Finding the ideals in a ring of fractions I am dealing with the ring $$R=\left\{\frac{a}{b} \mid a,b\in\mathbb{Z}\mbox{, $b$ is not divisible by 3}\right\}$$ with addition and multiplication as defined in $\mathbb{Q}$ and I'm trying to find all the ideals of the given ring.
My initial thought is to find all the additive subgroups of $(R,+)$, but I am having trouble reasoning through this step. I'm not sure how to classify all such subgroups, then to prove that these are all such subgroups. 
A prod in the right direction would be greatly appreciated! ~Dom
 A: Hints:
1) Show the set $\,M:=\{r\in R\;\;;\;r\,\,\,\text{is not invertible}\}$ is an ideal in $\,R\,$
2) Deduce $\,M\,$ is a maximal ideal (and, in fact, the only maximal ideal) of $\,R\,$
Your ring $\,R\,$ is what's called a local ring , a rather important class of rings in commutative algebra and some other mathematical realms. This is, apparently, what BenjaLim was aiming at in his comment, as local things appear as localizations wrt prime ideals in some rings...
A: Step 1: Let $I$ be an ideal in $R$.  Show that $\mathfrak{i} = I \cap \mathbb{Z}$ is an ideal of $\mathbb{Z}$ and that $I = \{ \frac{i}{s} \ | \ i \in \mathfrak{i}, s \in \mathbb{Z}^+, \ 3 \nmid s \}$.  Deduce that if $\mathfrak{i} = n \mathbb{Z}$, then $I = n R$.  Thus all ideals of $R$ are principal and generated by elements of $\mathbb{Z}$.
Step 2: Figure out for which $m,n \in \mathbb{Z}^+$ we have $m R = nR$.  The answer given by Math Gems is relevant here.
Remark: $\S 7.2$ of these notes contains a more general discussion along these lines.  It amounts to finding the ideals in a localization, as Benjalim mentioned.
A: Hint $\ $ Every prime $\rm\:p \ne 3\:$ becomes a unit in $\rm\,R\,$ since $\rm\:1/p\in R.\:$ But the prime $\rm\,p = 3\,$ is not a unit in $\rm\,R\,$ since $\rm\,1/3\not\in R.\:$ Hence $\rm\ (n) = (2^a 3^b 5^c\cdots) = (3^b)\:$ in $\rm\,R,\,$ and $\rm\,3\nmid 1\:\Rightarrow\:3^b\nmid 1.$
