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Thread: Associates

  1. #1
    Super Member Deadstar's Avatar
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    Associates

    R is an Integral Domain.
    Let c be a greatest common divisor of a and b in R and let f be an associate of c. Show that f is also a greatest common divisor of a and b in R.

    Thoughts...
    I think i should show that c|f and f|c. Then f would be the gcd as well.
    f = uc, where u is a unit, so c|f. Can find another unit v $\displaystyle \in$ R s.t. uv. = 1 (i.e. v is the inverse of u).
    Then c = vf, hence f|c. So f is gcd as well?

    On second thoughts, since R is an Int Domain maybe im not allowed to assume that uv = 1...
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Deadstar View Post
    R is an Integral Domain.
    Let c be a greatest common divisor of a and b in R and let f be an associate of c. Show that f is also a greatest common divisor of a and b in R.

    Thoughts...
    I think i should show that c|f and f|c. Then f would be the gcd as well.
    f = uc, where u is a unit, so c|f. Can find another unit v $\displaystyle \in$ R s.t. uv. = 1 (i.e. v is the inverse of u).
    Then c = vf, hence f|c. So f is gcd as well?

    On second thoughts, since R is an Int Domain maybe im not allowed to assume that uv = 1...
    Hi Deadstar.

    If $\displaystyle f=uc,$ suppose $\displaystyle uu'=u'u=1$ for some $\displaystyle u'.$ Now $\displaystyle a=a'c$ and $\displaystyle b=b'c$ for some $\displaystyle a',b'.$ Hence $\displaystyle a=a'u'f$ and $\displaystyle b=b'u'f,$ showing that $\displaystyle f$ divides both $\displaystyle a$ and $\displaystyle b$. Now if $\displaystyle g$ divides both $\displaystyle a$ and $\displaystyle b$ then $\displaystyle g$ divides $\displaystyle c$ as $\displaystyle c$ is a GCD. So $\displaystyle c=g'g$ for some g'. It follows that $\displaystyle f=ug'g$ and so $\displaystyle g$ divides $\displaystyle f.$ This shows that $\displaystyle f$ is a GCD.
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