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Let $R$ be a commutative ring with $1$ as identity and let $S$ be a multiplicative closed subset of $R$ containing $1$. Show that $S^{-1}R=0 \text{ iff } 0$ belongs to $S$.

My approach: Now $S^{-1}R=0$ implies if the $1$ in this ring is $zero$ i.e. $(1,1)~(0,1)$. This occurs if $x.1=0$ for some $x$ in $S$. That is if $0$ belongs to $S$. Now conversely let $0$ belongs to $S$. Then help me to procced in this direction. Thanks for all your help in advance.

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    $\begingroup$ You've done the tricky part. If $0$ belongs to $S$, then every pair $(r, s) \in S^{-1}R$ is equivalent to $(0, 1$), since $0(r \cdot 1 - s \cdot 0) = 0$. $\endgroup$ Commented Oct 15, 2017 at 1:05

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Recall that $(x,s)\simeq (x',s')$ iff $s'x=sx'$. The other way $(1,1)\simeq (0,0)$ since $1.0=0.1$

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    $\begingroup$ I believe that your relation on pairs is not quite correct; two pairs $(r, s), (r', s') \in R \times S$ belong to the same equivalence class in $S^{-1}R$ if there exists $s'' \in S$ such that $s''(rs' - sr') = 0$. $\endgroup$ Commented Oct 15, 2017 at 1:11

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