For an exercise in my book I have to find all the prime ideals of $R = \left.\left\{\frac{a}{b}\;\right|\; a \in \mathbb{Z}, b \in \mathbb{N}_0 \text{ odd}\right\}\leq (\mathbb{Q},+,\cdot)$
I proceeded as follows, for $\frac{a}{b}\in R$ where $a$ is odd we know that $\frac{b}{a} \in R$, thus $\frac{a}{b}$ is a unit and thus $\left(\frac{a}{b}\right) = R$. Thus $\left(\frac{a}{b}\right)$ cannot be a prime ideal.
Now, for $\frac{a}{b} \in R$ where $a$ is even, we know that $\left(\frac{a}{b}\right)$ only contains fractions with an even numerator. Thus $(\frac{a}{b})$ is a proper ideal of $R$. Now, \begin{align*} \frac{c}{d}\cdot\frac{e}{f} \in \left(\frac{a}{b}\right)&\Rightarrow \exists \frac{x}{y}\in R: \frac{c}{d}\cdot\frac{e}{f} = \frac{x}{y}\cdot\frac{a}{b}\\ & \Leftrightarrow \frac{c}{d} = \left(\frac{f}{e}\cdot\frac{x}{y}\right)\frac{a}{b}\\ &\Leftrightarrow \frac{e}{f} = \left(\frac{d}{c}\cdot\frac{x}{y}\right)\frac{a}{b}\end{align*}
If either $c$ or $e$ are odd, then the fraction with the even numerator is certainly in $\left(\frac{a}{b}\right)$, since suppose that $c$ odd, then $c \cdot y$ and thus $\frac{d}{c}\cdot\frac{x}{y} \in R$.
However, when both $c$ and $e$ are even, I'm stuck, since then both $c \cdot y$ and $e \cdot y$ are both even, so the method I used above does not work anymore.
Furthermore, how can I identify whether an ideal generated by more than one element is prime without brute force checking? I tried showing that every ideal is generated by one element, but I got stuck there. Could anyone give me some tips (No complete solutions please)?
EDIT:
As a response to Arturo's answer: Suppose $\frac{x}{y}\frac{c}{d}\in R$ and suppose that $\frac{x}{y}\cdot \frac{c}{d}\in (\frac{a}{1})$ with $a \in 2\mathbb{Z}$. Thus $a=2^nr$ for some $n\geq 1$ and $r$ odd. Hence $(\frac{xc}{1})\subseteq (\frac{a}{1})$, which implies that $a\mid xy$ and thus $xy=2^ms$ with $m\geq n$. But now I do not know how to proceed.