# Isomorphisms of free modules and extension of scalars

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 $\phi:F\to F'$ be a morphism of free $A$-modules such that $B\otimes_A\phi:B\otimes_AF\to B\otimes_AF'$ is an isomorphism.

Does this imply that $\phi$ is an isomorphism?

The answer is of course yes if $F$ and $F'$ have finite ranks, because we can use determinants.

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

• can you give some examples of A is proper subring of B? – Sky Sep 2 '17 at 12:49
• @Sky - Let $K$ be a field and $X$ an indeterminate, set $B:=K[X]/(X^3)$, let $x\in B$ be the image of $X$ and set $A:=K[x^2]\subset K[x]=B$. Then the units of $A$ and $B$ coincide, and $B$ is not $A$-flat. – Pierre-Yves Gaillard Sep 2 '17 at 13:11
• you mean B=$K[X]/(X^4)$?how do you show B is not flat A-module?is there some good way?we can also consider A=k[x,y],B=k[x,y,z].but if B is flat A-modue? – Sky Sep 2 '17 at 13:27
• Choose bases for $F$, $F'$, and write $[\phi]$ (resp. $[\phi^{-1}]$) for the matrix of $\phi$ (resp. $(B\otimes\phi)^{-1}$), 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 $[\phi^{-1}]$. If two ring homomorphisms $f,g:B\to C$ agree on $A$, both $f([\phi^{-1}])$ and $g([\phi^{-1}])$ are inverses of $f([\phi])=g([\phi])$, so $f([\phi^{-1}])=g([\phi^{-1}])$ and hence $f=g$ (here functions act on matrices entry-wise). This means $A\to B$ is an epimorphism of rings. – Julian Rosen Sep 4 '17 at 19:55
• (cont.) An example of an injective ring epimorphism introducing no new units is $k[x^2,x(x^2-1)]\subset k[x,(x-1)^{-1}]$ (see this MO answer), so if a counterexample exists this might be a place to start. – Julian Rosen Sep 4 '17 at 19:55