$\def\truesubset{\subsetneq}$I took grad school algebra many eons ago and recently decided to re-learn it just because. I have been unable to figure out the following question that is the key to a different problem I was working on (that problem described at the bottom).
Let $R$ be a commutative ring with unity. Let $I_1, I_2$ be two ideals. Are the following two equivalent?
If $I_1, I_2$ are co-prime (e.g. $I_1+I_2=R$), then there is no prime ideal $P$ such that $I_1+I_2 \truesubset P \truesubset R$.
If there is no prime ideal $P$ such that $I_1+I_2 \truesubset P \truesubset R$, then $I_1, I_2$ are co-prime.
In other words, being co-prime and the sum not being contained in any prime ideal are equivalent characterizations.
1 above is obvious from the definition of co-prime. Since $I_1 + I_2=R$, it is impossible to squeeze any SET between $I_1+I_2$ and $R$.
I can see why 2 would be true in a principal ideal ring, but I cannot prove it as stated. I cannot disprove either because I don't have fresh in my mind enough examples of non-principal ideal rings.
FYI, the characterization in 2 was used in this website some years ago to show that if two ideals are co-prime, so are the sums of any powers of the two ideals (e.g. $I_1+I_2=R$ implies $\forall n \in \mathbb N^+, I^n_1+I^n_2=R$).