Lemma 10.33.1. 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. Then $S^{-1}R \cong R/I$ for some ideal $I \subset R$.

Proof. Let $I = \mathop{\mathrm{Ker}}(R \to S^{-1}R)$ so that $V(I)$ contains the image. Say the image is the closed subset $V(I') \subset \mathop{\mathrm{Spec}}(R)$ for some ideal $I' \subset R$. So $V(I') \subset V(I)$. For $f \in I'$ we see that $f/1 \in S^{-1}R$ is contained in every prime ideal. Hence $f^ n$ maps to zero in $S^{-1}R$ for some $n \geq 1$ (Lemma 10.17.2). Hence $V(I') = V(I)$. Then this implies every $g \in S$ is invertible mod $I$. Hence we get ring maps $R/I \to S^{-1}R$ and $S^{-1}R \to R/I$. The first map is injective by choice of $I$. The second is the map $S^{-1}R \to S^{-1}(R/I) = R/I$ which has kernel $S^{-1}I$ because localization is exact. Since $S^{-1}I = 0$ we see also the second map is injective. Hence $S^{-1}R \cong R/I$. $\square$

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