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I've decided to study some number theory in my free time this summer, and have started reading Marcus' "Number Fields" and working through the exercises, but I've got myself stuck on one. This is the problem:

For $R$ an integral domain, prove that two ideals $A$ and $B$ of $R$ are isomorphic as $R$-modules if and only if $\alpha A = \beta B$ for some $\alpha,\beta \in R$.

If $\alpha A = \beta B$, then showing that the map $A \rightarrow B$ defined by $x \mapsto y$ if $\alpha x = \beta y$ is an isomorphism is easy, but I'm having trouble showing the reverse implication. Anyone care to give me a hint?

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up vote 4 down vote accepted

If $f:A\rightarrow B$ is an $R$-module isomorphism, then for any $\alpha\in A$, we have $f(\alpha)A=\alpha B$ because $$f(\alpha)A=\{f(\alpha)a:a\in A\}=\{f(\alpha a):a\in A\}=\{\alpha f(a):a\in A\}=\{\alpha b:b\in B\}=\alpha B$$ (we know every $b\in B$ is some $f(a)$ because $f$ is an isomorphism).

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