Let $R$ be a ring and let $S$ be a submonoid of $R$. The localization $R_S$ consists of the equivalence classes (fractions) $\frac{a}{b}$ such that $a\in{R}$, $b\in{S}$ defined by the relation $\sim$ in $R_S$ such that $(a,b)\sim(c,d)$ if and only if there exists an element $s\in{S}$ such that $s(ad-bc)=0$.
It is proved that $R_S$ is zero if and only if $S$ contains zero.
I wonder to know if $a$ in $S$ is a zero-divisor what is the quotient $\frac{a}{b}$ equal to?
I think that since $a$ is a zero-divisor in $S$ then there exists $s\in{S}$ such that $as=0$. Hence $\frac{a}{b}=\frac{as}{bs}=0$.
Moreover, what can we say about the quotient $\frac{0}{0}$? If $\frac{0}{0}=\frac{a}{b}$ then there exists $s\in{S}$ such that $s(0b-0a)=s0=0$. Does it follow that $\frac{0}{0}$ gives all $R_S$?
Would you help me, please? Thank you in advance.
If $0\in S$, then $R_S$ is the zero ring, yes, because $\frac00$ is equivalent to any other fraction, as you have seen, so the ring necessarily contains only one element.
The map $R\to R_S$ given by $x\mapsto \frac x1$ is injective iff $S$ contains no zero divisors. If $a\in S$ is a zero divisor, and $b\in R$ is such that $ab=0$, then $\frac b1=\frac01$. But if $b\notin S$, $R_S$ is not necessarily the zero ring.
Your $\frac ab$ doesn't make sense because we don't know that $b\in S$.
I think I covered most of it. Let me know if there is something I missed.
Finally, the $s(ad-bc)=0$ requirement I like to think of as "you can expand both fractions so that numerators are equal and denominators are equal". It's a lot more intuitive in my opinion. Just remember that you're only allowed to use elements of $S$ when expanding.