EDIT 4. [Version of Nov. 26, 2011, UTC.] The following result is implicit in Rentschler's paper, and solves Bass's Exercise (c):
Theorem. Let $T$ be ana nonempty ordered set $ ( * ) $ with no maximum. Then there is a domain $A$ and an increasing $ ( * ) $ map fromwhich has the following property. If $T$ to$P$ denotes the poset $ ( * ) $ of proper sub-$A$-modules of the field of fractions of $A$, then there is an increasing $ ( * ) $ map $f:T\to P$ such that $f(T)$ is cofinal in $P$.
This follows immediately from Example 6, Section V.3.4, and Proposition 7, Section V.3.5, in Bourbaki's Algèbre commutative, coupled with Section 5 of Rentschler's paper. --- This also solves Bass's Exercise (c).
$ ( * ) $ Since I'm using references written in French while writing in English (or at least trying to), I adhere strictly to linguistic conventions. In particular:
increasing = strictement croisssant.
Proof. Let $T_0$ be the ordered set opposite to $T$, let $\mathbb Z^{(T_0)}$ be the free $\mathbb Z$-module over $T_0$ equipped with the lexicographic order. Then $\mathbb Z^{(T_0)}$ is an abelian ordered group (groupe abélien totalement ordonné). By Example 6 in Section V.3.4 of Bourbaki's Algèbre commutative, there is a field $K$ and a surjective valuation $$
v:K\to\mathbb Z^{(T_0)}\cup \{ \infty \}.
$$
Say that a subset $F$ of $\mathbb Z^{(T_0)}$ is a final segment if
$$F\ni x < y\in\mathbb Z^{(T_0)}
$$
implies $y\in F$. Attach to each such $F$ the subset
$$
S(F):=v^{-1}(F)\cup \{ 0 \}
$$
of $K$. Then $A:=S(F_0)$, where $F_0$ is the set of nonnegative elements of $\mathbb Z^{(T_0)}$, is a subring of $K$. Moreover, by Proposition 7 in Section V.3.5 of the book quoted above, $F\mapsto S(F)$ is an increasing bijection from the final segments of $\mathbb Z^{(T_0)}$ to the sub-$A$-modules of $K$.
Write $e_{t_0}$ for the basis element of $\mathbb Z^{(T_0)}$ corresponding to $t_0\in T_0$. Then the intervals
$$
I_{t_0}:=[-e_{t_0},\infty)
$$
are cofinal in the set of all proper final segements of $\mathbb Z^{(T_0)}$, and we have $I_{t_0}\subset I_{u_0}$ if and only if $t\le u$. [We denote an element $t$ of $T$ by $t_0$ when we view it as an element of $T_0$.]
