The square degree of a Mumford map

Put $H=K(\mathcal{L})$, finite by hypothesis. The morphism $\phi ={\phi }_{\mathcal{L}}:A\to A^{\vee }$ is proper by [F6]. After any algebraically closed field extension, each nonempty fibre is a translate of $H$ as a scheme, hence finite. Thus $\phi$ is quasi-finite and therefore finite by [F6]. Its geometric image is closed and irreducible. The image-dimension formula, with zero-dimensional generic fibre, gives image dimension $g$; since the geometrically integral target has dimension $g$, this closed image is the whole target. Surjectivity descends to $k$. Form the fppf quotient $q:A\to Q=A\mathrm{/}H$ of [F6]. The induced homomorphism $u:Q\to A^{\vee }$ has trivial scheme-theoretic kernel: any kernel section lifts fppf-locally to a section $a$ of $A$, and $\phi (a)=0$ means $a\in H$, whose quotient class is zero. Hence $u$ is a closed immersion by [F6]; it is surjective because $\phi =uq$ is surjective. The target $A^{\vee }$ is reduced, so the ideal of this surjective closed immersion is zero and $u$ is an isomorphism. Consequently $\phi =q$ is faithfully flat. Together with its already proved finiteness, this makes ${\phi }_{\ast }{\mathcal{O}}_A$ a finite flat module on each Noetherian affine target chart. It is free at each local ring by [F6]; a local basis and its inverse spread to a neighbourhood because the module is finitely presented. Thus $\phi$ is finite locally free. Its rank is constant on the connected target and its fibre at zero is $H$, so that rank is ${\dim }_k\mathcal{O}(H)=\mathrm{deg}\phi$. This proves the exact flatness needed for the following base change, including nonreduced $H$.

Consider the Cartesian square with $\phi ={\phi }_{\mathcal{L}}$, the morphism $(\mathrm{id}\times \phi ):A{\times }_{k}A\to A{\times }_k{A}^{\vee }$, and projections $p_2:A{\times }_{k}A\to A$ and $p_2:A{\times }_k{A}^{\vee }\to A^{\vee }$. By flat base change [F4] applied to [F1] we have $R{p}_{2,\ast }((\mathrm{id}\times \phi {)}^{\ast }\mathcal{P})\cong {\phi }^{\ast }(R{p}_{2,\ast }\mathcal{P})\cong {\phi }^{\ast }(k(0)[-g])\cong {\mathcal{O}}_{K(\mathcal{L})}[-g]$, where ${\phi }^{-1}(0)=K(\mathcal{L})$ is finite of length $\mathrm{deg}\phi$ by [F2]. Since $(\mathrm{id}\times \phi {)}^{\ast }\mathcal{P}\cong \mathrm{\Lambda }(\mathcal{L})$ by [F2], the direct images of $\mathrm{\Lambda }(\mathcal{L})$ under $p_2$ vanish except in degree $g$, where they equal ${\mathcal{O}}_{K(\mathcal{L})}$.

The Leray spectral sequence of $p_2$ for $\mathrm{\Lambda }(\mathcal{L})$ has only the $q=g$ row, supported on the finite scheme $K(\mathcal{L})$; hence $H^n(A{\times }_{k}A,\mathrm{\Lambda }(\mathcal{L}))\cong H^{n-g}(K(\mathcal{L}),{\mathcal{O}}_{K(\mathcal{L})})$, which is $k^{\mathrm{deg}\phi }$ for $n=g$ and zero otherwise. Therefore $\chi (A{\times }_{k}A,\mathrm{\Lambda }(\mathcal{L}))=(-1{)}^g\mathrm{deg}{\phi }_{\mathcal{L}}$, with the full scheme length, including any characteristic-dividing part.

We compute the same Euler characteristic by an automorphism. Let $\sigma =(m,p_1):A{\times }_{k}A\to A{\times }_{k}A$ be the automorphism $(x,y)\mapsto (x+y,x)$, with inverse $(x,y)\mapsto (y,x-y)$. Then $\mathrm{\Lambda }(\mathcal{L})\otimes p_2^{\ast }\mathcal{L}={\sigma }^{\ast }(\mathcal{L}⊠{\mathcal{L}}^{-1})\otimes {\pi }^{\ast }{e}^{\ast }\mathcal{L}$, where $\mathcal{L}⊠{\mathcal{L}}^{-1}=p_1^{\ast }\mathcal{L}\otimes p_2^{\ast }{\mathcal{L}}^{-1}$ and ${\pi }^{\ast }{e}^{\ast }\mathcal{L}$ is the constant pullback of a line bundle from the base field; this is immediate from ${\sigma }^{\ast }(p_1^{\ast }\mathcal{L}\otimes p_2^{\ast }{\mathcal{L}}^{-1})=m^{\ast }\mathcal{L}\otimes p_1^{\ast }{\mathcal{L}}^{-1}$. Because $R{p}_{2,\ast }\mathrm{\Lambda }(\mathcal{L})={\mathcal{O}}_{K(\mathcal{L})}[-g]$ is supported on the finite scheme $K(\mathcal{L})$ [step 2.1], tensoring by the base-pulled line bundle $p_2^{\ast }\mathcal{L}$ does not change the Euler characteristic: it tensors the finite-length cohomology by the rank-one bundle $\mathcal{L}{\mathrm{\mid }}_{K(\mathcal{L})}$, preserving all lengths. Hence $\chi (\mathrm{\Lambda }(\mathcal{L})\otimes p_2^{\ast }\mathcal{L})=\chi (\mathrm{\Lambda }(\mathcal{L}))$.

Since $\sigma$ is an automorphism and ${\pi }^{\ast }{e}^{\ast }\mathcal{L}$ is pulled back from the base, $\chi ({\sigma }^{\ast }(\mathcal{L}⊠{\mathcal{L}}^{-1})\otimes {\pi }^{\ast }{e}^{\ast }\mathcal{L})=\chi (\mathcal{L}⊠{\mathcal{L}}^{-1})$, and by Kunneth multiplicativity for external tensor products [F3] this is $\chi (A,\mathcal{L})\chi (A,{\mathcal{L}}^{-1})=(-1{)}^g\chi (A,\mathcal{L}{)}^2$, the last equality by Serre duality [F5]. Combining with steps 3.1 and 3.2 gives $(-1{)}^g\mathrm{deg}{\phi }_{\mathcal{L}}=(-1{)}^g\chi (A,\mathcal{L}{)}^2$, hence $\mathrm{deg}{\phi }_{\mathcal{L}}=\chi (A,\mathcal{L}{)}^2$.

Finally let $\lambda :A\to A^{\vee }$ be a polarization. By definition ${\lambda }_{\bar{k}}={\phi }_{\mathcal{L}}$ for an ample invertible sheaf $\mathcal{L}$ on $A_{\bar{k}}$, and ${\phi }_{\mathcal{L}}$ is an isogeny with $\mathrm{deg}{\phi }_{\mathcal{L}}=\chi (A_{\bar{k}},\mathcal{L}{)}^2$ by step 4.1. Degree and Euler characteristic are unchanged by field extension (the degree is the rank of a finite locally free morphism, and coherent cohomology is compatible with flat field base change [F4]), so $\mathrm{deg}\lambda =\chi (A_{\bar{k}},\mathcal{L}{)}^2$ is a perfect square in $\mathbb{Z}$.