Endomorphisms of free modules and extension of scalars

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:FF be an endomorphism of a free A-modules F such that BAϕ:BAFBAF 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. [ϕ1]) for the matrix of ϕ (resp. (Bϕ)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 [ϕ1]. If two ring homomorphisms f,g:BC agree on A, both f([ϕ1]) and g([ϕ1]) are inverses of f([ϕ])=g([ϕ]), so f([ϕ1])=g([ϕ1]) and hence f=g (here functions act on matrices entry-wise). This means AB is an epimorphism of rings. An example of an injective ring epimorphism introducing no new units is k[x2,x(x21)]k[x,(x1)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(x21)]k[x,(x1)1] is an epimorphism, see this text.)

Answer

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

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