I am trying to construct a proper non-projective surface following the indications in section III.5 in Hartshorne's 'Alebraic geometry'.

In $X=\mathbb{P}_k^2$ consider the sheaf of differential 2-forms $\omega_X$ and let $(X',\mathcal{I})$ be a nontrivial extension of $X$ by $\omega_X$,namely a scheme $X'$ together with a sheaf of ideals $\mathcal{I}$ with $\mathcal{I}^2=0$ such that $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$. This gives a short exact sequence of sheaves $0\rightarrow \omega_X \rightarrow \mathcal{O}_{X'}^{\ast} \rightarrow \mathcal{O}_X^{\ast} \rightarrow 0$ inducing a long exact cohomology sequence $\cdots \rightarrow \underbrace{H^1(X,\omega_X)}_0 \rightarrow \underbrace{H^1(X',\mathcal{O}_{X'}^{\ast})}_{Pic(X')} \rightarrow \underbrace{H^1(X,\mathcal{O}_X^{\ast})}_{Pic(X)} \stackrel{\delta}{\longrightarrow} \underbrace{H^2(X,\omega_X)}_k \rightarrow \cdots$

Non-projectivity of $X'$ shall follow from the fact that $Pic(X')=0$ and in order to see this 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

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

where $U$ is the standard cover of $\mathbb{P}^2$.

Let $X'$ be the non-trivial extension given by the cocyle $\xi\in H^1(X,\Omega_X^1)$ given by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$.

The 1-cocycle $\alpha$ corresponding to $\mathcal{O}_X(1)$ in $\check{C}^1(U,\mathcal{O}_X^{\ast})$ is $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ and we only have to prove that it maps to some nonzero element in $\check{C}^2(U,\omega_X)$.

In order to lift $\alpha$ to $\beta=(\beta_0,\beta_1,\beta_2)\in \check{C}^1(U,\mathcal{O}^{\ast}_{X'})$ we first need a description of $\check{C}^1(U,\mathcal{O}^{\ast}_{X'})=\bigoplus_{i<j} \Gamma(U_{ij},\mathcal{O}_{X'}^{\ast})$

I recall having read somewhere that

$\Gamma(U_{ij},\mathcal{O}_{X'})\cong \Gamma(U_{ij},\mathcal{O}_X)[\eta_{ij}]=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$

where $\eta_{ij}^2=0$. How does this isomorphism follow? What is the description of the lift $\beta$?

Thanks in advance for any insight.

  • 3
    $\begingroup$ In case you want a different example (perhaps easier) of a proper nonprojective surface, see section 17.4.8 of the May 2 version of the notes here: math216.wordpress.com/2011-12-course $\endgroup$
    – Ravi Vakil
    May 10, 2012 at 23:53
  • $\begingroup$ This comment is not directly relevant to your question, but it is an interesting fact that every smooth proper surface is projective. $\endgroup$
    – Rex
    Jun 5, 2012 at 18:55

2 Answers 2


If I understand what you are writing correctly, this comes from the facts that:

  1. $\omega_X$ is trivial on $U_{ij}$

  2. The differential lifting property of affine schemes makes the extension of $\mathcal O_X$ trivial on $U_{ij}$

Together they imply that the extension is the trivial extension of $\mathcal O_X$ by $\mathcal O_X$, which of course comes from adjoining $\eta_{ij}$.

In regards to your second question, I don't think it's the lift where the subtlety lies. Since the map from $\mathcal O_{X'}$ to $\mathcal O_X$ is canonically the quotient by the radical, you can just take the lift to be given by the same symbols as the cocycle. I think you will find subtlety in evaluating symbolically the restriction maps from $\Gamma(\mathcal O_{X'}, U_{ij})$ to $\Gamma(\mathcal O_{X'}, U_{ijk})$. I will think about how to do that.


Thank you for your answer: regarding the second question, I just found some notes by prof. Mckernan in which he addresses this issue although there are some details that escape me.

We have $\mathcal{O}_X(1)$ represented by the 1-cocyle $\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ on $U_{01}$, $U_{12}$ and $U_{20}$ respectively, where $U_{ij}=U_i\cap U_j$, which we want to lift to $\check{C}^1(U,\mathcal{O}_{X'}^{\ast})=\bigoplus_{i<j} \Gamma(U_{ij},\mathcal{O}_{X'})$ in the first place, where $\Gamma(U_{ij},\mathcal{O}_{X'})=k\left[\frac{x_i}{x_j},\frac{x_j}{x_i},\frac{x_k}{x_i}\right][\eta_{ij}]$.

The 1-cocycle $\xi\in H^1(X,\omega_X^1)$ defining $X'$ determines a derivation $D\in Hom(\Omega_X^1,\omega_X)$ on $U_{ij}$ as follows: we have an identification $\Omega_X \rightarrow Hom(\Omega_X,\omega_X)$ given by $v \mapsto (w\mapsto w\wedge v)$. The 1-cocycle $\xi=(\xi_{ij})$ hence determines the derivation: $$d\left(\frac{x_i}{x_j}\right)\wedge \xi_{ij}=\frac{x_i}{x_j}\xi_{ij}\wedge \xi_{ij}=0, \quad d\left(\frac{x_j}{x_i}\right)\wedge \xi_{ij}=-\frac{x_j}{x_i}\xi_{ji}\wedge \xi_{ji}=0$$ $$d\left(\frac{x_k}{x_j}\right)\wedge \xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_k}{x_j}\right) \wedge d\left(\frac{x_i}{x_j}\right)$$

Then, computing in terms of the isomorphism $\phi:\Gamma(U_{12},\mathcal{O}_{X'}^{\ast})\rightarrow k\left[\frac{x_1}{x_2},\frac{x_2}{x_1},\frac{x_0}{x_1}\right]$, the lift $\beta=(\beta_2,\beta_0,\beta_1)\in \check{C}(U,\mathcal{O}_{X'}^{\ast})$ of $\alpha=\left(\frac{x_1}{x_0},\frac{x_2}{x_1},\frac{x_0}{x_2}\right)$ satisfies $$\phi(\beta_2)=\frac{x_1}{x_0}, \quad \phi(\beta_0)=\frac{x_2}{x_1}+\eta_{12}\frac{x_0}{x_1}d\left(\frac{x_2}{x_0}\right) \wedge d\left(\frac{x_1}{x_0}\right), \quad \phi(\beta_1)=\frac{x_0}{x_2}$$ I am not sure about why this holds yet, but then he concludes that the image of $\beta$ in $\check{C}^2(U,\mathcal{O}_{X'}^{\ast})=\Gamma(U_{012},\mathcal{O}_{X'}^{\ast})$ is represented by $$\frac{x_1}{x_0}\left(\frac{x_2}{x_1}+\eta_{12}\frac{x_0}{x_1}d\left(\frac{x_2}{x_0}\right)\wedge d\left(\frac{x_1}{x_0}\right) \right) \frac{x_0}{x_2}=1+ \eta_{12}\frac{x_0}{x_2}d\left(\frac{x_2}{x_0}\right)\wedge d\left(\frac{x_1}{x_0}\right)$$ which lifts to the element $\frac{x_0}{x_2}d\left(\frac{x_2}{x_0}\right)\wedge d\left(\frac{x_1}{x_0}\right)$ of $\check{C}^2)U,\omega)$.


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