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Here's what I have so far:

I know that $\mathbb{Z}$ is an integral domain, and from the formal definition of an ideal generated by the single element $p$ where $p$ is a prime,

$$ (p) = \{r_1ps_1 + ... + r_nps_n: n \in \mathbb{N}_0, r_1, ..., r_n, s_1, ..., s_n \in \mathbb{Z}\} = \{pz : z \in \mathbb{Z}\} = p\mathbb{Z}. $$

From the definition of localization of the ring $\mathbb{Z}$ with respect to the subset $(p)$,

$$ \mathbb{Z}_{(p)} = \{\frac{a}{b} : a \in \mathbb{Z}, b \in \mathbb{Z} \setminus (p)\} = \{\frac{a}{b} : a,b \in \mathbb{Z}, p \nmid b\}. $$

I am a little confused with showing that for $n \geq 0$ the set $(p^n) = p^n\mathbb{Z}$ is in fact an ideal of $\mathbb{Z}_{(p)}$, since for any $x \in \mathbb{Z}_{(p)}$, $x$ is of the form

$$ x = \frac{a}{b}, \quad a,b \in \mathbb{Z}, p\nmid b, $$

and for any $w \in (p^n)$, $w$ is of the form

$$ w = mp^n, \quad m \in \mathbb{Z}, n \geq 0 $$

so that

$$ xw = \frac{a}{b} \cdot \frac{mp^n}{1} = \frac{amp^n}{b} \in \mathbb{Z}_{(p)}, \quad wx = \frac{mp^n}{1} \cdot \frac{a}{b} = \frac{amp^n}{b} \in \mathbb{Z}_{(p)} $$

provided that I associate the element $mp^n \in (p^n)$ with the element $\frac{mp^n}{1} \in \mathbb{Z}_{(p)}$. It doesn't seem to be that these elements are necessarily elements of $(p^n)$ however (i.e., how do I know that $\frac{am}{b}$ will be an integer?). Am I misinterpreting the ideal $(p^n)$ or the localization $\mathbb{Z}_{(p)}$?

Furthermore I have no idea on where to start with proving that the converse is true; that every ideal of $\mathbb{Z}_{(p)}$ is in fact of this form. Right now I am working with the idea that if in fact $(p^n)$ is an ideal of $\mathbb{Z}_{(p)}$, then there exists a ring $S$ and a ring homomorphism $f: \mathbb{Z}_{(p)} \rightarrow S$ such that $\ker(f) = (p^n)$, but I'm not sure if this is this path will lead to any meaningful results.

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Please note that here $(p^n)$ means $p^n\mathbb Z_{(p)}$, not $p^n\mathbb Z$.

In $\mathbb Z_{(p)}$ $x$ is a non-unit if and only if $x$ is of the form $p^k\frac{a}{b}$ where neither $a$ nor $b$ are in $(p)$ with $k>0$ (clearly this is only the case if $x=p\frac{x'}{b}$. By factoring out $p$s one reaches this form).

Thus any proper Ideal $I$ in $\mathbb Z_{(p)}$ has to consist only of Elements of the form $p^k\frac{a}{b}$ with some $k>0$ and any $a,b\not\in(p)$ (and it does in fact contain $p^k\frac{a'}{b'}$ for any $a'/b'$, and thus $p^k$).

Now set $n$ the minimal $k$ so that $p^k\in I$. Then obviously any other Element of $I$ is of the form $p^l \frac{a}{b}$ with $l\geq n$, so $p^l\frac{a}{b} = p^n (p^{l-n}\frac{a}{b})$. Thus $I=(p^n)$.

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  • $\begingroup$ Since $\mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $(p)$, we have that $\mathbb{Z}_{(p)}$ is a local ring and therefore the non-units do in fact form a unique maximal ideal of $\mathbb{Z}_{(p)}$, but this doesn't necessarily prove that every non-zero ideal of $\mathbb{Z}_{(p)}$ will consist of elements which are all non-units does it? $\endgroup$
    – Oderus
    Aug 28 at 7:29
  • $\begingroup$ @OderusUrungus Note that if an ideal contains a unit, it also contains $1=p^0$, so it is in fact the whole ring. That’s why I specified $I$ to be a proper Ideal. Although that is not really nescessary, as for $(1)$ the proof works the same. The point is that every Element of $\mathbb Z_{(p)}$ is of the form $p^ke$ with some unit $e$. $\endgroup$
    – Lazy
    Aug 28 at 10:21
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First to clarify some confusion: $(p^n)$ as and ideal of $\Bbb Z_{(p)}$ is not $p^n\Bbb Z$, but $p^n\Bbb Z_{(p)}$ (as more generally, the ideal $(a)$ of a ring $R$ is $aR$ and hence quite clearly an ideal).

Next let $\mathfrak a$ be any non-zero ideal of $\Bbb Z_{(p)}$. For each non-zero $x\in\mathfrak a$, its valuation $v_p(x)$ (i.e., the maximal exponent $m$ such that $x=p^my$ for some $y\in\Bbb Z_{(p)}$) is a natural number. We can let $n=\min\{\,v_p(x)\mid 0\ne x\in\mathfrak a\,\}$, and with this show that $p^n\in\mathfrak a$ and in fact that $\mathfrak a=(p^n)$.

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  • $\begingroup$ For formality, shouldn't we say then that $(p^n)$ as an ideal of $\mathbb{Z}_{(p)}$ is $\frac{p^n}{1}\mathbb{Z}_{(p)}$ and not simply $p^n\mathbb{Z}_{(p)}$? $\endgroup$
    – Oderus
    Aug 28 at 6:09
  • $\begingroup$ Yes and no. For one thing you can define $\mathbb Z_{I}$ as well as $\mathbb Q$ as equivalence classes of pairs. You can also axiomatically start with $\mathbb R$, have $\mathbb Q$ the smallest subfield, $\mathbb Z$ the smallest ring. Then you can have $\Z_{I\$ as the subset of $\mathbb Q$ where the denominator is not in $I$. This way $\mathbb Z\subset\mathbb Z_I$. On the other hand you can also define a group action of $\mathbb Z$ on $\mathbb Z_I$ and thus also legitimize the notation $p^n\mathbb Z_{(p)}$. But you are more or less correct. $\endgroup$
    – Lazy
    Aug 28 at 10:34

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