Irreducible components

Question

Why the set of irreducible components of a scheme is locally finite? Actually,we know that a noetherian space has only a finitely many irreducible components,but I want to know something more

Answer 1

This is not true in general. For instance, let $k$ be a field and let $A$ be the ring of sequences of elements of $k$ that are eventually constant. Then the prime ideals of $A$ are the following: for each $n$, the ideal $P_n$ of sequences whose $n$th term is $0$, and the ideal $Q$ of sequences which are eventually $0$. All of these ideals are maximal, so each irreducible component of $\operatorname{Spec} A$ is just a single point. But every open neighborhood of $Q$ contains infinitely many of the $P_n$, and so there is no neighborhood of $Q$ with only finitely many irreducible components. (In fact, $\operatorname{Spec} A$ is homeomorphic to $\mathbb{N}\cup\{\infty\}$, with $Q$ corresponding to $\infty$.)

Answer 2

Sorry,I have known,since the scheme is covered by affine space,and affine space has only finitely many irreducible components,and the intersection of the affine with an irreducible component is an irreducible component of the affine space,so the set of irreducible components is locally finite!