If a multiplicative set contains zero, then $S^{-1}A$ is trivial

Question

Let $A$ be a commutative ring with unity, and $S\in A$ a multiplicative subset. One then defines the localization of $A$ at $S$,$\,\,S^{-1}A, $ by means of an equivalence relation: $$(a,s)\sim (a',s')\, \Leftrightarrow \exists s_1\in S\,\, \text{such that}\,\, s_1(s'a-sa')=0, \,\, \text{where}\,\,a,a' \in A,\, \text{and}\,\, s,s'\in S.$$

The elements of $S^{-1}A$ are then the equivalence classes, which are usually denoted $a/s.$

In the Algebra book is then stated that if $0\in S, $ then $S^{-1}A$ has precisely one element, namely $0/1.$ (To avoid this trivial case one thus assumes $0\notin S).$ It is this statement that I dont understand.

Let $0\in S$. Suppose that there exists un equivalence class $a_1/s_1 \in S^{-1}A, \,\,a_1/s_1\neq 0 \,\,$such that $\exists \,s\in S,\, \, s(s_1a_2-s_2a_1)=0.$ One can distinguish two cases: $1.\,\, s=0, \,\,$ in which case the expression in parenthesis must not be zero. $2.\,\, s\neq0, $ in which case the expression in parenthesis is a zero divisor. I can't go further then this. Can somebody provide the additional steps or give me some hint ?

Many thanks.

Answer 1

$(a,s)\simeq (0,1)$ since $0(a1-0s)=0$.

Answer 2

Directly, without contradiction:

$$\forall\,a\in A\;,\;\;(a,0)\in S^{-1}A\implies\text{ if we take an arbitrary element}\;\;(a',s')\in S^{-1}A$$

then

$$(a',s')=(a',0)\;,\;\;\text{because}\;\;0\cdot(0\cdot a'-s'a')=0\;\ldots$$

Answer 3

It is a general fact that one can assume, without loss of generality, that $1\in S$. The ring homomorphism $\lambda_S\colon A\to S^{-1}A$ defined by $\lambda_S(a)=a/1$ maps elements of $S$ to invertible elements in $S^{-1}A$.

The zero element in $S^{-1}A$ is $0/1=\lambda_S(0)$. What are the rings in which the zero element is invertible?