MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ be the spectrum of a discrete valuation ring with generic point $\eta$. Let $C/\eta$ be a smooth connected curve with an $\eta$-valued point, and let $\mathcal{C}/S$ be the smooth locus of the minimal proper regular model of $C$ over $S$. Let $N/S$ denote the Neron model of the Jacobian of $C_\eta$, and let $\alpha:C \rightarrow N_\eta$ denote the Abel-Jacobi map.

Now by the Neron mapping property, we obtain a (unique) extension of $\alpha$ to $\overline{\alpha}:\mathcal{C} \rightarrow N$.

Question: Is the map $\overline{\alpha}$ necessarily an immersion?

If the genus is 1, we are OK! In general, I cannot think of a reason why this should be true, but I also cannot think of a counterexample.

I started thinking about this question from the point of view of Serre's book `Algebraic groups and class fields', ie as a question about line bundles on singular curves. I wanted to use such a result in my thesis, but in the end found an alternative approach. However, I am still curious...

Thank you for your time.

share|cite|improve this question

See my article:

Best regards, Bas Edixhoven.

P.S. Thanks to Liu Qing who drew my attention to this.

share|cite|improve this answer
Thanks very much, I am really enjoying the paper. Regarding open immersions, in the final corollary you show that the morphism discussed above is proper iff all double points of the geometric special fibre are non-disconnecting. I guess an easy example of (one direction of) this is to take a genus 2 curve whose special fibre is a pair of elliptic curves meeting transversely at a single point; then the Neron model of the Jacobian is proper. Is there a reason this map cannot be an immersion (ie an open immersion followed by a closed immersion)? I am sorry if this is clear from your paper, I... – David Holmes Nov 27 '11 at 10:07
...haven't had time to read all of it yet. Thanks also to Liu Qing! – David Holmes Nov 27 '11 at 10:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.