For u transcendental over the field of ℒ̃ with uū algebraic: if S ⊆ ℒ + ℚu + ℚū and ℚ[S] has transcendence degree ≤ 1, then S lies in ℒ, on a line ℚe, or in 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 $\mathbb{Q}$. Then one of the following holds:

• $S \subseteq \mathcal{L}$;
• there is $e \in \mathbb{C}$ such that every $z \in S$ is $re$ for some $r \in \mathbb{Q}$;
• every $z \in S$ is $pu + q\bar u$ for some $p, q \in \mathbb{Q}$.

Proof. By DiazModulus.mem_logAlg_adjoin_circle_point_iff every $z \in S$ is $c + pu + q\bar u$ with $c \in \mathcal{L}$ and $p, q \in \mathbb{Q}$, and the same node at the point $0$ shows that $\mathcal{L}$ is closed under rational combinations. If $S \subseteq \mathcal{L}$ there is nothing to prove; otherwise fix $z_0 = c_0 + p_0u + q_0\bar u \in S$ outside $\mathcal{L}$, so $(p_0, q_0) \neq (0, 0)$. If some $z_1 = c_1 + p_1u + q_1\bar u \in S$ has $D = p_1q_0 - q_1p_0 \neq 0$, then $(q_0z_1 - q_1z_0)/D = c_1' + u$ and $(p_1z_0 - p_0z_1)/D = c_2' + \bar u$ lie in $\mathbb{Q}[S]$ with $c_1', c_2' \in \mathcal{L}$, and DiazModulus.logAlg_adjoin_circle_point_trdeg_le_one_hyperbola_case gives $S \subseteq \mathbb{Q}u + \mathbb{Q}\bar u$. Otherwise every $(p, q)$ is $r(p_0, q_0)$ with $r = (pp_0 + qq_0)/(p_0^2 + q_0^2)$, so $z - rz_0 \in \mathcal{L} \cap \mathbb{Q}[S]$, and DiazModulus.logAlg_adjoin_circle_point_trdeg_le_one_logAlg_part_eq_zero with the point $z_0$ makes it $0$: $S$ lies on the line $\mathbb{Q}z_0$.

Novelty. Not found in the sources read; routine.