I've just learned the definitions of Artinian and Noetherian module and I'm now trying to think of examples. Can you tell me if the following example is correct:
An example of a $\mathbb Z$-module $M$ that is not Noetherian: Let $G_{1/2}$ be the additive subgroup of $\mathbb Q$ generated by $\frac12$. Then $G_{1/2} \subset G_{1/4} \subset G_{1/8} \subset \dotsb$ is a chain with no upper bound hence $M = G_{1/2}$ as a $\mathbb Z$-module is not Noetherian.
But $M$ is Artinian: $G_{1/2^n}$ are the only subgroups of $G_{1/2}$. So every decreasing chain of submodules $G_i$ is bounded from below by $G_{1/2^{\min i}}$.
Edit In Atiyah-MacDonald they give the following example:
Let $G$ be the subgroups of $\mathbb{Q}/\mathbb{Z}$ consisting of all elements whose order is a power of $p$, where $p$ is a fixed prime.
Then $G$ has exactly one subgroup $G_n$ of order $p^n$ for each $n \geq 0$, and $G_0 \subset G_1 \subset \dotsb \subset G_n \subset \dotsb$ (strict inclusions) so that $G$ does not satisfy the a.c.c.
On the other hand the only proper subgroups of $G$ are the $G_n$, so that $G$ does satisfy d.c.c.
Does one have to take the quotient $\mathbb{Q}/\mathbb{Z}$?
Best Answer
Fix a prime $p$ and let $M_p={\Bbb Z}(\frac1p)/{\Bbb Z}$.
It is not difficult to see that the only submodules of $M_p$ are those generated by $\frac1{p^k}+{\Bbb Z}$ for $k\geq0$. From this it follows that $M_p$ is Artinian but not Noetherian.