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Why does it follow that if $d_1 = \gcd(a,b)$, $d_2 = \gcd(a,b)$ and $d_1\mid d_2$ then $d_1$ and $d_2$ are associates?

JavaMan
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user61835
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    in which ring are you working? namely, where do $a,b$ come from? (are they integers or what?) – Ittay Weiss Feb 23 '13 at 04:15
  • They are in an integral domain D – user61835 Feb 23 '13 at 04:17
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    Any two gcds $d_1,d_2$ of $a,b$ are associate (since $d_1\mid d_2$ and $d_2\mid d_1$, both following from the universal property of the gcd). However if you are prescribing some sort of well-defined gcd function and define $d_1$ and $d_2$ to both be gcd(a,b), then not only are $d_1,d_2$ associate they are equal by definition. The wording seems strange to me. – anon Feb 23 '13 at 04:23

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In a commutative ring, if $d=gcd(a,b)$ and $e$ divides both $a$ and $b$ then, by definition, $e$ divides $d$. Now, given that $d_1=gcd(a,b)$ and $d_2=gcd(a,b)$ it follows that $d_1$ divides $d_2$ (by taking $e=d_2$) and it also follows that $d_2$ divides $d_1$ (by taking $e=d_1$), so that $d_1$ and $d_2$ are associates (so that the extra condition that $d_1$ divides $d_2$ is superfluous.

Ittay Weiss
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  • Thanks, I was confused if $d_1$ divides $d_2$ and $d_2$ divides $d_1$ then $d_1 = d_2$ but this theorem says that gcd is not unique . Could you clarify that? – user61835 Feb 23 '13 at 04:29
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    In an arbitrary ring, $x$ may divide $y$ and $y$ may divide $x$ without them being equal. They are then called companions. For an extreme example, in a field any two non-zero elements are companions. – Ittay Weiss Feb 23 '13 at 04:34
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    @user61835 gcd is unique up to multiplication by a unit. For example, the gcd in $\mathbb{Z}$ is determined up to sign, and we normally take it to be the positive choice by convention. In other rings, there may not be so natural a choice. – Stahl Feb 23 '13 at 04:35
  • In an arbitrary ring (even with unit) not any two companions defer by a unit (this does hold in an integral domain). – Ittay Weiss Feb 23 '13 at 04:55
  • Remark that in non-domains the notion of associate, irreducible etc bifurcate into motley inequivalent notions, e.g. see the papers cited here. So perhaps it would be better to write "domain" instead of "ring". – Math Gems Feb 23 '13 at 14:47