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The Stacks project

Lemma 110.65.1. The properties

  1. \mathcal{P}(f) =f is projective”, and

  2. \mathcal{P}(f) =f is quasi-projective”

are not Zariski local on the base. A fortiori, they are not fpqc local on the base.

Proof. Following Hironaka [Example B.3.4.1, H], we define a proper morphism of smooth complex 3-folds f:V_ Y\to Y which is Zariski-locally projective, but not projective. Since f is proper and not projective, it is also not quasi-projective.

Let Y be projective 3-space over the complex numbers {\mathbf C}. Let C and D be smooth conics in Y such that the closed subscheme C\cap D is reduced and consists of two complex points P and Q. (For example, let C=\{ [x,y,z,w]: xy=z^2, w=0\} , D=\{ [x,y,z,w]: xy=w^2, z=0\} , P=[1,0,0,0], and Q=[0,1,0,0].) On Y-Q, first blow up the curve C, and then blow up the strict transform of the curve D (Divisors, Definition 31.33.1). On Y-P, first blow up the curve D, and then blow up the strict transform of the curve C. Over Y-P-Q, the two varieties we have constructed are canonically isomorphic, and so we can glue them over Y-P-Q. The result is a smooth proper 3-fold V_ Y over {\mathbf C}. The morphism f:V_ Y\to Y is proper and Zariski-locally projective (since it is a blowup over Y-P and over Y-Q), by Divisors, Lemma 31.32.13. We will show that V_ Y is not projective over {\mathbf C}. That will imply that f is not projective.

To do this, let L be the inverse image in V_ Y of a complex point of C-P-Q, and M the inverse image of a complex point of D-P-Q. Then L and M are isomorphic to the projective line {\mathbf P}^1_{{\mathbf C}}. Next, let E be the inverse image in V_ Y of C\cup D\subset Y in V_ Y; thus E\rightarrow C\cup D is a proper morphism, with fibers isomorphic to {\mathbf P}^1 over (C\cup D)-\{ P,Q\} . The inverse image of P in E is a union of two lines L_0 and M_0, and we have rational equivalences of cycles L\sim L_0+M_0 and M\sim M_0 on E (using that C and D are isomorphic to {\mathbf P}^1). Note the asymmetry resulting from the order in which we blew up the two curves. Near Q, the opposite happens. So the inverse image of Q is the union of two lines L_0' and M_0', and we have rational equivalences L\sim L_0' and M\sim L_0'+M_0' on E. Combining these equivalences, we find that L_0+M_0'\sim 0 on E and hence on V_ Y. If V_ Y were projective over {\mathbf C}, it would have an ample line bundle H, which would have degree > 0 on all curves in V_ Y. In particular H would have positive degree on L_0+M_0', contradicting that the degree of a line bundle is well-defined on 1-cycles modulo rational equivalence on a proper scheme over a field (Chow Homology, Lemma 42.20.3 and Lemma 42.28.2). So V_ Y is not projective over {\mathbf C}. \square


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