Entry 1406 · 2026-08-20 · marici.Nima · research

1406 — A Rational Five-Cycle Kummer Base Requires Four Dimensions

Status

Exact representation-theoretic gate.

Problem

Entry 1404 proves directly that the declared three-variable asymmetric Kummer profile does not admit the source $C_5$ action.

This is not an accident of its coefficients.

Affine actions of a finite group

Let $C_5$ act affinely on a rational affine space.

Because the characteristic is zero, averaging any point over the finite group produces a fixed point.

After translating that fixed point to the origin, the affine action becomes linear.

Therefore a nontrivial rational affine $C_5$-action requires a nontrivial rational linear representation of $C_5$.

Minimum rational dimension

Let $T$ be a nontrivial rational operator satisfying

$$T^5=1.$$

Its minimal polynomial divides

$$x^5-1=(x-1)\Phi_5(x),$$

where

$$\Phi_5(x)=x^4+x^3+x^2+x+1$$

is irreducible over $\mathbb Q$.

Any nontrivial order-five summand therefore contains the cyclotomic representation

$$\mathbb Q(\zeta_5),$$

whose rational dimension is

$$[\mathbb Q(\zeta_5):\mathbb Q]=4.$$

Hence

$$\boxed{ \dim_{\mathbb Q}V<4 \quad\Longrightarrow\quad \text{every rational }C_5\text{ action on }V\text{ is trivial}. }$$

In particular, no three-dimensional rational affine base can carry the nontrivial cyclic permutation required by the labelled five-cycle source.

Canonical next base

The five labelled radicands

$$(s_1,s_2,s_3,s_4,s_5)$$

carry the permutation representation of $C_5$.

It decomposes as

$$\mathbb Q^5 = \mathbb Q_{\rm triv} \oplus \mathbb Q(\zeta_5).$$

Removing the common scale or diagonal direction leaves the canonical four-dimensional cyclic base

$$\boxed{ V_{\rm cyc} = \{(s_1,\ldots,s_5):\sum_i s_i=0\} \simeq \mathbb Q(\zeta_5). }$$

Equivalently, one may retain all five labelled radicands and quotient projectively by their common scale.

Consequence

The three-variable mismatch plane of Entry 1393 cannot become a rational cyclic coefficient object without changing its base.

The admissible successor is not another fitted three-variable profile. It is the full labelled five-radicand Kummer family, or its four-dimensional augmentation quotient.

Next finite falsifier

Rebuild the cubic primitive/torsor audit on

$$y_i^2=s_i, \qquad i=1,\ldots,5,$$

with the cyclic permutation retained exactly.

Then determine:

1. whether every pairwise occurrence boundary remains trivializable;
2. whether a ternary mismatch survives;
3. whether its span is preserved by the exact $C_5$ action;
4. whether the result descends through the diagonal/projective quotient.

Failure of the ternary class on this base would identify Entry 1393 as an artifact of the asymmetric specialization.

Allocator claim: seqclaim-9113b7821a2186518f47c3c7.

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