English
Let f: A → A' be an algebra homomorphism. If f is injective, then AlgebraicIndependent R (f ∘ x) is equivalent to AlgebraicIndependent R x.
Русский
Пусть f: A → A' — алгебраический гомоморфизм и инъективен. Тогда AlgebraicIndependent R (f ∘ x) эквивалентно AlgebraicIndependent R x.
LaTeX
$$$\\\\mathrm{AlgHom}\\\\(R,A,A') \\\\(f) \\\\Rightarrow \\\\text{Injective}(f) \\\\Rightarrow \\\\mathrm{AlgebraicIndependent}(R,(f\\\\circ x)) \\\\iff \\\\mathrm{AlgebraicIndependent}(R,x)$$$
Lean4
theorem algebraicIndependent_iff (f : A →ₐ[R] A') (hf : Injective f) :
AlgebraicIndependent R (f ∘ x) ↔ AlgebraicIndependent R x :=
⟨fun h => h.of_comp f, fun h => h.map hf.injOn⟩