Let $A$ and $B$ be abelian varieties, and consider the natural map $$f:(A\times B)^\vee\to A^\vee\times B^\vee$$ sending a line bundle on $A\times B$ to the restrictions to $A\times\{0\}$ and $\{0\}\times B$. I would like to check that this is an isomorphism.
I had some old notes from a course on abelian varieties giving a hint which I'm trying to decypher. They say, first check that $f$ induces an isomorphism between the tangent spaces, hence deduce it finite etale. Then show that the only geometric points in the kernel is the identity, and then the claim follows since finite etale isogenies satisfying this property are isomorphisms.
Would anyone be able to enlighten me with a more detailed proof, or give a reference where I could read a proof?
So, this was fun to track down! (If you don’t know a result that I use and don’t give a reference for, you can search [keywords]+stacks project).
We will show that for any finite type $k$-scheme, $(A\times B)^{\vee}(S) \rightarrow A^{\vee}(S) \times B^{\vee}(S)$ is injective.
In particular, this means that $f: (A \times B)^{\vee} \rightarrow A^{\vee} \times B^{\vee}$ is proper with finite fibres hence finite. Moreover, since $A \times B)^{\vee}$ and $A^{\vee} \times B^{\vee}$ are abelian varieties of the same dimension, they are regular (integral) schemes of the same dimension, so the miracle flatness theorem applies and $f$ is flat.
Thus $f$ is finite flat to an integral base, so it is locally free of constant rank $d \geq 1$, where $d$ is the degree of the kernel of $f$ as a finite $k$-scheme. But since $f$ is injective on $S$-points for any $S$-scheme of finite type, one must have $d=1$ and we are done.
First, we show that $f$ is injective on geometric points: we can thus assume $k$ algebraically closed.
What we need to prove is that if $L \in \mathrm{Pic}^0(A \times B)$ is trivial when restricted to $A \times \{0\}$ and $\{0\} \times B$, then $L$ is trivial.
Since $L$ is in $\mathrm{Pic}^0$, it is stable under translation (Milne, Abelian varieties), so $L_{|A \times \{t\}}$ is trivial for all $t \in B(k)$.
The morphism $q: A \times B \rightarrow B$ is proper smooth with geometrically integral fibres, so there is, by Stacks 0BDP (aka Milne’s “seesaw principle”), a “greatest” locally closed subscheme $Z \subset B$ on which $L_{|A \times Z}$ comes from a line bundle on $Z$.
But the above shows that $Z$ contains all the $k$-points of $B$. Since $k$ is algebraically closed and $B$ is reduced, it follows that $L=q^{\ast}M$ for a line bundle $M$ on $B$. But then $L_{|\{0\} \times B} \cong M$ is trivial, hence $L$ is trivial.
Now, we discuss the general case. We have a line bundle $L$ on $A \times B \times S$ (for some $k$-scheme $S$ of finite type) such that $L_{|A \times B \times \{t\}} \in \mathrm{Pic}^0(A \times B \times \{t\})$ for all $t \in S$. By assumption, we know that $L_{|A \times \{0\} \times S}$ and $L_{|\{0\} \times B \times S}$ come from line bundles $L^A$ and $L^B$ on $S$. The goal is to prove that $L$ comes from a line bundle on $S$.
Note that then $L^A=L^B=L_{|\{0\} \times \{0\} \times S}$, so we can assume, after tensoring $L$ with the pullback of $L^A$, that $L_{|A \times \{0\} \times S}$ and $L_{|\{0\} \times B \times S}$ are trivial.
By our study of the geometric point case, we see that for any geometric point $s$ of $S$, $L_{|A \times B \times\{s\}}$ is trivial.
We then apply the theorem of the cube (Stacks 0BF4) to $X=A, Y=B$, $Z$ being any connected component of $S$ and the line bundle $L$.