CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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An ordinary irreducible representation can have reducible reduction modulo p

Statement refuted

Reducing an ordinary irreducible lattice modulo $p$ always preserves irreducibility.

Facts & Assumptions

Given: The splitting $3$ -modular system $(K, O, k) = (Q_{3} (ζ_{3}), Z_{3} [ζ_{3}], F_{3})$ and lattice $L = {(a, b, c) \in O^{3} : a + b + c = 0}$ from Reducing a standard integral lattice for S3 modulo 3 produces a reducible kS3-module, where $ζ_{3}$ is a primitive cube root of unity. The maximal ideal is $m = (1 - ζ_{3})$ ; reduction is modulo $m$ , not the ideal $(3)$ of this ramified valuation ring.

[F1]

Reduction modulo the maximal ideal produces a $k G$ -module (An OG-lattice is a finite free module over the valuation ring with G-action, and reduction modulo the maximal ideal produces a kG-module).

[L1]

The defining-characteristic page route allows reducibility after reduction (When the characteristic divides the group order, Maschke can fail and kG need not be semisimple).

[L2]

The specified triple is a splitting $3$ -modular system; $L$ has basis $(1, 0, - 1), (0, 1, - 1)$ and its reduction is the coordinate-sum-zero plane in $k^{3}$ , containing the nonzero invariant vector $(1, 1, 1)$ (Reducing a standard integral lattice for S3 modulo 3 produces a reducible kS3-module).

Counterexample

technique · direct

1.1givenL2choosealgebra

Extending scalars gives $V = K \otimes_{O} L = {(a, b, c) \in K^{3} : a + b + c = 0}$ with the same two basis vectors. If $W \subseteq V$ is a nonzero invariant subspace, choose $0 \neq v \in W$ . Its coordinates cannot all agree, since $3 a = 0$ implies $a = 0$ in characteristic zero. For unequal coordinates $v_{i}, v_{j}$ , subtracting the image of $v$ under the transposition $(i j)$ gives $(v_{i} - v_{j}) (e_{i} - e_{j}) \in W$ . Thus $e_{i} - e_{j} \in W$ , and permuting coordinates gives all differences, which span $V$ . Hence $W = V$ and $V$ is irreducible.

2.1F1L1L2step 1.1

By [F1] and [L2], reducing $L$ modulo $m$ gives a two-dimensional $k S_{3}$ -module $\overline{L}$ . The vector $(1, 1, 1)$ is the reduction of $(1, 1, - 2) \in L$ and spans a nonzero proper invariant line, so $\overline{L}$ is reducible. The residue characteristic is $3$ , which divides $∣ S_{3} ∣ = 6$ , consistent with [L1].

3.1step 1.1step 2.1∎

Therefore an ordinary irreducible representation can have reducible reduction modulo $p$ , refuting the statement.

Depends on

Used by

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Sources

Leonard Tomczak, Local Fields lecture notes (2022), cyclotomic extension example, p.31 (standard reference, not scraped)
Peter Webb, A Course in Finite Group Representation Theory (23 Feb 2016 draft) (standard reference, not scraped)