Why rings of fraction is defined as that? Let $R$ be the commutative ring, S be the multiplicative set. As we know in $S^{-1}R$,  $a/b\sim c/d$ iff there is a $t \in S$ such that $adt=bct$. The question is that why we need $t$? Why not just defining $a/b\sim c/d$ iff $ad=bc$? Thank you!
 A: If you have an integral domain, then the $t$ doesn't matter because $adt=bct$ iff $(ad-bc)t=0$, and if $t\not=0$, $ad-bc=0$.  The problem comes up when rings have zero divisors.
Suppose $a\in S$ and $a$ is a zero divisor with $ab=0$.  Let $d\in S$ such that $bd\not=0$.  In this case, $b/c=0/a$ since $ab=0c$.  Also, $0/a=0/d$ since $0d=0a$.  The problem is that without multiplying by $t$, you would have a problem with $b/c=0/d$ since $bd\not=0c$, but, by multiplying both sides by $a$ you get $abd=0ac$, which is true.
A: Hint $\ $ If $\,sr=0\,$ in $\,R\,$ then in $\,S^{-1}R$ we have $\,r=0\,$ because $\,s\,$ is a unit so cancellable.  Hence, necessarily, all elements of that form will be killed in $\,S^{-1} R.\,$ This necessary condition turns out to be sufficient, as the proof you are reading shows.
Remark $\ $ A simpler and more natural way to construct localizations is to universally adjoin inverses of all $\rm\,s\in S\,$ to $\rm\,R.\,$ via $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations. In particular, the calculation of the kernel is quite simple and natural from that viewpoint - see the two-line proof I give here. 
