Let $A$ be a commutative domain and $K = \text{Frac}(K)$ its field of fractions. $A$ is called a valuation ring if its ideals are totally ordered by inclusion and one can easily show that it is sufficient that all principal ideals are ordered by inclusion.

My course notes in Field Arithmetic claim that

$A$ is a valuation ring if and only if the $A$-submodules of $K$ are totally ordered by inclusion.

This seems to me like a strictly stronger statement (it is not weaker since ideals of $A$ are also $A$-submodules of $K$) and I cannot find a proof of the non-trivial implication in this equivalence.


$\newcommand{\Frac}{\operatorname{Frac}}$ Let $A$ be a valuation ring, $K = \Frac(A)$. For $x\in K$ we have $x\notin A \implies x^{-1}\in A$ (this follows from the fact that principal ideals are totally ordered by inclusion: if $a,b\in A$, then at least one of $a/b\in A$ or $b/a\in A$ holds true).

So given $A$-submodules $I,J\subseteq K$, we assume $J\not\subseteq I$. Take $x\in J\setminus I$. We want to show $I\subseteq J$. So let $y\in I$ be arbitrary. If we had $x/y\in A$, then $x = x/y\cdot y\in I$, contradicting $x\notin I$. Hence we have $y/x\in A$. Therefore $y = y/x\cdot x\in J$. This shows $I\subseteq J$.

As you can see, the proof is the same as “principal ideals are totally ordered by inclusion implies all ideals are totally ordered by inclusion”.

  • $\begingroup$ Elegant and simple, thanks a lot! $\endgroup$ – Bib-lost Oct 12 '16 at 12:23
  • $\begingroup$ Could someone say more about how this is true: if $a,b\in A$, then at least one of $a/b\in A$ or $b/a\in A$. Struggling to see it from the definition of valuation ring. $\endgroup$ – Brofessor Oct 21 '20 at 2:49
  • $\begingroup$ Given $a,b\in A$, you either have $Aa \subseteq Ab$ or $Ab\subseteq Aa$ by definition of a valuation ring. In the first case you have $\frac ab\in A$ and in the second case you have $\frac ba \in A$. $\endgroup$ – Claudius Oct 21 '20 at 6:04

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