$\gcd$ in fields

Question

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.

Answer 1

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 $\{0\}$.

Answer 2

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 $\gcd(A)=0$.