The answer is negative. Suppose for contradiction that $S$ is such a surface, and let me first assume that it is smooth and projective. Fix $g\geq 24$. Then the coarse moduli space of genus $g$ ...
Let $C$ be a smooth projective connected complex curve of genus $\geq 2$. Let me show that $C^{(d)}$ is of general type if $1\leq d\leq g-1$. Equivalently, one needs to show that the image $W_d$ of $... View answer 23 votes Here is a partial answer. In a UFD, the following statement is true : "either an element is prime or you can write it as a nontrivial product". In a Dedekind ring with finite class number h, it is ... View answer Accepted answer 22 votes It has been conjectured by Euler that this equation has no solutions in positive integers when$n\geq 4$. When$n=4$, this was disproved by Elkies in the paper [Elkies, On A4+B4+C4=D4] in a very ... View answer Accepted answer 17 votes Let me give an example showing that the normality hypothesis is necessary. Let$Y=\mathbb{P}^1$with natural$G=\mathbb{G}_m$-action. Let$X$be the$G$-variety obtained by glueing transversally the ... View answer Accepted answer 16 votes A much more general result is proven in Martin-Deschamps, Lewin-Menegaux : Applications rationnelles séparables dominantes sur une variété de type général. Théorème 2 : if$X$and$Y$are smooth and ... View answer Accepted answer 16 votes The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22]. Let me ... View answer Accepted answer 15 votes I believe that this statement is not true. Take$Y=Spec(k[t]/t^2)$and$X=\mathbb{P}^1_Y$. On$X$, extensions of$\mathcal{O}$by$\mathcal{O}(-2)$are parametrized by : $$Ext^1_X(\mathcal{O},\mathcal{... View answer Accepted answer 15 votes Here are two examples. The moduli space of dimension g principally polarized abelian varieties A_g contains complete codimension g subvarieties in any positive characteristic p (for instance, ... View answer Accepted answer 15 votes It is a theorem of Sumihiro (Equivariant completion, Corollary 2) that a normal variety over an algebraically closed field with an action of a torus is covered by invariant affine open subsets. Here, ... View answer Accepted answer 14 votes I think it's not true : Let X=Spec(A) with A=k[x,y,z]/(x^2-y^2-z^2) be a quadratic cone. Let Y be a line through the origin of the cone : its ideal is I=(z,x-y). We calculate :$$X'=Proj_{... View answer Accepted answer 13 votes Let$f:X\to Y$be a finite surjective map of degree$d$between smooth projective varieties of dimension$n$. Then$f$is flat (apply EGA IV 2 Prop 6.1.5), so Example 1.7.4 of Fulton's Intersection ... View answer Accepted answer 12 votes I think the following should be a counter-example. Let$X=\text{Spec}(\mathbb{C}[T])$be the affine line and$X'=\bigsqcup_{x\in\mathbb{C}}\text{Spec}(\mathbb{C}[T]_{(T-x)})$be its faithfully flat ... View answer Accepted answer 12 votes The new edition of SGA3 by Philippe Gille et Patrick Polo provides a connected example due to Raynaud. It is in SGA3 Exposé VIA Exemples 1.3.2 (2) and you may find it at http://www.math.jussieu.fr/~... View answer 11 votes It is true if$Y$is normal. Let me explain why. I will say that a variety has the Chevalley-Kleiman (CK) property if every finite subset is contained in an affine open. By Corollary 48 of Kollar's "... View answer 10 votes If$f:X\to Y$is separated of finite type between noetherian schemes and$f_*$preserves coherence, then$f$is proper. Here is a proof that follows the geometric idea (given in the comments of Piotr ... View answer Accepted answer 10 votes These$2$-spheres are called 'twistor lines'. They indeed cover the moduli space (in the non-polarized case) : more precisely, any two points of the moduli space may be linked by a chain of twistor ... View answer Accepted answer 10 votes A prime ideal of$S^{-1}R[X]$is the extension of a unique prime ideal of$R$, so that the morphism$Spec(S^{-1}R[X])\to Spec(R)$is a bijection, and even an homeomorphism. All the extensions of ... View answer Accepted answer 9 votes An example of a compact$16$-dimensional submanifold of$\mathbb{R}^{30}$with trivial normal bundle that is not defined by such global equations may be found in [Akbulut-King, Submanifolds and ... View answer Accepted answer 9 votes Let me explain why this statement cannot be true in general. I will give a counterexample where$X$is a complex K3 surface. By a result of [Beauville and Voisin, On the Chow ring of a K3 surface], ... View answer Accepted answer 9 votes No, it is not ! For instance, if$k=\bar{\mathbb{Q}}$,$K=\mathbb{C}$and$X$has dimension$\geq 1$, there is exactly one closed point of$X_K$above each closed point of$X$. But there is only ... View answer 9 votes What follows does not answer your precise question, but is very related to it and may be of interest to you. I consider the case where$G$is finite, but not necessarily abelian. Then there are ... View answer 9 votes No, the smoothness is really needed ! To construct a counter-example, let us choose$X=\mathbb{A}^2$. Consider the action of$\mathbb{Z}/2\mathbb{Z}$by$(x,y)\mapsto (-x,-y)$. The quotient of$X$... View answer Accepted answer 9 votes The description of the Hilbert scheme of complete intersections (obtained by taking in an iterative way open subsets of grassmannian bundles, as explained in the answer above) may be found in part 2.2 ... View answer Accepted answer 9 votes This locus is indeed open. I will explain why using Kollar's "Hulls and Husks" (arXiv:0805.0576). More generally, this article studies in great detail when taking the double dual commutes with base ... View answer Accepted answer 8 votes Let me give a counter-example, where$f$is moreover proper (see Piotr Achinger's comment), but$Y$is not normal. Let$X$be the affine plane over a field$k$and let$\tilde{Y}$be the blow-up of ... View answer Accepted answer 8 votes EDIT : Olivier Wittenberg pointed out to me that a positive answer to the question follows from Theorem 1 of [Altman-Kleiman, Bertini theorems for hypersurface sections containing a subscheme]. I keep ... View answer 8 votes I think this is a consequence of (variants of) Hilbert's irreducibility theorem. Let me explain why. Suppose that$C$is a geometrically integral curve defined over a number field$K$. Let$K'/K$be a ... View answer Accepted answer 7 votes Suppose that the secant variety of your curve is a surface$S$. This implies that$S$is covered by lines in such a way that there is a$1$-dimensional family of such lines through every general ... View answer Accepted answer 7 votes Let me work over an algebraically closed field$k$, let$X$be a reduced$k$-scheme of finite type and let$G$be a smooth connected$k$-group scheme acting on$X\$, and acting trivially on an open ...