Two questions on free modules over a PID

While reading about free modules over a PID I had two natural questions come up. To provide some context, here are two standard facts:

1. every submodule of a free module over a PID is free, and the proof involves the well-ordering principle,
2. given $N \leq M$ where $M$ has finite rank, there exists a basis $\{m_1, \dots, m_k\}$ of $M$ and ring elements $r_1 \mid \dots \mid r_j$ such that $\{r_1 m_1, \dots, r_j m_j\}$ is a basis of $N$. (Of course $N$ and $M$ are over a PID.) The standard proof is over the Smith Normal Form and requires us to rely on the fact that the required row/column operations we do to perform a change of basis are finite.

Relationship to the Axiom of Choice

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What is the relationship of the first fact to the Axiom of Choice? For context we know “every vector space has a basis” is equivalent to Choice. So it does not seem implausible that the first fact is equivalent to Axiom of Choice too.

Nice bases of a submodule of a free module of infinite rank

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Is the second fact true if $M$ has infinite rank? If not, which $M$ disprove it? Alternatively what extra conditions do we need to put on the underlying ring of $M$? (For instance we know making it a field is sufficient.) So on and so forth.