TITLE: Proof of transitivity of the equivalence relation on the localization of a ring
QUESTION [1 upvotes]: I'm having a bit of trouble proving that the relation defined with localization is in fact a transitive relation. Let $R$ be a commutative ring and let $S\subset R$ be closed under multiplication and suppose that $1\in S$. We define the relation $\sim$ on $R\times S$ by $(a,b)\sim (x,y)\iff (ay-bx)s=0$ for some $s\in S$. Can anyone help me prove that this relation is transitive?
What I've tried so far:
Suppose we have $(a,b),(c,d),(e,f)\in R\times S$, such that $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$. Then we have $(ad-bc)s_1=0$ and $(cf-de)s_2=0$ for some $s_1,s_2\in S$. We want to find some $s\in S$ such that $(af-be)s=0$. It's probably a good idea to use the first two equations, to be able to factor them out together I multiplied both sides of the first equation by $s_2$ and the second equation by $s_1$. Since they are both still equal to $0$, we get the equation: $$(ad-bc)s_1s_2=(cf-de)s_1s_2.$$ This leads to the equation $$(ad+de-bc-cf)s_1s_2=0.$$ From this equation, I tried to get to $(af-be)s=0$, but I couldn't find a way to get rid of any elements $c$ and $d$. My other attempts were mostly guessing, so I couldn't really figure anything out. What am I missing?

REPLY [2 votes]: Try multiplying $(ad - bc)s_1 = 0$ by $fs_2$ and $(cf - de)s_2 = 0$ by $bs_1$ and see what happens.
{"set_name": "stack_exchange", "score": 1, "question_id": 3783278}