Open set in the spectrum of a ring?

Question

Consider $Spec(K[X])$ where $K$ is an algebraically closed field. Is $0$ open in the Zariski topology on spectrum? Does the spectrum have points which are neither open nor closed?

Answer 1

Closed points correspond to maximal ideals, hence are all those of the form $[(X-a)]$ for $a\in K$. The point $\eta=[(0)]$ corresponding to the $(0)$ ideal, which is prime but not maximal, is neither closed nor open. A closed subset of $\textrm{Spec }K[X]$ is of the form $V(f_1,\dots,f_r)$, thus (if non-empty) consists of finitely many closed points. Any non-empty open set is thus infinite (and contains $\eta$).