Consider a commutative ring R with unity and consider and element $f\in R$. I wish to show that there is a homeomorphism between the two sets $\{\mathcal{p}\in Spec(R),f\notin \mathcal{p}\}$ where $p$ are the primary ideals in $R$ and $Spec(R_{f})$, the prime spectrum of the localisation of $R$ at $f$.
I already know there is a bijection between the two sets, but I am really struggling to show it is continuous. My attempt has been, consider an ideal $I\in R_{f}$. Then $V(I)$ is the set of all prime ideals $p$ which contain $I$. I want to now show that the preimage is also a closed set so I need to find an $I'\in R$ such that $V(I')$ is the preimage of $V(I)$. At this stage I keep getting lost. Hint rather than answer would be appreciated
Denote $U_f=\{p\in Spec(R),$ $f$ is not in $p\}$. Let $i:R\rightarrow R_f$ the canonical map, it induces a continuous map $u:Spec(R_f)\rightarrow Spec(R)$ whose image is the open subset $U_f$, Since it is bijective, you have only to show that $u$ is open, here $U_f$ is endowed with the topology induced by the topology of $Spec(R)$ which is equivalent to saying that an open subset of $U_f$ is $V-f$ where $V$ is an open subset of $V$.
Remark that the topology of $Spec(R_f)$ is generated by $V=Spec(R_f)-V(I)$ where $I$ is an ideal of $R_f$ which is the image of an ideal $J$ of $R$ which does not contains $f$. This implies that $u(V)$ is $U_f-V(J)$ thus is open, done.