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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

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Hint: Your ring is the localization of $\Bbb{Z}$ at the prime ideal $3$. –  fpqc Feb 17 '13 at 21:08
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@BenjaLim , I very much doubt that hint will help as this seems to be a beginners exercise... –  DonAntonio Feb 17 '13 at 21:09
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@DonAntonio: BenjaLim's hint is a great one for someone who knows about localization. Maybe the OP does, and maybe she doesn't, but some other readers will, so it seems like a worthwhile comment. –  Pete L. Clark Feb 17 '13 at 22:13
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My answer here might be of help: math.stackexchange.com/questions/157685/… –  Robert Feb 17 '13 at 22:25
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@Pete, I know it is a very good hint, yet someone with the necessary level to understand would hardly ask such an elementary and, imo, straightforward question. This is why I thought it is unlikely the OP would understand the hint, and I agree that anyway it is a good hint for everybody. –  DonAntonio Feb 18 '13 at 2:32
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3 Answers

up vote 4 down vote accepted

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...

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The OP said he wanted to find all ideals in $R$, not only the maximal / prime ideals. If you're willing to explain why $R$ is a PID, the full answer follows quickly from what you've done. –  Pete L. Clark Feb 17 '13 at 22:04
    
I think that as a hint the answer fulfills its goal, in particular taking into account that one of the deep (or "deep", as you will) theorems that must beginners know after a very short while is that in a commutative unitary ring every ideal is contained in a maximal one. After all, this question's tagged as "homework"... –  DonAntonio Feb 18 '13 at 2:34
    
$@$Don: sure, okay. I'm probably being too critical: sorry for that. –  Pete L. Clark Feb 18 '13 at 2:59
    
Thanks! As a quick side question, is there any way to determine all the ideals of a given ring generally, or is this a classification that is done case by case? –  Domonic Mei Feb 19 '13 at 2:23
    
I think it is the second one (in particular if we don't restrict ourselves to commutative rings and then we must talk about left or right ideals), but in some cases we can do quite much, for example: if the ring is a Principal Ideal Ring, then we can try to do some work with non-invertible elements and see what the ideals generated by them look like: zero divisors, non-zero divisors, prime elements, etc... –  DonAntonio Feb 19 '13 at 2:28
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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.

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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 $\rm\,3\nmid 1\:\Rightarrow\:3^b\nmid 1.$

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