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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 $A$.

Pierre-Yves Gaillard
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  • can you give some examples of A is proper subring of B? – Jian Sep 02 '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 02 '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? – Jian Sep 02 '17 at 13:27
  • @Sky - I mean $B:=K[X]/(X^3)$. Why do you think I mean $B:=K[X]/(X^4)$? – Pierre-Yves Gaillard Sep 02 '17 at 13:35
  • $(x^2)^2=x$ if you take B in your way. – Jian Sep 02 '17 at 13:47
  • @Sky - No, $(x^2)^2=x^4=x^3x=0x=0$. – Pierre-Yves Gaillard Sep 02 '17 at 13:51
  • yeah,you are right.I will be careful.thank you! is there someway to show if B is not A-flat? – Jian Sep 02 '17 at 13:55
  • It suffices to show that the multiplication $K[x]\otimes_{K[x^2]}(x^2)\to K[x]$ is not injective. Thus, it suffices to check $x\otimes x^2\ne0$ in $K[x]\otimes_{K[x^2]}(x^2)$. To prove this rigorously, you can use this answer. – Pierre-Yves Gaillard Sep 02 '17 at 14:06
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    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 04 '17 at 19:55
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    (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 04 '17 at 19:55
  • @JulianRosenn - Thanks for this awesome comment! Sorry for answering so late. I don't know enough algebraic geometry to follow the argument in the MO answer. I tried to find an elementary argument, and it took me a huge amount of time. I came up with this computation: http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Julian_MSE/ - I hope it's correct. – Pierre-Yves Gaillard Sep 10 '17 at 13:32
  • @JulianRosen - This is just to tell you that I replaced the $3\times4$ matrix with a $2\times3$ matrix in http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Julian_MSE/ – Pierre-Yves Gaillard Sep 11 '17 at 11:39

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