I asked this question on Mathematics Stackexchange, but got no answer.

Let B be a commutative ring with 1, let A be a subring such that any unit of B which belongs to A is a unit of A, and let ϕ:F→F be an endomorphism of a free A-modules F such that B⊗Aϕ:B⊗AF→B⊗AF is an automorphism.

Does this imply that ϕ is an automorphism?

The answer is of course yes if F has finite rank, by determinant theory.

The answer is also yes if B is faithfully flat over A.

Julian Rosen made the following interesting comment on Mathematics Stackexchange:

“Choose bases for F,F′, and write [ϕ] (resp. [ϕ−1]) for the matrix of ϕ (resp. (B⊗ϕ)−1), which is a column-finite (possibly) infinite with entries in A (resp. B). WLOG we may assume B is generated as an A-algebra by the entries of [ϕ−1]. If two ring homomorphisms f,g:B→C agree on A, both f([ϕ−1]) and g([ϕ−1]) are inverses of f([ϕ])=g([ϕ]), so f([ϕ−1])=g([ϕ−1]) and hence f=g (here functions act on matrices entry-wise). This means A→B is an epimorphism of rings. An example of an injective ring epimorphism introducing no new units is k[x2,x(x2−1)]⊂k[x,(x−1)−1] (see this MO answer), so if a counterexample exists this might be a place to start.”

Here k is a field of characteristic ≠2. (For an elementary proof of the fact that the inclusion k[x2,x(x2−1)]⊂k[x,(x−1)−1] is an epimorphism, see this text.)

**Answer**

**Attribution***Source : Link , Question Author : Pierre-Yves Gaillard , Answer Author : Community*