TITLE: Why Zariski topology?
QUESTION [75 upvotes]: Why in algebraic geometry we usually consider the Zariski topology  on $\mathbb A^n_k$? Ultimately it seems a not very interesting topology, infact the open sets are very large and it doesn't satisfy the Hausdorff separation axiom. Ok the basis is very simple, but what are the advantages?

REPLY [6 votes]: The following answer has a similar spirit to Zhen Lin's. Like his answer, it has a strong logical flavor; unlike his, no toposes appear (though they are lurking in the background).
That said, allow me a slight switch in terminology: Let's define $\operatorname{Spec} A$ not as the set of prime ideals of $A$, but as the set of filters in $A$. The axioms for a filter are precisely dual to the axioms of a prime ideal, so that a subset of $A$ is a prime ideal if and only if its complement is a filter. For instance, while prime ideals have the axiom "$x \in \mathfrak{p} \wedge y \in \mathfrak{p} \Rightarrow x+y \in \mathfrak{p}$", filters have the axiom "$x + y \in F \Rightarrow x \in F \vee y \in F$".
It happens that the axioms of a filter have a certain logical form; they form a so-called "geometric theory". For any geometric theory, there is an associated space of its models, which will be automatically endowed with a suitable topology. In the case of the geometric theory of filters, a model is precisely a filter, so the space of models coincides with the spectrum; and the automatically given topology is precisely the Zariski topology.
A very readable introduction to this point of view are notes by Steve Vickers ("Continuity and Geometric Logic").