Ring properties of $\mathbb Z_{(p)}\subset\mathbb Q$ I stumbled across the example $$\mathbb Z_{(p)}=\{\frac{a}{b}\mid p\nmid b\text{ and }\gcd(a,b)=1\}$$ for a localization of a ring and don't find it trivial to see that this indeed is a subring of $\mathbb Q$. How does one show the properties mathematically rigorous? If I take $\frac{a}{b}$ and $\frac{c}{d}$, then surely $p\nmid bd$ and also I can "cancel" the fraction $\frac{ad+bc}{bd}$ to a point where I get a coprime denominator and numerator. The same goes for $\frac{ac}{bd}$. But that doesn't sound rigorous to me.
 A: Just prove the required properties.
First, show that $0\in\mathbb{Z}_{(p)}$ and $1\in\mathbb{Z}_{(p)}$. Indeed 
$$0=\frac{0}{1}
\qquad
0=\frac{1}{1}
$$
and $p\nmid 1$.
Now, suppose $x,y\in\mathbb{Z}_{(p)}$. By definition, we can write $x=a/b$ and $y=c/d$ with $p\nmid b$ and $p\nmid d$. Then
$$
x-y=\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}
\qquad
xy=\frac{a}{b}\,\frac{c}{d}=\frac{ac}{bd}
$$
which belong to $\mathbb{Z}_{(p)}$ because $p\nmid bd$.
No need to “reduce” fractions.

An important point. Suppose $p=3$ just by way of example. Then
$$
x=\frac{9}{6}\in\mathbb{Z}_{(p)}
$$
notwithstanding $3\mid 6$, because we can express
$$
x=\frac{3}{2}
$$
and $2\nmid 3$.
In the arguments above I just assume $x=a/b$ with $p\nmid b$, because this is possible by definition of $\mathbb{Z}_{(p)}$.
You're probably mystified by the definition you've been given. The added condition $\gcd(a,b)=1$ is completely useless and can happily be removed. Defining
$$
\mathbb Z_{(p)}=\left\{\frac{a}{b}: p\nmid b\right\}
$$
would be exactly the same, because if $p\nmid b$, then also $p\nmid b'$ and
$$
\frac{a}{b}=\frac{a'}{b'}
$$
where $d=\gcd(a,b)$, $a=da'$ and $b=db'$. This frees you from “simplifying fractions”.
A: Recall that $\mathbb Q$ can be seen as the localization of $\Bbb Z$ at the prime ideal $(0)$.
By universal property of localization, there exists one and only one ring homomorphism $\varphi:\Bbb Z_{(p)}\to\Bbb Q$ making the following diagram commutative:

This ring homomorphism is, in fact, injective, hence $\Bbb Z_{(p)}$ can be identified with a subring of $\Bbb Q$, namely the image of $\varphi$:
$$\operatorname{Im}\varphi=\left\{\frac ab\in\Bbb Q:p\nmid b\right\}$$
Note that your requirement $\gcd(a,b)=1$ is unnecessary.
