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3 votes
0 answers
171 views

Nice blowups are universal algebraic fiber spaces?

We say that a proper (maybe projective) morphism $f:X \to Y$ is a universal algebraic fiber space if $f_* O_X = O_Y$ holds universally. (This means: for any morphism $Y' \to Y$, if $X' = Y' \times_Y X$...
iteo's user avatar
  • 39
6 votes
1 answer
1k views

Reference request: Generic k3 surface has Picard number 1

I keep running into the statement that "the generic k3 surface has Picard rank 1". For instance the answer of this question (end) and this paper (following Example 1.1) or this paper (proof ...
user147163's user avatar
2 votes
1 answer
324 views

Very general quartic hypersurface in $\mathbb{P}^3$ has Picard group $\mathbb{Z}$

I am looking for a reference from which I can cite the following statement: The Picard group of a very general quartic hypersurface $X\subset\mathbb{P}^3$ is generated by the class of a hyperplane ...
Hans's user avatar
  • 3,031
10 votes
2 answers
1k views

Picard group of a finite type $\mathbb{Z}$-algebra

Let $A$ be a finitely generated $\mathbb{Z}$-algebra. Is $\operatorname{Pic}(A)$ finitely generated (as an abelian group)? Thoughts: We may assume that $A$ is reduced since $\operatorname{Pic}(A) = \...
Minseon Shin's user avatar
  • 2,017
5 votes
0 answers
679 views

Picard group of normalization

Let $X$ be a projective variety with at worst (analytic) normal crossings singularities and $\pi:\tilde{X} \to X$ be the normalisation. Is there a "nice" description relating the picard group of $X$ ...
Ron's user avatar
  • 2,126
2 votes
0 answers
507 views

Fiber of the specialization map of Picard groups

Let $R$ be a Henselian discrete valuation ring with residue field $k$ of positive characteristic and fraction field $K$ of characteristic zero. Let $\pi:X_R \to \mathrm{Spec}(R)$ be flat, projective ...
user45397's user avatar
  • 2,323
2 votes
0 answers
188 views

Is Pic( G((z)) ) = $\mathbb{Z}$?

There are a fair number of papers by Beauville, Laszlo, Sorger, Kumar and others on the geometry of $LG/L^+G = G((z))/G[[z]]$ where $G$ is a simply connected and simple group over $\mathbb{C}$. In ...
solbap's user avatar
  • 3,968