Let $A, B$ be $\mathbb{C}$-algebras, which are also integral domains. Suppose there is an injective ring homomorphism $f:A \to B$. Assume further than $f$ is a finite morphism in the sense that $f$ induces a finite $A$-module structure on $B$. Let $M$ be a finitely generated $A$-module. Let $m \in M$ such that there exists $m' \in M$ and $b \in B$ for which $m \otimes 1=m' \otimes b$ in $M \otimes\_A B$. Does this imply that $b$ belongs to (the image of) $A$?



Naive question on tensor product
A trivial counterexample is $m=m'=0$. Perhaps more interesting is a situation where $b$ does not belong to the image of $A$, but some multiple or power of $b$ does. Say $M = B = \mathbb{C}[x]$ and $A = \mathbb{C}[x^2,x^3]$. Take $m' = x^3$ and $b = x$. Then


$$ m' \otimes b = x^3 \otimes x = x \otimes x^3 = x^4 \otimes 1 $$


because $x^2 \in A$, and then because $x^3 \in A$.


