Question: Let $R$ be a commutative ring and $S$ be a closed multiplication in $R$. Let $f \colon R \rightarrow S^{-1}R$ be the ring homomorphism defined by $f(x) = x/1$ for all $x \in R$. Prove that $f$ is injective iff $S$ is a subset of the set all elements which are not zero-divisors of $R$.
I didn't understand what the question say? Could you give me some explain to solve this problem. Thank all! Sorry for my poor English.
My attempt: Suppose $S$ is a subset of the set all elements which are not zero-divisors of $R$. So let $a,b \in R$. $\frac{a}{1} = \frac{b}{1} \ \Leftrightarrow (a-b)u = 0$ for some $u\in S$. Hence, $a-b = 0 \Rightarrow a=b$. So $f$ injective.
Let $f$ be injective and $s\in S$. If $as=0$, then $f(a)=\frac{a}{1}=\frac{sa}{s1}=\frac{0}{s}=0_{S^{-1}R}$. Now since $f$ is injective $a=0_R$, that is, $S$ is a subset of the set all elements which are not zero-divisors of $R$. Conversly, Let $S$ be a subset of the set all elements which are not zero-divisors of $R$. If $f(a)=0_{S^{-1}R}$, then $\frac{a}{1}=0_{S^{-1}R}$. Hence, there exists $s\in S$ such that $sa=0_R$ and so $a=0_R$ by our assumption, that if $f$ is injective.