Lemma 10.16.5. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. The map $R \to S^{-1}R$ induces via the functoriality of $\mathop{\mathrm{Spec}}$ a homeomorphism

\[ \mathop{\mathrm{Spec}}(S^{-1}R) \longrightarrow \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid S \cap \mathfrak p = \emptyset \} \]

where the topology on the right hand side is that induced from the Zariski topology on $\mathop{\mathrm{Spec}}(R)$. The inverse map is given by $\mathfrak p \mapsto S^{-1}\mathfrak p$.

**Proof.**
Denote the right hand side of the arrow of the lemma by $D$. Choose a prime $\mathfrak p' \subset S^{-1}R$ and let $\mathfrak p$ the inverse image of $\mathfrak p'$ in $R$. Since $\mathfrak p'$ does not contain $1$ we see that $\mathfrak p$ does not contain any element of $S$. Hence $\mathfrak p \in D$ and we see that the image is contained in $D$. Let $\mathfrak p \in D$. By assumption the image $\overline{S}$ does not contain $0$. By basic notion (54) $\overline{S}^{-1}(R/\mathfrak p)$ is not the zero ring. By basic notion (62) we see $S^{-1}R / S^{-1}\mathfrak p = \overline{S}^{-1}(R/\mathfrak p)$ is a domain, and hence $S^{-1}\mathfrak p$ is a prime. The equality of rings also shows that the inverse image of $S^{-1}\mathfrak p$ in $R$ is equal to $\mathfrak p$, because $R/\mathfrak p \to \overline{S}^{-1}(R/\mathfrak p)$ is injective by basic notion (55). This proves that the map $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ is bijective onto $D$ with inverse as given. It is continuous by Lemma 10.16.4. Finally, let $D(g) \subset \mathop{\mathrm{Spec}}(S^{-1}R)$ be a standard open. Write $g = h/s$ for some $h\in R$ and $s\in S$. Since $g$ and $h/1$ differ by a unit we have $D(g) = D(h/1)$ in $\mathop{\mathrm{Spec}}(S^{-1}R)$. Hence by Lemma 10.16.4 and the bijectivity above the image of $D(g) = D(h/1)$ is $D \cap D(h)$. This proves the map is open as well.
$\square$

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