let A=($a1, a2, .... ,an$) be a non-zero finitely generated ideal of R. Prove that there is an ideal B which is maximal with respect to property that it does not contain A.
I don't know how to proceed. Maybe by Zorn's lemma. If I define order by containment of ideals. Then in a chain what will be the maximal element. If ∪J { J ∈ Chain } how to check that A is not contained inside ∪J.
You are right, the idea, is the Zorn, consider the set $J$ of ideals which does not contain $A$, let $(B_i)$ be a family of well ordered (by the inclusion) element of $J$, $B=\cup_{i\in I}B_i$ is the sup. Suppose that $B$ contains $A$, $a_j\in B_{i_j}$ let $k=sup(i_1,...,i_n)$, since $B_{i_j}\subset B_k$, $a_1,...,a_n\in B_k$ contradiction. Thus $J$ has a maximum element.