Let $R$ be a subring of a field $F$ such that for all $x\in F^\times$, either $x\in R$ or $x^{-1}\in R$. Prove that if $I$ and $J$ are ideals of $R$, either $I\subseteq J$ or $J\subseteq I$. I am preparing for my qualifying exam and this is from one of our previous exams. I can tell that $R$ is an integral domain, but I couldn't go any further. Can anyone help? Thanks a lot. Here is the proof after reading the hints in the link below. Let $I$ and $J$ be two ideals of $R$. Suppose $I\not\subseteq J$. Then there exists $a\in I\backslash J$. Since $0\in J$, $a\neq0$. Now let $b\in J\backslash\{0\}$. Then in $F$, we have $b=(ba^{-1})a$ with $ba^{-1}\in F$. Claim: $ba^{-1}\in R$. If $b^{-1}a=(ba^{-1})^{-1}\in R$, then $a=b^{-1}ab\in J$ which is a contradiction. So by the definition of $R$, we have $ba^{-1}\in R$. Then $b=(ba^{-1})a\in I$. So all nonzero elements of $J$ are in $I$. Since $0\in I$, we have $J\subseteq I$.
There is the next characterization for valuation domains:  Let $D$ be an integral domain and $K$ its field of fractions. The following are equivalent:   For every nonzero $x$ in $K$, either $x$ in $D$ or $x^{−1}$ in $D$. The ideals of $D$ are totally ordered by inclusion. The principal ideals of $D$ are totally ordered by inclusion (i.e., the elements in D are totally ordered by divisibility.) There is a totally ordered abelian group $Γ$ (called the value group) and a surjective group homomorphism (called the valuation) $v :$ $K^×$ $→ Γ$ with $D = \{ x ∈$ $K^×$ | $v(x)$ ≥ $0 \}$ $∪$ {0}.   The equivalence between 1, 2 and 3 are clear, and I understand that the fourth can be found in Krull's article (1936), but I don't speak german and I can't find a proof anywhere. Any link or hint will be much appreciated.
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