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 3folds 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?

There is a construction of a proper normal nonprojective 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. 


A simple example of a proper nonprojective surface can be found in Vakil's AGnotes: 


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:
Then Hartshorne suggests that we perform the following computation: let $X=P_k^2$ and consider the sheaf of differential 2forms $\omega_X$; 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 nonprojective surface and in order to see that it is indeed nonprojective 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. 

