Lemma 10.33.2. Let R be a ring. Let S \subset R be a multiplicative subset. Assume the image of the map \mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R) is closed. If R is Noetherian, or \mathop{\mathrm{Spec}}(R) is a Noetherian topological space, or S is finitely generated as a monoid, then R \cong S^{-1}R \times R' for some ring R'.
Proof. By Lemma 10.33.1 we have S^{-1}R \cong R/I for some ideal I \subset R. By Lemma 10.24.3 it suffices to show that V(I) is open. If R is Noetherian then \mathop{\mathrm{Spec}}(R) is a Noetherian topological space, see Lemma 10.31.5. If \mathop{\mathrm{Spec}}(R) is a Noetherian topological space, then the complement \mathop{\mathrm{Spec}}(R) \setminus V(I) is quasi-compact, see Topology, Lemma 5.12.13. Hence there exist finitely many f_1, \ldots , f_ n \in I such that V(I) = V(f_1, \ldots , f_ n). Since each f_ i maps to zero in S^{-1}R there exists a g \in S such that gf_ i = 0 for i = 1, \ldots , n. Hence D(g) = V(I) as desired. In case S is finitely generated as a monoid, say S is generated by g_1, \ldots , g_ m, then S^{-1}R \cong R_{g_1 \ldots g_ m} and we conclude that V(I) = D(g_1 \ldots g_ m). \square
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