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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.

Buh
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    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"? – rschwieb Jan 11 '18 at 19:30
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    What do you mean? A $\gcd$ is always defined as a set, since $R$ has usually units. – Buh Jan 11 '18 at 19:31
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    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. – rschwieb Jan 11 '18 at 19:33
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    It's a little silly to talk about associates in a field anyway since there are only two classes: ${0}$ and $\mathbb F\setminus{0}$. – rschwieb Jan 11 '18 at 19:33
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    How do you define the $\gcd$ then? - I have never seen a definition that only produces one element. – Buh Jan 11 '18 at 19:35
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    You don't define gcd in fields as it is usually defined in rings which are not fields, because that'd be boring... – DonAntonio Jan 11 '18 at 19:39
  • @Buh boggle you have never seen something like $\gcd(6,10)=2$? You have always seen it written $\gcd(6,10)={-2, 2}$? – rschwieb Jan 11 '18 at 19:40
  • @rschwieb Yes, I've seen that and I think it's wrong. What is your definition? – Buh Jan 11 '18 at 19:40
  • @Buh I"m not here to regurgitate and/or apologize for definitions that appear in most books. I was just trying to understand what you wrote, that's all. Like I said, it is standard practice to write and refer to the associates of an element as a single element when discussing factorization in domains. There is (afaict) zero benefit derived from insisting on referring to them as sets. – rschwieb Jan 11 '18 at 19:44
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    For some reason you keep arguing with me instead of just giving me your definition. This is pointless indeed. – Buh Jan 11 '18 at 19:48
  • @rschwieb Maybe you want to have a look in here: https://math.stackexchange.com/questions/85565/motivation-behind-the-definition-of-gcd-and-lcm – Buh Jan 11 '18 at 20:01
  • @buh Rather, I am refusing to participate in the argument which you are referring to (you are currently the only one signed up.) The information you ask for is here and I am not going to waste my time typing it. If you are not interested in my information about the standard way these things are discussed, then I'm afraid it's your loss. Good day. – rschwieb Jan 11 '18 at 20:37
  • No reason to get passively aggressive here. It is just that I don't see a point in your "standard way" since I am sure your definition (which you still haven't provided) isn't applicable in many rings. – Buh Jan 11 '18 at 20:50
  • So, the way I'm interpreting this disagreement is: $\gcd$ in either interpretation really lives in the quotient of $R$ by the equivalence relation of $x \sim y$ if $x = ay, a \in R^*$. Then you can think of the quotient either as a setoid, whose elements are elements of $R$ but with a different notion of equality, or as the set of equivalence classes of $\sim$. – Daniel Schepler Jan 11 '18 at 21:53
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    @DanielSchepler That's how I interpret the disagreement too--in which case it's basically a semantic disagreement, and really doesn't impact the question asked here. The question is clear and should not have been closed. – BallBoy Jan 14 '18 at 00:09

2 Answers2

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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 $\{0\}$.

BallBoy
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  • 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? – Buh Jan 11 '18 at 20:35
  • @Buh Yes, I suppose so! I'll add that to the answer. – 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 $\gcd(A)=0$.

jgon
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