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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