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Let $S\subset R$ be a multiplicative set and given a ring homomorphism $\phi:R\rightarrow S^{-1}R$ by $r\rightarrow \frac{r}{1}$ to be the localization map. We define $\frac{r}{s}=\frac{r'}{s'}$ with $r,r'\in R, s, s'\in S$ only if $\exists k\in S$ such that $k(rs'-r's)=0$

By definition we have $\ker(\phi)=\{a\in R: \exists s\in S\text{ such that } sa=0\}$. If we add some restrictions on $R$ and $S$ such that $R$ is a domain and $S$ contains no zero-divisors or zero, then $\ker(\phi)=0$.Or $0\in S$ then $\ker(\phi)=R$.

However, when in general, I can't think of a general form of such kernel (in order to "determine" it). (the only thing I can say is $\ker(\phi)$ is an ideal) Thanks in advance.

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There is not much you can do in general, it really depends on the particular case. Consider, for instance, $R= \mathbb{Z} / 8 \mathbb{Z}$, with $S:= \{f^n \mid n \ge 0\}$ for some $f \in R$. You can then have a look at some possibilities:

  • $f=2, S= \{0,1,2,4\}, \ker(\phi)=R$.
  • $f=3, S= \{1,3\}, \ker(\phi)=\{0\}$.
  • $f=4, S= \{0,1,4\}, \ker(\phi)=R$.
  • $f=6, S= \{0,1,4,6\}, \ker(\phi)=R$.

Every element is either a nilpotent or invertible, hence $\ker(\phi)$ can only be $R$ or $\{0\}$. But for instance, if $R= \mathbb{Z} / 6 \mathbb{Z}$, the element $2$ is a zero-divisor but is not nilpotent, and you get

  • $f=2, S= \{1,2,4\}, \ker(\phi)=\{0,3\}$.
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I think this answers what you are asking: one can show that $S$ contains no zero divisors if and only if $a\mapsto a/1$ is injective.

Indeed, say $S\subset R$ is a multiplicative set, and $S$ contains no zero divisors. Say that $f(a)=f(b)=a/1=b/1$ then for some $s\in S$ there is $$s(a-b)=0,$$ where $s$ is not a zero divisor, and hence $(a-b)=0\implies a=b$.

Conversely, let $f(a)=f(b)\implies a=b$. If $S$ has a left zero divisor $s$ such that $ss'=0$, then $f(s')=s'/1$, but $s'/1=0/1$ since $s(s'-0)=0$. By injectivity, $s'=0$.

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