Intuition for fractions of the localization of a non integral domain

Consider the localization of $R = mathbb Z/10mathbb Z$ at $6$. We use the multiplicative subset $S = { 1, 6 }$. we build fractions of the form $r/s$ with $rin R, s in S$, with the equivalence relation: $r/s sim r’/s’ $ iff there exists an $alpha in S$ such that $alpha (rs’ – r’s) = 0$.

Crunching the equivalence relations, we find:

ring: Ring of integers modulo 10
localizing at s0=6 | S = [1, 6]
===Equivalence classes===
((0, 1), set([(0, 1), (5, 6), (5, 1), (0, 6)]))
((8, 1), set([(8, 1), (8, 6), (3, 1), (3, 6)]))
((2, 1), set([(7, 6), (2, 6), (2, 1), (7, 1)]))
((4, 1), set([(4, 1), (9, 6), (4, 6), (9, 1)]))
((6, 1), set([(6, 1), (1, 6), (1, 1), (6, 6)]))
****0/1****
    -  0/ 1 (~  1) [mod 10]  |  ( 0* 1 -  1* 0) *  1 = ( 0 -  0) * 1 =  0 *  1 = 0
    -  5/ 6 (~  6) [mod 10]  |  ( 0* 6 -  1* 5) *  6 = ( 0 -  5) * 6 =  5 *  6 = 0
    -  5/ 1 (~  6) [mod 10]  |  ( 0* 1 -  1* 5) *  6 = ( 0 -  5) * 6 =  5 *  6 = 0
    -  0/ 6 (~  1) [mod 10]  |  ( 0* 6 -  1* 0) *  1 = ( 0 -  0) * 1 =  0 *  1 = 0
****8/1****
    -  8/ 1 (~  1) [mod 10]  |  ( 8* 1 -  1* 8) *  1 = ( 8 -  8) * 1 =  0 *  1 = 0
    -  8/ 6 (~  1) [mod 10]  |  ( 8* 6 -  1* 8) *  1 = ( 8 -  8) * 1 =  0 *  1 = 0
    -  3/ 1 (~  6) [mod 10]  |  ( 8* 1 -  1* 3) *  6 = ( 8 -  3) * 6 =  5 *  6 = 0
    -  3/ 6 (~  6) [mod 10]  |  ( 8* 6 -  1* 3) *  6 = ( 8 -  3) * 6 =  5 *  6 = 0
****2/1****
    -  7/ 6 (~  6) [mod 10]  |  ( 2* 6 -  1* 7) *  6 = ( 2 -  7) * 6 =  5 *  6 = 0
    -  2/ 6 (~  1) [mod 10]  |  ( 2* 6 -  1* 2) *  1 = ( 2 -  2) * 1 =  0 *  1 = 0
    -  2/ 1 (~  1) [mod 10]  |  ( 2* 1 -  1* 2) *  1 = ( 2 -  2) * 1 =  0 *  1 = 0
    -  7/ 1 (~  6) [mod 10]  |  ( 2* 1 -  1* 7) *  6 = ( 2 -  7) * 6 =  5 *  6 = 0
****4/1****
    -  4/ 1 (~  1) [mod 10]  |  ( 4* 1 -  1* 4) *  1 = ( 4 -  4) * 1 =  0 *  1 = 0
    -  9/ 6 (~  6) [mod 10]  |  ( 4* 6 -  1* 9) *  6 = ( 4 -  9) * 6 =  5 *  6 = 0
    -  4/ 6 (~  1) [mod 10]  |  ( 4* 6 -  1* 4) *  1 = ( 4 -  4) * 1 =  0 *  1 = 0
    -  9/ 1 (~  6) [mod 10]  |  ( 4* 1 -  1* 9) *  6 = ( 4 -  9) * 6 =  5 *  6 = 0
****6/1****
    -  6/ 1 (~  1) [mod 10]  |  ( 6* 1 -  1* 6) *  1 = ( 6 -  6) * 1 =  0 *  1 = 0
    -  1/ 6 (~  6) [mod 10]  |  ( 6* 6 -  1* 1) *  6 = ( 6 -  1) * 6 =  5 *  6 = 0
    -  1/ 1 (~  6) [mod 10]  |  ( 6* 1 -  1* 1) *  6 = ( 6 -  1) * 6 =  5 *  6 = 0
    -  6/ 6 (~  1) [mod 10]  |  ( 6* 6 -  1* 6) *  1 = ( 6 -  6) * 1 =  0 *  1 = 0

1. I find it exremely counter-intuitive that $0/1 sim 5/6 ~(operatorname{mod} 10)$. How should I think of the fraction $5/6$ becoming equal to $0$ when localizing at $6$, without trying to find an element $alpha in S$ such that $(6*0 + 5*1) alpha = 0$. How can I "instantly see" which fractions are equivalent?

2. Similarly, I find it really counter intuitive that somehow, $3/1 = 8/1$ on performing this localization. What is actually "going on" here?

3. In general, giving $a/b$ and $c/d$ in $mathbb Z/nmathbb Z$ localized at $s_0$, how do we know if $a/b sim c/d$ by "just looking", without having to cross multiply and find an element $alpha in S$ that kills of the cross ratio?

4. Even more generally, how do I gain an intuition for the general case in a ring $R$ with multiplicative subset $S$?

• SAGE code to generate the above table