# fiberwise-quasi-compact implies quasi-compact?

Let $$f\colon X\to \mathbb{A}^n_{\mathbb{C}}$$ be a morphism of $$\mathbb{C}$$-schemes. Suppose $$f$$ is (a) separated, (b) flat, (c) locally of finite type, (d) all fibers are quasi-compact, is $$X$$ necessarily quasi-compact?

• @crispr Ok, fixed, thanks! – Qixiao Jul 12 at 23:41

Example. We will define $$X$$ as a union of affine varieties $$U_0 \subseteq U_1 \subseteq \ldots$$ as follows: start with $$U_0 = \mathbf A^1 \times (\mathbf A^1 \setminus 0) \subseteq \mathbf A^2 = V_0$$ with its natural projection to $$\mathbf A^1$$, and let $$Z_0 = \mathbf A^1 \times 0$$ be the complement of $$U_0$$ in $$V_0$$.
Choose a sequence of points $$x_1,x_2,\ldots$$ on $$\mathbf A^1$$. Define $$V_i$$ as the blowup of $$V_0$$ in the points $$(x_1,0), \ldots, (x_i,0)$$, so we have maps $$\ldots \to V_i \to V_{i-1} \to \ldots \to V_0.$$ Let $$E_i$$ be the exceptional divisor for $$V_i \to V_{i-1}$$, let $$Z_i$$ be the strict transform of $$Z_0$$ in $$V_i$$, and let $$U_i$$ be its complement in $$V_i$$. For each $$i$$, the centre of the blowup $$V_i \to V_{i-1}$$ is contained in $$Z_{i-1}$$, giving an isomorphism $$V_i\setminus(E_i \cup Z_i) \stackrel\sim\longrightarrow V_{i-1}\setminus Z_{i-1},$$ hence an open immersion $$U_{i-1} = V_{i-1}\setminus Z_{i-1} \cong V_i\setminus (E_i \cup Z_i) \hookrightarrow V_i \setminus Z_i = U_i.$$ Define $$X$$ as the union. The maps $$U_i \to \mathbf A^1$$ are compatible, so they give a map $$X \to \mathbf A^1$$. It is flat since $$X$$ is integral and dominant over the Dedekind scheme $$\mathbf A^1$$. It is separated and locally of finite type since $$X \to \operatorname{Spec} k$$ is. Finally, the fibres are quasi-compact: each step $$U_{i-1} \hookrightarrow U_i$$ only modifies the fibre over $$x_i$$. But $$X$$ itself is not quasi-compact. $$\square$$
• Thanks! Fixed a very small typo $U_{i-i}\to U_{i-1}$ – Qixiao Jul 13 at 2:52
Let $$X$$ be the scheme obtained by gluing the generic points of all $$\operatorname{Spec}\mathcal{O}_p$$ for all closed points $$p$$ of $$\mathbb{A}^1_{\mathbb C}$$. The obvious morphism $$X \to \mathbb{A}^1_{\mathbb C}$$ is a bijection, but $$X$$ is not quasi-compact.