# Show that every non-zero ideal of the ring of rationals with odd denominator is generated by $2^{n}$ [duplicate]

Let $$R \subset \mathbb{Q}$$ be the subring $$\left\{\frac{a}{b} \mid a, b \in \mathbb{Z}, b \text { odd }\right\}$$. Prove that the ideals of $$R$$ are the zero ideal $$\{0\}$$ and $$2^{n} R$$ for $$n \geq 0$$.

Hint: if $$\{0\} \neq I \subset R$$, the element $$2^{n}\frac{\text { odd }}{\text { odd }}$$ with the smallest $$n$$ generates $$I$$.

I have seen similar questions asked but I am stuck on the specific step which shows that $$I=2^{n} R$$. I can show that there exists $$n$$ such that for every element $$x$$ of a non-zero ideal of R, it can be expressed as $$x = 2^{n} \frac{a}{b}$$ with $$a, b \in \mathbb{Z}, b \text { odd }$$. This gives me that $$I \subset 2^{n}R$$, however I am having difficulties proving it in the other direction.

Apologies if this still counts as a duplicate question.

## marked as duplicate by Bill Dubuque abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 6 at 21:09

• See here for an insightful way to view it (Dedekind-Hasse criterion for a PID) – Bill Dubuque Mar 6 at 21:14

If $$a/b \in I$$ with $$a/b \gt 0$$, then $$a \in I$$. Thus, $$I$$ contains some positive integer. Let $$a$$ be the smallest positive integer in $$I$$. Then $$a = 2^nk$$ with $$k$$ odd. If $$k \gt 1$$, then $$a/k \in I$$, contradicting the choice of $$a$$ as the smallest positive integer in $$I$$. Thus, $$a=2^n \in I$$ for some $$n \in \Bbb Z^+$$ so $$2^nR \subseteq I$$.
Alternatively, just take the element you've already found and multiply by $$b/a \in R$$ to show that $$2^n \in I$$.
• Oh so the key is showing that $2^{n} \in I$ and then using the closed under multiplication from the ring condition of an ideal to show the final inequality? – Albert B Mar 6 at 19:19
Let $$\dfrac{a}{b}\in I$$ then $$a=2^{m}t$$ with $$t$$ odd. We can choose $$x=2^n\dfrac{t}{u}\in I$$ with $$t,u$$ odd and $$n$$ the smallest possible. Then $$Rx\subset I$$ and if $$\dfrac{a}{b}\in I$$ then $$\dfrac{a}{b}=2^l\dfrac{u'}{t'}$$ with $$u',t'$$ odd and $$l\ge n$$ so $$\dfrac{a}{b}=rx$$ so $$I \subset Rx$$
Hence $$I=Rx$$
If you know a bit more of commutative algebra, you can also view the ring $$R$$ as the localization of $$\mathbb{Z}$$ at the multiplicatively closed set $$S = \mathbb{Z} \setminus (2) = \{n \in \mathbb{Z}\mid n \text{ odd} \}$$, where $$(2)$$ is the ideal generated by $$2$$, so $$R = S^{-1}\mathbb{Z}$$. Then there is a correspondence of ideals, where every ideal $$I \subset R$$ comes from an ideal $$J \subset \mathbb{Z}$$, where $$J \cap S = \emptyset$$, in the way that $$I$$ is generated by elements of the form $$\frac{j}{1}$$ for $$j \in J$$. Other way to write this is $$I = R \cdot J = S^{-1}J$$. But then, the only ideals in $$\mathbb{Z}$$ which do not contain odd numbers, are of the form $$(2^n)$$ for $$n \in \mathbb{N}$$, or $$0$$.