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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.2012-01-31
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    @ArturoMagidin thanks:)2012-01-31
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    @ArturoMagidin I do know about it, however, this exercise comes way before that chapter. So I would rather do it without it.2012-01-31

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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 $\frac{c}{d} = \frac{qa}{d}$.2012-01-31
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    @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 $\frac{qa}{d}$.2012-01-31
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    True, thank you very much.2012-01-31
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    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?2012-01-31
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    @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 $c=0$.2012-01-31
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    Oh right i was simply looking to far.2012-01-31
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    Sorry for the load of questions, however I still seem to get nowhere. 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 2Z. 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 $m \geq n$. However, now I am stuck.2012-01-31
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    @DimitriSurinx: I'm sorry, I don't know what it is you are trying to prove. Are you trying to determine which ideals are prime? Only $(2)$ is prime, since for any $m\gt 1$, the ideal $(\frac{2^m}{1})$ contains the product of $2^{m-1}$ and $2$, but neither one is in the ideal.2012-01-31
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    Yes I'm indeed trying to show which are prime. But how about the ideal $(\frac{2^mr}{1})$ with $m \geq 1$ and $r$ odd?.2012-01-31
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    @Dimitri: That ideal contains $\frac{1}{r}\times\frac{2^mr}{1} = \frac{2^m}{1}$; conversely, the ideal generated by $\frac{2^m}{1}$ contains $\frac{2^mr}{1}$. Note that $r$ is a *unit* in this ring, and in any ring, if $u$ is a unit, then $(au) = (a)$. This is a consequence of my point 2 in the post: every ideal can be written in the form $\left(\frac{2^m}{1}\right)$ for some $m\geq 1$.2012-01-31
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    Oh right how stupid of me. I don't know why it is but when it comes down to working with ideals I have a hard time. While I find other topics in ring/group/field theory a lot easier. Is this a normal feeling among students?2012-01-31
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    @Dimitri: In my experience, it takes practice to work with ideals; while subgroups, subfields, etc. are things in and of themselves, normal subgroups of ideals are *contextual*: you need to keep in mind the overgroup/overring and how it affects things.2012-01-31
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    Thanks for your patience. It's always nice to see the elegance in your solutions! Your students are lucky!2012-01-31