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Lets $A\subset \mathbb F$ where $\mathbb F$ is an arbitrary field. Is is correct, that there are two possibilities?

  1. $A=\{0\}$, then $\gcd(A)=A$
  2. $A\neq\{0\}$, then $\gcd(A)=A\backslash\{0\}$

My reasoning here would be: If $0$ is a common divisor of all elements in $A$, then $\gcd$ must be divisible by $0$ which is only the case for $0$ itself. If any nonzereo element is in $A$, then a common divisor can only be another nonzero element. This yields $A\backslash\{0\}$ because those are all associated.

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closed as unclear what you're asking by J.-E. Pin, rschwieb, Brandon Carter, John_dydx, kimchi lover Jan 11 '18 at 23:33

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    $\begingroup$ What does it mean for gcd(A)=A? how can a gcd be a subset? Do you mean the set of common divisors of $A$ rather than "the greatest common divisor"? $\endgroup$ – rschwieb Jan 11 '18 at 19:30
  • $\begingroup$ What do you mean? A $\gcd$ is always defined as a set, since $R$ has usually units. $\endgroup$ – Buh Jan 11 '18 at 19:31
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    $\begingroup$ Perhaps you are reading strange books. Typically a gcd is an element of the ring and we confuse that with the class of associate elements. I've never seen a resource that handles a gcd as a set of associates. $\endgroup$ – rschwieb Jan 11 '18 at 19:33
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    $\begingroup$ It's a little silly to talk about associates in a field anyway since there are only two classes: $\{0\}$ and $\mathbb F\setminus\{0\}$. $\endgroup$ – rschwieb Jan 11 '18 at 19:33
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    $\begingroup$ You don't define gcd in fields as it is usually defined in rings which are not fields, because that'd be boring... $\endgroup$ – DonAntonio Jan 11 '18 at 19:39
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In the first case, the gcd set would be $\{0\}$; no nonzero element would be in the gcd set because the nonzero elements all divide $0$, so $0$ is a "greater" common divisor.

In the case where there is some $x \in A$ s.t. $x \neq 0$, any nonzero element of $\mathbb{F}$ (not only those of $A$) would be in the gcd set because they all divide every element of $A$ (since they all divide $1$, as they are all invertible), and they are all associates. So the gcd set would be $\mathbb{F}\setminus\{0\}$.

As Buh pointed out in the comments, there is a third case -- $A = \varnothing$ -- which has a gcd set $\{0\}$.

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  • $\begingroup$ Ah, of course I meant $\mathbb F\backslash\{0\}$ in the second case. Even wrote it in my reasoning. But there is a third case that just came to mind: If $A=\emptyset$, then $\gcd(A)=\{0\}$ too, right? $\endgroup$ – Buh Jan 11 '18 at 20:35
  • $\begingroup$ @Buh Yes, I suppose so! I'll add that to the answer. $\endgroup$ – BallBoy Jan 11 '18 at 20:38
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Another way to think about what's going on here is to think of the divisibility poset of the field.

Let $R$ be a commutative domain, then the divisibility poset, which I'll call $D(R)$ is defined as a set to be the quotient $R/\sim$ where $a\sim b$ if $a$ and $b$ are associates. The order on the quotient is that induced by the usual divisibility preorder on $R$, $a\le b$ if $a\mid b$.

Then if $F$ is a field $D(F)$ has two elements, which I will denote by their representatives, $1$ and $0$, and $1\le 0$. The gcd of a subset of $F$ is the infimum of the corresponding subset of $D(F)$. Hence $\gcd(A)=1$ if $x\ne 0\in A$, otherwise $\gcd(A)=0$.

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