This question seems quite trivial, but I don't seem to be able to avoid it.
Let $R$ be a commutative ring with unity. Then trivially $R$ is a multiplicative subset of itself, so we can localize, obtaining $R^{-1} R$. This seems like a silly thing to do, but it came up when I was trying to compute something, so it's not just a silly question for the sake of a silly question.
My question is, does $R^{-1} R$ make sense? I can't see any formal reason why it wouldn't. However, this means we have a multiplicative inverse for zero, which my instincts say is problematic.
Suppose $S$ contains $0$. Then the natural map $R \to S^{-1}R$ sends $0$ to a unit in $S^{-1}R$. Ring homomorphisms always send $0$ to $0$, so $S^{-1}R$ must be a ring where $0$ is a unit. The only such ring is the zero ring, i.e. the ring with one element.