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So my question is what does it mean to be $0$ in $S^{-1} M$, where $S$ is a multi-closed subset of a ring $A$, $M$, lets assume to be a finitely generated $A$ module.

I was reading Atiyah Macdonalds book on commutative algebra. From what I gather, $S^{-1} M$ is a set the fractions of the form $\frac{m}{s}$. So I was wondering whats does the $0$ fraction, $``\frac{0}{s}"$ looks like. I tried going back to the definition of his construction, but cant really get a good idea.

Any help or insight is deeply appreciated.

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  • $\begingroup$ For some conceptual motivation see the links I posted on egreg's answer. $\endgroup$ – Gone Nov 2 '19 at 13:59
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In $S^{-1}M$ and element $m/s$ (with $s\in S$ and $m\in M$) is zero iff $tm=0$ for some $t\in S$, that is iff $S\cap\text{Ann}(m)$ is non-empty.

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The idea is to make equivalence classes from pairs $(m,s)$ with $m\in M$ and $s\in S$ and denote the equivalence class of $(m,s)$ by $m/s$. We need to ensure that multiplying numerator and denominator by the same element of $S$ doesn't change the equivalence class, so $(mt)/(st)$ should be the same as $m/s$.

But when should $m/s=n/t$? It should be so when $mt=ns$, but it turns out that this is insufficient to ensure an equivalence relation: this is just a sufficient condition to put $(m,s)$ and $(n,t)$ in the same equivalence class. On the other hand, we should have $$ \frac{m}{s}=\frac{mu}{su},\qquad \frac{n}{s}=\frac{nu}{tu} $$ for every $u\in S$. It turns out that defining $$ (m,s)\sim(n,t) \quad\text{if and only if}\quad mtu=nsu \text{ for some } u\in S $$ makes $\sim$ into an equivalence relation. Defining $$ \frac{m}{s}+\frac{n}{t}=\frac{mt+ns}{st},\qquad \frac{m}{s}\frac{r}{t}=\frac{mr}{st} $$ does not depend on the representatives of the equivalence classes and makes $S^{-1}M$ (the quotient set) into a module over $S^{-1}R$ (with the similar definitions for the ring structure).

Clearly, for every $s\in S$, we need $0/s$ to be the zero element in $S^{-1}M$. By the very definition, then $$ \frac{m}{s}=\frac{0}{t} $$ if and only if $mtu=0su=0$, for some $u\in S$. But then we see that it's equivalent to say that $mu=0$, for some $u\in S$. One direction has been shown, as $tu\in S$; for the other direction $$ \frac{m}{s}=\frac{mu}{su}=\frac{0}{su} $$ is the zero element.

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