A remark on Cohen's theorem It is well known as Cohen's theorem that a commutative ring is Noetherian if all its
prime ideals are finitely generated. Is this statement true or false when prime ideals are replaced by maximal ideals? 
 A: See the following paper (and search for its citations for related work)
Gilmer, R; Heinzer W.
A non-Noetherian two-dimensional Hilbert domain with principal maximal ideals,
Michigan J. Math. 23 (1976), 353-362
Link
A: This is an old question, but Julian Rosen and I observed an example that is a bit simpler than those written above. I am posting it in case it's helpful to others.
It is a bit closer to algebraic geometry, in particular a one-dimensional subquotient of $k[x,y]$ where $k$ is an algebraically closed field, basically a thickening of $k[x]$ at the origin.
Let $\displaystyle{R = \frac{k[x,y,\tfrac{y}{x}, \tfrac{y}{x^2}, \ldots]}{(y)}.}$
Claim 1. The maximal ideals of $R$ are $(x-c)$ for $c \in k$.
Claim 2. The only other prime ideal is the nilradical of $R$, which is $\displaystyle{(\tfrac{y}{x}, \tfrac{y}{x^2}, \ldots)}$.
Proof: By the description of the nilradical, $R / \mathrm{rad}(R) \cong k[x]$, which makes it easy to list all the prime ideals. It suffices to check their generators in $R$, i.e. each $(x-c) \subset R$ actually is maximal and that the nilradical is as described.
For the nilradical claim, $(\tfrac{y}{x^n})^2 = y \cdot \tfrac{y}{x^{2n}} = 0$ for all $n$. Quotienting by them leaves $k[x]$ which is reduced.
For the maximal ideals claim, set $x=c$. If $c \ne 0$, then $\tfrac{y}{x^n} \cdot c^n = \tfrac{y}{x^n} \cdot x^n = y = 0$, so $\tfrac{y}{x^n} = 0$ for all $n$, which leaves a field. On the other hand if $c = 0$, then $\tfrac{y}{x^n} = \tfrac{y}{x^{n+1}} \cdot x = 0$ for all $n$. Either way, we get a field.
A: Here is another nice ring which shows that the answer is: "false". It is due to B. Osofsky and can serve as (counter)example in other situations.
Start with the $p$-adics $\mathbb{Z}_p$ and $\mathbb{Q}_p$. (Any other complete DVR with its fraction field would do.) Our ring, call it $R$, is the additive group $\mathbb{Z}_p \oplus \mathbb{Q}_p/\mathbb{Z}_p$ with multiplication $(a,b) \cdot (c,d) = (ac, ad + cb)$. (This is called the "trivial extension" of the ring $\mathbb{Z}_p$ by its module $\mathbb{Q}_p/\mathbb{Z}_p$.) The ideals in $R$ form the chain
$R \supsetneq pR \supsetneq p^2R \supsetneq \; ... \; (0, \mathbb{Q}_p/\mathbb{Z}_p) \; ... \; \supsetneq (0, p^{-2}\mathbb{Z}_p/\mathbb{Z}_p) \supsetneq (0, p^{-1}\mathbb{Z}_p/\mathbb{Z}_p) \supsetneq 0$
Among other interesting properties that $R$ has (like being a cogenerator ring, in particular self-injective), it is local with principal maximal ideal $pR$; but it has exactly one non-maximal prime ideal -- the lonely one there in the middle of the chain, having no neighbours --, and this sad thing is not finitely generated.
