Let $R$ be a UFD, and let $S$ be a multiplicative subset of $R$ containing the unity of $R$ then $R_S$ is also UFD.
There are several answer on M.SE like here, here. But none of them resolve the problem I am facing with the solution provided by the book.
solution: First they show if $a\in R$ is irreducible in $R$, then $a/1$ is irreducible in $R_S$. I understand this part. Second, they claim if $a/s\in R_S$ is irreducible then it is prime.
First we show that if $a \in R$ is irreducible in $R$, then $a / 1$ is irreducible in $R_s$. For, let $a / 1=\left(b / s_1\right)\left(c / s_2\right)$, where $b / s_1$ and $c / s_2$ are nonunits in $\boldsymbol{R}_s$. Then $a s_1 s_2=b c$. Because $a$ is irreducible and, hence, prime, it follows that $a \mid b$ or $a \mid c$. For definiteness let $a \mid b$; so $a b_1=b$ for some $b_1 \in R$. Then $a / 1=\left(b / s_1\right)\left(c / s_2\right) \Rightarrow 1=\left(b_1 / s_1\right)\left(c / s_2\right)$, a contradiction, because $c / s_2$ is not a unit. Now let $a / s \in R_s$. Write $a=a_1 a_2 \cdots a_r$ as a product of irreducible elements in $R$. Then $a / s=(1 / s)\left(a_1 / 1\right)\left(a_2 / 1\right) \cdots\left(a_r / 1\right)$ is a product of irreducible elements in $R_s$. This proves one of the conditions for $\boldsymbol{R}_{\boldsymbol{s}}$ to be a UFD.
We proceed to prove the second condition, that if $a / s \in \boldsymbol{R}_{\boldsymbol{s}}$ is irreducible then it is prime. Now $a / s$ irreducible implies $a$ is irreducible in $R$, so $a$ is prime in $R$. We prove $a / 1$ is prime in $R_s$. Let $a / 1$ divide $\left(b / s_1\right)\left(c / s_2\right)$. Then $(a / 1)\left(d / s_3\right)=\left(b / s_1\right)\left(c / s_2\right)$ for some $d / s_3 \in R_s$. This implies $a d s_1 s_2=b c s_3$, so $a \mid b c s_3$. But then $a$ divides $b, c$, or $s_3$. If $a \mid s_3$, it follows that $a / s \mid 1$, so $a / s$ is a unit in $R_s$, a contradiction. Thus, $a \mid b$ or $a \mid c$. This implies $a / s$ divides $b / s_1$ or $c / s_2$ as desired.
I couldn't understand the following lines, "If $a \mid s_3$, it follows that $a / s \mid 1$, so $a / s$ is a unit in $R_s$. Thus, $a \mid b$ or $a \mid c$. This implies $a / s$ divides $b / s_1$ or $c / s_2$ as desired." Like isn't we are working with $a/1$ then why $a/s$ was came in?
