equivalence relation of a localization of a ring
1. Proposition
Let \(R\) be a commutative ring, \(S \subseteq R\) a multiplicatively closed set. Then the relation
\begin{align*} (a,s_1) \sim (b,s_2) :\Leftrightarrow \exists s : s(a s_2 - b s_1) = 0 \end{align*}is an equivalence relation
2. Proof
2.1. reflexive
for \(1 \in S\), we conclude
\begin{align*} 1 \cdot (a s - a s) =& 1 \cdot 0 \\ =& 0 \end{align*}2.2. symmetric
follows from Multiplication with \(-1\):
\begin{align*} sas_2 - sbs_1 =& 0 && \vert \cdot -1 \\ sbs_1 - sas_2 =& 0 \end{align*}2.3. transitive
It holds true:
\begin{align*} s(as_2 - bs_1) =& 0 \\ s'(bs_3 - cs_2) =& 0 \\ \end{align*}there we conclude
\begin{align*} s(as_2 - bs_1) \cdot s's_3 =& 0 \\ ss' (as_2s_3 - bs_1s_3) =& 0 \\ s'(bs_3 - cs_2) \cdot ss_1 =& 0 \\ ss' (bs_1s_3 - cs_1s_2) =& 0 \end{align*}and adding them together results in
\begin{align*} ss'( as_2 s_3 - bs_1s_3 + bs_1s_3 - cs_1s_2) =& 0 \\ ss'( as_2 s_3 - cs_1s_2) =& 0 \\ ss's_2( a s_3 - cs_1) =& 0 \end{align*}