Let $A$ be a commutative ring with $1$ and $S$ be a multiplicatively closed subset of $A$. Can $S$ contain zero-divisors of $A$ and still we get $S^{-1}A \ne \{0\}$? My confusion is in the following point: For all $s \in S \ ; \ \frac s1$ is a unit in $S^{-1}A$. Also if $a \in A$ is a zero divisor then $\frac a1$ is a zero divisor in $S^{-1}A$. So if $S$ contains a zero divisor $a$ then $\frac a1$ is both a unit and a zero divisor which is a contradiction. Where am I going wrong?
Let $R$ be a commutative ring with identity, and $S$ a multiplicative subset of $R$ not containing $0$. Which step in my reasoning is wrong?   Every element of $S$ is a unit in $S^{-1}R$. Every element of $S$ that's a zero divisor in $R$ is a zero divisor in $S^{-1}R$. But zero divisors can't be units. Therefore it's not possible to localize $R$ at $S$.
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