Keep in mind, I'm strictly an amateur, though a very old one. I learned about imaginary numbers barely two years ago and ideals a year ago, and I'm still decidedly a novice in both topics.

In the university library, I was looking at Modules over Non-Noetherian Domains by Fuchs and Salce and I couldn't really understand anything. I'm also looking at the "Questions that may already have your answer," but if they do, it's not in a way that I can understand.

Then I thought what about a finite ring, like maybe $\mathbb{Z}_{10}$, but that's only created more questions, like: can a finite ring be non-Noetherian? Although $5 = 5^n$ for any $n \in \mathbb{Z}_{10}$ besides $0$, we're still dealing with only the ideal $\langle 5 \rangle$, right? There's no ascending chain of ideals even though some numbers in this domain have infinitely many factorizations, right? It is a Noetherian ring after all, right?

My question, it seems, has then become if it's possible for a non-Noetherian ring to be within the comprehension of a dilettante such as myself, or must it necessarily be esoteric and exotic?

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    $\begingroup$ Typically when people say "domain", they mean an integral domain which means no zero divisors. So $\mathbb{Z}_{10}$ is not a good place to start since it is not a domain. A very useful fact is also the following - if a ring $S$ is a quotient $R/I$ of a noetherian ring $R$ by an ideal $I$ then $S$ is also noetherian. So all of the finite rings $\mathbb{Z}/(n)$ will be noetherian because $\mathbb{Z}$ is. You might try things like $k[x_1,x_2,x_3,\ldots]$ where you adjoin countably infinitely many indeterminates to a field... $\endgroup$ Jun 9, 2016 at 22:39
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    $\begingroup$ @JohnMartin I think I've understood everything in your comment except the last sentence. Would $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6} \ldots)$ be the sort of thing you're talking about, or would that be too determinate? $\endgroup$
    – Mr. Brooks
    Jun 9, 2016 at 22:47
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    $\begingroup$ What you mention would be a field, and a field is always noetherian (since it has only two ideals). $\endgroup$ Jun 9, 2016 at 22:51
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    $\begingroup$ @Mr.Brooks: Something related to yours. Let $R$ be the ring of all algebraic integers. Then the ideals generated by $\sqrt{2}$, $\sqrt[4]{2}$, $\sqrt[8]{2}$, and so on are strictly ascending. $\endgroup$ Jun 9, 2016 at 23:07
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    $\begingroup$ Observe that sqrt(6) is already included if you have both sqrt(2) and sqrt(3). You only need the roots of primes in this field extension. $\endgroup$
    – Nij
    Jun 10, 2016 at 4:22

5 Answers 5


For a field $k$ the ring $k[x_1,x_2,\dots]$ of polynomials with infinite indeterminates is non-Noetherian because you can take the ascending chain

$$(x_1)\subset (x_1,x_2)\subset(x_1,x_2,x_3)\subset\cdots$$

And also to answer one of your questions, a finite ring must be Noetherian because an equivalent definition of Noetherian is "every ideal is finitely generated", so if $R$ is finite then each ideal $I$ is finite and in particular it's generated by itself, so every ideal is finitely generated.

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    $\begingroup$ Alternately (for the second part): a finite ring must be Noetherian because it can't have an infinite ascending chain of ideals, because it has only finitely many ideals at all, because it has only finitely many subsets! $\endgroup$ Jun 10, 2016 at 7:48

Being "Noetherian" can be read as a ring for which any ascending chain of ideals has a "biggest ideal", one that contains all the others but is only contained by ideals which are equal to itself.
To be non-Noetherian, the ring simply needs to have an infinite ascending chain of ideals. The ring of algebraic integers, for example, has the infinite chain of ideals generated by $2^{1/{2^{n}}}$.
That is, $$\langle \sqrt{2} \rangle \subset \langle \sqrt[4]{2} \rangle \subset \langle \sqrt[8]{2} \rangle \subset \langle \sqrt[16]{2} \rangle \subset \dots$$ forms a chain without a "biggest link".


The ring of integer-valued polynomials (the subring of $\mathbf Q[x]$ of polynomials which take integer values at integers) is another example of a non-noetherian integral domain.

The ring of continuous functions on $[a, b]$ is yet another example (it's not an integral domain).

  • $\begingroup$ why is your first example non-Noetherian? $\endgroup$ Jun 11, 2016 at 21:16
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    $\begingroup$ I don't know how it's proved, and I think it's not trivial . You'll probably find a proof or a reference to a proof in Cahen, P-J.; Chabert, J-L. , Integer-valued polynomials, Mathematical Surveys and Monographs 48, American Mathematical Society(1997). They were first introduced by G. Pólya in a paper from 1919 (in German). $\endgroup$
    – Bernard
    Jun 11, 2016 at 21:40
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    $\begingroup$ @AlexMathers It is not noetherian because the ascending chain of ideals of the form $I_j:=I(\binom{X}{1},\binom{X}{p_1},\ldots,\binom{X}{p_j})$ where $p_j$ is the $j$th prime does not stabilize. This is found in What are rings of integer-valued polynomials? by M. Steward. $\endgroup$
    – Jose Brox
    Nov 17, 2020 at 12:04

You should check out Richard Borcherds' very accessible (at least to begin with) and entertaining online lecture series on Commutative Algebra. In his lecture on Noetherian Rings he gives several very instructive examples of rings which are all somewhat similar, and shows that some are Noetherian while others aren't.

Here are his examples:

  • polynomials over the reals (Noetherian)
  • analytic functions over the reals (Non-Noetherian)
  • analytic funtions on [-1,1] (Noetherian)
  • analytic functions on (-1,1) (Non-Noetherian)
  • functions analytic in a neighbourhood of zero (Noetherian)
  • functions that are smooth in a neighbourhood of zero, i.e. germs of smooth functions (Non-Noetherian)
  • formal power series over the reals (Noetherian)

As Marc van Leeuwen answered in here, sub $\mathbb{Z}$ module $R=X\mathbb{Q}[X]+\mathbb{Z}$ of $\mathbb{Q}[X]$ forms a ring, and this is not Noetherian! I commented the reason there.


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