# On the natural homomorphism $\nu:R\longrightarrow S^{-1}R$

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.

## 1 Answer

Note that, if $$0\in S$$, then $$S^{-1}R$$ is a zero ring where everything trivially is a unit.

• Thank you for your answer. Indeed, $S^{-1}R=\{0_R\} \iff 1_{S^{-1}R}=0_{S^{-1}R} \iff \frac{1_R}{1_R}=\frac{0_R}{1_R} \iff \exists s\in S:\ s(1_R0_R-0_R1_R)=0_R \iff \exists s\in S:\ s=0_R \iff 0_R\in S$. Apr 17, 2019 at 1:19
• @Chris exactly. You might want to accept the answer if it did solve your question. Apr 17, 2019 at 15:36
• Of course. Thank you very much again :) Apr 17, 2019 at 22:38