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Let $R$ be a valuation ring with fraction field $K$ and let $A$ be a subring of $K$ with $R \subseteq A \subsetneq K$ . With $S := \{ a \in R : 1/a \in A \}$ , I can show that $S^{-1} R =A$ (actually to show this, I only need that $R$ is a Bezout domain i.e. every finitely generated ideal is principal) .
Now if $A$ is also a valuation ring , then is it true that $A=R_P$ for some prime ideal $P$ of $R$ ?
Yes.
The reason is that we may always replace $S$ with its saturation $\tilde{S}$ without changing the localization. Now every saturated multiplicatively closed set is the complement of a union of prime ideals, but since $R$ is a valuation ring, the ideals are totally ordered by inclusion, so any union of prime ideals is itself a prime ideal.
To give more details on the last point, let $(P_i)_{i \in I}$ be a collection of prime ideals in a valuation ring $R$. Then I claim that $P=\displaystyle \bigcup_{i \in I} P_i$ is again a prime ideal. $0 \in P$, $1 \not \in P$. Let $r \in R$ and $p \in P$, then $p \in P_i$ for some $i$, so $rp \in P_i \subset P$. Let $x,y \in P$, then $x \in P_i, y \in P_j$ for some $i$ and $j$. Then because $R$ is a valuation ring, we have $P_i \subset P_j$ or $P_j \subset P_i$, so depending on the case $x+y \in P_i$ or $x+y \in P_j$, thus $x+y \in P$. (This step fails horribly in most cases when $R$ is not a valuation ring: consider for example $\Bbb Z$, then $2,3 \in (2) \cup (3)$, but $2+3=5 \notin (2) \cup (3)$) Thus $P$ is an ideal. Suppose that $a,b \in R$, such that $ab \in P$, then for some $i$, $ab \in P_i$, so $a \in P_i$ or $b \in P_i$ because $P_i$ is a prime ideal. Thus $a \in P$ or $b \in P$, so $P$ is a prime ideal.
(Compare Atiyah-Macdonald Exercise 7 in chapter 3).
