TITLE: Commutative non Noetherian rings in which all maximal ideals are finitely generated
QUESTION [38 upvotes]: 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

REPLY [5 votes]: 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 E. 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 $\mathfrak{m}^{\infty}\ne 0$, see the Cauchy's example of a function with infinite order of vanishing at $a$.