Q:

Find all the prime ideals of $\{\frac{a}{b}| a \in \mathbb{Z}, b \in \mathbb{N}_0 \text{ odd}\}$

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. 

A:

If you know about localizations of rings, then this ring is the localization of $\mathbb{Z}$ away from (or "at"; I'm never sure of the correct terminology) the ideal $(2)$. From here there are general theorems that tell you exactly what the ideals and prime ideals of your ring are.
If you don't know about localizations (or, as you indicated in your comment, don't want to use them):

To show that every ideal is principal, let $I$ be an ideal of the ring, and let $a$ be the smallest positive integer (if one exists) for which there exists an odd $b$ with $\frac{a}{b}\in I$. Show that $I=\left(\frac{a}{1}\right)$. If no such $a$ exists, then show $I=(0)$. 
Let $a=2^nr$ with $r$ odd, and $b=2^ms$ with $s$ odd, $n,m\geq 0$. Show that $\left(\frac{a}{1}\right) = \left(\frac{b}{1}\right)$ if and only if $n=m$. That is: the only thing that really matters are the powers of $2$ (Intuition: anything else can be cancelled by multiplying by a suitable unit of the ring).
Go from there.

