# Prime and maximal ideals of this ring?

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

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.

• 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 $2+4=6 \notin S$. – Robert Shore Mar 7 at 1:39
• @RobertShore oops - it was meant to say gcd(a,3)=3. I have now edited to fix. – user651483 Mar 7 at 1:50

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 $$R$$).
• 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. – user651483 Mar 7 at 1:56
• 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. – Robert Shore Mar 7 at 2:01