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
Best Answer
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:
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.