Let $R$ be a commutative ring with unity, with at least one ideal which is non-finitely generated. Let $S$ be the family of such ideals. Show $S$ has a maximal element.
I need to show every chain in $S$ has an upper bound and use Zorn's lemma. I can take a chain $\mathcal{C}: I_{1} \subset I_{2} \subset I_{3} \subset \cdots$ in $S$ and let $I := \bigcup_{n} I_{n}$. If $I = (a_{1}, \ldots, a_{t})$ is finitely generated, then I want to arrive at a contradiction. From here I can say that $I_{n} = (b_{n, 1}, b_{n, 2}, \ldots) \subset (a_{1}, \ldots, a_{t})$ for each $n \in \mathbb{N}$, whence $b_{n, i} = r_{n, 1}^{i}a_{1} + \cdots + r_{n, t}^{i}a_{t}$ (some linear combination). However, I don't know where to go from here, or if this is even the right direction. I suppose I should use the inclusion $I_{1} \subset I_{2} \subset \cdots$ somehow, seeing as subideals of finitely-generated ideals need not be finitely generated.
Since this looks like it might be a homework problem, I'll just give a hint.
Let's consider the extreme case where $I = (a)$, i.e., $I$ is principal. Since $I = \bigcup_n I_n$, there must be some $k$ such that $a \in I_k$. Then it follows (why?) that $I = I_k$, which means $I_k$ was finitely generated, contradiction.
Now suppose $I = (a_1, a_2)$. Then there is a $k_1$ with $a_1 \in I_{k_1}$ and a $k_2$ with $a_2 \in I_{k_2}$. See if you can use the fact that the $I_n$'s form a chain to again conclude that one of the $I_n$'s must have been finitely generated.
Once you see how to do the case where $I$ has two generators, you'll probably also see how a similar proof takes care of the general case.