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Let $mathbb{Z}[1/2] = {a/(2^{k}) : text{$a$ is odd}, k in mathbb{Z} }$ be a ring,
I would like to classify the prime ideals.

I am having a hard time thinking about this problem, I have no ideal how to start thinking about it.

Any insight to the problem will be highly appreciated.

Sketch: Observe that $mathbb Z[1/2]cong mathbb Z[x]/(2x-1)$. Let $eta:mathbb Z[x]to mathbb Z/(2x-1)$ be the usual quotient map. Each prime ideal $Isubseteq mathbb Z[x]/(2x-1)$ is taken to a prime ideal in $mathbb Z[x]$ via its preimage under $eta$. So it suffices to classify the proper prime ideals of $mathbb Z[x]$ properly containing $(2x-1)$. One can show that these are exactly those $Jsubseteqmathbb Z[x]$ of the form $(p,2x-1)$ for prime $pneq 2$ and furthermore that $eta(p,2x-1)=eta(p',2x-1)iff p=p'$.

We conclude that the prime ideals of $mathbb Z[1/2]$ are exactly those of the form $(text{loc}(p))subseteq mathbb Z[1/2]$ where $pneq 2$ is prime and $text{loc}:mathbb Ztomathbb Z[1/2]$ is the localization map $Box$

Let $p$ be a prime different from $2$; we want to show that $pmathbb{Z}[1/2]$ is a prime ideal.

Note that every element of $pmathbb{Z}[1/2]$ can be written as $px/2^k$, for some $xinmathbb{Z}$ and $kinmathbb{N}$. Similarly, elements in $mathbb{Z}[1/2]$ can be written as $y/2^k$.

Suppose $a,binmathbb{Z}$ and $m,ninmathbb{N}$; then
$$
frac{a}{2^m}frac{b}{2^n}=frac{px}{2^k}
$$
implies that $pmid ab$, because we have $2^{m+n}px=2^kab$, so $pmid 2^kab$. Since $pne2$, we have $pmid ab$ and the conclusion easily follows.

Suppose $I$ is a prime ideal in $mathbb{Z}[1/2]$. Then
$$
I^c={xinmathbb{Z}:xin I}=Icapmathbb{Z}
$$
is a prime ideal in $mathbb{Z}$, so it is of the form $pmathbb{Z}$. Note that $pmathbb{Z}[1/2]subseteq I$. On the other hand, if $y/2^kin I$, then $yin I$ and so $y=px$ for some $x$. This proves the reverse inclusion.

Note that $pne2$, because $2mathbb{Z}[1/2]=mathbb{Z}[1/2]$ is not a prime ideal.

More generally, given a multiplicative set $S$ in $R$:

1. if $P$ is a prime ideal of $R$ such that $Pcap S=emptyset$, then $S^{-1}P$ is a prime ideal of $S^{-1}R$;


2. if $Q$ is a prime ideal of $S^{-1}R$, then $Q^c={xin R:x/1in Q}$ is a prime ideal in $R$ and $Q^ccap S=emptyset$;


3. if $P$ is a prime ideal of $R$ and $Pcap S=emptyset$, then $P=(S^{-1}P)^c$.

Proof of 1. Suppose $(a/s)(b/t)=x/u$, with $a,bin R$, $xin P$ and $s,t,uin S$. Then $abuv=xstuv$, for some $vin S$. Therefore $abuvin P$; since $uvnotin P$, we conclude $abin P$ and therefore $a/sin S^{-1}P$ or $b/tin S^{-1}P$.

Proof of 2. The fact that $Q^c$ is a prime ideal is obvious. If $sin Q^ccap S$, then $s/sin Q$: contradiction.

Proof of 3. Let $xin P$; then $x/1in S^{-1}P$, so $xin(S^{-1}P)^c$. Let $xin(S^{-1}P)^c$; this means $x/1=y/s$ for some $yin P$ and $sin S$; therefore, for some $tin S$, $xst=yt$; but $ytin P$ and $stnotin P$; therefore $xin P$.

Conclusion: the map $Pmapsto S^{-1}P$ defines a bijection between the prime ideals of $R$ not intersecting $S$ and the prime ideals of $S^{-1}R$.

In your case $R=mathbb{Z}$ and $S={2^n:ninmathbb{N}}$. You can see the steps in the proof of the general fact as mirrored in the special case above.