Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: reduction mod p of an ordinary irreducible is always irreducible

Statement

If χ is an ordinary irreducible character, then its reduction modulo p is always irreducible.

Facts & Assumptions

Given: A primitive cube root ζ3, the local cyclotomic triple (K,O,k)=(Q3(ζ3),Z3[ζ3],F3), and the standard OS3-lattice L={(a,b,c)O3:a+b+c=0}, whose scalar extension to K affords the ordinary standard irreducible representation of S3.

[F1]

Reduction modulo p is recorded by the decomposition map (Decomposition map from ordinary to modular Grothendieck groups).

[F2]

Decomposition numbers describe the simple factors of that reduction (Decomposition numbers and the decomposition matrix).

Refutation

technique · direct
1.1

By the given realization, the ordinary character afforded by KOL is irreducible.

given
2.1

By [F1], reducing modulo the maximal ideal gives the kS3-module L=L/(1ζ3)L. The nonzero vector (1,1,1)L is fixed by every permutation matrix and belongs to L because it is the reduction of (1,1,2)L. Hence L has a nontrivial proper invariant line and is reducible.

F1step 1.1algebra
3.1

Thus an ordinary irreducible representation can have reducible reduction modulo p. In the decomposition process recorded by [F2], this reduction is therefore not irreducible. The statement is false.

F2step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources