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$E/\mathbb{C}$ and $E'/\mathbb{C}$ are isomorphic elliptic curves. Then if
$$E :\ y^{2} = x^3 + Ax + B$$ then $$E': \ y^{2} = x^3 + \mu ^4 Ax + \mu ^6 B$$ and the isomorphism map $\phi : E \to E'$ is $$\phi (x, y) = (\mu^2x, \mu^3y)$$

Except this isomorphism, is there any other isomorphism between the two curves $E$ and $E'$?

alpha
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1 Answers1

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If you have two isomorphisms $E\to E'$ then you have an automorphism of $E$, those are found from the isomorphism $E\to \Bbb{C}/L,L=\Bbb{w_1Z+w_2Z}$, there is always $z\to -z$ which corresponds to $(x,y)\to (x,-y)$, there are more automorphisms only in two kind of elliptic curves with complex multiplication

  • either $L=r(\Bbb{Z+iZ})$ which corresponds to $j(E)= 1728$ and $B=0$ (the additional automorphism is $z\to iz,(x,y)\to (-x,iy)$)

  • or $L=r(\Bbb{Z+e^{2i\pi /3}Z})$ which corresponds to $j(E)=0,A=0$ (the additional automorphism is $z\to e^{2i\pi /3}z,(x,y)\to (e^{2i\pi /3}x,y)$)

reuns
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  • Where does the magical number 1728 occur? – kelalaka Dec 24 '19 at 15:57
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    The $j$-invariant of $E:y^2=x^3+ax+b$ is $C \frac{a^3}{4a^3+27b^2}$ where $C$ is a normalization constant that we don't really care about, it is finite when the denominator (the discriminant) is non-zero ie. when $x^3+ax+b$ has distinct roots ie. when $E$ is an elliptic curve, it is invariant under the $\mu$ transformation of the OP, it remains to show the isomorphism with a complex torus, the $\mu$ transformation corresponds to $\Bbb{C}/L\mapsto \Bbb{C}/\mu^{-1} L$ and two complex tori are isomorphic iff they are related this way, thus $j(E)$ depends only on the isomorphism class of $E$. – reuns Dec 26 '19 at 06:00