Prime ideals in $R= \left\{ \frac ab\in \mathbb Q\mid a,b\in \mathbb Z,p\nmid b \right\}$

Question

Let $p\in \mathbb Z$ be a prime. Define $R= \left\{ \frac ab\in \mathbb Q\mid a,b\in \mathbb Z,p\nmid b \right\}$. I'm supposed to prove $pR$ is the only prime ideal in $R$, and the $R/pR\cong \mathbb Z/p\mathbb Z$. I proved it's a prime ideal, but I don't see how to prove it's the only one. I think $R$ is a pid and every ideal generated by an integers, so all prime ideals are generated by primes in $\mathbb Z$. I tried supposing there's a prime $q\in \mathbb Z$ not associate to $p$ which is a prime element in $\mathbb Z[\frac 1n]$, but I don't see how to get a contradiction.. Clueless about the isomorphism too..

Also, is $R$ a localization of $\mathbb Z$?

Update I think $R$ is basically inverting all integers except powers of $p$, so it looks like we're localizing at the complement of $p\mathbb Z$. Then, since localization commutes with quotients we and $\mathbb Z/p\mathbb Z$ is already a field, we get the isomorphism we wanted. Is this a correct solution for the last part?

Update 2 Does this work for the first part? All primes except $p$ are inverted in $R$, so they're units and hence not prime elements.

Answer 1

Yes, $pR$ contains all the non-invertible elements of $R$ so it is the unique maximal ideal of a local ring (in this case indeed the localisation of $\mathbb Z$ at $p$).

Any hypothetical prime ideal $Q$ must be contained in this maximal ideal. If it contains any element $p^na/b=p.p^{n-1}a/b$ ($a$ not a multiple of $p$) then by primeness it also contains $p^{n-1}a/b,$ and so on until we have $Q=pR.$

The isomorphism is induced by the map $R\to \mathbb Z/p\mathbb Z$ that sends $a/b$ to $\overline a{\overline b}^{-1}$.

Answer 2

Just a nitpick, but (depending on your definitions) $(0)$ is also a prime ideal of $R = \mathbb{Z}_{(p)}$. But besides that, yes, the proof you gave in the update is correct.

Here's an equivalent but slightly different way to see the result. Let $R$ be a commutative ring and $S$ be a multiplicative subset with $1 \in S$. The correspondence theorem for localization states that there is a bijection as below

$\hspace{2.7cm}$

where $e$ and $c$ are the extension and contraction maps, respectively. (Cf., Proposition 38, \S 15.4, p. 709 of Dummit and Foote.) In the case where $S = R \setminus Q$ for some prime ideal $Q \trianglelefteq R$, then we get a bijection between prime ideals of $R_Q = S^{-1}R$ and prime ideals of $R$ contained in $Q$. If $Q$ is a minimal prime, i.e., does not properly contain any nonzero primes, then the correspondence shows that the only primes of $R_Q$ are $0$ and ${^e Q} = Q R_Q$. Since $(p)$ is a minimal prime for any nonzero prime $p \in \mathbb{Z}$, then this is true in your case. (A fancy way of saying this is that $\mathbb{Z}$ has Krull dimension $1$.)