Suppose we have commutative rings $A$ and $B$, a (maybe injective) ring homomorphism $f: A \rightarrow B$ and an ideal $I \subseteq A$. Is it true that $I^e \cong I \otimes_A B$, where $I^e$ denotes the extension of $I$ into $B$?
In other words, does the extended ideal have the same module structure as the module obtained through extension of scalars?
Consider the quotient map $\pi :\mathbb{Z}\longrightarrow \mathbb{Z}/2\mathbb{Z}$. Let $I=6\mathbb{Z}$. Then $I^e=0$. Now $I\otimes _{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \cong I/2I=6\mathbb{Z}/12\mathbb{Z}$ (as $\mathbb{Z}$-modules). Note that the underlying abelian groups are non isomorphic. So this gives a counterexample.