To anyone encountering this problem: As the comments and answer correctly point out, if $R$ is a commutative ring with identity, and $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$, then $d$ is a gcd of $a_1, ..., a_n$. If, additionally, $R$ is Noetherian (for instance if $R$ is a PID), then this becomes an equivalence. However, if $R$ is not Noetherian this may fail (as in the example given by Exodd).
Furthermore, in any commutative ring $R$ with identity, the following stronger statement is true: $d\in R$ is a gcd of $a_1, ..., a_n$ and is equal to $r_1 a_1 + \cdots + r_n a_n$ for some $r_i\in R$ iff $\langle d\rangle = \langle a_1 \rangle + \cdots + \langle a_n\rangle$.
Let $R$ be a commutative ring with identity, and let $a_1, ..., a_n\in R$. I would like to prove that $d\in R$ is a gcd of $a_1, ..., a_n$ iff $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$.
If $\langle d\rangle = \langle a_1\rangle + \cdots + \langle a_n\rangle$, then $\langle a_i\rangle \subseteq\langle d\rangle$, so $d\mid a_i$ for all $i$. Furthermore, if $c\mid a_i$ for all $i$, then $\langle a_i\rangle\subseteq\langle c\rangle$, so $\langle a_1 \rangle + \cdots + \langle a_n\rangle\subseteq\langle c\rangle$, or $\langle d\rangle\subseteq\langle c\rangle$, so $c\mid d$, implying $d$ is a gcd of $a_1, ..., a_n$.
I'm finding the other direction trickier. I thought I had an argument but I can't quite make it work. First, if $d \mid a_i$ then $\langle a_i\rangle \subseteq\langle d\rangle$ for all $i$, so $\langle a_1\rangle + \cdots +\langle a_n\rangle \subseteq \langle d\rangle$.
My idea was this: let $J = \{I\subseteq \langle a_1\rangle + \cdots + \langle a_n\rangle\ |\ I\ \text{principal ideal}\}$, ordered by inclusion. $\color{red}{\text{Then $J$ has a maximal element by Zorn's lemma.}}$ This is not actually true, as only finite chains have upper bounds in $J$, unless $R$ is a PID. That is, if we have an arbitrary collection of principal ideals $\{\langle a\rangle\}_{a\in A}$, totally ordered by inclusion, then we cannot guarantee that $\bigcup_{a\in A}\langle a\rangle$ is also a principal ideal (it is an ideal however).
If we roll with it, letting $\langle d\rangle$ be a maximal element of $J$, clearly as $\langle a_i\rangle\subseteq \langle a_1\rangle + \cdots + \langle a_n\rangle$ we must have $\langle a_i\rangle\subseteq\langle d\rangle$, so $d \mid a_i$. Furthermore, if $c \mid a_i$, then $\langle a_1\rangle + \cdots + \langle a_n\rangle \subseteq\langle c\rangle$, so $\langle d\rangle\subseteq\langle c\rangle$, and thus $c \mid d$. So $d$ is a gcd of $a_1, ..., a_n$.
Can I modify my argument to make this work? Is there a simpler route I've missed?
