$R$ is a subring of a field $F$ with $x\in R$ or $x^{-1}$ for all $x\in F^\times$. If $I,J$ are ideals of $R$, then $I\subseteq J$ or $J\subseteq I$.

Asked Aug 10 '20 at 19:21

Active Jun 27 '22 at 00:47

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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 $J\subseteq I$.

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 $J\subseteq I$.

edited Jun 27 '22 at 00:47

Clemens Bartholdy

14,222

asked Aug 10 '20 at 19:21

Collier Gaiser

2

Such rings are called valuation rings. Here is a post with the same question: https://math.stackexchange.com/questions/3002778/ideals-in-valuation-rings – Viktor Vaughn Aug 10 '20 at 19:45
Why is $0 \in J$ – Clemens Bartholdy Jun 27 '22 at 01:05

$R$ is a subring of a field $F$ with $x\in R$ or $x^{-1}$ for all $x\in F^\times$. If $I,J$ are ideals of $R$, then $I\subseteq J$ or $J\subseteq I$.

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