Good example teaches sometimes more than couple of theorems. I wonder what are your favourite examples in complex algebraic geometry, the ones that were astonishing for you, the simpler (at least available for people without PhD in AG) the better.

I will give an example which I was looking for a long time by myself. The question concerns CONTRACTIBILITY OF CURVES ON SMOOTH SURFACES: Having a smooth (hence projective) surface $X$ and a divisor $E$ when can $E$ be contracted into an algebraic singularity (i.e. such that the quotient $X/E$ is an algebraic surface)? Background: It is known that a divisor on a smooth projective surface can be contracted in the analytic category if and only if its intersection matrix is negative definite (Grauert). However, the question in the algebraic category is much more subtle. There are criteria (numerical) by Artin for contractibility of bunch of rational curves to a rational singularity but no general ones. The following two examples show that it is not possible to give a general numerical criterion for contractibility (people refer to Zariski, The theorem of RiemannRoch for high multiples of an effective divisor on an algebraic surface. Ann. of Math. (2) 76, 1962, 560615). Background for the example: Let $C$ be an elliptic curve and let $P_0$ be a point in $C$. The linear system $3P_0$ gives an embedding of $C$ into $\mathbb{P}^2$, in particular for some hyperplane section $H$ we have $H\cap C=3P_0$. The bijection $P\to \mathcal{O}_C(PP_0)$ gives $C$ the group structure from $Pic^0(C)$. Example 1: Take 12 points $P_1,\ldots,P_{12}\in C$. Let $p:X\to \mathbb{P}^2$ be the blowup in these points, denote the proper transform of $C$ by $C'$. Since $C'^2=912=3$, it can be contracted analytically (to a singular point), let $\pi:X\to Y$ be the contraction. We show that for a general choice of $P_i$'s each divisor on $Y$ meets the singular point, so $Y$ cannot be algebraic. Suppose $D$ is a divisor on $Y$ which does not meet the singular point. The divisor $B=p_*\pi^*D$ meets the cubic only in the chosen points, so schemetheoretically for some integers $k_i$ we have $\sum k_i P_i=B\cap C\sim deg B\cdot H\cap C=3deg B\cdot P_0$. we see that $\sum k_iP_i$ is the zero of the group, which does not happen for a general choice of $P_i$'s. (Note that since $C$ is uncountable, one can easily find $12$ $\mathbb{Z}$independent points on it). Example 2: In the example above take for $P_i$'s the points of intersection of some quartic $Q$ with our cubic. The linear system $\mathfrak{s}$ spanned by $Q$ and quartics of type $C+line$ is transformed birationally by $p$ to a free linear system $p^{1}\mathfrak{s}$. We see easily that for an irreducible $U\subseteq X$ we have $U\cdot (C'+p^{1}line)=0$ if and only if $U=C'$, so the new system realizes the contraction $X\to Y$ algebraically. 

