TITLE: Example of non-noetherian ring whose spectrum is noetherian
QUESTION [6 upvotes]: Since spectrum of noetherian ring is a noetherian topological space, I am finding an example s.t. a non-noetherian ring whose spectrum is noetherian.
Since most nice rings are noetherian, actually I do not have many examples to start, does any one can help? Thanks!

REPLY [14 votes]: The standard example here is $A=k[x_1,x_2,\dots]/(x_1^2,x_2^2,\dots)$, for $k$ a field.  Since each variable $x_n$ is nilpotent, every prime must contain $I=(x_1,x_2,\dots)$.  But $A/I$ is just $k$, so $I$ is already a maximal ideal.  So $I$ is the only prime, and so $\operatorname{Spec}(A)$ has only one point and is obviously Noetherian.  But $I$ is not finitely generated, so $A$ is not Noetherian.