# Show that every non-zero ideal of the localization $\mathbb{Z}_{(p)}$ of $\mathbb{Z}$ at a prime $p$ is of the form $(p^n)$ for some $n \geq 0$

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.

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 $$I=(p^n)$$.

• 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? Aug 28 at 7:29
• @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 $e$.
– Lazy
Aug 28 at 10:21

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 $$\mathfrak a=(p^n)$$.

• 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 $p^n\mathbb{Z}_{(p)}$? Aug 28 at 6:09
• 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.
– Lazy
Aug 28 at 10:34