16
votes
Accepted
Relative Picard functor for the Zariski topology
Let $Y$ be Cayley's nodal cubic surface over the complex numbers, given in $\mathbb{P}^3$ by
$X_0X_1X_2 + X_0X_1X_3 + X_0X_2X_3 + X_1X_2X_3 = 0.$
This surface has four simple double points. It also ...
14
votes
Restriction of the Picard group of a surface to a curve
Edit. I edited the argument below to make it work in all characteristics. By SGA $7_{II}$ Exposé XVII, this requires working with a sufficiently general pencil of divisors in $\mathcal{O}_{\...
Community wiki
12
votes
Accepted
Pic^0 and H^0(K,Pic^0)
By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of ...
Community wiki
11
votes
Accepted
Picard group of a finite type $\mathbb{Z}$-algebra
This is false. A counterexample is given in [Kahn06, Rmq. 1 (6)]. The example uses the cuspidal cubic $B = A[x^2,x^3]$ over a finite type $\mathbb Z$-algebra $A$ that is not a finitely generated $\...
11
votes
Accepted
Picard group of connected linear algebraic group
$\DeclareMathOperator\Pic{Pic}$The statement is false over most imperfect fields, even for smooth affine group schemes.
In particular, it is false over any separably closed imperfect field $k$. I will ...
11
votes
Accepted
p-torsion in the Picard group of a regular projective curve
Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, putting $L:=K(t^{1/3})$, $C_L$ is isomorphic to the usual cuspidal cubic ...
11
votes
Accepted
Reference request: Generic k3 surface has Picard number 1
Welcome new contributor. I am just writing my comment as an answer, and expanding on the observation of Prof. Arapura. For a smooth, projective scheme $X$ over a field $k$, the space of first order ...
Community wiki
10
votes
Accepted
Does every relative curve have a Picard scheme?
The answer is yes: $\mathbf{Pic}_{X/S}$ is representable by a scheme. I will argue that this follows from the SGA 3 result mentioned by user27920 and from Theorem 2 (c) in section 6.6 of Neron models (...
10
votes
Accepted
Are there varieties with non finitely generated Picard group and vanishing irregularity?
As pointed out by Jason Starr, the answer to your last question is yes, so the answer to the question in your title is no. Let me give a quick (and straightforward) proof in the case $k= \mathbb{C}$.
...
9
votes
Accepted
Is $Pic^0(X)$ of a curve of genus $\geq 1$ over a non-algebraically closed field still non-finitely generated?
As Will Sawin says, the key phrase here is Mordell-Weil theorem. Also, this isn't really a theorem about Picard groups of curves, it's a theorem about abelian varieties. Here is a fairly general ...
9
votes
Accepted
Del Pezzo surfaces and Picard-Lefschetz theory
Indeed you can see it this way. This is my symplectic geometer's
perspective on it (I blame Paul Seidel's Lecture notes on four-dimensional Dehn twists).
Consider the $n$-point blow-up of $\mathbf{CP}...
8
votes
Accepted
Picard group of infinite direct product of DVRs trivial
We aim to show that every $\mathbb{G}_m$-torsor over $R$ is trivial. By descent, such a torsor is represented by an affine $R$-scheme $X$. Due to affineness, $X(R) = \prod_n X(R_n)$, so it remains to (...
8
votes
Accepted
Extension of line bundle defined over an open subscheme
As explained here Extending vector bundles on a given open subscheme the only possible such extension is
$$
\tilde{L} = (i_*L)^{\vee\vee},
$$
where $i \colon U \to X$ is the embedding. The sheaf $\...
8
votes
Accepted
On a morphism from the Brauer group to the Picard group
It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.
Here is the ...
8
votes
Accepted
Picard-surjectivity and Morita-equivalence
Yes, the basic algebra of $A$ will be Picard-surjective.
The basic algebra is the endomorphism algebra $\operatorname{End}_A(\bigoplus_{i=1}^{n}P_i)$ of the direct sum of indecomposable projective (...
7
votes
What is the right definition of the Picard group of a commutative ring?
1) About the second definition:
$\alpha$) It is not true that for an arbitrary ring a) is equivalent to c):
Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring ...
7
votes
Accepted
Injectivity under flat base change of the Picard group on smooth projective curves
This map is injective. There is a Hochschild-Serre spectral sequence with $E^{pq}_2=H^p(\mathrm{Gal}(\bar{K}/K), H^q(X_{\bar{K}},\mathbb{G}_m))$ converging towards $H^{p+q}(X_{K},\mathbb{G}_m)$. This ...
7
votes
Injectivity under flat base change of the Picard group on smooth projective curves
Yes this is true. It can be proved using the Hochschild-Serre spectral sequence plus Hilbert's theorem 90 (it is true more generally for any geometrically connected projective variety $X$).
The ...
7
votes
Picard group of symplectic group modulo orthogonal group
With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $\mathbb{C}^{2n}$ into the sum of two Lagrangian (with respect to the both forms) subspaces:
$$
...
7
votes
Accepted
Picard group of a cubic hypersurface
It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{...
6
votes
Accepted
Picard group of classifying stack
$\mathrm{Pic}([S/G])$ is the group of line bundles on $S$ together with a $G$-linearization. When $S=\mathrm{Spec}(k)$, any line bundle on $S$ is trivial, and a $G$-linearization is given by a ...
6
votes
Are Picard stacks group objects in the category of algebraic stacks
Converting my comment into an answer: stacks form a 2-category, not a category. If a stack takes values in groupoids, then a "group stack" takes values in 2-groups, or equivalently in monoidal ...
6
votes
Accepted
Fiberwise vanishing of $H^2$ and formal smoothness of the Picard functor
I'll write $p = f \times_S Z$, and ${\cal O}^\times$ (resp. $\bar{{\cal O}}^\times$) for the units in the structure sheaf of $X \times_S Z$ (resp. $X \times_S Z_0$, viewed as a sheaf on the same space)...
6
votes
Accepted
Cohomological interpretation of G-equivariant line bundles
See Theorem 4.2.2 in https://www-fourier.ujf-grenoble.fr/~mbrion/lin.pdf
In particular, in your example properness of $X=G/B$ simplifies the left part of the sequence, turning it into $$0\to \hat{G}\...
6
votes
Accepted
Why is it useful for the (relative) Picard functor to be representable?
Let me give one cute example: the Theorem of the Cube (cf. e.g. Mumford's "Abelian varieties") can be proved quite easily if we have Picard schemes at our disposal. In contrast, the proof in Mumford's ...
6
votes
Accepted
Picard group of derived category of sheaves
Thanks to Drew Heard's comment I was able to find answers to my questions. In his paper "Picard groups of derived categories" H. Fausk proves the following theorem (see Theorem 4.2).
Theorem: Let $(\...
6
votes
Accepted
Galois invariant Picard group elements
Updated. The example by @Lucifer is completely correct. Thanks to @Count Dracula who explained the example proposed by @Lucifer.
That example is fine. I am keeping the counterexamples below, since ...
Community wiki
6
votes
Examples of smooth projective varieties with "nice" Picard group
Here is a simple example. Let $X = P^1 \times Q^3$ and $H = O(1,1)$. The Hilbert polynomial of the line bundle $O(a,b)$ is
$$
P(t) = (t+a+1)(t+b+1)(t+b+2)(t+b+3/2)/3.
$$
It has three integral roots $-...
6
votes
Picard group and reduced schemes
Let $f : X \to Y$ be a universal homeomorphism of schemes. Then as you noted above, the pullback functor on (small) étale sites
$$\begin{eqnarray} Y_{\text{ét}} &\to& X_{\text{ét}}\\
U &\...
6
votes
Accepted
Picard group modulo codimension 2
The group $G_X$ can be identified with the group of rank 1 reflexive sheaves on $X$ ($F$ is reflexive if the canonical morphism $F \to F^{\vee\vee}$ is an isomorphism) by taking a sheaf $F$ to the ...
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