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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.

12 votes
Accepted

When two k-varieties with the same underlying topological spaces isomorphic?

The condition you are looking for is seminormality. A variety (or a reduced scheme) $Y$ is seminormal if any proper bijective morphism $f:X\to Y$, with $X$ reduced, inducing isomorphisms on residue fi …
VA.'s user avatar
  • 13.1k
6 votes

Proving if fibres are reduced or not.

How about: $Y$ is the cubic curve $y^2=x^2+x^3$ in $\mathbb A^2$ minus the point $(-1,0)$, $g:Y\to Z$ is the projection to the $x$-axis, and $X$ is the normalization of $Y$, which would be $\mathbb P^ …
VA.'s user avatar
  • 13.1k
8 votes

Intersection numbers and theorem of the cube

The reason for both is that the third derivative of a (nonhomogeneous) quadratic function is zero. $\chi(O(D))$ on a surface is a quadratic function of $D$, by Riemann-Roch. Theorem of cube on an ab …
VA.'s user avatar
  • 13.1k
7 votes

Flatness over non-reduced schemes

(1) A family of hypersurfaces of degree $d$ over an affine scheme $Spec\ R$ means the following: it is a closed subscheme of $\mathbb P^n_R$ given by a homogeneous polynomial $f(x_0,\dots,x_n)$ of deg …
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  • 13.1k
12 votes

Why a subvariety of a variety of general type is of general type

You need to be careful about what you mean by "a general point". Usually, this means "a point in a certain Zariski open set". So in particular, this statement would say that on a surface of general ty …
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  • 13.1k
21 votes

Varieties cut by quadrics

As Pete already indicated, Mumford's theorem says that for any projective variety $X\subset \mathbb P^n$, its Veronese emberdding $v_d(X)\subset \mathbb P^N$ is cut out by quadrics, for $d\gg0$. So a …
VA.'s user avatar
  • 13.1k
25 votes
Accepted

Ample line bundles, sections, morphisms to projective space

1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally gene …
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13 votes

The importance of EGA and SGA for "students of today"

I think EGAs and SGAs are not useful for "students of today" but they are indispensable for "researchers of today", and "tomorrow". There is just so much stuff there that is not available anywhere els …
11 votes
Accepted

Union of closed subschemes with the structure sheaf over it

Because the first one is the right answer in the case of affine varieties, and the second one is not. Indeed, $R/I$, $R/J$ are nilpotent-free implies $R/I\cap J$ is nilpotent-free, but not so for $R/I …
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9 votes

Two-dimensional quotient singularities are rational: why?

In positive characteristic, quotient singularities need not be rational. For an example, see Artin's paper "Wildly ramified Z/2-actions in dimension two". In characteristic zero, quotient singulariti …
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8 votes
Accepted

When is the canonical divisor of an algebraic surface smooth?

Any smooth projective surface with nonempty $K_X$ is obtained by blowing up finitely many points on its unique minimal model. From the formula $K_X=f^*K_Y+E$ for the blowup, you see that the exception …
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  • 13.1k
7 votes

Classification of simply connected smooth projective varieties?

That is way too ambitious, I think, considering what is known about classification of algebraic varieties. Ignoring the trivial, for these purposes, case of curves, the next and best studied case is a …
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  • 13.1k
6 votes

Degrees of etale covers of stacks

Make a base change $f':Y'\to Y$ so that $f':X'=X\times_Y Y' \to Y'$ is a morphism of schemes. That should be possible if $f:X\to Y$ is representable. Then define $\deg(f):=\deg(f')$. Most, if not all …
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  • 13.1k
7 votes

Moduli space of K3 surfaces

Why aren't you happy with the moduli space of polarized K3's, i.e. pairs $(X,L)$ where $L$ is an ample line bundle? This is standard, and this at least makes sense over any field or $\mathbb Z$. And t …
VA.'s user avatar
  • 13.1k
4 votes
Accepted

Automorphism group of bi-elliptic surface

The answer is always "no". By classification, a bielliptic surface over $\mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, $G=\subset Aut(E,0)$ is an abelian group acting by com …
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