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Let $X$ be a stable curve over a complete DVR $R$ with smooth generic fiber. If $X$ has a $R$-rational point, by the universal properties of Neron models, we obtain a morphism from $f: X-X_{s}^{\rm Sing} \rightarrow J_{X}$, where $X_{s}^{\rm Sing}$ denotes the singular set of the special fiber.

If $X$ is smooth, then $f$ is an embedding. How about the general case ?

Remark: If the special fiber of the scheme-theoretic image of $f$ is reduced, then $f$ is an embedding. I think this is not true in general case....

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  • $\begingroup$ "Remark: If the special fiber of the scheme-theoretic image of $f$ is reduced, then $f$ is an embedding." Why is that true? Consider the case when $X_s$ contains a genus $0$ irreducible $Y$ component that meets the other components in three points (that are ordinary double points of $X_s$. It seems to me that $f$ factors through the contraction $X'$ of $Y$ in $X$. Then $X'$ is a local complete intersection scheme (the image of $Y$ is an $A_2$-singularity of $X'$), so the fiber $X'_s$ is $S1$, hence reduced (it is clearly generically reduced). $\endgroup$
    – Jason Starr
    Commented Jan 16, 2016 at 2:00
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    $\begingroup$ Perhaps I should have specified that in my example $X_s$ equals $Y_0\cup Z_1\cup Z_2\cup Z_3$ for disjoint smooth curves $Z_i$ that intersect $Y_0$ in ordinary double points $p_i$ of $X_s$. Thus $p_a(Z_1) + p_a(Z_2) + p_a(Z_3)$ equals $p_a(X_\eta)$, and the closed fiber of the Neron model of $J_X$ equals $J_{Z_1}\times J_{Z_2}\times J_{Z_3}$. The restriction of $f$ to $Y_0$ is constant. $\endgroup$
    – Jason Starr
    Commented Jan 16, 2016 at 2:28

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I am just posting as an answer my comments above, so that this question will not remain unanswered.

One case where the morphism $f$ is non-immersive, in fact non-injective, is when the special fiber $X_s$ contains an irreducible component $Y$ that is a genus $0$ curve such that every connected component $Z_1,\dots,Z_m$ of the residual curve $Z$ to $Y$ in $X_s$ intersects $Y$ in a single ordinary double point $q_i$. For this to be stable, $m$ must be at least $3$. Also, the sum of the arithmetic genera, $p_a(Z_1)+\dots +p_a(Z_m)$, equals the arithmetic genus $p_a(X_\eta)$ of the generic fiber $X_\eta$.

For simplicity, assume that every curve $Z_i$ is a curve of compact type, so that the "Jacobian" of $Z_i$ is a product of Abelian varieties that has dimension equal to $p_a(Z_i)$. Then the Neron model of the Jacobian $\text{Jac}(X_\eta)$ of the generic fiber is an Abelian scheme over $S$. Since all morphisms from a genus $0$ curve to an Abelian variety are constant, $f$ must be constant on $Y\setminus \{q_1,\dots,q_r\}$. Thus $f$ is not an immersion; it is not even injective.

This seems to contradict the OP's assertion that $f$ is an immersion whenever the scheme-theoretic image of $f$ has a reduced special fiber. It is straightforward to arrange a special fiber as above and such that $X$ is regular. By the Weil extension theorem, the rational transformation $f$ extends to a regular morphism on all of $X$, not just on $X\setminus X_s^{\text{Sing}}$. This morphism necessarily factors through the contraction $X'$ of $Y$ in $X$. To prove that the induced morphism is a closed immersion, it suffices to prove that the morphism is unramified at the image point $y'$ of $Y$ in $X'$. The Zariski tangent spaces of the curves $Z_i$ at $q_i$ together with the Zariski tangent space of a transversal give a direct sum decomposition of the Zariski tangent space to $X'$ at $y'$. It seems to me that these subspaces map to linearly independent subspaces of the Zariski tangent space of $J_X$, so that the morphism is unramified. Thus the special fiber of the scheme theoretic image of $f$ equals the special fiber of $X'$. Since $X'$ is normal, the special fiber is $S1$. Since the special fiber is generically reduced, it is everywhere reduced.

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