Fractions: cross-multiply equivalence is least with $\frac{a}b\approx \frac{ac}{bc}$

Question

Background

The following is a paraphrasing of E. B. Vinberg's A Course of Algebra, page 129, discussing the construction of quotient fields.

Let $A$ be an integral domain. Consider the set of pairs $(a,b)$ where $a,b \in A, b\neq0$. Define an equivalence relation [on this set of pairs] by the following rule: $$(a_1,b_1) \sim (a_2,b_2) \overset{\text{def}}{\Longleftrightarrow} a_1b_2=a_2b_1.$$

The above definition is reasonable. I also understand this statement, $(3.34)$,

The above definition implies that $$(a,b) \sim (ac,bc) \tag{3.34}$$ for any $c\neq0$.

Now next comes the part that is unclear to me. Especially the equality marked by $(*)$; the asterisk is not present in the book.

On the other hand, any equivalence $(a_1,b_1) \sim (a_2,b_2)$ is a corollary of equivalences of the form $(3.34)$, as the following chain of equivalences demostrates: $$(a_1,b_1) \sim (a_1b_2,b_1b_2) \overset{(*)}{=} (a_2b_1,b_1b_2) \sim (a_2,b_2).$$ (We first multiplied both entries in $(a_1,b_1)$ by $b_2$ and then cancelled $b_1$ in both entries of the resulting pair.)

The emphasis is also added by me.

Question

As I understand the passage, we are trying to show that we can prove any equivalence $(a_1,b_1) \sim (a_2,b_2)$ by using some other equivalence of the type $(3.34)$. It is clear to me why $(a_1,b_1) \sim (a_1b_2,b_1b_2)$ and $(a_2b_1,b_1b_2) \sim (a_2,b_2)$ hold, as they directly use relationship $(3.34)$ and multiplication commutativity in $A$.

However, I fail to understand the equality denoted by $(*)$: $(a_1b_2,b_1b_2) \overset{(*)}{=} (a_2b_1,b_1b_2)$. It is supposed to be an equality of ordered pairs. That is, $(*)$ is true by definition iff $a_1b_2 = a_2b_1 \land b_1b_2 = b_1b_2$. The latter part of the conjunction is clear but the first half $a_1b_2 = a_2b_1$ is equivalent to our definition of $(a_1,b_1) \sim (a_2,b_2)$. Yet this is what we wish to show (cf. corollary), and hence one cannot assume $(a_1,b_1) \sim (a_2,b_2)$ is true when it is exactly what we are trying to demonstrate.

Q: How does $(a_1b_2,b_1b_2) \overset{(*)}{=} (a_2b_1,b_1b_2)$ when $(a_1,b_1) \sim (a_2,b_2)$ is not yet known? Wherein does my misunderstanding lie?

Answer 1

Intuitively, the point is to show that the cross-multiplication rule for fraction equivalence $(\sim)$ is the smallest equivalence relation $(\approx)$ equating $\,a/b\,$ and $\,ad/(bd)\,$ for all $\,d\neq 0$, i.e. satisfying $(3.34)$.

The unclear part shows $(a,b)\sim (c,d) \Rightarrow \, (a,b)\approx (c,d),\,$ i.e. any $\rm\color{#c00}{equiv}$. relation $\approx$ satisfying $(3.34)$ includes all relations in $\,\sim.\,$ So, being an equiv. relation satisfying $(3.34),\,$ $\sim\,$ is the smallest such.

Below we give a very detailed presentation of the argument. Recall that the relation $(3.34)$ is $$(a,b)\, \approx\, (ad,bd)\ \ \ {\rm for\ any}\,\ d\neq 0\qquad\qquad \tag{3.34}$$

To show that $\,\sim\,$ is the smallest $\rm\color{#c00}{equivalence}$ relation satisfying $(3.34)$ it suffices to show that any such equivalence relation $\,\approx\,$ includes all elements of $\,\sim,\,$ i.e. if $\,(f,g)\,$ is in $\,\sim\,$ then $\,(f,g)\,$ is in $\,\approx,\,$ i.e. $\,f\sim g\,\Rightarrow\, f\approx g.\ $ The Lemma below proves this. The proof outline, in common notation, is

$$\dfrac{a}b\sim \dfrac{c}d\,\Rightarrow\,\color{#0a0}{ad = cb}\,\Rightarrow\, \dfrac{a}{b}\,\approx\, \dfrac{\color{#0a0}{a\,d}}{b\,d}\,\approx\,\dfrac{\color{#0a0}{c\,b}}{d\,b}\,\approx\, \dfrac{c}d\qquad\qquad $$

Lemma $\,\ (a,b)\,\sim\, (c,d)\, \Rightarrow \, (a,b)\,\approx\, (c,d)\ $ for any $\rm\color{#c00}{equivalence}$ relation $\,\approx\,$ satisfying $(3.34)$

$\!\begin{align}{\bf Proof}\:\ \ \ \ (a,b)\, &\approx\, (\color{#0a0}{ad},bd)\ \ \ {\rm by}\ \approx\ {\rm satisfies}\ (3.34) \ {\rm and}\ \, d\neq 0 \\[.2em] &\approx\, (\color{#0a0}{cb},\,db)\ \ \ {\rm by}\ \ \color{#0a0}{ad=cb}\ \ {\rm by\ definition\ of}\,\ (a,b)\sim (c,d)\ \ {\rm and}\ \approx\ \color{#c00}{\rm reflexive}\\[.2em] &\approx\ (c,d) \ \ \ \ \ \ \ {\rm by}\ \approx\ {\rm satisfies}\ (3.34)\ {\rm and}\ \approx\, {\rm\color{#c00}{symmetric}\ and}\,\ b\neq 0\\[.2em] \Rightarrow\ \ (a,b)\, &\approx\, (c,d)\ \ \ \ \ \ \ \, {\rm by}\ \approx\ \rm \color{#c00}{transitive} \end{align}$

Note that above we (implicitly) used commutativity of multiplication: $\, bd = db$.

Answer 2

The full statement of what is being proved here is that the equivalence relation [on this set of pairs] that is described in your first box is "generated by" the relation in your second box (3.34), or to say this more formally, the first is the reflexive-symmetric-transitive closure of the second.

The logic of the argument goes like this:

• Given $a_1,b_1,a_2,b_2 \in A$, if $b_1 \ne 0$, if $b_2 \ne 0$, and if $a_1 b_2 = b_1 a_2$, then the ordered pair $\bigl((a_1,b_1),(a_2,b_2)\bigr)$ is an element of the reflexive-symmetric-transitive closure of the relation (3.34). Or to put this more informally, the relation $(a_1,b_1) \sim (a_2,b_2)$ may be deduced by a finite chain of relations in the reflexive-symmetric-transitive closure of the relation (3.34).

Notice: We are not trying to show that $a_1 b_2 = b_1 a_2$. Instead we are assuming that equation to be true in the integral domain $A$, and you may use this equation in your calculations. Equation (*) is exactly where this equation is being used.