Showing that equivalence in ring of fractions is well defined

Question

I'm reading this answer, in which, one of the step involves showing that:

$$s_3(r_1s_2-r_2s_1)=0,\quad s_1(r_2s_3-r_3s_2)=0\implies s_2(r_1s_3-r_3s_1)=0$$

I am utterly confused on how this implication was arrived at. How does having the "s" prefactor make it any more possible to show the implication then if it doesn't?

Answer 1

I think there is a problem of notation on the answer that you cited. You have to prove that if $(r_1,s_1)\sim (r_2,s_2)$ and $(r_2,s_2)\sim (r_3,s_3)$ then $(r_1,s_1)\sim (r_3,s_3)$.

$(r_1,s_1)\sim (r_2,s_2)$ implies there exists $x\in S$ such that $x(r_1s_2-s_1r_2)=0$;

$(r_2,s_2)\sim (r_3,s_3)$ implies there exists $y\in S$ such that $y(r_2s_3-s_2r_3)=0$.

Now consider $(r_1s_3-s_1r_3)$. By using the previous equations, you can observe that

$xys_2s_3(r_1s_3-s_1r_3)=xys_3^2s_1r_2-xys_1r_2s_3^2=0,$

where $xys_2s_3\in S$ since $S$ is a multiplicative set.

Therefore $(r_1,s_1)\sim (r_3,s_3)$ and you are done.

Answer 2

The implication is quite easy to see when you remove the subtractions and write it as $$\color{blue}{s_3 r_1 s_2} = \color{green}{s_3 r_2 s_1}, \quad \color{green}{s_1 r_2 s_3} = \color{red}{s_1 r_3 s_2} \implies \color{blue}{s_2 r_1 s_3} = \color{red}{s_2 r_3 s_1},$$ which is clear. But: this is not a complete proof of transitivity. After all, in the definition of the equivalence relation an arbitrary elements of $S$ appears, not one of the denominators.

If $(r_1,s_1) \sim (r_2,s_2) \sim (r_3,s_3)$, there are $u,v \in S$ with $$u \cdot s_2 \cdot r_1 = u \cdot s_1 \cdot r_2$$ $$v \cdot s_3 \cdot r_2 = v \cdot s_2 \cdot r_3$$ Let us assume for the moment that $u=v=1$. (The general case will be reduced to this later.) So we have $s_2 \cdot r_1 = s_1 \cdot r_2$ and $s_3 \cdot r_2 = s_2 \cdot r_3$. Then: $$s_2 \cdot (s_3 \cdot r_1) = s_3 \cdot (s_2 \cdot r_1) = s_3 \cdot (s_1 \cdot r_2) = s_1 \cdot (s_3 \cdot r_2) = s_1 \cdot (s_2 \cdot r_3) = s_2 \cdot (s_1 \cdot r_3)$$ Thus, $(r_1,s_1) \sim (r_3,s_3)$.

Now, for the general case, notice that we can write the given equations as $$s_2 \cdot r'_1 = s'_1 \cdot r_2$$ $$s'_3 \cdot r_2 = s_2 \cdot r'_3$$ with $r'_1 := u \cdot r_1$, $s'_1 := u \cdot s_1$, $r'_3 := v \cdot r_3$, $s'_3 := v \cdot s_3$. So the special case gives us some $t \in S$ with $$t \cdot (s'_3 \cdot r'_1) = t \cdot (s'_1 \cdot r'_3).$$ But then $$u \cdot v \cdot t \cdot (s_3 \cdot r_1) = u \cdot v \cdot t \cdot (s_1 \cdot r_3)$$ and we are done.

Of course, you could also verify directly that $$u \cdot v \cdot s_2 \cdot (s_3 \cdot r_1) = u \cdot v \cdot s_2 \cdot (s_1 \cdot r_3)$$ holds. But I think the approach above is more conceptual and explains how to find this equation.

Notice that we don't need any subtraction or addition here. In fact, the localization works perfectly fine for commutative monoids.