Exercise 2.2

Proof. Let $R$ the set of such rationals. Simplifying these fractions, we obtain

$\text{∙}$ $1=1∕1\in R$.

$\text{∙}$ if $r,r^{\prime }\in R$, $r=p∕q,r^{\prime }=p^{\prime }∕q^{\prime }$, with $q\wedge p_1{p}_2\cdots p_t=1,q^{\prime }\wedge p_1{p}_2\cdots p_t=1$. then $q{q}^{\prime }\wedge p_1{p}_2\cdots p_t=1$, and $r-r^{\prime }=\frac{p{q}^{\prime }-q{p}^{\prime }}{q{q}^{\prime }}$, $r{r}^{\prime }=\frac{p{p}^{\prime }}{q{q}^{\prime }}$, so $r-r^{\prime },r{r}^{\prime }\in R$.

Thus $R$ is a subring of $\mathbb{Q}$.

If $r=a∕b\in R$ is an unit of $R$, then $b∕a\in R$, so ${\mathrm{ord}}_{p_i}a={\mathrm{ord}}_{p_i}(b),i=1,\dots ,t$. After simplification, $r=p∕q$, with $p\wedge p_1\cdots p_t=1,q\wedge p_1\cdots p_t=1$, and such rationals are all units.

Note that $p_i,1\le i\le t$, is a prime: if $p_i\mid \mathit{rs}$ in $R$, where $r=a∕b,s=c∕d\in R$, then there exists $u=e∕f\in R$ such that $\mathit{rs}=p_{i}u$, with $b,d,f$ relatively prime with $p_1,\dots ,p_t$. Then $\mathit{acf}=p_i\mathit{bde}$. As $p_i\wedge f=1$, $p_i$ divides $a$ or $c$ in $\mathbb{Z}$, so $p_i$ divides $r$ or $s$ in $R$.

If $r=a∕b\in R$, with $b\wedge p_1\cdots p_r=1$, $a=p_1^{k_1}\cdots p_t^{k_t}v,v\in \mathbb{Z},k_i\ge 0,i=1,\dots ,t$. So $r=u{p}_1^{k_1}\cdots p_t^{k_t}$, where $u=v∕b$ is an unit.

Let $\pi$ be any prime in $R$. As any element in $R$, $\pi =u{p}_1^{k_1}\cdots p_t^{k_t},k_i\ge 0,u=a∕b$ an unit. $u^{-1}\pi =p_1^{k_1}\cdots p_t^{k_t}$, so $\pi \mid p_1^{k_1}\cdots p_t^{k_t}$ (in $R$). As $\pi$ is a prime in $R$, $\pi \mid p_i$ for an index $i=1,\dots ,t$. Thus $p_i=\mathit{q\pi }$, where $q\in R$. Since $p_i$ is irreducible, $q$ is a unit, so $p_i$ and $\pi$ are associate.

Conclusion: the primes in $R$ are the associates of $p_1,\dots ,p_t$. □