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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Conjugate norms from the dyadic eighth-cyclotomic ring are real dyadics

Statement

Let $D = Z [ω, 1 / 2]$ and $D_{+} = Z [\sqrt{2}, 1 / 2] \subseteq R$ . For every $z \in D$ , its complex-conjugate norm satisfies $z^{†} z \in D_{+}$ . More precisely, if $z = (a_{0} + a_{1} ω + a_{2} ω^{2} + a_{3} ω^{3}) / 2^{r}$ is its canonical encoding, then $z^{†} z = \frac{P + \sqrt{2} Q}{2^{2 r}}, P = \sum_{j = 0}^{3} a_{j}^{2}, Q = a_{0} (a_{1} - a_{3}) + a_{2} (a_{1} + a_{3}) .$

Facts & Assumptions

Given: $ω = e^{i π / 4}$ , $\sqrt{2} = ω - ω^{3}$ , $z \in D$ , and scalar adjoint means complex conjugation: $z^{†} = \overset{ˉ}{z}$ .

[F1]

Canonical dyadic coefficient encoding: every $z \in D$ has a canonical power-basis numerator over a power $2^{r}$ .

[F2]

Square root of two divisibility criterion in the eighth cyclotomic ring: for $a = a_{0} + a_{1} ω + a_{2} ω^{2} + a_{3} ω^{3} \in O$ , $a^{†} a = P + \sqrt{2} Q$ with the displayed integer forms $P, Q$ .

Proof

1.1F1givenalgebra

Write $z = a / 2^{r}$ using [F1]. Since $2^{r}$ is real, $z^{†} z = (a^{†} a) / 2^{2 r}$ .

2.1F2step 1.1algebra∎

Substituting the exact norm formula from [F2] gives $z^{†} z = (P + \sqrt{2} Q) / 2^{2 r}$ . Both coefficients are dyadic rationals, so this element lies in $D_{+} = Z [\sqrt{2}, 1 / 2]$ and has the claimed form.

Depends on

Used by

Dependency tree · two levels

5 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

Kliuchnikov, Maslov, and Mosca, Practical approximation of single-qubit unitaries by single-qubit quantum circuits (standard reference, not scraped)