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

  1. Feb 26, 2017 #1
    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:



    ?temp_hash=4cad108bde00c38f1acae6d92c8cf1dc.png



    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
     

    Attached Files:

  2. jcsd
  3. Feb 26, 2017 #2
    First off, you should note that the author doesn't mean to write $m>r$ and $s>1$. What he means is that $m>r>1$ and $m>s>1$.

    Now, there is nothing fancy going on here. If $d=\gcd(x,m)$, then $d$ divides both $x$ and $m$, so we can write $m=rd$ and $x=sd$ for some integers $r,s$. Now, since we're assuming $d\neq1$, i.e. $d>1$, then we must have $r<m$ because if $r\ge m$, then we'd have $m=rd\ge md>m$, which is a contradiction. Similarly, $s<x$, and $x<m$ by assumption so $s<m$.

    Now actually, I believe there is a typo here from the author. You don't necessarily need $s>1$, you could have $s=1$. Note the fact that $s>1$ isn't actually used in the rest of the proof, so it doesn't matter.

    Edit: sorry for the poor formatting, this is my first time using this forum to answer a math question and I assumed it would use TeX commands.
     
  4. Feb 26, 2017 #3

    fresh_42

    Staff: Mentor

    It does, but the syntax varies a bit from editor to editor. Here you have to use [itex][\itex] or double # # for inline tex and [tex][\tex] or double $ $ for single line tex. See https://www.physicsforums.com/help/latexhelp/
     
  5. Feb 28, 2017 #4
    Thanks mathers101 ... just worked through what you have said ...

    Your argument is VERY clear and helpful ...

    Thanks again ...

    Peter
     
  6. Feb 28, 2017 #5
    Hi mathers101 ...

    Just realised that I am unsure of where m > r > 1 is used in the rest of the proof ...

    Can you help with this matter ...

    Peter
     
  7. Feb 28, 2017 #6
    r > 1 is again not used at all, and r < m tells you that [r] is a nonzero element of Z_m, so since [x][r] = 0, we see [x] is a zero divisor.
     
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