Four exponentials in transcendence degree one for ℒ_u = ℒ + ℚu + ℚū (u transcendental over the field of ℒ̃, uū algebraic): a singular 2×2 matrix over ℒ_u∖{0} has ℚ-dependent rows or columns

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 $\ell_{11}, \ell_{12}, \ell_{21}, \ell_{22} \in \mathcal{L}_u$ be non-zero with $\ell_{11}\ell_{22} = \ell_{12}\ell_{21}$, and suppose the $\mathbb{Q}$-subalgebra of $\mathbb{C}$ they generate has transcendence degree at most $1$ over $\mathbb{Q}$. Then there are $a, b \in \mathbb{Q}$, not both zero, with $a\ell_{11} + b\ell_{21} = a\ell_{12} + b\ell_{22} = 0$ (the rows of the matrix $(\ell_{ij})$ are linearly dependent over $\mathbb{Q}$), or with $a\ell_{11} + b\ell_{12} = a\ell_{21} + b\ell_{22} = 0$ (its columns are).

Proof. Apply the trichotomy DiazModulus.logAlg_adjoin_circle_point_trdeg_le_one_trichotomy to the four entries. If all lie in $\mathcal{L}$, their exponentials are algebraic and the four exponentials theorem in transcendence degree one (DiazModulus.four_exponentials_trdeg_one) gives the conclusion verbatim. If all are $r_{ij}e$ with $r_{ij} \in \mathbb{Q}$, then $e \neq 0$, $r_{11}r_{22} = r_{12}r_{21}$, and $(a, b) = (r_{21}, -r_{11}) \neq (0, 0)$ makes the rows dependent. If all are $p_{ij}u + q_{ij}\bar u$, DiazModulus.four_exp_barrier_of_no_quadratic_relation with $e = (u, \bar u)$ applies once $u, \bar u$ satisfy no rational quadratic relation: $au^2 + bu\bar u + c\bar u^2 = 0$ times $u^2$ reads $a(u^2)^2 + b\rho u^2 + c\rho^2 = 0$ with $\rho = u\bar u$ algebraic and non-zero, and as $u^2$ is transcendental, Transcendence.quadratic_coeffs_eq_zero_of_transcendental gives $a = b = c = 0$.

Novelty. With $\mathcal{L}_u$ in place of $\mathcal{L}$, this is the four exponentials theorem in transcendence degree one (Waldschmidt 1973, Brownawell 1974), from which it is derived. Not found in the sources read; routine.