Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In commutative rings we have the following

Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated.

From this Theorem I am looking for commutative rings $R$ in which every maximal ideal is finitely generated but $R$ is non Noetherian.

Question: Is there a straightforward example of a commutative ring $R$ so that each maximal ideal is finitely generated, but $R$ is non Noetherian?

Thank You

share|improve this question

4 Answers 4

Let $A = C^\infty(S^1)$ be the ring of smooth functions on the circle (if you prefer, you can see it as the ring of smooth $2\pi$-periodic functions $\mathbb R \to \mathbb R$).

First, $A$ isn't Noetherian : the ideal $I_{\mathscr V(0)}$ of functions vanishing on a neighbourhood of $0$ isn't finitely generated.

But the maximal ideals of $A$ are exactly the $$\mathfrak m_p = \left\{ f \in A\, \Big |\, f(p) = 0 \right \},$$ for $p \in S^1$, which are generated by the two functions $(x,y) \mapsto x-x_p$ and $(x,y) \mapsto y - y_p$. (If you think of $A$ as a set of trigonometric functions, $x$ is $\cos$ and $y$ is $\sin$).

Proof of the various claims:

  1. $I_{\mathscr V(0)}$ isn't f.g. : Suppose ad absurdum that $I_{\mathscr V(0)} = (f_1, \ldots, f_r)$ where each $f_i$ vanishes on a neighbourhood $V_i$ of $0$. Then any function of $(f_1, \ldots, f_r)$ vanishes on $V = \bigcap V_i$, which is a fixed neighbourhood of 0. But it is easy to construct functions of $A$ vanishing on a neighbourhood of $0$ as small as desired (in particular, strictly smaller than $V$), a contradiction.
  2. $\mathrm{Max}(A) = \left\{ \mathfrak m_p \, \Big | \, p \in S^1 \right\}$ : Let $I$ be an ideal of $A$. We are going to prove that either $I$ is contained in some $\mathfrak m_p$ or $I = A$. The negation of “$I$ is contained in some $\mathfrak m_p$” is “forall $p \in S^1$, there is a function $f$ s.t. $f(p) \neq 0$”. Since the set on which a function doesn't vanish is open and $S^1$ is compact, that implies the existence of finitely many functions $f_1, \ldots, f_r \in I$ such that $\forall p \in S^1, \exists i : f_i(p) \neq 0$. Then, $f = f_1^2 + \cdots + f_r^2 \in I$ is everywhere nonzero, so it is invertible in $A$ and $I = A$.
  3. $\mathfrak m_p = (x-x_p, y-y_p)$ : The inclusion $\supseteq$ is clear. Let $f \in \mathfrak m_p$. By definition of a smooth function on a submanifold, $f$ is the restriction of a smooth function $F \in C^1(V)$ for some neighbourhood $V$ of $p$ in $\mathbb R^2$. Of course, $F$ still vanishes on $p$. The claim then follows from Hadamard's lemma.

PS : All this seems to indicate that $A$ has some strange (in particular non f.g.) prime ideals. I must confess I cannot really understand who they are.

share|improve this answer
An enthusiastic +1 for this great answer, PseudoNeo! –  Georges Elencwajg Aug 16 '12 at 14:33
+1 for a nice, natural and geometric answer! –  Dedalus Feb 20 '13 at 12:03

As I say in this MO answer, a valuation ring with value group $\mathbb{Z} \times \mathbb{Z}$ (ordered lexicographically) is a non-Noetherian domain whose unique maximal ideal is principal.

share|improve this answer

In this topic I've constructed a ring $R\times M$ which is called the idealization of the $R$-module $M$ or the trivial extension of $R$ by $M$. In the special case $R=\mathbb{Z}_{(2)}$ (the localization of $\mathbb{Z}$ at the prime ideal $2\mathbb{Z}$) and $M=\mathbb{Q}$ one obtains a local ring which is not noetherian and its maximal ideal $2\mathbb{Z}_{(2)}\times\mathbb{Q}$ is principal.

share|improve this answer

Based on @PseudoNeo: answer. Take the ring of germs of smooth functions $\mathcal{C}_a$ at a point $a$ in some manifold, a local algebra. Hadamard's lemma says that the unique maximal $\mathfrak{m}$ ideal is finitely generated. The ideal $$\mathfrak{m}^{\infty} \colon = \bigcap_{n \ge 0} \mathfrak{m}^n$$ is the set of germs of functions all whose derivative vanish at the base point. This coincides with the kernel of the map germ of function $\mapsto$ Taylor expansion $$\mathcal{C}_a \to \mathbb{R}[t_1, \ldots, t_m]$$ to the domain of power series, so this ideal is prime. (by the way, a surjective morphism, by A. Borel)

Now $\mathfrak{m}^{\infty}$ is not finitely generated. Indeed, we have $$\mathfrak{m}\cdot \mathfrak{m}^{\infty} = \mathfrak{m}^{\infty}$$ If $\mathfrak{m}^{\infty}$ were finitely generated, then from Nakayama's lemma we would conclude that $\mathfrak{m}^{\infty}=0$. But $\mathfrak{m}^{\infty}\ne 0$, for instance the Cauchy example of a function with infinite order of vanishing at $a$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.