A transcendence-degree-one field as quotients of integer polynomials in ω and ω₁

Presenting the field. Let $x_1, x_2$ be $\mathbb{Q}$-linearly independent, and likewise $y_1, y_2$. Suppose every $e^{x_iy_j}$ is algebraic and $\operatorname{trdeg}_{\mathbb{Q}}\mathbb{Q}[x_1,x_2,y_1,y_2] \le 1$. Then there are $\omega, \omega_1 \in \mathbb{C}$ such that $\omega$ is transcendental; $\omega_1$ is a root of $Q \in \mathbb{Z}[X][Y]$, monic in $Y$ of degree $d \ge 1$ and minimal, in the sense that no non-zero $A \in \mathbb{Z}[X][Y]$ with $\deg_Y A < d$ vanishes at $(\omega, \omega_1)$; and $D, E_i, G_j, H_{ij} \in \mathbb{Z}[X][Y]$ with $D(\omega,\omega_1) \ne 0$, $x_i D = E_i$, $y_j D = G_j$ and $e^{x_iy_j} D = H_{ij}$ at $(\omega, \omega_1)$. Evaluation is $A \mapsto A(\omega)(\omega_1)$ via eval₂.

Proof sketch. By Hermite–Lindemann $\omega = x_1y_1$ is transcendental, since it is non-zero with algebraic exponential. The field generated by $x, y$ and the exponentials is finite over $\mathbb{Q}(\omega)$. A primitive element, scaled by a polynomial in $\omega$, becomes $\omega_1$, integral over $\mathbb{Z}[\omega]$ with a monic minimal relation $Q$. Clearing denominators gives $D$, $E$, $G$ and $H$.

Role. It is the opening of the 1973 §III, in integral form, for FourExp.construction_core_1973.

About the hypotheses. They are contradictory: that is the four exponentials theorem in transcendence degree one, which this subtree proves. This node is a step of that proof by contradiction, so its proof must not use the theorem.