Let $R$ be a commutative ring with $1_R$ and $S$ an multiplicatively closed set. We define the natural homomorphism \begin{align*} \nu:R &\longrightarrow S^{-1}R, \\ a&\longmapsto \nu(a):=\frac{a}{1_R}. \end{align*} where $ S^{-1}R$ is the localization of $R$ on $S$.
In a proposition, I found that $\forall s\in S\subseteq R\implies \nu(s)\in U( S^{-1}R)$, because $$\nu(s)\cdot \frac{1_R}{s}=\frac{s}{1_R}\cdot \frac{1_R}{s}=\frac{s}{s}=1_{ S^{-1}R} $$ But should it be $\forall s\in S^*:=S\backslash \{0_R\}$?
I mean that if we take $s=0_R$, then $\nu(0_R)=\frac{0_R}{1_R}=0_{ S^{-1}R}$. So, it is not unit.
Thank you.
Note that, if $0\in S$, then $S^{-1}R$ is a zero ring where everything trivially is a unit.