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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

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It's spelled noetherian. –  Rasmus Aug 16 '12 at 13:17
It's even often spelled Noetherian :-) –  PseudoNeo Aug 16 '12 at 13:56

3 Answers 3

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.

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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.

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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.

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great and simple answer. –  messi May 8 '13 at 9:31

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