Let $ R$ be a a commutative ring with identity. Let $ M$ be an $ R $-module and let $N$ be an $R$-submodule of $M$. $N$ is called a small submodule in $ M$ if it satisfies the following condition:
the fact that $M =T +N$ for some $R$-submodule $T$ implies $T = M$. If every proper submodule of $M$ is small, we call $M$ a hollow.
We know that any ring $R $ has a maxima ideal. Now I want to know that if every commutative ring has at least a non-zero hollow ideal as an $R-module? If not under wath condition a hollow ideal exists?
This is only a partial answer, too long for a comment.
Proof: If $R$ is not local, then let $\mathfrak{m}$ and $\mathfrak{n}$ be distinct maximal ideals, then $\mathfrak{m} + \mathfrak{n} = R$, but $\mathfrak{m},\mathfrak{n} \neq R$, so $R$ is not hollow.
If $(R,\mathfrak{m})$ is local and $I$ and $J$ are proper ideals of $R$, then $I, J \subset \mathfrak{m}$, so $I+J \subset \mathfrak{m} \subsetneq R$. By contraposition, this implies that $R$ is hollow.
This allows us to give examples of rings with no hollow ideals:
Proof: Any ideal $I=(a)$ of $R$ is isomorphic as an $R$-module to $R$ itself via $r \mapsto ar$, so the result follows from the previous lemma.
With more effort, we can strengthen this result.
Proof Let $X$ be a maximal submodule of $M$. If $Y$ is another maximal submodule, then $X+Y=M$, so $M$ is not hollow. Thus $X$ is the unique maximal submodule. If we choose $a \in M \setminus X$, then $M=X+\langle a \rangle$, so $M=\langle a \rangle$
Using this, we get the following result
Only one direction is left to prove. Let $I$ be a non-zero hollow ideal, then $I$ is a Noetherian module, so $I$ has a maximal submodule. Thus $I$ is cyclic by the previous lemma. As $R$ is an integral domain, principal ideals are isomorphic to $R$, so $R$ is hollow as a module over itself. Thus $R$ is local by the first lemma.
This gives already plenty of examples of rings that don't have nonzero hollow ideals.