What does it mean to be $0$ in $S^{-1} M$

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.

• For some conceptual motivation see the links I posted on egreg's answer. – Gone Nov 2 '19 at 13:59

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.
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.