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

Peter

 

[Evgeny.Makarov]

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

$m>r$ 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