Having read this (which is now technically correct), it makes sense to fill in some missing details, which could help other readers.
To be completely correct, one needs:
- Assume $I \neq R$
- Define $P=\{I \subset A : A \text{ is a proper ideal in } R\}$
Assumption 1 is necessary because the result is not true if $I=R$. A maximal ideal is by definition a proper subset of the ring.
For $P$ in assumption 2 above to exist, $P$ must contain at least one element. This follows #1 because $I$ is contained $P$.
To prove that $P$ contains a maximal ideal (which would clearly contain $I$) requires invoking the Zorn's lemma. Since set inclusion is a partial ordering on any collection of sets, we use the set inclusion predicate as the partial ordering on $P$.
To apply Zorn's lemma we take an arbitrary chain $C$ in $P$ and show it has an upper bound in $P$. Define $U_C = \cup C$ (which exists by the axiom of union of set theory). You can show that $U_C$ is an ideal and that it contains $I$. This takes a little bit of work, but it is routine. To show that $U_C$ is in $P$, one notices that $1 \neq U_C$ because none of the elements in $C$ contain $1$ (else they would not be proper ideals in $R$ and would not belong to $P$). Finally, it obvious that $U_C$ is an upper bound for the chain $C$.
Since we have shown that every chain $C$ in $P$ has an upper bound in $P$, Zorn's lemma states that $P$ has a maximal element. The maximal element in in $P$. Hence, by the definition of $P$, it contains $I$.
One more note, the ring $R$ must have unity. The unity requirement is essential to show that $U_C$ is an ideal not equal to $R$. Without assuming $1 \in R$, one CANNOT show that $U_C$ is a proper ideal in $P$.
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.
Best Answer
This depends heavily on what your definition of "Noetherian" is. Are you working with the ascending chain condition on ideals?
If you start with an infinite nondecreasing chain $A_i\subseteq A_{i+1}$ of proper ideals (in a ring with identity), then $\bigcup_{i=1}^\infty A_i=(a)$ is a proper ideal containing all the $A_i$'s.
But $a\in A_j$ for some $j$, and that would mean that $(a)\subseteq A_j$. In that case, $A_j=\bigcup A_i=A_k$ for every $k\geq j$.
This says that all such chains stabilize, that is, the ring satisfies the ascending chain condition on ideals.
The same reasoning shows you that if you just suppose all ideals are finitely generated, then the ascending chain condition holds.
Are you using the maximum condition on ideals?
I think, perhaps, it was intended for you to use Zorn's Lemma to prove that ACC on ideals implies this. For if you are given a nonempty set of ideals of $R$, Zorn's lemma and the ACC together imply that every chain in the set is bounded, and hence the whole set has a maximal element.
The converse implication (maximal condition implies ACC) always holds, of course: given any chain, the maximal condition implies the chain has a greatest element, and that would make the chain stable.
Despite the name, Zorn's Lemma is just the Axiom of Choice in another guise. You are free to assume it holds, or not, and we often assume it. Using it is easy: if its hypotheses are satisfied, then its conclusion holds. To do this, you need to validate that ascending chains in a poset are bounded in that set, and then you magically get a maximal element somewhere in the poset.