2
$\begingroup$

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.

$\endgroup$

marked as duplicate by Bill Dubuque abstract-algebra Mar 6 at 21:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See here for an insightful way to view it (Dedekind-Hasse criterion for a PID) $\endgroup$ – Bill Dubuque Mar 6 at 21:14
2
$\begingroup$

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 $2^n \in I$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Albert B Mar 6 at 19:19
  • $\begingroup$ Correct. That will show the inclusion relationship you need. $\endgroup$ – Robert Shore Mar 6 at 19:30
0
$\begingroup$

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 $I=Rx$

$\endgroup$
0
$\begingroup$

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 $0$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.