English
If e is an algebra equivalence, IsTranscendenceBasis R x implies IsTranscendenceBasis R (e ∘ x).
Русский
Если e — алгебраическое эквивалентство, то IsTranscendenceBasis R x влечёт IsTranscendenceBasis R (e ∘ x).
LaTeX
$$$\\mathrm{AlgEquivIsTB}(e)\\_R(x) \\Rightarrow \\mathrm{IsTB}_R(e \\circ x)$$$
Lean4
/-- Also see `IsTranscendenceBasis.algebraMap_comp`
for the composition with a algebraic extension. -/
theorem isTranscendenceBasis (e : A ≃ₐ[R] A') (hx : IsTranscendenceBasis R x) : IsTranscendenceBasis R (e ∘ x) :=
.of_comp e.symm.toAlgHom e.symm.injective (by convert hx; ext; simp)