by K Math
Last Updated August 10, 2018 12:20 PM

$A$ is an integral domain, $S=\{(a,b)\mid a,b \in A \text{ and } b\neq 0\}$. If $(a,b)\sim (c,d)$ means $ad=bc$, prove that $\sim$ is an equivalence relation on $S$.

I have already proven that $\sim$ is reflexive and symmetric. I am stuck on transitivity. I have:

If $(a,b)\sim(c,d)\Leftrightarrow ad=bc$ and $(c,d)\sim(e,f)\Leftrightarrow cf=de$ then $(a,b)\sim (e,f)\Leftrightarrow af=be$. Should I be solving with subsitution?

Let be $(a,b)\sim (a', b')$ and $(a',b')\sim (a'', b'')$.

Then:

$b'(ab''-a''b)=\underbrace{ab'b''-a'bb''}_{b''(ab'-a'b)}+\underbrace{a'bb''-a''bb'}_{b(a'b''-a''b')}=0$, because $A$ is an integer domain and $b'\neq 0$ it is $ab''=a''b$

You can't make substitution such as $a=\frac{bc}{d}$ because you only know that $A$ is an integral domain, but not necessarily a field. Instead, what you do is try to prove that $afx=bex$ for some well-chosen $x$; in an integral domain, this is equivalent to $af=be$.

Particularly interesting examples of $x$ would be $x=c$ or $x=d$, so that you can use the different relations you already know.

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