Homeomorphism between spectra of two rings

Question

Given a ring $R$ and a multiplicative system $S\subseteq R$, take the inclusion $j:R\hookrightarrow S^{-1} R$. I want to prove that $\operatorname{Spec} j:\operatorname{Spec} S^{-1}R\to \operatorname{Spec} R$ induces a homeomorphism between $\operatorname{Spec}S^{-1}R$ and $\{p\in \operatorname{Spec}R:p\cap S=\emptyset\}$. I already proved that $\operatorname{Spec}j $ induces a bijection, and I know that is continuous since $\operatorname{Spec} $ is a functor. My problem is that I don't see why it's a closed map: in other words, if I take an ideal $I$ of $R$ and that doesn't have any element in common with $S$, and I consider the prime ideals containing $I$, I don't see why I recover only ideals that don't have any element in common with $S$. Can you clarify my ideas? Thanks in advance

Answer 1

In general the map might not be either open nor closed. For example, for $p \in Spec(R)$, $Spec(R_p)$ is the set of all prime ideals that are contained in $p$. So $(2) \in Spec(\mathbb{Z})$ is the set $\{(0), (2)\}$ which is neither open nor closed.

if $f : X \rightarrow Y$ is a continuous bijection between topological spaces, then $f$ is a homeomorphism onto its range iff $f^{-1} : f(X) \rightarrow X$ is continuous. This requires that for any closed set $Z$ in $X$, $f^{-1}(Z)$ is closed in $f(X)$ in the subspace topology. Note that this is not the same as requiring that $f^{-1}(Z)$ is closed in $Y$.

Now suppose $I$ is an ideal in $S^{-1}R$. Let $J \subseteq R$ be the set $\{x \in R : x/1 \in I\}$. Note that since $I$ is an ideal, $J$ is also an ideal. Furthermore, for any $a/b \in I$, $a/b * b/1 \in I$ as well, so $a/1 \in I$, so $a \in J$.

Now, let $\phi : R \rightarrow S^{-1}R$ be the localization homomorphism, and let $p \in Spec(S^{-1}R)$ be arbitrary. $p \in V(I)$ iff $I \subseteq p$, and $\phi^{-1}(p) \in V(J)$ iff $J \subseteq \phi^{-1}(p)$.

Therefore, let's prove that $I \subseteq p$ iff $J \subseteq \phi^{-1}(p)$. Suppose $I \subseteq p$, and let $a \in J$ be arbitrary. Then, $\phi(a) = a/1 \in I$, so $a/1 \in p$. This implies that $a \in \phi^{-1}(p)$. Conversely, suppose $J \subseteq \phi^{-1}(p)$, and let $a/b \in I$ be arbitrary. From above, we have $a \in J$, so $a \in \phi^{-1}(p)$. This implies that $\phi(a) = a/1 \in p$, so $a/b = a/1 * 1/b \in p$ as well.