If t is transcendental over R and t, z are not algebraically independent over R, then z is algebraic over R[t]
ProvedTranscendence.isAlgebraic_adjoin_of_not_algebraicIndependent_pairdiaz-modulus-leannumber-theorytranscendence
Let be a commutative ring, a commutative -algebra and transcendental over . If and are not algebraically independent over , then is algebraic over the subalgebra generated by .
The pair is algebraically independent exactly when is transcendental over and is transcendental over (Mathlib's AlgebraicIndependent.option_iff_transcendental). Over a field, together with Transcendence.trdeg_adjoin_le_one_of_isAlgebraic_adjoin, this shows that elements of which no two are algebraically independent, one of them transcendental, generate an algebra of transcendence degree at most one. DiazModulus.two_algebraically_independent_of_exp_column and DiazModulus.algebraicIndependent_e_pi_of_exp_pi_sq_algebraic use it in this way.
Preamble
import Mathlib
Formal statement
namespace Transcendence
/-- Pair dependence. Let `t` be transcendental over `R`. If the pair `t, z` is not algebraically
independent over `R`, then `z` is algebraic over `R[t]`.
Indeed `t` alone is algebraically independent, and a pair `t, z` with `z` transcendental over
`R[t]` is algebraically independent. Together with
`Transcendence.trdeg_adjoin_le_one_of_isAlgebraic_adjoin`, this turns "no two of these numbers are
algebraically independent, and one of them is transcendental" into "they generate an algebra of
transcendence degree at most one". -/
theorem isAlgebraic_adjoin_of_not_algebraicIndependent_pair {R A : Type*} [CommRing R]
[CommRing A] [Algebra R A] {t z : A} (ht : Transcendental R t)
(h : ¬ AlgebraicIndependent R ![t, z]) :
IsAlgebraic ↥(Algebra.adjoin R ({t} : Set A)) z := by
sorry
end Transcendence
Source
Standard. Formal proof: Diaz modulus mission, 30 September 2026 (C. Perassi).