_\ell})(w-w_{r_\ell})-4

is irreducible in

K_0[z,w]

, the polynomials

Q_\ell

are pairwise distinct, and no

Q_\ell

divides any

L_m=(z-z_{r_m})(w-w_{r_m})

. Hence the exponent of

Q_\ell

in a rational function is an integer-valued valuation

v_\ell

with

v_\ell(D_\ell)=1

and

v_\ell(D_m)=0

for

m\ne\ell

In plain English

When the left locus is the plane, each

Q_\ell=(z-z_{r_\ell})(w-w_{r_\ell})-4

is irreducible in

K_0[z,w]

, the

Q_\ell

are pairwise distinct, and none divides any

L_m=(z-z_{r_m})(w-w_{r_m})

. So the order of vanishing along

Q_\ell

is an integer-valued valuation

v_\ell

with

v_\ell(D_\ell)=1

and

v_\ell(D_m)=0

for

m\ne\ell

Why it's here

This supplies the separate odd valuation for each radicand

D_\ell

in the plane case: each

D_\ell

vanishes to order one along its own irreducible curve

Q_\ell

and is a unit along the others. This is what lets the parent lemma conclude that the classes of

D_1,\ldots,D_j

are independent modulo squares.

Classified routine:

Direct algebraic check that a polynomial of the form (z-a)(w-b)-4 is irreducible over a field and that the order of vanishing along an irreducible factor of a UFD is a discrete valuation. This is standard bookkeeping in K_0[z,w].

Uses

Nothing else in the paper

Used by

claimSubset products of radicands are non-squares

Sources

PaperOpen

Section: §8.4 A separate odd valuation for each radicand
Page: —
Label: —
Line: sections/07-algebra.tex:454

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