MHB Units in Z_m .... Anderson and Feil, Theorem 8.6 .... ....

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 8: Integral Domains and Fields ...

I need some help with an aspect of the proof of Theorem 8.6 ...

Theorem 8.6 and its proof read as follows:
View attachment 6434

In the above text, Anderson and Feil write the following:

" ... ... Conversely, if $$gcd(x,m) = d$$ and $$d \neq 1$$, then $$m = rd$$ and $$x = sd$$, where $$r$$ and $$s$$ are integers with $$m \gt r, s \gt 1$$. ... ... "I cannot see exactly why/how $$m \gt r, s \gt 1$$ ... can someone help me to prove that $$m \gt r $$ and $$s \gt 1$$ ... ... ?
Help will be appreciated ...

Peter

 

[Evgeny.Makarov]

I cannot see exactly why/how $$m \gt r, s \gt 1$$ ... can someone help me to prove that $$m \gt r $$ and $$s \gt 1$$

$m>r$ holds because by definition of $r$, $m=rd$ and by assumption $d>1$. But I don't see where $s>1$ is used in the proof. It can happen, for example, that $m=12$ and $x=3$; then $d=3$, $r=4$ and $s=1$, but $xr=3\cdot4$ still gives a 0, so $x$ is a zero divisor.

 

[Math Amateur]

$m>r$ holds because by definition of $r$, $m=rd$ and by assumption $d>1$. But I don't see where $s>1$ is used in the proof. It can happen, for example, that $m=12$ and $x=3$; then $d=3$, $r=4$ and $s=1$, but $xr=3\cdot4$ still gives a 0, so $x$ is a zero divisor.

Thanks Evgeny ... appreciate your help ...

Peter