# Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module?

I'm confused.

Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module?

We know that $\Bbb Z_{p^{\infty}} \subset \Bbb Q/\Bbb Z$ is artinian. The following argument is true or not ?

$\mathbb Q / \mathbb Z$ (as $\mathbb Z$-module): subgroups of $\mathbb Q / \mathbb Z$ look like $(\frac{1}{n})$, the subgroup generated by $\frac1n$. If we have $(\frac1n) \supset (\frac1m)$ we know that $m$ divides $n$ so if $(\frac{1}{n_1}) \supset (\frac{1}{n_2}) \supset \dots$ is a decreasing chain it eventually becomes stationary because there are only a finite number of divisors of $n_1$. Hence $\mathbb Q / \mathbb Z$ is Artinian.

• Not all subgroups of $\Bbb Q/\Bbb Z$ are generated by a single element. The whole group for instance is not. Mar 4 '14 at 21:13
• No, it's an infinite direct sum of non-zero indecomposable summands (the Prüfer groups). Mar 4 '14 at 21:17

Starting from all of $\Bbb Q/\Bbb Z$, you can successively forbid all fractions whose denominator contains a factor $2$, then factors $3$, $5$, and so forth. This gives an infinite decreasing chain, and the module is not Artinian.
$\mathbb Q/\mathbb Z$ is not artinian. Let $L_i$ be the subgroup of $\mathbb Q$ consisting by all fractions $m/n$ with $\gcd(m,n)=1$ and $n$ doesn't contain any of the first $i$ prime numbers. Prove that $(L_i/\mathbb Z)_{i\ge 0}$ is a strictly descending chain of subgroups.