Show that a map is continuous in the Zariski topology Let $R,S$ be two commutative rings with unity and $\alpha :R\to S$ be a ring homomorphism, for $f\in R$ is a non nilpotent element let $R_f$ denote the localization of $R$ with respect to the multiplicative closed set $\{1,f,f^2,f^3,\dots\}$, and $q:R\to R_f,r\mapsto\frac{r}{1}$ be the canonical homomorphism of rings. I want to show that (1) the map$$\alpha ^*:\text{Spec}(S)\to\text{Spec}(R),Q\mapsto \alpha^{-1}(Q)$$is continuous with respect to Zariski topology.(2) the map$$q^*:\text{Spec}(R_f)\to \text{Spec}(R),Q\mapsto q^{-1}(Q)$$is a topological isomorphism onto the open subset $D(f)$ of $\text{Spec}(R)$. where $D(f):=\{P\in \text{Spec}(R) \mid f\notin P\}.$I don't know how to show continuity of maps between rings, need your help, thank you.
 A: Hints:
For (1): To show that a map is continuous you show that the preimage of a closed set is closed.  The closed sets in $\mathrm{Spec} \ R$ are all of the form $V(I) = \{\mathfrak p \in \mathrm{Spec} \ R \ | \ I \subseteq \mathfrak p\}$ where $I \subseteq R$ is an ideal.  So you show that $\alpha^\ast$ is continuous let $I \subseteq R$ be an ideal and let $\alpha(I)S$ be the ideal in $S$ generated by the image of $I$.  You must show that $(\alpha^\ast)^{-1}(V(I)) = V(\alpha(I)S)$.
There's quite a bit of unpacking of definitions to be done here, but it's a very healthy exercise.  If you want to work with Spec it really helps to be fluid at unpacking those definitions and that's something that will only come with practice.
For (2): You have to show that it's bijective and you have to show that it's a closed map.  For both of these you will heavily use the correspondence theorem for ideals in a localization.  Note that an ideal intersects $\{1, f, f^2, \ldots\}$ if and only if it contains $f^n$ for some $n$ and a prime ideal contains $f^n$ if and only if it contains $f$.
