classification of prime ideals $\mathbb{Z}[1/2]$

Question

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.

Answer 1

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 $I\subseteq \mathbb Z[x]/(2x-1)$ 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 $J\subseteq\mathbb Z[x]$ of the form $(p,2x-1)$ for prime $p\neq 2$ 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 $p\neq 2$ is prime and $\text{loc}:\mathbb Z\to\mathbb Z[1/2]$ is the localization map $\Box$

Answer 2

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

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

Suppose $a,b\in\mathbb{Z}$ and $m,n\in\mathbb{N}$; then $$ \frac{a}{2^m}\frac{b}{2^n}=\frac{px}{2^k} $$ implies that $p\mid ab$, because we have $2^{m+n}px=2^kab$, so $p\mid 2^kab$. Since $p\ne2$, we have $p\mid ab$ and the conclusion easily follows.

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

Note that $p\ne2$, because $2\mathbb{Z}[1/2]=\mathbb{Z}[1/2]$ is not a prime ideal.

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

1. if $P$ is a prime ideal of $R$ such that $P\cap S=\emptyset$, then $S^{-1}P$ is a prime ideal of $S^{-1}R$;
2. if $Q$ is a prime ideal of $S^{-1}R$, then $Q^c=\{x\in R:x/1\in Q\}$ is a prime ideal in $R$ and $Q^c\cap S=\emptyset$;
3. if $P$ is a prime ideal of $R$ and $P\cap S=\emptyset$, then $P=(S^{-1}P)^c$.

Proof of 1. Suppose $(a/s)(b/t)=x/u$, with $a,b\in R$, $x\in P$ and $s,t,u\in S$. Then $abuv=xstuv$, for some $v\in S$. Therefore $abuv\in P$; since $uv\notin P$, we conclude $ab\in P$ and therefore $a/s\in S^{-1}P$ or $b/t\in S^{-1}P$.

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

Proof of 3. Let $x\in P$; then $x/1\in S^{-1}P$, so $x\in(S^{-1}P)^c$. Let $x\in(S^{-1}P)^c$; this means $x/1=y/s$ for some $y\in P$ and $s\in S$; therefore, for some $t\in S$, $xst=yt$; but $yt\in P$ and $st\notin P$; therefore $x\in P$.

Conclusion: the map $P\mapsto S^{-1}P$ 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 $S=\{2^n:n\in\mathbb{N}\}$. You can see the steps in the proof of the general fact as mirrored in the special case above.