English
If f is an injective AlgHom and H asserts IsTranscendenceBasis R (f ∘ x), then IsTranscendenceBasis R x.
Русский
Если f — инъективный гомоморфизм Алгебра, и H имеет IsTranscendenceBasis R (f ∘ x), тогда IsTranscendenceBasis R x.
LaTeX
$$$(\\text{H}) \\Rightarrow \\text{IsTB}_R x$$$
Lean4
theorem of_comp {x : ι → A} (f : A →ₐ[R] A') (h : Function.Injective f) (H : IsTranscendenceBasis R (f ∘ x)) :
IsTranscendenceBasis R x :=
by
refine ⟨(AlgHom.algebraicIndependent_iff f h).mp H.1, ?_⟩
intro s hs hs'
have :=
H.2 (f '' s) ((algebraicIndependent_image h.injOn).mp ((AlgHom.algebraicIndependent_iff f h).mpr hs))
(by rw [Set.range_comp]; exact Set.image_mono hs')
rwa [Set.range_comp, (Set.image_injective.mpr h).eq_iff] at this