1

Proposition 2.3.5 in Artin says the following:

Let $a$ and $b$ be integers, not both zero, and let $d$ be their greatest common divisor, the positive integer that generates the subgroup $S = \mathbb{Z} a + \mathbb{Z} b$. So $\mathbb{Z} d = \mathbb{Z} a + \mathbb{Z} b$. Then

a. $d$ divides $a$ and $b$.

b. If an integer $e$ divides both $a$ and $b$, it also divides $d$.

c. There are integers $r$ and $s$ such that $d = ra + sb$.

I didn't have a problem in proving this statement, but Artin goes on to comment that (c) implies (b), but (b) does not imply (c). I'm struggling to find a counterexample which establishes this statement, as every example I've tried seems to work. I attempted a "proof" of this fact in order to find a problem with the assertion. So suppose that $e \mid a$ and $e \mid b$. Then there exist $r,s \in \mathbb{Z}$ such that $er = a$ and $es = b$. By the implication, $e$ also divides $d$, so there exists $k \in \mathbb{Z}$ such that $ek = d$. If $d$ divides $a$ and $d$ divides $b$, I believe the result is proved, but that would defeat the purpose, so it must be the case that $d$ either fails to divide $a$ or fails to divide $b$.

Any guidance on how to find such a counterexample, or intuition on why this result is false, would be greatly appreciated.

Bill Dubuque
  • 272,048
Brad G.
  • 2,238
  • They are equivalent in $\Bbb Z$ so you need to consider other GCD domains for a counterexample, e.g. as mentioned here generally there is no Bezout identity for gcds, e.g. in the polynomial ring UFDs $,\Bbb Z[x],$ and $,\Bbb Q[x,y],$ we have $,\gcd(2,x)=1,$ and $,\gcd(x,y)=1,$ but there is no Bezout identity for those gcds (if it existed then evaluating at $,x,y=0,$ yields a contradiction). – Bill Dubuque Jul 14 '23 at 06:42
  • The statement "(b) does not imply (c)" could be true if you add in (c) "where GCD of $r$ and $s$ is $1$".

    Indeed, if (b) is true for all $e$, then it is true in particular for $e=a\wedge b$ (GCD of $a,b$) then $a\wedge b$ divides $d$d, and with Bezout there exists $u,v$ such as $a\wedge b=au+bv$

     – hdci Jul 14 '23 at 07:20
  • "If $d$ divides $a$ and $d$ divides $b$, I believe the result is proved": not at all. Anyway, bad start. What is the point of taking some $e$ dividing both $a$ and $b$, when you want to prove that (b) implies (c)? However, (b) implies (c) indeed, because $$(b)\iff\gcd(a,b)\mid d\iff d\in\gcd(a,b)\Bbb Z\iff(c).$$ – Anne Bauval Jul 14 '23 at 21:57
  • @AnneBauval So the result is true? I'm a bit confused now, as Artin claims that (b) does not imply (c). – Brad G. Jul 14 '23 at 22:16
  • Yes, see the first line of Bill's comment above. Can you include in your post a scrrenshot of that page of Artin? – Anne Bauval Jul 14 '23 at 22:24

0 Answers0