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Apr 12, 2014 at 6:54 comment added Conder Indeed you are right, I was typing before thinking, not taking twists into account. Maybe I was muddling it to "integral j-invariant" rather than squarefree conductor.
Apr 12, 2014 at 4:35 comment added Will Sawin @Conder: Are you sure that's true? What about a quadratic twist of a curve of semistable reduction, so a fiber of type $I_n^*$?
Apr 12, 2014 at 3:49 vote accept 352506
Apr 12, 2014 at 2:14 answer added Joe Silverman timeline score: 4
Apr 12, 2014 at 0:26 comment added Conder Usually the conductor is arithmetic, not geometric in meaning. Square conductor in particular means that it has everywhere good reduction after a field extension, as opposed to the case where there is a prime of multiplicative reduction, when this is not possible.
Apr 11, 2014 at 21:03 history asked 352506 CC BY-SA 3.0