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 $I \subset R$.

## 10.33 Curiosity

Lemma 10.24.3 explains what happens if $V(I)$ is open for some ideal $I \subset R$. But what if $\mathop{\mathrm{Spec}}(S^{-1}R)$ is closed in $\mathop{\mathrm{Spec}}(R)$? The next two lemmas give a partial answer. For more information see Section 10.108.

**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 $S^{-1}R \cong R/I$.
$\square$

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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