Show that the following localization ring has one maximal ideal

Question

I was given the following question:

$R=\mathbb{Z}$ and $S$ contains all elements not divisible by $p$ (which is a given prime)

Prove that $S^{-1}R$ has only one maximal ideal

I tried assuming it has two and looking at one element in each that does not exist in the other but I didn't manage to come to any meaningful result so I'm pretty much stuck now, any thoughts on what to try next?

This subject is pretty new to me and I think I lack some intuition as to what the ring $S^{-1}R$ even is so any kind of explanation would be very much appreciated

Thank you

Answer 1

$S^{-1}R$ is the localization of $R$ at the prime ideal $p\mathbb{Z}$. Since $\mathbb{Z}$ is an integral domain, we can describe $S^{-1}R$ as a subring of its field of fractions, i.e. $\mathbb{Q}$. In this case $S^{-1}R$ consists of all those rational numbers which can be written as $\frac{a}{s}$ where $a\in R,s\in S$ that is $a$ and $s$ are integers with $s$ not divisible by $p$.
Now for the actual question regarding the maximal ideals:
Show that the units in $S^{-1}R$ are precisely those elements in $S^{-1}R\setminus pS^{-1}R$. This then shows that $pS^{-1}R$ is the unique maximal ideal of $S^{-1}R$ (why?).

Answer 2

$S^{-1}R:=\{a/s:a\in R,p\not\mid s\}$ is a local ring with a unique maximal ideal $(p)$, since one may check that $S^{-1}R\setminus(p)$ is the group of units of $S^{-1}R$.

Suppose there is a prime ideal $0\ne\mathfrak p\subseteq (p)$. Let $a/s\in\mathfrak p$ be nonzero. Then $a/1\in\mathfrak p$. Let $a=up^k$ for $p\not\mid u$. Since $u/1\in S^{-1}A$ is a unit, we must have $p^k\in\mathfrak p$, so that since $\mathfrak p$ is prime, $p\in\mathfrak p$. Thus, in fact $\mathfrak p=(p)$.