ac.commutative algebra - Are submodules of free modules free?

To answer ding8191's question: let $A$ be any valuation ring of height $1$ which is not a discrete valuation ring. That is to say, let $F$ be a field and suppose that there is a function $v : F to mathbb{R} cup { infty }$ such that $v(xy) = v(x) + v(y), v(1) = 0, v(0) = infty$ and $v(x + y) geq min $ ${ v(x), v(y) }$ (thus $v$ is a valuation), such that $v(F)$ is not a discrete subgroup of $mathbb{R}$, and let $A = { x in F :v(x)geq 0}$ (the valuation ring of $F$).

Then the maximal ideal $mathfrak{m}$ in this ring is a direct limit of principal ideals and is therefore flat. If $mathfrak{m}$ was projective, then it would have to be free by Kaplansky's Theorem (all projectives over any local ring are free), but since $mathfrak{m}$ is contained in $F$ it has to have rank at most $1$, hence $mathfrak{m}$ is principal. But this would force the valuation on $A$ to be discrete.

You can find this material in Chapter VI of Bourbaki's "Commutative Algebra" (Hermann, 1972). In particular, Lemma $1$ of $S 3.6$ says that if $A$ is a valuation ring (in a more general sense than I defined above), then every torsion-free $A$-module is flat, and Proposition $9$ of the same subsection implies that $A$ is a discrete valuation ring whenever $A$ is a valuation ring of height $1$ with principal maximal ideal.

To give a concrete example, consider the field $F$ of Puiseux series and let $A$ be its valuation ring.