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Result: Pic^0 is always trivial, and Pic = NS is always free abelian, with rank increasing by 1 each time you blow up. For surfaces in $\mathbb {P}^3$ of degree at most 3 the calculation of $Pic (X)$ is relatively easy: In this case $X$ is rational, hence $NS (X)$ modulo torsion equals $H^2 (X,\mathbb {Z})$.

2021年6月13日 · If $X$ is a non singular plane curve, what is the relationship between the Jacobian variety of $X$ $J(X)$ and the Picard group of divisors of degree 0 i.e. Pic $^0(X)$? I read somewhere that $$ J(X) = \text{Pic}^0(X) $$ but I can't quite make sense of this.

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By definition, $L$ is in $Pic^0$ if the homomorphism $\phi_L$ is identically zero, where $\phi: X \rightarrow Pic(X)$ is defined by sending $x$ to the isomorphism class of $T_x^*L \otimes L^{-1} $. The map $\phi_L$ being identically zero means precisely that $T_x^*L \otimes L^{-1} \cong \mathcal O_X$ for every $x \in X$, i.e. that $T_x^*L \cong L$.

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2018年8月14日 · $\mathrm{Pic}^0$ is meaningful only for complete varieties. If the variety is not normal, this is defined as an abelian group, but not as a scheme (see Variete du Picard, by CS Seshadri). To define as a group, one can do the following.

2015年12月20日 · Let $Pic^0(X)$ denote the subgroup of $Pic(X)$ consisting of divisors of degree $0$. Let $n\in \mathbb{Z}$ be non-zero and $Pic^0(X)[n]$ denote the kernel of the multiplication by $n$ map. If $(n,\text{char} (k))=1$, then $Pic^0(X)[n]\cong (\mathbb{Z}/n\mathbb{Z})^{2g}$;

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