Localization of $\mathbb{Z}$ at the powers of $15$ I am trying to determine the maximal ideals of $\mathbb{Z}[1/15] =\big\{\frac{a}{15^n}\;|\; a \in \mathbb{Z},\; n\geq 0 \big\}$. This is the integers localized at the powers of $15$. My attempt was to use a similar approach as that used for finding the maximal ideals of the integers localized at a prime number, but that did not work.
Any hints or suggestions would be greatly appreciated.
 A: Let's look at this from a more general point of view. Let $S$ be a multiplicative set in the domain $A$. We can assume $0\notin S$ (or the case would be trivial) and that $1\in S$ (the localization doesn't change if we add $1$ to $S$ anyway).
For the canonical morphism $\lambda\colon A\to S^{-1}A$, $\lambda(a)=a/1$ we have (like for any ring morphism) that $\lambda^\gets(P)$ (inverse image of $P$) is a prime ideal for every prime ideal $P$ of $S^{-1}P$.
We can also consider, for an ideal $I$ of $A$, the ideal $I^e$ generated by $\lambda^\to(I)$ in $S^{-1}A$, which is nothing else than
$$
I^e=\{a/s:a\in I,s\in S\}
$$
Can we say that $I^e$ is a prime ideal when $I$ is a prime ideal? Suppose
$(x/t)(y/u)\in I^e$. Then there exist $a\in I$ and $s\in S$ such that
$$
\frac{a}{s}=\frac{xy}{tu}
$$
which means that $atu=xys$. In particular $xys\in I$, so at least one among $x,y,s$ must belong to $I$. In case $I\cap S=\emptyset$ we have the desired conclusion, because $x\in I$ or $y\in I$ and so $x/t\in I^e$ or $y/u\in I^e$. Hence, under $I\cap S=\emptyset$, we have that $I^e$ is a prime ideal in $S^{-1}A$. Well, we also need $1/1\notin I^e$, but this is clearly true, for the same reason that $I\cap S=\emptyset$.
For $P$ a prime ideal of $S^{-1}A$ we have $\lambda^\gets(P)\cap S=\emptyset$, because $1/1\notin P$.
Now let's prove that $(\lambda^\gets(P))^e=P$. Suppose $a\in\lambda^{\gets}(P)$; then $a/1\in P$, so for every $s\in S$ we have $a/s\in P$. Suppose $a/s\in P$. Then also $(s/1)(a/s)=a/1\in P$ and so $a\in\lambda^{\gets}(P)$ and therefore $a/s\in(\lambda^\gets(P))^e$.
You can apply this to your particular case.

 If $P$ is a prime ideal of $\mathbb{Z}[1/15]=S^{-1}\mathbb{Z}$, where $S=\{15^n:n\ge0\}$, then $\lambda^\gets(P)$ is a prime ideal of $\mathbb{Z}$ that doesn't meet $S$, so it has the form $p\mathbb{Z}$ for some prime integer $p$, where $p\ne 3$ and $p\ne 5$. Or it is the zero ideal, of course.


 And conversely, any prime ideal has the form $(p\mathbb{Z})^e$ for $p\notin\{3,5\}$ a prime, or it is the zero ideal.


 Thus the prime ideals are the zero ideal and those of the form $p\mathbb{Z}[1/15]$ for $p$ a prime different from $3$ and $5$.

