What is an appropriate generalization of this module/ring result Consider $R=\mathbb{Z}$ as a $\mathbb{Z}$-module, i.e. an abelian group.
Then, every (nonzero) submodule of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ (as $\mathbb{Z}$-modules).
I am curious what is the appropriate generalization to rings $R$?
For instance, if $R$ is a PID, do we have this result:
Every nonzero submodule of $R$ is isomorphic to $R$ (as $R$-modules)?
In other words, I am curious what is the "essential property" of $\mathbb{Z}$ that makes the statement true?
Thanks.
 A: I'll make a try ...
For $R$ a PID we have that a nonzero $R$-submodule is a principal ideal $I=(a)$ of $R$. So indeed $I\cong R$ as $R$-modules with $a\mapsto1$ being the isomorphism.
A: Yes, the result is true for PIDs in general: see this for example.
What you describe is equivalent to a PID. Obviously such a ring is a principal ideal ring (you just said the ideals are isomorphic to $R$, which is cyclic.) Furthermore, there could be no zero divisors, because for nonzero $r$, the homomorphism $(r)\cong R$ implies that the annihilator of $r$ is zero.
To slightly generalize we could look at rings such that


*

*$R$ is hereditary, meaning submodules of projective modules are projective, and 

*$R$ has the property that projective modules are free, so now submodules of free modules are free modules as well. I have seen an eponymous name for such rings in the past, but it is escaping me at the moment.
These are related to another concept: Cohn's notion of a free ideal ring.  A ring is called a right free ideal ring if every right ideal is a free module with well-defined rank.  Among commutative rings, the free ideal rings are exactly the PIDs.
So among commutative rings, the kind of ring you describe is precisely a PID, since all ideals are clearly free.  But I believe there are hereditary local rings which aren't PID's as well, and they would have the two properties above.
