curves with complex multiplication (CM curves) of prime power conductor $p^r$. The reason is, first, that there are finitely many imaginary quadratic orders ${{\mathcal{O}}}$ of class number $1$, and second, that if $p$ divides the discriminant of ${{\mathcal{O}}}$, then all curves $E$ with endomorphism ring isomorphic to ${{\mathcal{O}}}$ have additive reduction at $p$, and the field where $E$ attains good reduction is not abelian. In particular, any other rational curve isomorphic to $E$ (that is, a twist of $E$) will also have bad reduction at $p$, so the prime power discriminant condition gives at most one curve per order.

Over an imaginary quadratic field $K$, this finiteness statement no longer holds true. If $E/K$ has CM by ${{\mathcal{O}}}_K$, the ring of integers of $K$, and $p$ ramifies in $K$, the curve attains good reduction over an abelian extension of the completion $K_p$. In particular, some twist of $E$ (quadratic, quartic or sextic, depending on the number of roots of unity in $K$) attains good reduction at $p$. However the required local extension does not come from a global one, so we do not get a curve over $K$ of conductor $1$ (which do not exist), but an infinite family, all 