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).
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http://math.stackexchange.com/questions/7392/a-non-noetherian-ring-with-textspecr-noetherian?rq=1 – Youngsu Jun 08 '14 at 03:04
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Thanks for the link. The rings there seem to all be local. If $R$ is local then Spec($R$) is a Noetherian scheme only if $R$ is a Noetherian ring. – user51197 Jun 08 '14 at 03:34
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You are right. Thanks for pointing it out. – Youngsu Jun 08 '14 at 07:08
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$\DeclareMathOperator{\Spec}{Spec}$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.
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