I got a little confused by the different definitions of valuation rings while reading Atiyah and Macdonald's introduction to commutative algebra.
Let $A$ be an integral domain and $K$ its field of fractions. We say $A$ is a valuation ring of $K$ if for any $x\in K^*$ either $x$ or $x^{-1}$ lies in $A$.
However, if we specify a valuation $v$ on a field $K$, we can define the valuation ring of $v$ as the ring $A=\{x\in K^*:v(x)\geq0\}\cup\{0\}$.
My questions:
(1) In the book, it is said that a valuation ring of $v$ (2nd definition) is a valuation ring of $K$ (1st definition). But if we stick to the two definitions above, we would find that "the valuation ring $A$ of $K$" is an undefined concept if $K$ is not the field of fractions of $A$ (the 2nd definition does not require $K$ to be the field of fractions of $A$). However, this concept frequently appears in the book. How should I understand it?
(2) Is the converse also true? Suppose $A$ is a valuation ring of $K$, does there necessarily exists a valuation $v$ such that $A$ is the valuation ring of $v$?
(1) But $K$ is the field of fractions of $A$ in Definition 2. To see this, note that valuation takes value in a totally ordered group, so for every $x\in K^\times$ either $v(x)\geq 0$ or $v(x^{-1})\geq 0$, i.e., $x$ or $x^{-1}$ is in $A$. So $\operatorname{Frac}A\supseteq K$. But $A\subseteq K$ by definition.
(2) Yes. There is a totally ordered group $\Gamma$ and a map $v\colon K^\times\to\Gamma$ such that $A=v^{-1}(\Gamma_{\geq 0})\cup\{0\}$. Just take $\Gamma:=K^\times/A^\times$ (where $A^\times$ is the units of $A$). The ordering is by looking at whether $xy^{-1}\in A$.