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Let $C: z^2 = f(x,y)$ be a hyperelliptic curve defined over $\mathbb{Q}$, with $f$ a binary form with integer coefficients and non-zero discriminant. Let $A_C$ denote the Jacobian variety of $C$, and for a non-zero integer $d$, let $C_d$ denote the quadratic twist given by $dz^2 = f(x,y)$.

Suppose that $A_C$ does not have complex multiplication. What can be said about whether $A_{C_d}$ has complex multiplication as $d$ varies over the non-zero integers?

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  • $\begingroup$ $A_C$ and $A_{C_d}$ are isomorphic over $\bar{\mathbb{Q}}$, thus they either both have complex multiplication over $\bar{\mathbb{Q}}$ or neither does. It would help if you clarified your definition of complex multiplication, e.g. whether it takes into account the ground field; for example for elliptic curves the extra endomorphisms are never defined over $\mathbb{Q}$, but only over an imaginary quadratic field. $\endgroup$
    – Daniel Loughran
    Commented Apr 24, 2018 at 11:34
  • $\begingroup$ That's what I thought ought to be true (that they are isomorphic over $\overline{\mathbb{Q}}$, but it was only clear to me in the case of elliptic curves. $\endgroup$
    – Stanley Yao Xiao
    Commented Apr 24, 2018 at 12:10

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