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 1


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

  • $\begingroup$ Thank you for your answer. Indeed, $S^{-1}R=\{0_R\} \iff 1_{S^{-1}R}=0_{S^{-1}R} \iff \frac{1_R}{1_R}=\frac{0_R}{1_R} \iff \exists s\in S:\ s(1_R0_R-0_R1_R)=0_R \iff \exists s\in S:\ s=0_R \iff 0_R\in S$. $\endgroup$
    – Chris
    Apr 17, 2019 at 1:19
  • 2
    $\begingroup$ @Chris exactly. You might want to accept the answer if it did solve your question. $\endgroup$
    – Mihail
    Apr 17, 2019 at 15:36
  • $\begingroup$ Of course. Thank you very much again :) $\endgroup$
    – Chris
    Apr 17, 2019 at 22:38

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