Let $R$ be a commutative ring, I always thought that finitely generated ideals are "small" and the non finitely generated ideals are "large" but this is not quite correct...

Lemma. In a non-noetherian ring $R$, there exists a maximal non finitely generated ideal w.r.t. set inclusion.

The proof is an application of Zorn's lemma.

Given the non empty collection $\mathcal{C}$ of non finitely generated ideals, if we have any ascending chain $$I_1\subset I_2\subset \cdots$$ we claim that the union $\bigcup_k I_k$ is an upper bound in $\mathcal{C}$.

$\bigcup_k I_k$ is an ideal because the union is nested, and if it is finitely generated (i.e. not in $\mathcal{C}$), then the set of finite generators must be contained in some $I_{n^*}$, and this finite set of generators would generate $I_{n^*}$ which contradicts $I_{n^*}\in \mathcal{C}$. By Zorn's lemma, there exists a maximal element in $\mathcal{C}$.

We call this maximal element $P$, then given any element $x\in R-P$, then we see that $P+\langle x\rangle$ must be finitely generated! Because $P\subsetneq P+\langle x\rangle$ and $P$ is the maximal non finitely generated ideal in $R$. So in a sense, a lot of these "large" ideals will be finitely generated.

  • 1
    $\begingroup$ What exactly is your question? Have you worked through any examples? E.g., rings of polynomials in infinitely many variables. $\endgroup$
    – Rob Arthan
    Sep 9, 2017 at 21:35
  • $\begingroup$ I know a few standard examples of non finitely generated ideals from Noetherian rings chapter we did. I do not really see a lot of finitely generated ideal examples it is like $\langle r_1, \cdots r_n \rangle$. I have seen polynomial ring with infinitely many variable, and for rings of formal power series, we did some exercise about it and it is a local ring. $\endgroup$
    – Xiao
    Sep 9, 2017 at 21:43
  • $\begingroup$ Perhaps you could add something like "where is my intuition going wrong?" to make your "question" into a question. I've tried to answer that question. $\endgroup$
    – Rob Arthan
    Sep 9, 2017 at 22:16

1 Answer 1


Your idea that finitely generated ideals are "small" and non-finitely generated ideals are "large" is misleading. Think about $R = \Bbb{Z}[X_1, X_2, \ldots]$, the ring of polynomials over the integers in countably many variables $X_1, X_2, \ldots$ If you define $J_k = \langle X_1, \ldots, X_k\rangle$, then $J_1 \subseteq J_2 \subseteq \ldots$ is a strictly increasing chain of ideals, so $R$ is not Noetherian. The finitely generated ideal $J = \langle 2 \rangle $ properly contains the non-finitely generated ideal $K = \langle 2X_1, 2X_2, \ldots \rangle$. For any prime $p$, the ideal $L_p$ comprising the polynomials with constant term a multiple of $p$ is maximal non-finitely generated.

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    $\begingroup$ Thank you for the answer, I think I know where my understanding is wrong now. When I see $\langle r_1, \cdots r_n\rangle$, I always think of it as "$\text{span}\{r_1, \cdots r_n\}$" because the elements are linear combinations, but here scalars are all of $R$, so a small set does not imply small ideal. $\endgroup$
    – Xiao
    Sep 9, 2017 at 22:23

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