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Let $K$ be a field which is complete with respect to a discrete valuation $v$ with ring of integers $R$ and residue field $k$. Let $A$ an abelian variety over $K$ and let $A^t$ be the dual abelian variety. Now we can take the Néron model $X$ of $A$ and also the Néron model $X^t$ of $A^t$. Is there any relation between $X^t(R)$ and line bundles on $X$?

Here are some immature thoughts:

Since $K$-points of $X^t$ are $K$-points of $A^t$, we have that $X^t(K)$ is in one-to-one correspondence with line bundles of degree zero on $X_K=A$. Now it is clear that the exact same statement cannot be true for the special fiber but there seems to be still some relation. For instance let $A$ be an elliptic curve with bad reduction and assume for simplicity that $k$ is algebraically closed. Then the identity component of $(X^t)_k$ is $(k^*,*)$ if $A$ has multiplicative reduction and $(k,+)$ if $A$ has additive reduction (since A=A^t). This is exactly the degree zero part of the Picard group of the nodal curve in the first case and the cuspidal curve in the second case.

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    $\begingroup$ This is not an answer, but a special case that's already quite interesting. If $A$ is the Jacobian of a smooth projective curve $C/K$ with minimal regular model $\mathcal{C}/R$, then $A = A^t$ and the identity component of $X_k$ is isomorphic to $\text{Pic}^0_{C_k/k}$, when the gcd of the multiplicities of $C_k$ is one; this is a theorem of Raynaud and explains your elliptic curve example. $\endgroup$
    – Jef
    Commented Jun 1, 2023 at 18:13
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    $\begingroup$ Of course $X^t(R) = A^t(K)$ just corresponds to degree zero line bundles on $A$, but there is a more subtle relationship too. The universal line bundle on $A \times A^t$ (Poincare bundle) gives rise to a so-called biextension of $A\times A^t$. Grothendieck showed that it extends to a biextension on $X^{0}\times (X^t)^{0}$. For (rather intricate) details, SGA7, Expose IX. $\endgroup$
    – Jef
    Commented Jun 1, 2023 at 18:23
  • $\begingroup$ Also this question seems very related: mathoverflow.net/questions/114337/… $\endgroup$
    – Jef
    Commented Jun 1, 2023 at 18:27
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    $\begingroup$ I read more about this in a survey article of Cedric PEPIN: math.univ-paris13.fr/~cpepin/neron-picard-final.pdf There is "almost" a functorial description. There is a "semi-normal", proper $R$-model $\overline{A}$ of $A_K$, and the N'eron model $A^\dagger$ is the unique "group smoothening" of the relative $\text{Pic}^0$ of $\overline{A}/R$. As noted by user @Jef, when the groups of connected components of the closed fibers of $A$ and $A^\dagger$ are nontrivial, the Poincar'e bundle almost never extends. So the group smoothening is nontrivial. $\endgroup$
    – Jason Starr
    Commented Jun 1, 2023 at 20:18
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    $\begingroup$ Thanks a lot for your comments and thoughts! I will have a look at these references. $\endgroup$
    – Hans
    Commented Jun 2, 2023 at 16:56

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