My homework asks:
Given the set $R$ of rationals $a/b$, where a prime $p$ does not divide $b$, is $R$ a principal ideal domain? Justify your claim.
I believe that $R$ is a principal ideal domain, but I'm lost as to how to follow through. Thanks for helping me.
Oh goody! I love homework questions.
Let $I \subseteq R$ be an ideal. Now elements of $R$ are of the form $\frac{a}{b}p^n$ where $p$ does not divide $a$ and $b$ and where $0\leq n$.
Of all such elements in $I$ take one with minimal $n$, say $\frac{a}{b}p^n\in I$ then we have that $\frac{b}{a}\in R$ since $a$ is not divisible by $p$. Thus $\frac{b}{a}\frac{a}{b}p^n \in I$, that is $p^n \in I$. We now claim $I=(p^n)$. Let $\frac{c}{d}p^m\in I$ with $p$ not dividing $c$ or $d$ and $n\leq m$ (since $n$ was minimal). This means that $\frac{c}{d}p^{m-n}\in R$ and $\frac{c}{d}p^m=\frac{c}{d}p^{m-n} \cdot p^n$ so $I \subseteq (p^n)$. The other direction is trivial so we are done.