Let $\Phi:A \rightarrow B$ be a flat morphism of commutative rings. Let $f \in A$, not a unit and $A/fA \cong B/fB$ induced by $\Phi$.


Let $M$ be an $A\_f$-module. Is it true that $M \otimes\_A B = 0 \Rightarrow M=0$?


Here's my way of thought so far: by flatness of $B$ it's enough to show this for $M$ a cyclic $A\_f$ module.


Let $M = A\_f/IA\_f \neq 0$. We have to see $B\_f/IB\_f \neq 0$. In other words, if $a \in A$ doesn't become a unit in $A\_f$, then $\Phi(a)$ doesn't become a unit in $B\_f$.


Sloppy spoken "for ruling out the case $f^n$ " I can use the isomorphism $A/fA \cong B/fB$. But maybe it can happen that already $\Phi(a) \in B$ is a unit?


I would like to either see a proof or counterexample. I'm not sure if the statement holds.



In this special situation, does $M \otimes B=0$ imply $M=0$?
No, that is not true. Let $A$ be $\mathbb{Z}[x]/\langle x(1-x) \rangle$. Let $B$ be $A/\langle x \rangle$ with the obvious quotient morphism, $\Phi$. Let $f$ be $x$. Then the natural $A$-algebra homomorphism, $$ A[y]/\langle yx-1 \rangle \to A/\langle 1-x \rangle, \ \ y \mapsto 1, $$ is an isomorphism. To see this, observe that in $A[y]/\langle yx-1 \rangle$ we have the congruences, $$ 1-x = 1(1-x) \equiv (yx)(1-x) = y(x(1-x)) = y\cdot 0 = 0.$$ So let $M$ be $A[y]/\langle xy-1 \rangle = A/\langle 1-x \rangle$. Then $M\otimes\_A B$ equals $A/\langle 1-x \rangle \otimes\_A A/\langle x \rangle$, which is $\{0\}$. But $M$ is not zero.


