"maximal condition" means if any non-empty collection of ideals in R has a maximal element (under set inclusion). And we define noetherian ring to be a ring such that any ascending chain of ideals is finite.

I don't see why the definition implies the maximal condition. Let's suppose $S=\{\mathfrak{A}_1, \mathfrak{A}_2\}$ such that $\mathfrak{A}_1\bigcap\mathfrak{A}_2=\{0_R\}$ and none of them is unit ideal. Then there is no such maximal element. There is no need for an ideal properly contains in another ideal if it is not a maximal element. Can anyone point out what's the problem here? Thanks in advance

Edit: What I am confusing is what will be the maximal element in my example.

My definition for maximal element is: given $(S, R)$ a poset and $T\subset S$, then $s\in S$ a maximal element of $T$ if $sRy$ with $y\in S$ implies $s=y$.

And I proved "noetherian ring satisfies the maximal condition" in this way: Since any non-empty collection of ideals is a poset, let $S$ be one and $T$ be a chain of $S$. As R is noetherian, there must be an "end" for the chain, namely $\mathfrak{A}\in T$. Then it is an upper bound of $T$ and hence $S$ is inductive and the result follow from Zorn's Lemma. But then I found I can't tell what is the maximal element in my example. So either my proof is false or something else goes wrong

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    $\begingroup$ I'm not entirely certain what your question is, but are you confusing "maximal" with "greatest"? $\endgroup$
    – Arthur
    Feb 1, 2018 at 11:10
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    $\begingroup$ maximal = there is no bigger than this; greatest = this is bigger than all others $\endgroup$ Feb 1, 2018 at 11:11
  • $\begingroup$ @HagenvonEitzen So both elements in my example are maximal? $\endgroup$ Feb 1, 2018 at 11:40
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    $\begingroup$ Yep, it looks like it. Neither of them is the greatest, because they are not comparable to each other, but both are maximal, because, for either of them, the 'other' one is not greater than that one... Hope it makes sense. $\endgroup$
    – user491874
    Feb 1, 2018 at 11:51
  • $\begingroup$ This subtlety is absent in e.g. Zorn’s Lemma which requires the subset in question to be totally ordered, meaning you can compare every element of the subset with each other. But in the Noetherian condition you are allowed to have subsets which are incomparable, as you’ve noticed. $\endgroup$
    – Ducky
    Feb 1, 2018 at 11:58

1 Answer 1


Suppose $S$ is a nonempty set of ideals with no maximal element.

If $I_0\in S$, there is $I_1\in S$ with $I_1\supsetneq I_0$. But neither $I_1$ is maximal, so…

This starts a recursion, and proves the ring is not noetherian.

More formally, suppose we have found a chain of ideals belonging to $S$, $I_0\subsetneq I_1\subsetneq \dots \subsetneq I_k$. Then, as $I_k$ is not maximal by assumption, we can further extend the chain. Thus $R$ is not noetherian.

Your argument is good as well: chains in $S$ are finite, so every chain has an upper bound, namely the terminal element. Zorn's lemma provides the required maximal element. But you have to remember that Zorn's lemma is a “purely existential” statement and offers no method/algorithm to determine a maximal element.


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