English
If A and B are linearly disjoint with A ⊔ B = S, then any R-basis of B induces an A-basis of S via the natural isomorphism.
Русский
Если A и B линейно раздельны и A ⊔ B = S, то любой R-базис B порождает базис S как модуль над A через естественное изоморфизм.
LaTeX
$$$H \\text{ is a basis transport: } \\text{basisOfBasisRight} : \\text{Basis}_R(B) \\to_B S$$$
Lean4
/-- If `A` and `B` are linearly disjoint subalgebras in a commutative algebra `S` over `R`
such that `A ⊔ B = S`, then any `R`-basis of `B` is also an `A`-basis of `S`.
-/
noncomputable def basisOfBasisRight (H' : A ⊔ B = ⊤) {ι : Type*} (b : Basis ι R B) : Basis ι A S :=
(b.baseChange A).map (H.mulMapLeftOfSupEqTop H').toLinearEquiv