Let $A = (a_{1}, a_{2}, a_{3}, \cdots, a_{n})$ be a non-zero finitely generated ideal of $R$. Prove that there is an ideal $B$ which is maximal w.r.t. the property that it doesn't contain $A$.
I can understand we have to use Zorn's Lemma, but I need some help.
Let $S$ be the set of all ideals not containing $A = (a_1, a_2, \ldots, a_n)$. It is partially ordered by set inclusion. Zorn's lemma can find us a maximal element in this set, that is, there can't be any other element that contains our element.
To apply Zorn's lemma we need to show that every chain $C$ of ideals in $S$ has an upper bound and that $S$ is non-empty.
For any such chain $C$ let $I = \bigcup_{I_i \in C} I_i$ be the union of all ideals in the chain. It is easy to see that it is an ideal that does not contain $A$. After all if it did we would have that each generator in $A$ is contained in some element of $C$. For some large enough $k$ all generators of $A$ would be in $I_k$ because $C$ is totally ordered. That would mean that $I_k$ contained $A$, which is a contradition. The union of those ideals is the upper bound of the chain.
By Zorn's lemma we have a maximal ideal out of the ideals not containing $A = (a_1, a_2, \ldots, a_n)$.