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7h

comment 
Complex manifold with subvarieties but no submanifolds
My suggestion will not work. Ljudmila Kamenova reminded me that for a generic deformation of $Y$, the subvariety $E$ does not deform. 
7h

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Complex manifold with subvarieties but no submanifolds
The following example does not quite work, but perhaps it can be made to work. Let $X$ be a generic Kaehler K3 surface. Let $n\geq 3$ be an integer. Consider the Douady space $Y=\text{Hilb}^n_{X/\mathbb{C}}$ parameterizing closed analytic subspaces of $X$ that are zerodimensional of length $n$. There is a singular closed analytic hypersurface $E$ in $Y$, namely the locus parameterizing closed subspaces that have at least one point that is nonreduced. Sadly, the deepest stratum of the singular locus of $E$ is a manifold! What if we deform $(Y,E)$ with one multiple point? 
13h

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What is the integral cohomology of an Enriques surface over a finite field?
If you are searching Google, you can also use the keywords "Cossec", "Dolgachev", and "Schoeer". 
Apr
29 
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SchwartzZippel lemma for an algebraic variety
Just to clarify, "Pr" means "probability"? 
Apr
29 
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Monodromy theorem of degeneration of smooth projective varieties to nonreduced central fiber
"My monodromy acts on the middle cohomology of a fixed general fiber." If $n=2m$, $m\geq 1$, in my example above, then the action of the monodromy on the middle cohomology decomposes into a trivial action on a onedimensional subspace, and cyclic actions of order $d_1,\dots,d_m$ on subspaces of dimensions $d_1,\dots,d_m$. This follows from the description of the (additive) cohomology of a blowing up of a smooth subvariety of a smooth variety in terms of the cohomology of the original ambient variety and the cohomology of the subvariety. 
Apr
29 
revised 
Monodromy theorem of degeneration of smooth projective varieties to nonreduced central fiber
added 8 characters in body 
Apr
29 
answered  Monodromy theorem of degeneration of smooth projective varieties to nonreduced central fiber 
Apr
29 
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Canonical model of Kontsevich's moduli space
I want to second what Jim Bryan said: the moduli space is almost always singular and, for $g>0$, even for $X=\mathbb{P}^n$, the moduli space is typically not even equidimensional. Often this is clear by considering the locus parameterizing stable maps with two components: a contracted genus$g$ component, and a noncontracted genus$0$ component. Nonetheless, for $g=0$ and for $X$ a Fano manifold, there are many good cases when the moduli space is equidimensional, normal, and even $\mathbb{Q}$factorial. There are a few cases where the moduli space is known to have canonical singularities. 
Apr
29 
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Which varieties are flat degenerations of projective space?
One famous theorem is that, if the special fiber is smooth, then the base change to the algebraic closure of the ground field is isomorphic to projective space, at least in characteristic 0 (I believe it is now known in all characteristics, but I am forgetting a reference). Siu first proved this over the complex numbers using differential geometry, but now it is a corollary of an algebraic geometry theorem of Cho, Miyaoka and ShepherdBarron. 
Apr
29 
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Canonical model of Kontsevich's moduli space
Did you delete a question that abx and I answered? 
Apr
28 
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Isomorphism vs. projective equivalence: the $10$dimensional spinor variety
There is an enumerative computation of the number of double points (when this set is finite or empty) on a smooth, projective variety $S$ of dimension $n$ associated to a pair $(\mathcal{L},V)$ of an invertible sheaf $\mathcal{L}$ on $S$ and a linear subspace $V\subset H^0(S,\mathcal{L})$ of dimension $2n+1$. The answer is a polynomial in $c_1(\mathcal{L})$ whose coefficients are monomials in the Chern classes of $T_S$. For $\mathcal{L}=\mathcal{O}_S(d)$, this will be positive for $d\geq d_0$. In your case, it might be positive for $d\geq 2$. Then you can follow Sasha's suggestion. 
Apr
27 
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When a ring is a polynomial ring?
In the result you cite, there is a slight generalization. To deduce the same result for every normal, onedimensional subring $A$ of a $k$algebra $B$, it suffices that (i) for the integral closure $k^*$ of $k$ in $B$, the fraction field $k^*(B)$ has vanishing irregularity, i.e., there is no nonconstant $k^*$rational transformation from $\text{Spec}(B)$ to an Abelian variety, and (ii) every invertible element of $B$ is integral over $k$. 
Apr
27 
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Isomorphism vs. projective equivalence: the $10$dimensional spinor variety
@Sasha. I doubt that is necessary. Based on parameter counts with secant loci, it seems unlikely that any Veronese reembedding of $S$ would have a linear projection into $\mathbb{P}^{15}$ that is an isomorphism (i.e., does not identify some pair of secant points or contract a tangent direction). 
Apr
26 
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Rationally connected spaces over nonalgebraicallyclosed fields
Maybe I should not have written "rinse, repeat". Anyway, this construction is in Koll'ar's article about rational curves and Requivalence over local fields (for these purposes, the field of real numbers behaves like a local field). 
Apr
25 
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Rationally connected spaces over nonalgebraicallyclosed fields
"Would you mind elaborating a little bit more?" Begin with a smooth fiber of the conic bundle over a real point and that contains a real point. Glue on a complex conjugate pair of very free sections. Deform. Rinse. Repeat. 
Apr
25 
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Rationally connected spaces over nonalgebraicallyclosed fields
There are counterexamples over $k=\mathbb{R}$ coming from the fact that $X(\mathbb{R})$ can be disconnected, yet $\mathbb{P}^1(\mathbb{R})$ is connected (for the analytic topology). For instance, inside $\mathbb{P}^1\times \mathbb{P}^2$ with homogeneous coordinates $([s,t],[u,v,w])$, consider the zero scheme of $s^3(u^2+v^2) = t(t^2s^2)w^2$. The image in $\mathbb{P}^1(\mathbb{R})$ of the real points are the open intervals $1\leq t/s\leq 0$ and $0\leq s/t \leq 1$. On the other hand, it is not difficult to construct a morphism $u:\mathbb{P}^1\times M \to X$ with $u^{(2)}$ dominant. 
Apr
17 
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Weil conjectures for higher dimensional cycles?
For projective space and for $n=\text{dim}(X)1$, there are summands of the form $\sum q^{d^n/n!} t^d$. That certainly is not the formal power series of any rational function. 
Apr
16 
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A question about canonical bundle of moduli space of Kahler Einstein metrics
The very first example is when $X$ is a smooth, genus $2$ curve. Then the moduli space $\mathcal{M}$ is a dense open subset of a rational threefold. What is your definition of "nef" if you allow the canonical bundle of an open subset of a rational threefold to be nef? 
Apr
13 
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Number of free summands of finite local extensions
There are no examples arising as homogeneous coordinate rings, $\Gamma_*(Y)\to \Gamma_*(X)$, (localized at the maximal ideal of the vertex) associated with a finite 'etale morphism $f:X\to Y$ and an ample invertible sheaf $\mathcal{O}(1)$ on $Y$ (use the pullback to $X$ to define the coordinate ring). Every $R$summand of $S/R$ corresponds to a direct summand of $f_*\mathcal{O}_X/\mathcal{O}_Y$ that is isomorphic to $\mathcal{O}(d)$. Via the perfect trace pairing, there is a summand $\mathcal{O}(d)$. One of $d$ and $d$ is nonnegative, so that $H^0(X,\mathcal{O}_X)$ is greater than $1$. 
Apr
12 
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A necessary condition for existence of Ricci flat metric on pair (X,D)
Just to make this explicit, are you asking about complete Kaehler metrics that are Ricci flat? If you do not demand completeness, then you can begin with a log CalabiYau open with a Ricciflat metric, and then you can restrict to a smaller open by removing additional prime divisors. This will preserve the Ricciflat condition, yet it destroys completeness. 