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Let $f:C\to S$ be a regular fibered surface where $S=Spec(R)$, $R=dvr$. Assume $C$ has smooth geometrically integral generic fibre $C_K$. We also assume the existence of a section $x\in C(S)$. Let $J_K$ be the Jacobian of $C_K$ and $u_K:C_K\hookrightarrow J_K$ the Abel-Jacobi map sending $x_K$ to $0_{J_K}$. Let $N$ be the NĂ©ron model of $J_K$. Can we find an Abel-Jacobi map for $C$, i.e. a morphism $u:C\to N$ extending $u_K$? We also assume $C$ is not smooth! What if $C$ is only normal? Thank you!

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No. For example, if $C_K$ is an elliptic curve and $C$ is its minimal regular proper model then $N$ is the $R$-smooth locus in $C$, so $N$ is an open subscheme of $C$ that is strictly smaller than $C$ if $C$ is not $R$-smooth (i.e., $C_K$ does not have "good reduction"). – BCnrd Sep 21 '10 at 22:17
Thank you very much! Your example is very clear and simple! Sorry if I didn't think about it. But I think that if $C_K$ has good reduction (but $C$ is still not smooth) then such a morphism $u:C\to N$ can finally be found: let $D$ be the smooth projective model of $C_K$ and $v:D\to N$ the unique natural morphism extending $u_K$. Then since $N=\mathbf{Pic^0_D}=\mathbf{Pic^0_C}$ from $v:D\to \mathbf{Pic^0_C}$ we finally obtain $u:C\to \mathbf{Pic^0_D}$. If $C$ is only normal this should also work. Am I wrong? – Fede Sep 23 '10 at 7:47
Sorry, $N^0=\mathbf{Pic}^0_D=\mathbf{Pic}^0_C$, ... – Fede Sep 23 '10 at 9:16
No, this still cannot work if $C$ is just normal. Again, think of the elliptic curve case with bad reduction. Then any Weierstrass model $C$ is normal (use Serre's normality criterion and facts about Weierstrass cubics over fields), but there's no $S$-map $C \rightarrow N$ extending the isomorphism on generic fibers. Indeed, such a map from a proper $S$-scheme to a a separated $S$-scheme is necessarily proper and thus surjective onto $N^0$ (since it clearly factors through $N^0$), forcing $N^0$ to be proper and thus an abelian scheme. That contradicts the bad reduction hypothesis. – BCnrd Sep 28 '10 at 3:06

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