Lemma 10.31.5. If R is a Noetherian ring then \mathop{\mathrm{Spec}}(R) is a Noetherian topological space, see Topology, Definition 5.9.1.
Proof. This is because any closed subset of \mathop{\mathrm{Spec}}(R) is uniquely of the form V(I) with I a radical ideal, see Lemma 10.17.2. And this correspondence is inclusion reversing. Thus the result follows from the definitions. \square
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