Example of non-noetherian ring whose spectrum is noetherian and infinite A topological space is noetherian if it satisfies the descending chain condition for its closed subsets. Let be $R$ a commutative ring and let $\mathrm{Spec}(R)$ its spectrum with Zariski topology.
I already know some examples of non-noetherian rings whose spectrum is noetherian, but in all these cases the spectrum is noetherian as it is finite. 

Can someone give me an example of a non-noetherian ring whose spectrum is noetherian and with infinite points ? 

Thanks!
 A: Theorem:  If Spec$(R)$ is Noetherian, then so is Spec$(R[X])$.  [Theorem 2.5 in ``Rings with Noetherian spectrum'' by Ohm and Pendleton]
So for the example, take $R[X]$, where $R$ is any non-Noetherian ring with Noetherian spectrum.
A: Commutative ring without unit.
Let $\mathbb{K}$ be a field, let $R=\mathbb{K}[x_n\mid n\in\mathbb{N}]_{\displaystyle/(x_n\mid n\in\mathbb{N})^2}$: we know that $R$ is not a Noetherian ring but $\operatorname{Spec}(R)=\{(x_n\mid n\in\mathbb{N})\}$ is a Noetherian topological space.
Let $S=\bigoplus_{k\in\mathbb{N}}R$, then $\operatorname{Spec}(S)=\coprod_{k\in\mathbb{N}}\operatorname{Spec}(R)$ where the Zariski topology is the cofinite topology: $S$ is not a Noetherian ring but $\operatorname{Spec}(S)$ is a Noetherian topological space with infinitely many points.
Commutative ring with unit.
Let
\begin{equation}
R=\{f\in\mathbb{Q}[t]\mid f(\mathbb{Z})\subseteq\mathbb{Z}\}
\end{equation}
the ring of numerical polynomial (over $\mathbb{Q}$); defined:
\begin{gather*}
f_0=1,\,\forall n\in\mathbb{N}_{\geq1}\equiv\mathbb{N}_1,f_n=\frac{1}{n!}t(t-1)\dots(t-n+1),
\end{gather*}
easily one can prove that $f_n\in R$ for any $n\in\mathbb{N}_0$.
Let
\begin{equation*}
\forall p\in\mathbb{P}_{\geq2},\,I_p=(f_1,\dots,f_{p-1});
\end{equation*}
by definition $f_p\notin I_p$, then $\{I_p\subset R\}_{p\in\mathbb{P}_{\geq2}}$ is a non stationary ascending chain of ideals in $R$, in other words $R$ is not a Noetherian ring.
Let $\mathbb{Z}[t]\stackrel{i}{\hookrightarrow}R$ the canonical inclusions, then one can consider the canonical continuous map $\operatorname{Spec}(R)\stackrel{i^{*}}{\to}\operatorname{Spec}(\mathbb{Z}[t])$.
$\operatorname{Spec}(R)$ has finitely many points; because any prime ideal of $\mathbb{Z}[t]$ of the form $(t-m)$ can be extended to a prime ideal of $R$ (via $i$). By proposition V.2.7.(iii) from Cahen, Chabert - Integer-Valued Polynomials one can state that $\dim_{Krull}R=2$ and therefore $\operatorname{Spec}(R)$ is Noetherian.
Is it all clear?
