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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
6
votes
Conic sections are to cones as quadric surfaces are to what?
The thing that makes quadric surfaces "3D analogs of conic sections" is just that they are defined by a single equation of degree 2. It's not a particularly helpful characterization though, I would sa …
8
votes
Picard groups of quartic K3 surfaces
The first example of a smooth quartic surface in $\mathbb{P}^3$ with (geometric) Picard number 1 was found by Ronald van Luijk (who, incidentally, was my PhD advisor). The following is cited from one …
3
votes
Quadratic twist of curve defined over finite field
[Edit: there was an obvious mistake in my original answer, which was noticed in the comments. Here's the amended statement.] Let $P$ be a rational point on $C$, then if $P$ is not a branch point then …
2
votes
Equidistribution of rational points on an algebraic variety
Without any restrictions on $X$, the answer is no. Consider the following setup. Suppose that $X=E$ is an elliptic curve over $\mathbb{Q}$ with $E(\mathbb{Q})\cong\mathbb{Z}$ generated by an element $ …
3
votes
Generalization of Weak Nullstellensatz?
Or see Proposition 2.4.6 in Bjorn Poonen's book Rational Points on Varieties (link). This is almost exactly the result you conjectured, just a bit more general:
Let $X$ be a $k$-variety. Then the map …
2
votes
Hasse principle and twists of $\mathbb{P}^n$
As Daniel mentioned, the answer is yes. This is Theorem 4.5.11 in Bjorn Poonen's Rational Points on Varieties.
5
votes
2
answers
1k
views
Singular models of K3 surfaces
Let us work over a ground field of characteristic zero. As is well-known, a K3 surface is a smooth projective geometrically integral surface $X$ whose canonical class $\omega_X$ is trivial and for whi …
5
votes
Singular models of K3 surfaces
For what it's worth, I wrote up a proof (pretty detailed) following the hints in Francesco Polizzi's answer. It's in an unpublished preprint found here (p. 38 onwards). I am not a geometer, so the exp …
10
votes
Accepted
Does GAGA hold over other topological fields?
If $k$ is a field that is complete with respect to some ultrametric valuation, then there is the "GAGR" (i.e. géométrie algébrique et géométrie rigide) theorem. A succinct explanation (in French, with …
11
votes
2
answers
773
views
Geometrically unirational varieties that are not unirational
By a variety over a field $k$, I mean a scheme that is separated and
of finite type over $k$. I indicate changes of the base ring by
subscripts.
Does there exist a smooth and projective variety $V$ …
8
votes
Accepted
Is any quadric birational to a product of Brauer-Severi varieties?
Consider the projective quadric $V$ given by
$$
2x^2+y^2+z^2+w^2=0
$$
over $\mathbb{Q}$. Inspired by Jason Starr's remark on splitting fields, I will prove that $V$ is not birational to a product of S …
9
votes
The boundedness of the rank of twists of a fixed curve
This is not a complete answer to your question (since I think it isn't known), but you might still find this interesting.
Mestre has shown in Rang de courbes elliptiques d'invariant donné (http://arx …
8
votes
Brauer group of projective space
Let $X$ be an $n$-dimensional projective space over a field $k$. Let $k_s$ be a separable closure of $k$, and $X_s$ the base change of $X$ to $k_s$. The algebraic part $\textrm{Br}_1(X)$ of the Brauer …
15
votes
Elliptic Curves over Rings?
Elliptic curves can be defined over arbitrary base schemes $S$. In particular, for every (commutative!) ring $R$ one can talk about elliptic curves over (the spectrum of) $R$. Loosely speaking, what o …
4
votes
A curve is proper iff the space of global sections is finite-dimensional
The answer to the first question is certainly yes, because if a curve is non-proper, it must be affine, and hence its ring of global sections is not finite as a $k$-module. See this link to a M.SE top …