Given an integral domain $R$ and its field of fractions $K$, let $M, M', N$ and $N'$ be $R$-modules. Then we can extend the $R$-module action to a $K$-action naturally. Then do I have the following implication?:
If ($M\otimes_{K}N$) $\cong$ ($M'\otimes_{K}N'$) then ($M\otimes_{R}N$) $\cong$ ($M'\otimes_{R}N'$).
I think the implication should hold because $R$ action is restriction of above defined $K$-action. I am not entirely sure.
Take $M = N = \mathbb{Z}/2\mathbb{Z}$, $M' = N' = \mathbb{Z}/3\mathbb{Z}$, $R = \mathbb{Z}$.
Lets denote the extensions by $M_\mathbb{Q}$, so $M_\mathbb{Q} = M \otimes_\mathbb{Z} \mathbb{Q}$.
Then one has $M_\mathbb{Q} = N_\mathbb{Q} = 0 = M'_\mathbb{Q} = N'_\mathbb{Q}$, so certainly $M_\mathbb{Q} \otimes_\mathbb{Q}N_\mathbb{Q} = M'_\mathbb{Q} \otimes_\mathbb{Q}N'_\mathbb{Q}$.
Now $\mathbb{Z}/n\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/m\mathbb{Z} \simeq \mathbb{Z}/(n,m)\mathbb{Z}$ where $(n,m)$ is the greatest common divisor of $n,m$.
Hence $M \otimes_\mathbb{Z} N = \mathbb{Z}/2\mathbb{Z} \neq \mathbb{Z}/3\mathbb{Z} \simeq M' \otimes_\mathbb{Z} N'$.
Basically tensoring with $K$ kills the torsion of $M$, so you lose information.