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.

  • 2
    $\begingroup$ Your proof of (<=) does not prove what you want. Showing there is a finite chain of ideals for $R$ like this does not prove $R$ is Noetherian. You need to start with an ascending chain and prove that it terminates. $\endgroup$ – rschwieb Feb 27 '17 at 0:09
  • $\begingroup$ @rschwieb I take any ideal $I$ of $R$ and show that for any ideal the nested chain is finite. What exactly do I need to show if this is insufficient? $\endgroup$ – sequence Feb 27 '17 at 0:49
  • $\begingroup$ The last sentence of my last comment explains exactly what you need to prove, assuming your definition is that ascending chains stabilize after finitely many steps. $\endgroup$ – rschwieb Feb 27 '17 at 1:52
  • $\begingroup$ @rschwieb Let $I_1\subset I_2 \subset \dots \subset I_n\subset \dots$. Since all ideals in $R$ are finitely generated, let $A_1, A_2, \dots, A_n, \dots$ be the generating sets of $I_i$. Then all these sets are finite, and so there exists integer $m<\infty$ such that $|A_i|=m$ for some $i$. This implies that the chain of nested ideals is finite. Do you think this looks OK? $\endgroup$ – sequence Feb 27 '17 at 2:07
  • $\begingroup$ No, nothing you wrote implies the chain is finite. In principle, the n'th item in the chain could have n generators. How does that imply anything about the finiteness of the chain? $\endgroup$ – rschwieb Feb 27 '17 at 2:38

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.)

  • $\begingroup$ If $I$ is generated by a finite set, and $I$ is a union of infinitely many sets, then, by exhaustion, there exists a $j$ such that $I=I_j$. However, what if $R$ is infinite? Can there be an exhaustion of finite sets? $\endgroup$ – sequence Feb 27 '17 at 4:37
  • $\begingroup$ @sequence The first sentence of your last comment is true, but you haven't given the details of the proof. You are totally skipping he step I boxed off above as being "very important." The second two sentences of your same comment are irrelevant concerns: $R$ can be infinite (it will often be) and "exhaustion of finite sets" appears to be an allusion to some odd technique you haven't explained fully.It doesn't look useful, though. $\endgroup$ – rschwieb Feb 27 '17 at 8:57
  • $\begingroup$ Well, if $\{a_1, \dots, a_n\}$ is finite then the preceding ideals must be generated by fewer elements of the set. This is why there exists this $j$. $\endgroup$ – sequence Feb 27 '17 at 15:14
  • $\begingroup$ @sequence That is patently not true. By that logic, since $R$ is generated by one element, every finitely generated ideal contained in $R$ is generated by one element or less. Containment simply does not say anything about relationships between generating sets or their sizes. $\endgroup$ – rschwieb Feb 27 '17 at 15:47
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    $\begingroup$ As I said, by that logic, if $R$ is a ring with identity, then $\{1\}$ generates $I=R$, and every ideal contained in $I$ is generated by one element. Therefore every ring with identity is a principal ideal ring. (this is of course nonsense.) It is an invalid deduction. $\endgroup$ – rschwieb Feb 27 '17 at 18:54

(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.

  • $\begingroup$ For (2), a generator is of the form $\langle a \rangle$, while an element is of the form $a$. I'm wondering if "finitely generated" also means "has finitely many generators"? Or can there be just one generator, but on a finite set? I'm somewhat confused in the definitions. For (3), what do you mean by "countably generated"? If an ideal is finitely generated, doesn't that mean that it is generated by a finite set? But a finite set contains finitely many elements, so it is countable. $\endgroup$ – sequence Feb 27 '17 at 4:33
  • $\begingroup$ You seem to be badly confused about what "generator" means. A set $A$ is said to generate an ideal $I$ if $I$ is the smallest ideal containing $I$. If we have picked such a specific set $A$ which generates $I$, we sometimes refer to the elements of $A$ as "generators" of $I$. But there is no intrinsic notion of a "generator" of an ideal: whether any particular element is a "generator" depends on which set $A$ we are are choosing to talk about which generates $I$. $\endgroup$ – Eric Wofsey Feb 27 '17 at 4:37
  • $\begingroup$ An ideal $I$ is said to be finitely generated if there exists a finite set $A$ which generates $I$. Similarly, an ideal is said to be countably generated if there exists a countable set $A$ which generates $I$. $\endgroup$ – Eric Wofsey Feb 27 '17 at 4:38
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    $\begingroup$ Yes, but in the forward direction of your proof you are assuming $I$ is not finitely generated. So you can't assume it is countably generated, as you have done. $\endgroup$ – Eric Wofsey Feb 27 '17 at 4:40
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    $\begingroup$ Right, you don't actually need all of the inclusions to be strict. But you need infinitely many of them to be strict, so that the sequence does not stabilize. And since you are choosing $a_n$, there's no point in choosing it so that the inclusion $\langle a_1,a_2,\dots,a_{n-1}\rangle\subseteq \langle a_1,a_2,\dots,a_n\rangle$ isn't strict. $\endgroup$ – Eric Wofsey Feb 27 '17 at 5:02

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