Example 29.51.2. Let $A = \prod _{n \in \mathbf{N}} \mathbf{F}_2$. Every element of $A$ is an idempotent. Hence every prime ideal is maximal with residue field $\mathbf{F}_2$. Thus the topology on $X = \mathop{\mathrm{Spec}}(A)$ is totally disconnected and quasi-compact. The projection maps $A \to \mathbf{F}_2$ define open points of $\mathop{\mathrm{Spec}}(A)$. It cannot be the case that all the points of $X$ are open since $X$ is quasi-compact. Let $x \in X$ be a closed point which is not open. Then we can form a scheme $Y$ which is two copies of $X$ glued along $X \setminus \{ x\}$. In other words, this is $X$ with $x$ doubled, compare Schemes, Example 26.14.3. The morphism $f : Y \to X$ is quasi-compact, finite type and has finite fibres but is not quasi-separated. The point $x \in X$ is a generic point of an irreducible component of $X$ (since $X$ is totally disconnected). But properties (3) and (4) of Lemma 29.51.1 do not hold. The reason is that for any open neighbourhood $x \in U \subset X$ the inverse image $f^{-1}(U)$ is not affine because functions on $f^{-1}(U)$ cannot separate the two points lying over $x$ (proof omitted; this is a nice exercise). Hence the condition that $f$ is quasi-separated is necessary in parts (3) and (4) of the lemma.

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