# Are these conditions sufficient for a $\mathbb{Z}$-module to be free?

There exist modules over the integers that, like $\mathbb{Q}$, manage to be torsion-free without being free. Ergo, its probably worth looking for conditions $P$ such that "$P$ + torsion-free" is equivalent to "free." Let try $P$ = "every cyclic submodule can be enlarged to a maximal cyclic submodule" since this obviously fails for $\mathbb{Q}$ and clearly holds for every free module over the integers, even on infinitely many generators.

Question.

Suppose $X$ is a module over the integers satisfying:

• $X$ is torsion-free
• Every cyclic submodule of $X$ can be enlarged to a maximal cyclic submodule.

Is $X$ necessarily free?

If not, does requiring enlargeability to a unique maximal cyclic submodule solve the problem?

• If a maximal cyclic enlargement exists, it is automatically unique (if $x$ is an element of an abelian group that is divisible by two integers $n$ and $m$, it is also divisible by their least common multiple). Commented May 18, 2016 at 6:41

$\prod_{i=1}^\infty \mathbb{Z}$ is a counterexample. It is torsion free, not free (Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?), and every nonzero element is only divisible by finitely many integers, so your extra hypotheses hold.
The group $\mathbb{Z}^\mathbb{N}$ satisfies your hypotheses but is not free.
For a much smaller counterexample, let $\hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$ and let $\alpha\in\hat{\mathbb{Z}}$ be an element such that $a\alpha+b$ is divisible by only finitely many integers whenever $a,b\in\mathbb{Z}$ with $a\neq 0$ (such an $\alpha$ can be constructed by a diagonalization argument, as in this answer). Define $X\subset\mathbb{Q}^2$ to be the subgroup consisting of pairs $(a/n,b/n)$ where $a,b,n\in\mathbb{Z}$ and $n$ divides $a\alpha+b$. Then $X$ is clearly torsion-free, and by our condition on $\alpha$ we know that every cyclic subgroup has a maximal cyclic enlargement (i.e. no nonzero element is divisible by infinitely many integers).
However, I claim $X$ is not free. Indeed, since $X\subset\mathbb{Q}^2$, if it were free then it would have rank at most $2$. But it is easy to see that $X$ is not even finitely generated. For instance, let $Y\subset X$ be the subgroup generated by $(0,1)$. Then the quotient $X/Y$ is divisible, since for any $a\in \mathbb{Z}$ and any nonzero $n\in\mathbb{Z}$, there exists $b\in\mathbb{Z}$ such that $a\alpha+b$ is divisible by $n$. In particular, $X/Y$ is not finitely generated, so $X$ is not finitely generated.