For a ring homomorphism, why does $f$ induces a homeomorphism from $SpecB$ onto the closed subset $V(\ker f)$ of $SpecA$.

263 Views Asked by At

Let $\varphi : A \rightarrow B$ be a ring homomorphism. Then we have a map of sets $Spec(\varphi):Spec(B) \rightarrow Spec(A)$ defined by $p \mapsto \varphi^{-1}(p)$ for every $p \in SpecB$. Let $f =Spec(\varphi)$ be the map associated to $\varphi$ as above.

a) If $\varphi$ is surjective, then $f$ induces a homeomorphism from $SpecB$ onto the closed subset $V(\ker \varphi)$ of $SpecA$.

b) If $\varphi$ is a localization morphism $A \longrightarrow S^{-1}A,$ then $f$ is a homomorphism from $Spec(S^{-1}A)$ onto the subspace $\{ p \in SpecA \mid p \cap S = \emptyset \}$ of $SpecA$.

Please tell me how to proof this two properties with details. It is about Qing Liu's book named Algebraic Geometry ang Arithmetic Curves (p. 28).