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Problem: Let $R$ be a ring such that $\Bbb Z \subset R \subset \Bbb Q$. Show that $R$ is a PID. Hint: If $I$ is an ideal of $R$ consider $A=\Bbb Z \cap I$.

Following the hint, if $I$ is an ideal of $R$, then $A$ is the set of rationals in $I$ where the denominator is $1$, and moreover since $\Bbb Z$ is an ideal of $R$, then $A$ is an ideal of $\Bbb Z$, and so is principle. This is about as far as I've gotten, insights appreciated.

IntegrateThis
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  • Try to show that a generator of $A$ as a $\Bbb Z$-ideal is also a generator of $I$ as an $R$-ideal. – Angina Seng Apr 20 '19 at 01:10
  • @LordSharktheUnknown if $a/b$ is in $I$, then $b (a/b)=a$ is in $I$, and so $a = pk$ for some $k$, where $p$ generates $A$, and now how can I show that $pt$ = $a/b$ for some $t \in \Bbb Q$ ? – IntegrateThis Apr 20 '19 at 01:18

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