Abstract Algebra - ideals of the ring $\{a/b : b\text{ odd}\}$

Question

Let us have a ring $R$ defined as $R=\{a/b:a,b \in \mathbb{Z} \text{ and } b \text{ is odd}\}$. Show that if an ideal is not the zero set, then it is generated by $2^n\cdot(\text{odd}/\text{odd})$.

Answer 1

Any nonzero element of $R$ can be written as $$ 2^n\frac{a}{b} $$ with $n\ge0$ and both $a$ and $b$ odd.

Let $I$ be a nonzero ideal of $R$ and consider $n$ minimal such that $2^n\frac{a}{b}\in I$ (with $a$ and $b$ odd).

Now prove that $2^n\frac{1}{1}$ generates $I$.

One inclusion is clear because $$ 2^n\frac{1}{1}=2^n\frac{a}{b}\frac{b}{a}\in I $$ Use the minimality of $n$ to prove the converse.