$Z(I)$ is homeomorphic to $\operatorname{Spec}(R/I)$ Let $R$ be a commutative ring. How do you prove that $Z(I)$ is homeomorphic to $\operatorname{Spec}(R/I)$, where $Z(I)$ is equipped with the topology induced from the Zariski topology on $\operatorname{Spec}R$?
It is clear that this is bijection. And the maps are defined in this question Correspondence theorem for rings. What is the easiest way to prove that they are continuous? I tried to show that both maps are closed but I didn't succeed.
 A: How do we define our map $f:Z(I)\to Spec(A/I)$? It is given by sending $\mathfrak p\mapsto \mathfrak p/I$ for a prime ideal $\mathfrak p$ containing $I$.
How do we see first that this is continuous? We can either check $(1)$ preimages of open sets are open, or $(2)$ preimages of closed sets are closed. Considering that we define things in the Zariski topology in terms of closed sets, $(2)$ is probably easier. 
What does a closed set of $Spec(A/I)$ look like? Well, since all ideals of $A/I$ are of the form $J/I$ for an ideal $J\subseteq A$ containing $I$, a closed subset of $Spec(A/I)$ has the form $Z(J/I)$. I encourage you to check that the preimage of this set under $f$ is given by $Z(J)$, which is closed in the subspace topology on $Z(I)$ because it is equal to $Z(J)\cap Z(I)$ (why?).
Now, the last thing you need to check is $f$ is a closed map. A closed subset of $Z(I)$ has the form $Z(I)\cap Z(J)$ for some ideal $J$ of $A$, which is equal to $Z(I+J)$. Using the same logic, it should be easy for you to check that the image of this set under $f$ is equal to $Z((I+J)/I)$, which is closed in $Spec(A/I)$.
