Denote the ring of $p$-adic integers by $\Bbb{Z}_p$. An abelian group is torsion-free iff every nonzero element has infinite order.
A subgroup is said to be pure if when divisibility by an integer can be performed in the group containing it, then it can be performed also in the subgroup).
For an abelian group, algebraic compactness means that such a group is a direct summand in every group containing it as pure subgroup.
I'm studying abelian groups and I found the following lemma: "A reduced, countably generated torsion-free $\Bbb{Z}_p$-module is free".
I don't understand many steps of the following proof. I tried to work out the details to test my understanding, but in most cases I failed.
Proof.
(1) A torsion-free $\Bbb{Z}_p$-module of rank $1$ is either cyclic or injective (the author says that it follows from the fact that all $\Bbb{Z}_p$-modules $\ne \Bbb{Q}_p$ between $\Bbb{Z}_p$ and $\Bbb{Q}_p$ are isomorphic to $\Bbb{Z}_p$). But both I don't see why and I was not able to prove the statement in parentheses.
Then he claims:
(2) As $\Bbb{Z}_p$ is algebraic compact, we easily deduce that pure torsion-free $\Bbb{Z}_p$-modules of rank $1$ are summands.(why?)
Consequently, finitely generated torsion-free $\Bbb{Z}_p$-modules are free (I got this)
(3) and the same holds for reduced $\Bbb{Z}_p$-modules of finite rank (how to prove it?).
Finally, the assertion follows then in the same way as in the proof of another result in the same book.
It's clear to me that what I have tried doesn't work. So, I need help. May you suggest how to show (1),(2),(3)?
For your comment between (2) and (3), we know that $\mathbb Z_p$ is a DVR, so torsion-free is the same as flat, and finitely generated flat modules are free.
For (1), if $M$ is torsion-free, then $M\to \mathbb Q_p\otimes_{\mathbb Z_p}M$ is injective, and if $M$ has rank one then $\mathbb Q_p\otimes_{\mathbb Z_p}M\cong\mathbb Q_p$. So we may assume that $M\subseteq\mathbb Q_p$. Now consider the set of $n$ such that $p^{-n}\in M$. If this set is finite, then it has a maximal element $n$, and $M=p^{-n}\mathbb Z_p\cong\mathbb Z_p$. Otherwise this set is infinite and $M=\mathbb Q_p$.
As an alternative approach to the whole question, note that every flat, countably generated Mittag-Leffler module is projective (eg Lemma 10.92.1 in the Stacks project). Also, for an hereditary ring, Kaplansky (for commutative, Albrecht in general with an easier proof) showed that every countably generated projective is isomorphic to a direct sum of finitely generated ideals. Finally, for a DVR, every finitely generated ideal is principal, hence free.
So, everything reduces to showing that for countably generated flat modules, reduced implies Mitag-Leffler.