[Math] Ideals of the ring of rational numbers with odd denominators

abstract-algebramaximal-and-prime-idealsring-theory

Consider the subring $R\subset\Bbb Q$ , $R=\{\frac ab\ | \ a,b\in\Bbb Z, b\text{ odd}\}$

I am struggling with the following questions:

(1) Prove that the ideals of $R$ are $\{0\}$ and $2^nR$ for $n\ge 0$

(2) Prove that $R$ has 2 prime ideals and 1 maximal ideal

For (1) I can see that $2^nR$ is an ideal for $n\geq0$ but I'm not sure how to show all ideals other than $\{0\}$ are of this form

With question (2),

$2^nR\subsetneq 2^{n-1}R\subsetneq…\subsetneq 2R\subsetneq R$

so $2R$ is clearly the only maximal ideal and is therefore prime, but I am not sure where to go from here.

Any suggestions?

Best Answer

To continue (1): The elements of $R$ look like elements of $\mathbb Z$ except that you also have inverses for all the odd numbers. Everything in $\mathbb Z$ factors into primes, but most of them are units, so the only thing that matters is the number of $2$'s in the factorization. So you should not have much difficulty in arguing this way: let $I\lhd R$ be a nonzero ideal. Among all elements in $I$, expressed in least terms $\frac{a}{b}$, pick one with the smallest power of $2$ dividing $a$. Then work to show $I=(a)\lhd R$.

To continue (2), you've seen that $(2)$ is prime, and obviously nothing of the form $(2^n)$ with $n>1$ could be prime, so the only thing left is $\{0\}$. Do you see it is prime as well?