We were given this defintion in our Algebra $2$ course:

Definition $4.1$: Let $R$ be a commutative ring and $a, b\in R$. An element $d \in R$ is called a greatest common divisor, $\operatorname{gcd}(a, b)$, of $a$ and $b$ if:
$d$ divides both $a$ and $b$ $($i.e., $\exists x$, $y \in R$ such that $a = dx$, $b = dy$$)$,
• If $e \in R$ divides both $a$ and $b$, then $e$ divides $d$.

I don't really understand how the second point could work with say the integers, for example $\operatorname{gcd}(10,12)$, if you pick $2$ to be the first $\operatorname{gcd}$, then for $e=1$ you have that it does divide both $10$ and $12$, and it does divide $2$. But then $2$ doesn't divide $1$ so $2$ is no longer a gcd according to this surely?

  • 1
    $\begingroup$ So then $\gcd(10, 12) \ne 1$. That seems fine to me! $\endgroup$ – Robert Lewis Nov 4 '18 at 19:02
  • $\begingroup$ It's not $2$ which has to divide $1$, according to this definition, but the reverse. $\endgroup$ – Bernard Nov 4 '18 at 19:06
  • $\begingroup$ Surely they'd both have to divide each other? $\endgroup$ – Andy010101 Nov 4 '18 at 19:23
  • $\begingroup$ Equivalently, $\, c\mid a,b \iff c\mid \gcd(a,b).\ $ This means the gcd is "greatest" w.r.t. divisibility partial order. $\endgroup$ – Bill Dubuque Nov 4 '18 at 19:44
  • $\begingroup$ In your example $ d\mid 10,12\iff d\mid 2\ $ so $\,\gcd(10,12) = 2\ $ $\endgroup$ – Bill Dubuque Nov 12 '18 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.