Multiplication of fractions is well-defined (Explanation)

Question

I am studying the construction of the field of fractions from an integral domain. The multiplication operation on this field work as follows

$$[a,b][c,d]=[ac,bd]$$

Also, $$[a,b]=[a_1,b_1]\Leftrightarrow ab_1=ba_1$$

To show that the multiplication operation is well-defined in my book appears the following:

Suppose that $[a_1,b_1]=[a,b]$ and $[c_1,d_1]=[c,d]$. We have $[a_1,b_1][c_1,d_1]=[a_1c_1,b_1d_1]$ and \begin{eqnarray*} acb_1 d_1 -a_1c_1bd&=&(acb_1d_1-a_1cbd_1)+(a_1cbd_1-a_1c_1bd) \\ &=& cd_1(ab_1-a_1b)+a_1b(cd_1-c_1d)\\ &=& cd_1(0)+a_1b(0)\\ &=& 0\end{eqnarray*}

My problem is that I do not know exactly where the term "$a_1cbd_1$" comes from. I have not been able to understand the proof at all for this detail, could you please help me?

they added and subtracted $a_1cbd_1$ (which has no net change) to simplify; cf. this question — J. W. Tanner, Jul 25 '21 at 13:51
Because it works. We know what we want the answer to be, and so we just have to do whatever weird manipulation that allows us to get there. After you fiddle for a bit you find this one, and then you write it down and move on. Such unmotivated "add zero" steps are rather common in elementary proofs, e.g. the product rules for limits and derivatives. — Eric Nathan Stucky, Jul 25 '21 at 14:42
For a deeper understanding that clarifies this see here and here. — Bill Dubuque, Jul 25 '21 at 14:56
Could have just as well used $ac_1b_1d$ in place of $a_1cbd_1$ and taken it from there — J. W. Tanner, Jul 25 '21 at 14:58

score 1 · Answer 1 · answered Jul 25 '21 at 15:10

My problem is that I do not know exactly where the term $a_1cbd_1$ comes from.

In order to find a relation between $acb_1d_1$ and $a_1c_1bd$, we find an "intermediate term" that can be related to both. To get an intermediate term, change subscripts on two of the terms at a time (either on $a$ and $b$, or on $c$ and $d$; changing from $ab_1$ to $a_1b$ has no effect when $ab_1=a_1b$, and likewise $cd_1=c_1d$). Thus the intermediate term is $a_1cbd_1$ or $ac_1b_1d$.

Multiplication of fractions is well-defined (Explanation)

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