I am reading Atiyah, MacDonald "Commutative Algebra" and there it is said that ring $A$ is Noetherian if three equivalent conditions are satisfied.

(1) Every nonempty set of ideals in $A$ has maximal element.

(2)Every ascending chain is stationary

(3) Every ideal is finitely-generated.

I see why $(1) \Longrightarrow (2)$ is correct, and I also see why $(2) \Longleftrightarrow (3)$ is correct. But my problems are the following:

(1) How to prove $(2) \Longrightarrow (1)$ or $(3) \Longrightarrow (1)$? I truly do not see.

(2) A corollary of Hilbert's Basis Theorem says that for any Noetherian ring $R$, $R[x,y]$ is Noetherian. Great. Take $\{<x>,<y>\}$. That is a nonempty set of ideals, right? What is its maximal element?


2 Answers 2


I think your confusion stems from the following fact: A maximal element isn't necessarily a greatest element. Maximal only means nothing is above it. It does not mean it's above everything else. You see this very clearly with, for instance, the concept of maximal ideals. There can be several of them, none contained in any other, and this is not an issue.

With that in mind, in your concrete example with $\{\langle x\rangle, \langle y\rangle\}$, they are both maximal. And the proof of $(2)\implies (1)$ is merely a matter of taking a set of ideals, picking an ideal, and then successively pick larger and larger ideals. $(2)$ says that this process necessarily stops after a finite number of steps, and the only way it can stop is if you have found a maximal element.


To prove $(2)\implies (1)$, let $\mathcal{F}$ be a nonempty set of ideals. Assume it has no maximal element. Take any $I_1\in\mathcal{F}$. It is not maximal, and so there is some $I_2\in\mathcal{F}$ that strictly contains it. Again, $I_2$ is not maximal, so there is some $I_3\in\mathcal{F}$ that strictly contains it. And so on. We get a strictly increasing sequence of ideals:

$I_1\subset I_2\subset I_3\subset...$

Which is a contradiction. Note that we do have to use some form of the axiom of choice here (to be precise, countable choice), but as you probably know, this axiom is always assumed in ring theory. (Zorn's lemma is used a lot in the book you are reading)

As for the second question, note that a maximal element is an element that is not strictly contained in any other element. (maybe you thought about the wrong definition, like an element that contains every other element in the set) So in your example, both elements are maximal.

  • $\begingroup$ Thankieees! So, what is wrong with my "counter-example"? $\endgroup$
    – Dibidus
    Apr 18 at 6:41
  • $\begingroup$ @Dibidus The element $\langle x\rangle$ is not strictly contained in any element of your set, right? So it is a maximal element. For the same reason, $\langle y\rangle$ is also maximal. So this is not a counter-example. $\endgroup$
    – Mark
    Apr 18 at 6:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .