I know this is probably a dumb question. Since a field is a PID then it should be a GCD domain according to the chain of domains. However, take $\mathbb{Q}$ for an example, I don't think it is a GCD because I can't find the greatest common divisor of $5$ and $7$. I mean, every nonzero element divides $5$ and $7$ since it is a field, and I can't find the largest one.
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Is it that take $a$ and $b$ which are not both zero, every non-zero element is a GCD of $a$ and $b$? – Coco Apr 04 '23 at 04:49
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In generic domains, GCDs are never meant to be unique (or canonically determined). They are greatest in the sense that they are $d$-s such that, for all $y$ such that $y\mid a$ and $y\mid b$, $y\mid d$. Note that $\mid$ is not an order relation. – Sassatelli Giulio Apr 04 '23 at 04:53
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4The "greatest common divisor" is "the greatest" in terms of *divisibility*, i.e. every other common divisor divides *it*. This has nothing to do with the _size_ of the number. The GCDs are not uniquely determined, but only up to _equivalence_ (two elements are equivalent if they divide each other). Thus, in $\mathbb Z$, the GCD's of $5$ and $7$ are $1$ and $-1$; in $\mathbb Q$, the GCDs of $5$ and $7$ are _all nonzero rational numbers_, and they are all equivalent to each other. – Stinking Bishop Apr 04 '23 at 04:54
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The gcd $d$ of any set $S$ containing a unit $u$ (invertible) is $1$, since $d\mid u\Rightarrow d\mid uu^{-1}=1,\,$ so $\,d\,$ is a unit (and unit gcds are normalized to $1)$. In a field every element $\neq 0$ is a unit, so every set $S\neq \{0\}$ contains a unit, so $\gcd S = 1.\ \ $ – Bill Dubuque Apr 04 '23 at 06:50
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@StinkingBishop Thank you!!! – Coco Apr 04 '23 at 20:02