If $a/s$ is irreducible in $R_{S}$, then $a$ is irreducible in $R$

Question

Let $R$ be a UFD, and let $S$ be a multiplicative subset of $R$ containing the unity of $R$. Show that if $a/s$ is irreducible in $R_{S}$, then $a$ is irreducible in $R$. (Bhattacharya, Basic Abstract Algebra, Example 1.7, page 227.)

I showed that if $a$ is irreducible in $R$, then $a/1$ is irreducible. (I don't know that it is useful.)

What I did :

Let $a=xy$. Since $a/s=xy/s=(x/1)(y/s)$ is irreducible, $x/1$ or $y/s$ is irreducible. If $x/1$ is unit, we done. Let $y/s$ be unit. It means that $y/1=s_{1}/s_{2}$, $ys_{2}=s_{1}$. We should show that $y$ is a unit.

Do $ys_{2}=s_{1}$ imply that $y$ is a unit?

Answer 1

Your claim is wrong: take $R=\mathbb Z$ and $S=\{1,2,\dots, 2^n,\dots\}$, $a=6$ and $s=2$.