Show that $\mathbb Z_{p\mathbb Z}$ is a subring of $\mathbb Q$ I have to show that $\mathbb Z_{p\mathbb Z} = \{\frac{a}{b} \ |\ a,b \in \mathbb Z, b \notin p \mathbb Z  \}$ is a subring of $\mathbb Q$.
What I know:

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*$\mathbb Z_{p\mathbb Z}$ is a subring of $\mathbb Q$ if:


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*$\mathbb Z_{p\mathbb Z} \subseteq \Bbb Q$

*$\mathbb Z_{p\mathbb Z}$ is a ring, so $(\mathbb Z_{p\mathbb Z}, +)$ is a subgroup and $(\mathbb Z_{p\mathbb Z}, \cdot)$ is a monoid


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*$p \Bbb Z = \{p \cdot z \ |\ p$ is some prime, $z \in \Bbb Z \}$ a subgroup of $\Bbb Z$
First I wan't to show the closure regarding $(\mathbb Z_{p\mathbb Z}, +)$.
Let $x, y \in \Bbb Z_{p\mathbb Z}$ such that $x = \frac{a}{b}$ and $y = \frac{c}{d}$. Then $x + y = \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$. Now I have to show that this is also in $\Bbb Z_{p\mathbb Z}$.
My reasoning: because $b,d \in \Bbb Z$ and $b,d \notin p\Bbb Z$, I concluded that they have to be of the form: $b = kz + r$ and $d = k'z' + r'$, where $k, k' \in \Bbb N$, $z \in \Bbb Z_{p\mathbb Z}$ and $0 \le r \lt k$, $0 \le r' \lt k'$. So, $b$ and $d$ cannot be divided by any $z \in p\Bbb Z$ without a remainder.
Then $b \cdot d = (kz + r) \cdot (k'z' + r') = (k k' z'+ kr')z + (k'z' + r')r = ez + r''$, so $b \cdot d \notin p\Bbb Z$.
Is my reasoning regarding $b,d$ correct? If not, a little help would be appreciated
 A: Your reasoning for $bd$ is neither complete nor correct.
You seem to be dividing by $k$ and $k'$ (note that conditions you are putting on the remainder); but you never tell us what $k$ and $k'$ are. That makes that division useless to detect whether something is a multiple of $p$.
Instead, you should be dividing by $p$, and writing $b=kp+r$ with $0\lt r\lt p$ (note the strict inequality, since you are assuming that $b$ is not a multiple of $p$), and $d=k'p+r'$, with $0\lt r'\lt p$. Then your calculations will end up with $bd = (kk'p + kr'+k'r)p + rr'$, and now you must show that $rr'$ is not a multiple of $p$... which just reduces to the original issue: showing that the product of two numbers, neither of which is a multiple of $p$, is also not a multiple of $p$.
Note that this claim is false when $p$ is not a prime (for example, $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so if you were trying to argue that if neither $b$ nor $d$ are multiples of $6$ then $bd$ is not multiple of $6$, the argument woudl fail). So you must somehow use the fact that $p$ is a prime number, which is something you have failed to do.
