On Spec of a localized ring Let $A$ be a commutative ring with unity, and $S$ a regular multiplicative subset of $A$ containing $1$. I know that given $f \in A$, Spec $A_f$ corresponds to an open subscheme of Spec $A$. I was wondering does Spec $S^{-1}A$ also correspond to an open subscheme of Spec $A$ as well (for any multiplicative subset $S$)? Thank you!
PS I guess what I am curious is that: when I consider the subset of Spec $A$ that does not meet $S$, is it isomorphic to Spec $S^{-1}A$ as schemes?
 A: By Bourbaki, Algèbre Commmutative, chapter II, paragraph 4, number 3, corollary of the proposition 13, the morphism of schemes $\textrm{Spec}(S^{-1} A)\to \textrm{Spec}(A)$ corresponding to the localisation map $A\to S^{-1} A$ induces an homeomorphism of $\textrm{Spec}(S^{-1} A)$ onto the subset of $\textrm{Spec}(A)$ consisting in points $x$ of $A$ such that $\mathfrak{p}_x \cap S = \varnothing$.
You can also see Grothendieck's Eléments de géométrie algébrique, I, corollaire 1.2.6. which results from the more general corollaire 1.2.4. Let me detail the latter.
corollaire 1.2.4. Let $\varphi : A' \to A$ a ring morphism such that each $f\in A$ can be written $f = h \varphi(f')$ for some $h\in A^{\times}$ and $f'\in A'$. Then on the spectra $\varphi$ induces an homeomorphism from $X = \textrm{Spec}(A)$ to its image.
Proof. Let's show that for each $P\subseteq A$ we can find a $P'\subseteq A'$ such that $V(E) = V(\varphi(E'))$. As $X$ is a $T_0$ topological space this will imply that the induced map is injective and is an homeomorphism, thanks to fact that ${\psi}^{-1}(V(E')) = V(\varphi(E'))$ where $\psi$ is the scheme morphism induced by $\varphi$. But for any $f \in E$ just take any $f'\in A'$ such that $f = h \varphi(f')$ for some $h\in A^{\times}$ (possible by hypothesis) and take $E'$ the set of such $f'$'s. $\square$
Note that a surjective $\varphi$ satisfies the hypothesis of this corollary. And that the localisation map $A\to S^{-1} A$ also satisfies it, giving an answer to your question.
