Non-Noetherian ring $R$ with Spec($R$) a Noetherian Scheme

Question

In looking at the examples of Non-Noetherian rings I knew/found I wasn't able to find one where I could conclude that Spec($R$) was a Noetherian scheme (not just merely a Noetherian topological space).

Answer 1

$\DeclareMathOperator{\Spec}{Spec}$In fact a ring $R$ is Noetherian iff $\Spec(R)$ is a Noetherian scheme. This is because the property of being (scheme-theoretically, locally) Noetherian is affine-local, in the sense that the following are equivalent for a scheme $X$:

i) $X$ can be covered by affines $\Spec(R_i)$ with $R_i$ Noetherian rings
ii) For every open affine $\Spec(R)$ of $X$, $R$ is a Noetherian ring

One way to check that a property $P$ of affine open subsets is affine local (i.e. satisfies (i) $\implies$ (ii)) is if the following two conditions are met:

a) if $P$ holds for $\Spec(A)$ and $f \in A$, then $P$ holds for $\Spec(A_f)$, and
b) if $f_1, \ldots, f_n \in A$ generate the unit ideal in $A$ and $P$ holds for $\Spec(A_{f_1}), \ldots, \Spec(A_{f_n})$, then $P$ holds for $\Spec(A)$

For a proof that this holds for $P = $ Noetherian, see e.g. Proposition 3.2 in Hartshorne.