Ideal in integral domain implies self-similar

Statement

Suppose $A$ is an integral domain, and $I$ is an ideal of $A$. Consider $I$ as an $A$-module. Then, any nonzero $A$-submodule of $I$ contains a submodule isomorphic to $I$ (as an $A$-module).

Proof

Given: $A$ is an integral domain, and $I$ is an ideal of $A$. $N$ is a nonzero $A$-submodule of $I$

To prove: $N$ contains a submodule $J$ isomorphic to $I$ as an $A$-module.

Proof: Pick $0\neq x\in N$. Consider the submodule $xI$. There is a natural homomorphism:

$a\mapsto xa$

from $I$ to $xI$. Since the multiplication is within an integral domain, the map is injective, and by definition, it is surjective. We thus have an isomorphism.