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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?

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    $\begingroup$ I added a complete argument to my answer. This elementary approach is almost surely what Vinberg intended since the book includes no prior discussion of general properties of equivalence relations. Vinberg's argument is far from clear so it is not surprising that it might prove perplexing to one first encountering these ideas. If anything remains unclear then let me know and I will be happy to elaborate. $\endgroup$ – Bill Dubuque Jun 29 '19 at 23:31
  • $\begingroup$ Note to future readers. The title of the question was edited (I assume) to make this question more easily found in the future. Do not get the impression that I knew the answer before asking. Because once it is clear what is being proven, the proof itself becomes much simpler. :) $\endgroup$ – Linear Christmas Jun 30 '19 at 18:59
  • $\begingroup$ Yes, that's why I edited the title. Unfortunately this key conceptual idea is rarely mentioned in algebra textbooks (and when it is often the essence of the matter is obfuscated as here). So anything we can do here to remedy this it time well spent. $\endgroup$ – Bill Dubuque Jun 30 '19 at 19:52
  • $\begingroup$ It is even more obscure in the case of general rings of fractions (localizations) where the cross multiply rule becomes $\, s ad = scb\,$ for some denominator $\,s,\,$ but a similar argument works as in my answer (here it is convenient to also use $\ (a,b) \simeq (c,d)\!\! \!\overset{\rm def}\iff (as,bs) = (ct,dt) \,$ for some $\,s,t\in S = $ denominator set. $\ \ $ $\endgroup$ – Bill Dubuque Jun 30 '19 at 19:52
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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: $\, bd = db$.

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  • $\begingroup$ Reply to your comment under the question as well as answer: Yes, I suppose this is what Vinberg had in mind. The word corollary threw me off entirely, and indeed, Vinberg does not exactly say that this smallness is what is being proven. Would you also agree that in the quote "On the other hand, any equivalence $(a_1,b_1) \sim (a_2,b_2)$ is a corollary of..." the symbol $\sim$ aqcuires a new meaning. That is, in that chain $\sim$ no longer designates the relation $a_1b_2 = a_2b_1$ but rather any equivalence relation satisfying $(3.34)$? $\endgroup$ – Linear Christmas Jun 30 '19 at 18:56
  • $\begingroup$ @Linear Yes, that's why I chose different symbols. Certainly the exposition there could be much better so it is not your fault that it was not immediately clear. This is one of the reasons that teaching is so difficult. To excel at teaching requires the capability to easily forget key ideas in order to revert to a prior (lower) knowledge level of the average reader in order to make the exposition accessible at that level. Alas, Vinberg didn't forget enough here. $\endgroup$ – Bill Dubuque Jun 30 '19 at 19:28
  • $\begingroup$ Agreed. Just to be sure not to be too rough on Vinberg: most of the time Vinberg isn't too chatty but the proofs are sometimes excellent. Also, I have been having some answer-accepting anxiety since there are two great and mostly equivalent answers to my question. I decided to use a random number generator (0 to 50 being Lee, and 51 to 100 being you). Congratulations are in order, as '68' popped out. $\endgroup$ – Linear Christmas Jun 30 '19 at 19:56
  • $\begingroup$ @Linear Pretty random! Usually I recommend that one accepts the answer that one thinks will be most helpful to others at your level. What Lee wrote is correct, but I doubt that's what Vinberg had in mind. In fact it is rare that any mention of generation of equivalence relations is explicitly mentioned in a first course in Abstract Algebra since it is not needed for the most common topics. It sometimes occurs in second courses, and almost always in courses on Universal Algebra (where congruences vs. ideals become more central). $\endgroup$ – Bill Dubuque Jun 30 '19 at 20:06
  • $\begingroup$ If you're willing, you may join me at a chatroom I created: Brief discussion: Quotient Fields (cont.). It should take about 15 minutes of your time, hopefully. If you don't have time, this is perfectly OK. $\endgroup$ – Linear Christmas Jul 2 '19 at 16:34
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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.

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  • $\begingroup$ Note that we don't need to invoke any knowledge of the structure of generated equivalence relations in order to prove (or comprehend) the claim - see my answer. $\endgroup$ – Bill Dubuque Jun 29 '19 at 16:49
  • $\begingroup$ This is just a point of language. One can decide to take for granted the existence of the "smallest" equivalence relation, if one so desires. But then that raises a mathematical question about why the "smallest" exists, and I chose language which suggests why that is so without committing to that decision. The "smallest" equivalence relation containing a given relation is simply the reflexive-symmetric-transitive closure of the given relation. $\endgroup$ – Lee Mosher Jun 29 '19 at 16:54
  • $\begingroup$ And, of course, instead of saying "the smallest equivalence relation..." one can use the alternate language of "the equivalence relation generated by...", which is what I chose. $\endgroup$ – Lee Mosher Jun 29 '19 at 16:57
  • $\begingroup$ In general yes, but the point is that we don't need such generality here since the argument in my answer proves that $\sim$ is smallest. $\endgroup$ – Bill Dubuque Jun 29 '19 at 16:58
  • $\begingroup$ @LeeMosher On its own, how do you define the relation $(3.34)$? Simply as $(a,b) \sim' (d, e) \overset{\text{def}}{\Longleftrightarrow} \exists c\in A\ (c\neq 0 \land d = ac \land e = bc)$, that is simply as the set $\left\{((a,b),(ca,cb))\mid(a,b)\in A^2, b\neq0, c\in A\setminus\{0\}\right\}$, I assume? $\endgroup$ – Linear Christmas Jun 29 '19 at 18:09

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