claimstandard

Wild inertia acts trivially over F_1

Shows wild inertia acts trivially over

F_1=F(A[p_0])

finiteness of fields for the small-prime cut

Claim·§4.8 A uniform bound for the modular level·inside lemModular level is polynomially bounded in M

Statement (paper)

Let

p_0\ge3

be a fixed prime different from

q

and put

F_1=F(A[p_0])

, so that

[F_1:F]\le|\mathrm{GL}_4(\F_{p_0})|

. Then inertia acts unipotently on

V_\ell(A)

over

F_1

, and its wild pro-

q

subgroup acts trivially on

V_\ell(A)

. Moreover, since

[F:\Q_q]

and

[F_1:F]

are bounded, the fields

F_1

and their normal closures over

\Q_q

range over finite sets.

In plain English

Fix a prime

p_0\ge3

different from

q

and let

F_1=F(A[p_0])

, an extension of

F

of degree at most

|\mathrm{GL}_4(\F_{p_0})|

. Over

F_1

, inertia acts unipotently on

V_\ell(A)

and wild inertia acts trivially. Also,

F_1

and its normal closure over

\Q_q

come from finite lists.

Why it's here

Adjoining the

p_0

-torsion kills the semistability obstruction (as in Serre–Tate/Grothendieck's criterion), so only unipotent tame inertia remains and the wild pro-

q

part is trivial. Because

[F:\Q_q]

and

[F_1:F]

are bounded, only finitely many fields

F_1

can occur, which lets the next claim choose a ramification cut

u_q

independent of

s

and bound the conductor at small primes.

Classified standard:

Raynaud's criterion with Grothendieck's semistable reduction theorem (SGA 7, 1972): an abelian variety acquires semistable reduction over the field generated by its m-torsion for m≥3 prime to the residue characteristic, so inertia acts unipotently and wild inertia trivially. Finiteness of extensions of Q_q of bounded degree is Krasner's lemma.

Uses

Nothing else in the paper

Used by

claimUniform conductor bound gives N ≤ CM²

Sources

PaperOpen

Section: §4.8 A uniform bound for the modular level
Page: —
Label: —
Line: sections/04-height.tex:1108

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