LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Fourier annihilator equation for cyclic discrete logarithm

Statement

Let $G = ⟨ g ⟩$ have order $q \geq 2$ , $x = g^{a}$ , and $H = ⟨ (- a, 1) ⟩ \leq Z_{q}^{2}$ as in Cyclic discrete logarithm hides a rank-one subgroup. Characters of $Z_{q}^{2}$ are labelled by $(c, d) \in Z_{q}^{2}$ as

χ_{c, d} (u, v) = \exp (2 π i (c u + d v) / q) .

Then

χ_{c, d} \in H^{⊥} ⟺ d \equiv a c (m o d q) .

The map $c \mapsto (c, a c)$ is a bijection from $Z_{q}$ to $H^{⊥}$ . Consequently, any sample that is uniform on $H^{⊥}$ has a uniform first coordinate $c$ .

Facts & Assumptions

Given: The cyclic discrete-log instance and the rank-one subgroup $H$ above.

[F1]

$H = ⟨ (- a, 1) ⟩$ is the fiber subgroup of $f (u, v) = g^{u} x^{v}$ (Cyclic discrete logarithm hides a rank-one subgroup).

[F2]

The annihilator is $H^{⊥} = {χ : χ (h) = 1 for every h \in H}$ (Character annihilator of a subgroup).

[F3]

Every character of the cyclic group $Z_{q}$ is uniquely labelled by $c \in Z_{q}$ with value $\exp (2 π i c / q)$ on $1$ ; characters of a direct product factor coordinatewise (A cyclic group of order n has exactly n additive characters, Characters of a finite product factor coordinatewise).

[F4]

$\exp (2 π i / q)$ is a primitive $q$ th root of unity, so its integer powers equal $1$ exactly at exponents divisible by $q$ (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ).

Proof

1.1givenF1F2F3F4algebra

By [F1] and [F3], every character label is a unique pair $(c, d)$ and its value on the generator $(- a, 1)$ of $H$ is $\exp (2 π i (- a c + d) / q)$ . By [F2], a character lies in $H^{⊥}$ exactly when it is trivial on that generator; [F4] makes this equivalent to $d - a c \equiv 0 (m o d q)$ .

2.1step 1.1algebra∎

For each $c \in Z_{q}$ , step 1.1 gives exactly one annihilator label, namely $(c, a c)$ . Every label in $H^{⊥}$ has this form, so the displayed map is a bijection. A uniform distribution on a finite set is carried by a bijection to the uniform distribution on its image, hence the first coordinate is uniform. This is a counting statement conditional on a uniform annihilator sample; the quantum sample distribution is established separately.

Depends on

Used by

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms (17 April 2025) (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)