CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-16

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A recurrence over $Q$ can require a proper splitting field for its exponential closed form

Statement refuted

Every constant-coefficient recurrence over $Q$ has its characteristic-root exponential closed form over $Q$ , without passing to a proper splitting field.

Facts & Assumptions

Given: The sequence defined by $a_{0} = 0$ , $a_{1} = 1$ , and $a_{n + 2} + a_{n} = 0$ .

[L1]

A recurrence beginning at zero is represented by its rational formal generating function (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L2]

The polynomial-times-exponential form is asserted over a named splitting field of the recurrence characteristic polynomial (Over a named splitting field in characteristic zero, repeated characteristic roots give polynomial-times-exponential closed forms).

Counterexample

technique · explicit splitting field

1.1givenL1algebra

Multiplying $A (x) = \sum_{n \geq 0} a_{n} x^{n}$ by $1 + x^{2}$ and using the recurrence leaves $x$ , so [L1] gives $A (x) = x / (1 + x^{2})$ over $Q$ .

1.2algebra

The characteristic polynomial $t^{2} + 1$ has no root in $Q$ , but in $Q (i)$ it factors as $(t - i) (t + i)$ . Thus a pure-exponential expression using its characteristic roots cannot be written with bases in $Q$ .

2.1step 1.2L2algebra

Applying [L2] in the splitting field and matching $a_{0} = 0, a_{1} = 1$ gives $a_{n} = (i^{n} - (- i)^{n}) / (2 i)$ . Its first four values are $0, 1, 0, - 1$ , which satisfy the given recurrence.

3.1step 1.1step 1.2step 2.1∎

This rational recurrence over $Q$ therefore requires the proper splitting field $Q (i)$ for its characteristic-root exponential form, refuting the claim.

Depends on

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A recurrence over QQ can require a proper splitting field for its exponential closed form