For u transcendental over the field of ℒ̃ with uū algebraic: if S ⊆ ℒ + ℚu + ℚū and ℚ[S], of transcendence degree ≤ 1, contains c₁ + u and c₂ + ū with c₁, c₂ ∈ ℒ, then S ⊆ ℚu + ℚū

Here $\mathcal{L}$ is the set of logarithms of algebraic numbers (the $z \in \mathbb{C}$ with $e^z$ algebraic), and $\widetilde{\mathcal{L}}$ is the $\overline{\mathbb{Q}}$-vector space spanned by $1$ and $\mathcal{L}$. Let $F$ be the subfield of $\mathbb{C}$ generated by $\widetilde{\mathcal{L}}$, that is, by $\overline{\mathbb{Q}}$ and $\mathcal{L}$. Let $u \in \mathbb{C}$ be transcendental over $F$, with $u\bar u$ algebraic, and put $\mathcal{L}_u = \mathrm{span}_{\mathbb{Q}}(\mathcal{L}) + \mathrm{span}_{\mathbb{Q}}(u, \bar u)$; as $\mathcal{L}$ is a $\mathbb{Q}$-vector space, this is $\mathcal{L} + \mathbb{Q}u + \mathbb{Q}\bar u$. Let $S \subseteq \mathcal{L}_u$ be such that $\mathbb{Q}[S]$, the $\mathbb{Q}$-subalgebra of $\mathbb{C}$ generated by $S$, has transcendence degree at most $1$ over $\mathbb{Q}$, and let $c_1, c_2 \in \mathcal{L}$ with $c_1 + u \in \mathbb{Q}[S]$ and $c_2 + \bar u \in \mathbb{Q}[S]$. Then every $z \in S$ is $pu + q\bar u$ for some $p, q \in \mathbb{Q}$.

Proof. An algebraically independent pair in $\mathbb{Q}[S]$ would exceed the bound on its transcendence degree, so $(c_1 + u, c_2 + \bar u)$ is dependent, and the hyperbola step DiazModulus.isAlgebraic_of_hyperbola_pair_not_algebraicIndependent makes $c_1$ and $c_2$ algebraic; Hermite–Lindemann (DiazModulus.hermite_lindemann_holds) then makes them $0$, so $u, \bar u \in \mathbb{Q}[S]$. Each $z \in S$ is $c + pu + q\bar u$ with $c \in \mathcal{L}$ (DiazModulus.mem_logAlg_adjoin_circle_point_iff), so $c = z - pu - q\bar u$ lies in $\mathcal{L} \cap \mathbb{Q}[S]$. Then DiazModulus.logAlg_adjoin_circle_point_trdeg_le_one_logAlg_part_eq_zero, applied with the point $u = 0 + 1 \cdot u + 0 \cdot \bar u$ of $\mathbb{Q}[S]$, gives $c = 0$, that is $z = pu + q\bar u$.

Novelty. Not found in the sources read; routine.