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 $$0 \ne x \in N$$. 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.