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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?

user26857
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beanshadow
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  • Is it possible that this provides a counter-example? https://math.stackexchange.com/questions/2220392/example-of-non-flat-modules The correspondence between your terminology and from that example: your A is their R, your B is their N, your I is their (t). It seems to me like that example is saying $I \otimes_A B \rightarrow B$ is not injective, so it shouldn't be isomorphic to $I^e \subseteq B$, or at least the isomorphism shouldn't be induced by the natural map from $I \otimes_A B \rightarrow B$. Does that seem right? (PS: If $A \rightarrow B$ is flat, see Matsumura CRT Theorem 7.7.) – CJD Nov 17 '18 at 04:39
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    For $A=k[x^2,x^3]$, $B=k[x]$, and $I=(x^2,x^3)$ we have $I^e=x^2B\simeq B$ (as $B$-modules). If $I\otimes_AB\simeq B$ then the canonical morphism $I\otimes_AB\to I^e$ is an isomorphism, a contradiction. – user26857 Nov 18 '18 at 23:25

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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.

Anirban Bose
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