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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.

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FYI: In this context, we usually use "odd" rather than "uneven"; and "numerator" rather than "nominator" is the correct term. –  Arturo Magidin Jan 31 '12 at 4:18
    
@ArturoMagidin thanks:) –  sxd Jan 31 '12 at 4:20
    
@ArturoMagidin I do know about it, however, this exercise comes way before that chapter. So I would rather do it without it. –  sxd Jan 31 '12 at 4:21
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1 Answer

up vote 5 down vote accepted

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):

  1. 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 $I=(0)$.

  2. 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).

  3. Go from there.

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Would the following approach be proper for 1? We only have to show that $I \subset (\frac{a}{1})$. Suppose that $\frac{c}{d} \in I$ in lowest terms, then $d$ is odd since $I$ is a subring of $R$. Due to the devision algorithm, $c = qa + r$ with $0 \leq r < a$. But then $\frac{c}{d} = \frac{qa+r}{d} = \frac{qa}{d}+\frac{r}{d}$ and thus $\frac{r}{d} = \frac{c}{d}-\frac{qa}{d}\in I$, thus $r = 0$ and thus $\frac{c}{d} = \frac{qa}{d}$. –  sxd Jan 31 '12 at 4:49
    
@DimitriSurinx: Yes, that works, though I don't think you need to assume $\frac{c}{d}$ is in lowest terms, just that $d$ is odd. And I would make it a bit more obvious by writing $(\frac{q}{d})\frac{a}{1}$ instead of $\frac{qa}{d}$. –  Arturo Magidin Jan 31 '12 at 4:51
    
True, thank you very much. –  sxd Jan 31 '12 at 4:55
    
Now I am trying to show that if such a does not exists that $I = (0)$. I'm trying this by contradiction. Suppose that I \neq (0). Then there exists $\frac{c}{d} \neq 0 \in I$ and thus $(\frac{c}{1}) \in I$ and thus $c \in I$. I think the clue is that showing that now this $c$ is this minimal positive integer. But now I have no clue how to even get to this, since $I$ could contain infinitely many elements. Could you point me in the right direction? –  sxd Jan 31 '12 at 5:35
    
@Dimitri: The only way there can be no smallest positive $a$ is if there are no positive $a$ at all (since any nonempty set of positive integers must have a smallest element). Since $\frac{c}{d}\in I$ implies $\frac{-c}{d}\in I$, the only way this can occur is if $\frac{c}{d}\in I$ with $d$ odd implies $c=0$. –  Arturo Magidin Jan 31 '12 at 5:46
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