The Stacks project

110.64 Being projective is not local on the base

In the chapter on descent we have seen that many properties of morphisms are local on the base, even in the fpqc topology. See Descent, Sections 35.22, 35.23, and 35.24. This is not true for projectivity of morphisms.

Lemma 110.64.1. 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 ${\mathbf C}$. $\square$

In different terminology, Hironaka's 3-fold $V_ Y$ is a small resolution of the blowup $Y'$ of $Y$ along the reduced subscheme $C\cup D$; here $Y'$ has two node singularities. If we define $Z$ by blowing up $Y$ along $C$ and then along the strict transform of $D$, then $Z$ is a smooth projective 3-fold, and the non-projective 3-fold $V_ Y$ differs from $Z$ by a “flop” over $Y-P$.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08J0. Beware of the difference between the letter 'O' and the digit '0'.