I've been trying to wrap my head around this and I feel like I'm getting stuck in circular logic.
The definition I'm using is the gcd of ideals $I_1, I_2$ of a ring $R$ is an ideal $J$ of $R$ such that $I_1 \subseteq J$ and $I_2 \subseteq J$.
Furthermore if $J'$ is any other ideal such that $I_1 \subseteq J'$ and $I_2 \subseteq J'$ then $J \subseteq J'$
If it helps I'm working in a Dedekind domain.
Okay so trivially there exists an ideal that contains $I_1, I_2$ that is the entire ring, so we know they have a common divisor.
Let $\gcd(I_1,I_2)=J$. It is clear that $J=I_1+I_2$ since if not, there exists $x\in J, x\notin I_1, I_2$. But we have $I_1, I_2 \subseteq I_1+I_2$ and $I_1+I_2\subset J$ and thus $J$ would not be the $\gcd$. Hence $\gcd(I_1,I_2)=I_1+I_2$.
Is there a better way to phrase this? I feel like I'm using the assumption there is a gcd to prove there is a gcd.
Let $K$ be the intersection of all ideals of $R$ which contain both $I_1$ and $I_2.$ Then $K$ is an ideal since any intersection of ideals is an ideal. [I'm assuming you're working with 2-sided ideals.]
This $K$ contains $I_1$ and $I_2,$ and if $J'$ is any ideal of $R$ containing each of $I_1$ and $I_2,$ then $K \subset J',$ since by definition $K$ is the intersection of all ideals of $R$ which contain each of $I_1$ and $I_2,$ and so must be a subset of $J'.$
So there is indeed a gcd for the two ideals, and that may coincide with your description, I just wanted to show how to see a gcd exists, even not knowing a description of it.