Let $(\mathbb{Z},+)$ be the additive group of integers, and $(\mathbb{Z}, +, \cdot)$ the ring of integers. By definition, every ideal of $(\mathbb{Z}, +, \cdot)$ is a subgroup of $(\mathbb{Z},+)$. Is the opposite true? Is every subgroup of $(\mathbb{Z},+)$ also an ideal of $(\mathbb{Z}, +, \cdot)$? If yes, how can it be proved? If not, is there any counterexample?
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Every subgroup of $(\Bbb Z, +)$ is generated by a single element (division algorithm), and gives a principal ideal of $(\Bbb Z, +, \cdot)$.
Kenny Lau
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The equivalence follows from the following two (easy to prove) results:
The ideals of a commutative ring $R$ are precisely the $R$-submodules of $R$ as a module over itself.
The Abelian groups are essentially $\mathbb Z$ modules, in the sense that the notions of group homomorphisms, subgroups, quotients, map to the appropriate notions of $\mathbb Z$-modules (module homomorphism, submodule, quotient).
I am sure someone with more knowledge of category theory would be able to formulate the second statement more precisely.
What I want to conclude is:
$$I\subseteq \mathbb Z\;\text{is an ideal of}\;\mathbb Z\Leftrightarrow I\;\text{is a}\;\mathbb Z\text{-submodule of}\;\mathbb Z\Leftrightarrow I\;\text{is a subgroup of}\;\mathbb Z$$
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Every subgroup of $(\mathbb{Z},+)$ is of the form $(n\mathbb{Z},+)$ with $n\in\mathbb{N}$. A proof uses division with remainder. I wrote it down here: Subgroups of $\mathbb{Z}$
Cornman
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