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Nov 10, 2015 at 22:47 vote accept Johnson-Leung
Oct 23, 2015 at 12:01 answer added Todd Trimble timeline score: 4
Oct 22, 2015 at 20:34 comment added David Roberts Eg you might want a textbook as a reference.
Oct 22, 2015 at 18:12 comment added eric (Edited) I don't want to post those references as an answer because they are overkill. Given an unramified representation of dimension $d$, if you twist it by a tamely ramified character the resulting representation has conductor $p^d$. It will be hard to find a reference for this because to the experts it's obvious. I've given some references to random places in the literature, but they are overkill because they prove a huge amount more stuff. For example the Serre-Tate Annals paper proves Neron-Ogg-Shaferevich, but you only need the easy direction, so I'd find this reference over the top.
Oct 22, 2015 at 15:22 comment added Johnson-Leung @eric Thanks! Could you post your comment as an answer so that I can accept it?
Oct 22, 2015 at 10:48 comment added eric (Edited) If the curve has good reduction then the Jacobian has good reduction. The standard reference for this is SGA7. Now you need that the ell-adic representation is unramified; this is one direction of Neron-Ogg-Shaferevich, and a reference is the Serre-Tate Annals paper. A quadratic twist is tamely ramified (for $p$ odd -- when $p=2$ the result you want is probably false in fact), and for this you can refer to the Artin-Tate notes on class field theory.
Oct 22, 2015 at 1:59 history edited Johnson-Leung CC BY-SA 3.0
changed question to reflect that the conductor of the Jacobian is the conductor of interest
Oct 22, 2015 at 1:58 comment added Johnson-Leung @eric and Felipe I am ultimately interested in the conductor of the Jacobian, so I can edit the question to reflect that. Again, though, I would really like to have a reference.
Oct 21, 2015 at 20:32 comment added eric In which case I don't know the definition we're using in this question.
Oct 21, 2015 at 1:48 comment added Felipe Voloch @eric There are curves of genus 2 (for example) that have bad reduction somewhere but whose Jacobian has good reduction, so the conductor of the curve is not the same as of its Jacobian, depending on the definition.
Oct 21, 2015 at 1:20 comment added Johnson-Leung That makes sense, but I'd still like a reference.
Oct 20, 2015 at 20:14 history edited Daniel Loughran
edited tags
Oct 20, 2015 at 19:41 comment added eric If the definition of the conductor is equal to the conductor of the Jacobian of the curve, then this is true because you can compute the conductor of the ell-adic Galois representation instead, and you have four tame characters each of which contributes $p^1$.
Oct 20, 2015 at 19:28 history asked Johnson-Leung CC BY-SA 3.0