Ideals of ring finitely generated if and only if ring Noetherian

Question

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.

Answer 1

1). Having identity does not factor into the proof. Using noncommutativity would require you to restate everything in terms of right or left ideals.

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? The second thing is more correct. Really you should probably be thinking in terms of "finitely generated" and "not finitely generated" (which is a little different from "infinitely generated".)

3) The proof of the on direction is not correct as we have been discussing in the comments. To get you started for the $\impliedby$ direction:

Assume all ideals are finitely generated and let $I_1\subseteq I_2\subseteq\ldots$ be an ascending chain. Let $I=\bigcup_{i=1}^\infty I_i$ be the union of these ideals (and prove it is an ideal, if you haven't already.

Now $I$ is finitely generated by $\{a_1, a_2,\ldots a_n\}$. The important step in logic takes place right here:

prove that there must be a $j\in \mathbb N$ such that $\{a_1, a_2,\ldots a_n\}\subseteq I_j$. Conclude that $I=I_j$, and that $I_k=I$ for all $k\geq j$.

The idea for $\implies$ is the right one, but it has a technical flaw: you do not establish that the chain is strictly increasing. You can do this inductively. (The other posts mentions a fault with assuming $I$ is countably generated; however, I only see that as a disagreement in notation, and what is written doesn't really say that. You can well-order whatever generating set and write the initial segment of countably many generators like that, IMO.)

Answer 2

(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.