Let $A$ be a ring, $S$ a multiplicatively closed subset. Is it true that $\frac a1\in S^{-1}A$ is a non zero-divisor if and only if $a\in A$ is?
I would say yes: for, if $\frac a1$ is a zero-divisor exist $b\in A,s\in S$ such that $abs=0$, while if $ab=0$ for some $b\in A$ automatically $\frac a1\frac b1=0$. Is it correct? I'm a bit suspicious because I've never seen this result mentioned anywhere, and also because it seems too general. Am I missing something?
Set $A=\mathbb Z_6$ and $S=A\setminus 3\mathbb Z_6$. Then $\hat 2$ is a zerodivisor in $A$ and $\frac{\hat 2}{\hat1}$ is not a zerodivisor in $S^{-1}A$. (Actually it is invertible in $S^{-1}A$.)