Let $A$ be a ring and $S$ be a multiplicatively closed subset. Can $S$ contain $0$? If so what will happen if we do $S^{-1}A$?
A concrete and easy example coming to my mind is $A = \Bbb Z$, and $S = \{0,1\} $. What will $S^{-1}A$ look like?
2026-03-27 14:11:16.1774620676
Can a multiplicatively closed subset contain zero?
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Yes, multiplicative subsets are just subsets which contain $1$ and are closed under $\cdot$. It is not required that they don't contain $0$. If $0 \in S$, then $S^{-1} A = 0$ is an easy consequence of the definition resp. the universal property (in fact, the elements of $S$ become invertible in $S^{-1} A$, so that $0$ is invertible in $S^{-1} A$, i.e. the ring must be trivial). In fact, $S^{-1} A = 0$ if and only if $0 \in S$.