Paper XI visual guide · Visual 26 of 47 · PART III / ACTUAL ARITHMETIC CONSTRUCTION

Odd towers retain native trace zero

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Original source locator: Block 166, recursive tower and full-character proofs

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For odd \(p\), \({\operatorname{Tr}}_{E/L}(\alpha)=0\). Choose \(b\) with \(qb\equiv-1\pmod p\), let \(a=(qb+1)/p\), and set \(z=t_L^a\alpha^b\). Then \(z\) is a uniformizer.

For its Eisenstein polynomial \(f\) and constant-coordinate projection \(P_0\), the normalized corestriction uses \(\kappa=-f'(z)/z^{p-1}\). Writing \(u_0=s(P_0)\),

\[ \operatorname{Lift}_E(C)=u_0^{-1}CP_0,\qquad {\operatorname{tr}}_L\operatorname{Lift}_E(C) ={\operatorname{Tr}}_{E/L}(\kappa^{-1}{\operatorname{tr}}_EC). \]

If \({\operatorname{tr}}_EC=0\), the lift has trace zero. Adding \(\alpha I_m\) preserves trace zero. Induction therefore gives trace zero at every native stage and every native correction.

Reading the diagram

This is stronger than the automatic ambient trace cancellation from a multiplicity divisible by \(p\). Native trace controls the scalar restriction used in full-character construction.

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