Polarizations and ampleness under Picard twists

By [F1], ${\phi }_{\mathcal{L},\bar{k}}$ is an isogeny with finite scheme-theoretic kernel. Every geometric fibre of ${\phi }_{\mathcal{L}}$ is, after choosing a point in it, a translate of that geometric kernel. In particular $K(\mathcal{L})\to \mathrm{Spec}k$ is proper and quasi-finite, since it is closed in $A$ and has a finite geometric fibre; [F1] makes it finite over $k$. Surjectivity follows from geometric surjectivity after the field extension, so ${\phi }_{\mathcal{L}}$ is an isogeny. By [F3], $(\mathrm{id},{\phi }_{\mathcal{L}}{)}^{\ast }\mathcal{P}$ has Mumford map $2{\phi }_{\mathcal{L}}$, the same as ${\mathcal{L}}^2$; since $\ker \phi ={\mathrm{Pic}}^0$ as a sheaf [F1], the two bundles differ by an algebraically trivial class: $(\mathrm{id},{\phi }_{\mathcal{L}}{)}^{\ast }\mathcal{P}\cong {\mathcal{L}}^2\otimes \mathcal{M}$ with $\mathcal{M}\in {\mathrm{Pic}}^0$. For symmetry, the family classifying ${\phi }_{\mathcal{L}}^{\vee }\circ {\kappa }_A$ on the second copy of $A$ is obtained by switching the Poincare factors and pulling back along ${\phi }_{\mathcal{L}}$ on the other factor. It is therefore the switched bundle ${\sigma }^{\ast }\mathrm{\Lambda }(\mathcal{L})$, where $\sigma (x,y)=(y,x)$. The family classifying ${\phi }_{\mathcal{L}}$ is $\mathrm{\Lambda }(\mathcal{L})$. Since multiplication is commutative, the two normalized bundles are isomorphic, including their rigidifications on both axes. The all-test universal property gives ${\phi }_{\mathcal{L}}^{\vee }\circ {\kappa }_A={\phi }_{\mathcal{L}}$, which proves symmetry rather than assuming it. For the bundle comparison in [F3], diagonal pullback of $\mathrm{\Lambda }(\mathcal{L})$ gives $[2{]}^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{-2}$ up to a constant line. For any homomorphism $f$, translation commutation and pullback of Picard classes give ${\phi }_{f^{\ast }\mathcal{L}}=f^{\vee }{\phi }_{\mathcal{L}}f$; with $f=[2]$ and additivity of duality this gives ${\phi }_{[2{]}^{\ast }\mathcal{L}}=4{\phi }_{\mathcal{L}}$ and hence the asserted $2{\phi }_{\mathcal{L}}$.

We first prove (c). Let $\mathcal{M}\in {\mathrm{Pic}}^0(A)$ and let $\mathcal{H}$ be ample. Over an algebraic closure, [F1] gives $\mathcal{M}\cong t_x^{\ast }\mathcal{H}\otimes {\mathcal{H}}^{-1}$ for some point $x$, so $\mathcal{H}\otimes \mathcal{M}\cong t_x^{\ast }\mathcal{H}$ is the pullback of an ample bundle under an isomorphism, hence ample. Ampleness is a geometric condition checked after faithfully flat field extension, so $\mathcal{L}\otimes \mathcal{M}$ is ample over $k$; this is (c).

Statement (a) now follows: since ${\mathcal{L}}^2$ is ample and $(\mathrm{id},{\phi }_{\mathcal{L}}{)}^{\ast }\mathcal{P}$ differs from it by the algebraically trivial class $\mathcal{M}$ of step 1.1, (c) gives that $(\mathrm{id},{\phi }_{\mathcal{L}}{)}^{\ast }\mathcal{P}$ is ample.

For (b), $A$ is projective by [F4], so there exists an ample invertible sheaf $\mathcal{L}$ on $A$; then ${\phi }_{\mathcal{L}}$ is a symmetric isogeny by (a) and $(\mathrm{id},{\phi }_{\mathcal{L}}{)}^{\ast }\mathcal{P}$ is ample. Since ${\phi }_{\mathcal{L}}$ is the Mumford map of an ample bundle already over $k$, it satisfies the geometric ample-realization definition of a polarization.

For (d), let $\lambda ={\phi }_{\mathcal{L}}$ be symmetric, now allowing $\mathcal{L}$ to be any invertible sheaf, and put $\mathcal{N}=(\mathrm{id},\lambda {)}^{\ast }\mathcal{P}$. The identity [F3], which holds without ampleness, gives ${\phi }_{\mathcal{N}}=2{\phi }_{\mathcal{L}}={\phi }_{{\mathcal{L}}^2}$; by [F1] this means $\mathcal{N}\cong {\mathcal{L}}^2\otimes \mathcal{M}$ for an algebraically trivial $\mathcal{M}$. If $\mathcal{L}$ is ample, then ${\mathcal{L}}^2$ is ample by [F4] and $\mathcal{N}$ is ample by step 1.2. Conversely, if $\mathcal{N}$ is ample, applying step 1.2 to $\mathcal{N}$ and ${\mathcal{M}}^{-1}$ makes ${\mathcal{L}}^2$ ample, and [F4] then makes $\mathcal{L}$ ample.