Let I be a f.g. ideal of R. Let S be the set of ideals J which do not include I. Show S has maximal element.

Question

Let I be a f.g. ideal of R. Let S be the set of ideals J which do not include I. Show S has maximal element.

My attempt is to construct a maximal element by taking the union of all elements in the set S. However a chain need not to be countable, and moreover I'm not sure this construction even stays in S. Any ideas to show this?

Answer 1

Use Zorn's lemma (assuming AC):

Write $I = Rx_1 + \cdots + Rx_n$.

Let $S = \{J \triangleleft R: I \not\subseteq J\}$. Partially order $S$ with inclusion. Note that an increasing union of ideals is an ideal. Let $J_1 \subset J_2 \subset\cdots$ be a chain in $S$.

Consider the ideal $J = \bigcup J_i$. If $I \subseteq J$, then for each $1 \leq i \leq n$ there is $j_i \in \mathbb{N}$ such that $x_i \in J_{j_i}$. Take $k =\max j_i$. Then $I \subseteq J_k$ - a contradiction. Zorn's lemma implies your result.