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$R=\bigl\{\frac{a}{b}\ | \gcd(3, b) = 1,\ a, b ∈ \Bbb Z\bigr\}$

I am trying to find all of the ideals in this ring. So far I have $\bigl\{0\bigr\}$, $R$, and $S=\bigl\{\frac{a}{b}\ | \gcd(3, b) = 1, \gcd(3,a)=3 \ a, b ∈ \Bbb Z\bigr\}$ . I'm not sure how to find any more or, if there are no more, show that this is the case. I am then interested in which of the ideals are prime and maximal.

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  • $\begingroup$ Which elements of this ring are not units? When you answer that question, you'll have an idea of some non-trivial ideals. I don't think your set $S$ is an ideal because it's not closed under addition: $2, 4 \in S$ but $2+4=6 \notin S$. $\endgroup$ – Robert Shore Mar 7 at 1:39
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    $\begingroup$ @RobertShore oops - it was meant to say gcd(a,3)=3. I have now edited to fix. $\endgroup$ – user651483 Mar 7 at 1:50
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This is the ring of fractions with denominator not divisible by $3$. In other words, it is the localisation of $\mathbf Z$ at the prime ideal $3\mathbf Z$, usually denoted as $\mathbf Z_{(3)}$.

It is known the only non-trivial ideals of this ring are generated by the powers of $3$, so the only prime ideals are $(0)$ and $3R$ (which is the single maximal ideal of $R$).

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    $\begingroup$ I wrote my $S$ down wrongly in my original post, but have now edited - is this equal to $3R$? I am quite new to ring theory and this is the first time I have met this ring. $\endgroup$ – user651483 Mar 7 at 1:56
  • $\begingroup$ Yes, your (corrected) ideal $S=3R$. It's not hard to prove the result stated in this answer. Prove that any non-trivial ideal $I$ must have some positive integer, and then consider the smallest positive integer in $I$. Prove that smallest positive integer must have the form $3^k$ for some $k \in \Bbb Z^+$. Then show that $I$ consists of $R$-multiples of that integer. $\endgroup$ – Robert Shore Mar 7 at 2:01

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