Intuition for fractions of the localization of a non integral domain

Question

Consider the localization of $R = \mathbb Z/10\mathbb Z$ at $6$. We use the multiplicative subset $S = \{ 1, 6 \}$. we build fractions of the form $r/s$ with $r\in R, s \in S$, 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 $\mathbb Z/n\mathbb Z$ 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

Answer 1

Intuition behind localization is better understood with other kind of rings. If you think of $R[x,y]$ and perform the same constructions you would get better ideas. The key idea is that you are selecting which polinomial can get a multiplicative inverse (unlike it happens if you take the whole field of fractions). You can develop a lot of intuition of commutative algebra with the ring $R[x,y]$, thinking of "functions-polynomials" defined on the plane. Then, you can think of ANY ring (for example, your $\mathbb{Z}/10\mathbb{Z}$) as functions defined in an special geometric object called Scheme, in this case, the spectrum of $\mathbb{Z}/10\mathbb{Z}$.

I recommend you the book: The geometry of Schemes, by Eisenbud and Harris.