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Let $X$ be a smooth, projective variety over an arbitrary ground field $k$.

I want to write $Pic^{[L]}(X)$ instead of $Pic^L(X)$ -- i.e., to make explicit that the variety depends only on the Neron-Severi class of $L$ -- for reasons which will become clear shortly.

Suppose first that $L$ is algebraically equivalent to $0$. Then $Pic^{[L]}(X) = Pic^0(X)$, so certainly it is an abelian variety.

Next suppose that $L$ is a $k$-rational line bundle on $X$. Then $Pic^{[L]}(X)$ is not literally an abelian variety, because it is a nonidentity coset of a group rather than a group itself. However, it is canonically isomorphic to the abelian variety $Pic^0(X)$ just by mapping a line bundle $M$ to $M - L$. So it might as well be an abelian variety, really.

Finally, supose that $L$ is not itself $k$-rational but that its Neron-Severi class $L$ is rational -- i.e., $L$ is given by a line bundle over the algebraic closure which is algebraically equivalent to each of its Galois conjugates. Then $Pic^{[L]}(X)$ is a well-defined principal homogenous space of the Picard variety $Pic^0(X)$ but need not have any $k$-rational points. For instance, suppose that $X$ is a curve. Then the Galois action on the Neron-Severi group is trivial, so taking $L/\overline{k}$ to be any degree $n$ line bundle, we get $Pic^{[L]}(X) = Pic^n(X) = Alb^n(X)$, a torsor whose $k$-rational points parameterize $k$-rational divisor classes of degree $n$. (Note that here when I write $Pic^0(X)$ I am talking about the Picard variety rather than the degree $0$ part of the Picard group. More careful notation would be $\underline{\operatorname{Pic}}^0(X)$.)

In particular, if $X$ is a genus one curve, then there is a canonical isomorphism $X \cong Pic^1(X)$, so $Pic^1(X)$ is can be endowed with the structure of an abelian variety iff $X$ has a $k$-rational point.

Some further material along these lines can be found in Section 4 of

http://math.uga.edu/~pete/wc2.pdf

1

Let $X$ be a smooth, projective variety over an arbitrary ground field $k$.

I want to write $Pic^{[L]}(X)$ instead of $Pic^L(X)$ -- i.e., to make explicit that the variety depends only on the Neron-Severi class of $L$ -- for reasons which will become clear shortly.

Suppose first that $L$ is algebraically equivalent to $0$. Then $Pic^{[L]}(X) = Pic^0(X)$, so certainly it is an abelian variety.

Next suppose that $L$ is a $k$-rational line bundle on $X$. Then $Pic^{[L]}(X)$ is not literally an abelian variety, because it is a nonidentity coset of a group rather than a group itself. However, it is canonically isomorphic to the abelian variety $Pic^0(X)$ just by mapping a line bundle $M$ to $M - L$. So it might as well be an abelian variety, really.

Finally, supose that $L$ is not itself $k$-rational but that its Neron-Severi class $L$ is rational -- i.e., $L$ is given by a line bundle over the algebraic closure which is algebraically equivalent to each of its Galois conjugates. Then $Pic^{[L]}(X)$ is a well-defined principal homogenous space of the Picard variety $Pic^0(X)$ but need not have any $k$-rational points. For instance, suppose that $X$ is a curve. Then the Galois action on the Neron-Severi group is trivial, so taking $L/\overline{k}$ to be any degree $n$ line bundle, we get $Pic^{[L]}(X) = Pic^n(X) = Alb^n(X)$, a torsor whose $k$-rational points parameterize $k$-rational divisor classes of degree $n$. (Note that here when I write $Pic^0(X)$ I am talking about the Picard variety rather than the degree $0$ part of the Picard group. More careful notation would be $\underline{\operatorname{Pic}}^0(X)$.)

In particular, if $X$ is a genus one curve, then there is a canonical isomorphism $X \cong Pic^1(X)$, so $Pic^1(X)$ is an abelian variety iff $X$ has a $k$-rational point.

Some further material along these lines can be found in Section 4 of

http://math.uga.edu/~pete/wc2.pdf