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Let k be an algebraically closed field. It's well known that every complete curve, period, is projective. Also, that every smooth surface is, and that there are smooth 3-folds which are not, and people go to reasonable lengths to include these examples all over the place, so they're easy to find. However, Hartshorne does say that singular complete surfaces are not all projective. Is there a simple example? A complete normal surface that is not projective? Is there some "least singular" possible such surface? I suspect that normality is too much to hope for, but I can't quite phrase why I think this, so is every normal complete surface projective?

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What if we drop completeness -- is it easier to write down examples of non-quasi-projective varieties is we don't require them to be proper? – David Zureick-Brown Nov 1 '09 at 0:56
@Igor: Kollar wrote an example for our paper (Example 34). The example is obtained by gluing two projective planes along three generic projective lines (with a twist). The result has no nontrivial line bundles, so it is not projective. – Misha Apr 21 '12 at 2:56
@Misha. But concerning this sort of twisted gluing, the result is non-algebraic, I believe. The main interest lies in algebraic examples. – Pelle Salomonsson Apr 21 '12 at 5:13
up vote 15 down vote accepted

There is a construction of a proper normal non-projective surface here .

There is an example given by Nagata in his paper "Existence theorems for nonprojective complete algebraic varieties" in the Illinois Journal, but I don't know where this is available on the web.

Over a finite field complete + normal implies projective for surfaces.

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A simple example of a proper non-projective surface can be found in Vakil's AG-notes:

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Very nice. This is essentially Hironaka's 3-dimensional example, which locally looks like the blow-up of ℙ² along two lines which intersect in two points (but the blow-up is done "in the wrong order" at one of them). Here, you just take the "exceptional locus" (the surface lying over the two lines) and use the same proof to show that it's not projective. – Anton Geraschenko Apr 19 '10 at 14:10
There are some typos in the construction in the notes, so read carefully. – Matt Aug 18 '10 at 22:08

There is also an example in an Exercise from Hartshorne's Algebraic geometry involving infinitessimal extensions which I am trying to understand.

Let me recall some definitions and properties in the first place:

  • Infinitessimal lifting property: given an algebraically closed field $k$, a finitely generated $k$-algebra $A$ with $X=\mbox{Spec } A$ non-singular, and an exact sequence $0\rightarrow \mathcal{I} \rightarrow B' \rightarrow B \rightarrow 0$, where $B,B'$ are k-agebras and $I\subset B'$ is an ideal such that $\mathcal{I}^2=0$, any k-algebra homomorphism $A\rightarrow B$ lifts to a h-algebra homomorphism $A\rightarrow B'$, and two such homomorphism differ by a k-derivation of A into $\mathcal{I}$, namely an element in $Hom_A(\Omega_{A/k},\mathcal{I})$.

  • An infinitessimal extension of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ is a pair $(X',\mathcal{I})$ where $X'$ is a k-scheme and $\mathcal{I}$ is a sheaf of ideals on $X'$ with $\mathcal{I}^2=0$ and such that we have isomorphisms $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$ (as k-schemes) and $\mathcal{I}\cong \mathcal{F}$ (as $\mathcal{O}_X$-modules). For instance, the trivial extension of $X$ by $\mathcal{F}$ is given by the pair $(X,\mathcal{F})$, where the $X$ has structure sheaf $\mathcal{O}_X'=\mathcal{O}_X\oplus \mathcal{F}$ with product $(a\oplus f)\cdot (a'\oplus f')=aa'\oplus (af'+a'f)$, so that $\mathcal{F}$ becomes an ideal sheaf in $X$.

  • If $X=\mbox{Spec }A$ is affine and $\mathcal{F}$ is a coherent sheaf, then any extension is isomorphic to the trivial one: we just use the previous lifting property to construct a splitting of an appropriate short exact sequence.

  • There is a correspondence between isomorphism classes of infinitessimal extensions of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ and the cohomology group $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ where $\mathcal{F}_X$ is the tangent sheaf of $X$. If $(X',\mathcal{I})$ is an infinitessimal extension of $X$ by $\mathcal{F}$ and and $\{U_i\}$ is an affine open cover of $X$ (so that sheaf cohomology is isomorphic to Cech cohomology) then on every open affine set the extension is trivial, namely of the form $(U_i,\mathcal{I}_{|U_i}=\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i})$. It is easy to see from the construction of the trivialisation (choosing a lift, and noting that the difference of two lifts gives an element of $Hom_A(\Omega_{A/k},\mathcal{I})$) that this gives a cocyle in $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$. Conversely, given a cocyle $\xi=(\xi_{ij})\in \check{H}^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ and an open affine cover $\{U_i\}$, on each $U_i$ we have a trvial extension $(U_i,\mathcal{F}_{|U_i|}$ with $\mathcal{O}_{|U_i}'\cong\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i}$ and we can glue them all via $\xi=(\xi_{ij})$ to give an extension of $X$ by $\mathcal{F}$.

Then Hartshorne suggests that we perform the following computation: let $X=P_k^2$ and consider the sheaf of differential 2-forms $\omega_X$; then $H^1(X,\Omega_X^1)\cong H^1(X,\omega_X\otimes \mathcal{T}_X)$ and a nontrivial extension $X'$ of $X$ by $\omega_X$ is given by the cocylce $\xi \in H^1(X,\omega_X^1)$ given over $U_{ij}=U_i\cap U_j$ (where the $\{U_i\}$ are the standard open subsets covering $P_k^2$) by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$. This is our target proper non-projective surface and in order to see that it is indeed non-projective we shall prove that it has no ample invertible sheafs (in fact no invertible sheaf at all, namely $Pic X'=0$). We have a short exact sequence

$0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast} \rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long exact cohomology sequence $\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$

and in order to see that $Pic X'==$ it suffices to prove that $\delta$ is injective and nonzero. Since $Pic X\cong \mathbb{Z}$, any invertible sheaf is of the form $\mathcal{L}=\mathcal{O}_X(d)\cong \mathcal{O}_X(1)^{\otimes d}$ and it suffices to see that $\delta(\mathcal{O}_X(1))\neq 0$. I am confused as to how to carry out this computation since I guess I still do not understand very well the correspondence between infinitessimal extensions and the cohomology group. What I intend to do is to compute $\delta$ explicitly in the standard way, namely via the diagram

$\begin{array}{ccccccccc} 0 & \rightarrow & \check{C}^1(U,\omega) & \rightarrow & \check{C}^1(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow & \check{C}^1(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \\ && \downarrow && \downarrow && \downarrow && \\ 0 & \rightarrow & \check{C}^2(U,\omega) & \rightarrow & \check{C}^2(U,\mathcal{O}_{X'}^{\ast}) & \rightarrow & \check{C}^2(U,\mathcal{O}_X^{\ast}) & \rightarrow & 0 \end{array}$

The cycle corresponding to $\mathcal{O}_X(1)$ in $\check{C}^1(U,\mathcal{O}_X^{\ast})$ is $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$. How does it map down to $\check{C}^2(U,\omega)$?

Thanks in advance for any insight.

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I don't undertand why it reads so bad. I have compiled this with a tex editor and there are no errors. – Marc Apr 20 '12 at 23:55
I cleaned up formatting a bit: tips: a. Do NOT use $$ b. DO use backquotes. – Igor Rivin Apr 21 '12 at 0:57

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