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Is there a characterization of elliptic curves over $\mathbb Q$ whose conductor is a square? Does this property have a geometric meaning?

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  • $\begingroup$ 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. $\endgroup$
    – Conder
    Apr 12, 2014 at 0:26
  • $\begingroup$ @Conder: Are you sure that's true? What about a quadratic twist of a curve of semistable reduction, so a fiber of type $I_n^*$? $\endgroup$
    – Will Sawin
    Apr 12, 2014 at 4:35
  • $\begingroup$ 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. $\endgroup$
    – Conder
    Apr 12, 2014 at 6:54

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Square conductor implies that all reduction is additive. (The converse is not quite true, since at 2 and 3, the conductor can have higher valuation, up to valuation 5 for $p=2$ and up to 3 for $p=3$. So maybe a better question is: what does it mean for the conductor to be powerful?) A geometric implication is that every fiber of the Neron model is a group of order 1, 2, 3, or 4. So given any point $P$ on a minimal Weierstrass equation, the point $12P$ goes through the identity component of the Neron model on every fiber. But I don't know if this is the sort of geometric implication you had in mind. As Conder notes, the conductor is more of an arithmetic invariant, but Ogg's formula relates it to the geometry of the special fiber of the Neron model.

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  • $\begingroup$ I think it is $2^8$ and $3^5$, but the powerful comment is along the correct lines IMO. $\endgroup$
    – Conder
    Apr 12, 2014 at 6:55
  • $\begingroup$ @Conder You're right, I misremembered. I was probably thinking of the exponent of the wild part, but even for that, the right numbers are 6 and 3, not 5 and 3. $\endgroup$ Apr 12, 2014 at 11:35

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