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Let $C$ be a smooth genus $g>1$ curve defined over a number field (say over $\mathbb Q$). Let $J(C)$ be its jacobian. Suppose that $J(C)$ has good reduction $J_p$ at a prime $p$ and moreover $J_p$ is absolutely simple (ie simple over any field extensions).

Is it true that $C$ has good reduction $C_p$ at $p$ (maybe after some finite extension) ? So far in the (few) examples I found in the literature of a curve that has bad reduction and for which the jacobian has good reduction, that reduction is isogeneous to a (non-trivial) product.

A question related to that problem is : what is the link between $C_p$ and $J_p$ ? Is there still a non-constant map $C_p \to J_p$ ? Does the image of that map generates $J_p$ ?

There is a criteria of Oda for good reduction of curves mentioned here: A curve with bad reduction for which the jacobian has good reduction but I do not see how to apply it to the hypothesis I make on the curve $C$...

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    $\begingroup$ If $J(C)$ has good reduction, then $J(C)$ has semi-abelian reduction and thus $C$ has stable reduction (by Grothendieck's theorem). In particular, extending the base field won't eliminate bad reduction of $C$ if the curve $C$ has bad reduction. (In other words: "stable + bad" implies "bad even after extending the base field".) $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 1, 2016 at 11:21
  • $\begingroup$ @AriyanJavanpeykar Thanks for that point. My main point is that the reduction mod $p$ of the Jacobian I consider is simple. I may reformulate the question as follows: is there an example of a curve of genus $>1$ with bad reduction but such that the reduction mod $p$ of the jacobian is a simple Abelian variety? $\endgroup$
    – Xavier Roulleau
    Commented Nov 1, 2016 at 11:39
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    $\begingroup$ What you ask is indeed true: your hypothesis implies that the Neron model of $J(C)$ is an abelian scheme (locally at $p$). The Neron model of a Jacobian is related to the relative Picard scheme of a semi-stable model of the curve, see for example Section 9.5 of the book "Neron models" by Bosch, Luetkebohmert and Raynaud. What you want follows immediately from this and the structure of $Pic^0$ of a semi-stable curve. $\endgroup$
    – naf
    Commented Nov 1, 2016 at 11:45
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    $\begingroup$ @XavierRoulleau Ulrich has already answered your question. Let me just rephrase slightly what he says below. If $X$ is a curve such that $J(X)$ has good reduction, then $X$ has stable reduction by Grothendieck's theorem. Now, the fact that $J(X)$ has good reduction implies that $X_p$ is a compact type curve. (You will find a definition of compact type curve in Brian Osserman's notes; see Section 3.1 math.ucdavis.edu/~osserman/classes/280-W13/llsbook.pdf.) However, the Jacobian of a compact type curve is absolutely simple if and only if the curve is smooth... $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 2, 2016 at 16:15
  • $\begingroup$ This finishes the proof, as the Jacobian is "functorial" in the sense that the Jacobian of the special fibre is the special fibre of the Jacobian (under your good reduction hypotheses). $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 2, 2016 at 16:16

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Yes, this is true. One can argue as follows:

Since $J(C)$ has good reduction, your other hypothsism implies that the special fibre of its Neron model at $p$ is an absolutely simple abelian variety. By Grothendieck's theorem, $C$ has semi-stable reduction at $p$, so it has a regular model whose fibre at $p$ is a semi-stable curve.

By the last line of Theorem 4, p. 267, of the book "Neron Models" by Bosch, Raynaud and Luetkebohmert, the identity component of the Neron model of $J(C)$ is $\mathrm{Pic}^0$ of the regular semi-stable model of $C$. (By a base change one can assume that all components of the special fibre are geometrically irreducible.)

By Example 8 on p. 246 of the same book, which gives the structure of $\mathrm{Pic}^0$ of a semi-stable curve over a field, and your assumptions on the reduction of $J(C)$, it follows that $C_p$, the special fibre of a regular semi-stable model of $C$, must have a unique smooth component of genus $g(C)$ and possibly some other smooth rational components. Since the arithemetic genus is constant in flat families it follows that any extra rational component must be a "tail" and can be contracted. We conclude that the special fibre of the stable model of $C$ is a smooth curve of the same genus, i.e., $C$ has good reduction at $p$.

(Note that no base change is required to get good reduction.)

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  • $\begingroup$ I thank you very much for these useful explanations ! $\endgroup$
    – Xavier Roulleau
    Commented Nov 2, 2016 at 19:41

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