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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 $v$?

trisct
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1 Answers1

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(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 $xy^{-1}\in A$.

user10354138
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  • Thank you very much for the answer. But I am still wondering: can we speak of a valuation ring of a field $K$ if there is no valuation previously specified? More specifically, let $A$ be a subring of a field $K$, is it reasonable to define $A$ to be a valuation ring of $K$ if $\forall x\in K^\implies x$ or $x^{-1}\in A$? (or alternatively, only for all $x\in Frac(A)^$?) – trisct May 25 '19 at 11:52
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    Yes you can, but if $K$ is not $\operatorname{Frac}A$ and you only demand it for $x\in(\operatorname{Frac}A)^*$ then many nice theorems won't work. – user10354138 May 25 '19 at 12:08