Let $R$ be a commutative ring, I always thought that finitely generated ideals are "small" and the non finitely generated ideals are "large" but this is not quite correct…
Lemma. In a non-noetherian ring $R$, there exists a maximal non finitely generated ideal w.r.t. set inclusion.
The proof is an application of Zorn's lemma.
Given the non empty collection $\mathcal{C}$ of non finitely generated
ideals, if we have any ascending chain $$I_1\subset I_2\subset
\cdots$$ we claim that the union $\bigcup_k I_k$ is an upper bound
in $\mathcal{C}$.$\bigcup_k I_k$ is an ideal because the union is nested, and if it is
finitely generated (i.e. not in $\mathcal{C}$), then the set of finite
generators must be contained in some $I_{n^*}$, and this finite set of
generators would generate $I_{n^*}$ which contradicts $I_{n^*}\in
\mathcal{C}$. By Zorn's lemma, there exists a maximal element in $\mathcal{C}$.
We call this maximal element $P$, then given any element $x\in R-P$, then we see that $P+\langle x\rangle$ must be finitely generated! Because $P\subsetneq P+\langle x\rangle$ and $P$ is the maximal non finitely generated ideal in $R$.
So in a sense, a lot of these "large" ideals will be finitely generated.
Best Answer
Your idea that finitely generated ideals are "small" and non-finitely generated ideals are "large" is misleading. Think about $R = \Bbb{Z}[X_1, X_2, \ldots]$, the ring of polynomials over the integers in countably many variables $X_1, X_2, \ldots$ If you define $J_k = \langle X_1, \ldots, X_k\rangle$, then $J_1 \subseteq J_2 \subseteq \ldots$ is a strictly increasing chain of ideals, so $R$ is not Noetherian. The finitely generated ideal $J = \langle 2 \rangle $ properly contains the non-finitely generated ideal $K = \langle 2X_1, 2X_2, \ldots \rangle$. For any prime $p$, the ideal $L_p$ comprising the polynomials with constant term a multiple of $p$ is maximal non-finitely generated.