TITLE: Ideals of ring finitely generated if and only if ring Noetherian
QUESTION [5 upvotes]: Let $R$ be a commutative unital ring. Then $R$ is Noetherian if and only if all ideals of $R$ are finitely generated.

My proof:

(=>): Suppose that $R$ is Noetherian, and $I$ is an ideal of $R$, which is not finitely generated. Then $\exists$ set $A\subseteq R$ which contains infinitely many elements, such that $I=\langle A\rangle$. Let $A=\{a_1, a_2,...,a_n,...\}$, then we $\langle a_1 \rangle\subset \langle a_1, a_2 \rangle\subset...\subset \langle a_1, a_1,...,a_n,... \rangle$, which is an infinite chain of nested ideals. This is a contradiction, thus $I$ must be finitely generated.
(<=): Suppose that all ideals of $R$ are finitely generated Let $I$ be an ideal of $R$, then $I=\langle A\rangle$, where $A\subseteq R$ is finite. Let $|A|=n$. Then, $\forall a_i\in A$, for $1\le i\le n$, $\langle a_1\rangle \subset \langle a_1, a_2 \rangle \subset... \langle a_1, a_2,..., a_n \rangle$, which is a finite chain of nested ideals. Hence, $R$ is Noetherian.
I would appreciate if someone could please clarify the following:
(1) Why do we need $R$ to be unital and commutative in this case?
(2) Is an ideal generated by an infinite set has in fact infinitely many generators, or does this just mean that such an ideal is simply generated by infinitely many elements, but not necessarily infinitely many generators?
(3) Is my proof OK?
Thank you very much.

REPLY [4 votes]: (1) You don't.  The theorem is true without these assumptions.
(2) The term "an ideal generated by an infinite set" is basically meaningless.  Literally, it means simply an ideal $I$ such that some infinite subset $A\subseteq I$ generates $I$.  But this is true of every infinite ideal $I$, since you can just take $A=I$.  In any case, I don't know why you're asking about the meaning of this term because it does not appear in the statement of what you are trying to prove.  I also don't understand what distinction you are making between "elements" and "generators".
(3) Your proofs of both directions are incorrect.  In the forward direction, you seem to be assuming that your ideal $I$ is countably generated, which need not be true.  If $I$ is countably generated, then your argument is correct: you know your ascending sequence of ideals cannot stabilize because then $I$ would actually be generated by finitely many of the $a_n$, contradicting the assumption that $I$ is not finitely generated.  However, there need not exist any countable set that generates $I$.
So if you do not know there exists a countable set $A$ that generates $I$, you must argue more carefully.  You might try choosing a sequence elements $a_n\in I$ by induction so that each $a_n$ is not in the ideal generated by $a_1,\dots,a_{n-1}$ (why is this possible?).
In the reverse direction, your argument is just completely incorrect: you are not proving the right thing.  To prove that $R$ is Noetherian, you need to start with an arbitrary ascending sequence of ideals $$I_1\subseteq I_2\subseteq I_3\subseteq\dots$$ and prove the sequence must stabilize.  To prove this, I suggest letting $I=\bigcup_n I_n$ and using the fact that $I$ is finitely generated.