In $\textit{A Classical Introduction to Modern Number Theory}$, the authors define $\mathbb{Z}_p$ as a set where $p$ is a prime number and $a,b$ form the rational number $a/b$ such that $p\nmid b$. After describing this set, the authors mention &quot;One easily checks using the remark following Corollary 1 to Proposition 1.1.1 that $\mathbb{Z}_p$ is a ring. That remark is &quot;If $p$ is a prime and $p\nmid b$ and $p\nmid c$, then $p\nmid bc$. I'm puzzled by the semantics of suggesting the use of the remark to the corollary. I understand how this applies: if I have elements $a/b,c/d$, where $p\nmid b$ and $p\nmid d$, then $a/b+c/d=(ad+bc)/bd\Rightarrow p\nmid bd$. A similar calculation is done with multiplication. This definitely shows $\mathbb{Z}_p$ is closed. For the sake of context, the next paragraph demonstrates $a/b$ is a unit implies $p\nmid a$ (and the converse) and all elements fo $\mathbb{Z}_p$ can be expressed as a power of $p$ times a unit. The conclusion is that all primes of $\mathbb{Z}_p$ are associate. (I'm totally fine with this part.) Does the phrasing indicate that I use this corollary remark to justify the work showing why the other ring properties are true, or is it an end-all statement saying &quot;because this is satisfied, it's a ring&quot;?
Let $p$ be a prime number. Let $R$ be the subset of all rational numbers $m / n$ such that $n \neq 0$ and $n$ is not divisible by $p$. Show that $R$ is a ring. My Proof-trying. $R1$. Associativity for under addition is clear. $R2$. $\dfrac {m} {n}+0=0+\dfrac {m} {n}$, so there exist the identitiy element for under addition. $R3$. $\dfrac {m} {n} + \dfrac {-m} {n}=\dfrac {-m} {n} +  \dfrac {m} {n}$, so there exist inverse element. $R4$. $\dfrac {m} {n} + \dfrac {m'} {n'}=\dfrac {m'} {n'}+\dfrac {m} {n}$, so commutativity. So $R$ is an additive abelian group. $R5$. $\dfrac {m} {n} ( \dfrac {r} {s} + \dfrac {t} {v})=\dfrac {mr} {ns} + \dfrac {mt} {nv}$ and $( \dfrac {r} {s} + \dfrac {t} {v})\dfrac {m} {n}=\dfrac {rm} {sn}+\dfrac {r} {s}+\dfrac {tm} {vn}$ $R6.$ $R$ is associativity for under multiplication, clear. $R7.$ $1 \dfrac {m} {n}=\dfrac {m} {n} 1 = \dfrac {m} {n}$, so there exist identity element for under multiplication. Can you check my answer? If false, can you write true answer?
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