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If we consider an elliptic curve $E/k$ given in Weierstrass form $y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}$, then I know that the translation maps $\tau_{P}$ with $P\in{E}$ fix the invariant differential $\omega_{E}=\frac{dx}{2y+a_{1}x+a_{3}}$, i.e. $\tau_{P}^{*}\omega_{E}=\omega_{E}$. But I'm wondering if the converse is true? Namely, if $\phi:E\rightarrow{E}$ is an automorphism of $E/k$ such that $\phi^{*}\omega_{E}=\omega_{E}$, then $\phi$ should be a translation map?

My only attempt is assuming that $\phi$ is of the form $(x,y)\mapsto{(u^{2}x+r,u^{3}y+u^{2}sx+t)}$ for some $u\in{k^{*}}$ and $r,s,t\in{k}$ (I think we can assume this) and applying the definition directly

$\phi^{*}\frac{dx}{2y+a_{1}+a_{3}}=\frac{d\phi^{*}x}{2\phi^{*}y+a_{1}\phi^{*}x+a_{3}}=\frac{d(u^{2}x+r)}{2(u^{3}y+u^{2}sx+t)+a_{1}(u^{2}x+r)+a_{3}}=\frac{u^{2}dx}{2u^{3}y+(2u^{2}s+a_{1}u^{2})x+(2t+a_{1}r+a_{3})}$

which by hypothesis is equal to $\omega_{E}$ again. This seems to suggest that $u=1$ and $s=0$, and also $2t+a_{1}r=0$. So $\phi$ is of the form $(x,y)\mapsto{(x+r,y+t)}$. Is it possible to conclude from this that $\phi$ is a translation map? How can I do that?

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[Corrected (mistake in characteristic 2: multiplication by $-1$ always fixes $\omega_E$)]

The converse is true, except in characteristic 2 and for some supersingular $E/k$ in characteristic 3.

Composing $\phi$ with translation by $-\phi(0)$ yields an automorphism $\psi: E \to E$ sending $0$ to $0$ and inducing the same action on $\omega_E$ as the $\phi$. Let $A$ be the group of automoprhisms sending $0$ to $0$.

Assume first that $E$ is not a supersingular curve in characteristic $2$ or $3$. Then $A$ is cyclic, usually of order $2$ except that curves with $j=1728$ and $j=0$ have an automorphism of order $4$ or $6$ if $k$ contains the correpsonding roots of unity. In this case the homomorphism $A \to k^*$ obtained from the action on $\omega_E$ is injective, except in characteristic 2 where $-1=1$ so the multiplication-by-$(-1)$ map also fixed $\omega_E$.

If $k$ is algebraically closed of characteristic $p=2$ or $3$, and $E$ is supersingular, then $A$ is non-abelian of order $24$ or $12$ respectively, so the map to $k^*$ cannot be injective because a finite subgroup of $k^*$ is cyclic. For $p=2$ the kernel still contains $-1$ as before, but also six further automorphisms, for a total of $8$, which constitute a quaternion group; the image of $A$ in $k^*$ consists of the cube roots of unity. For $p=3$ the kernel is cyclic of order $3$, and the image consists of the 4th roots of unity; but if $k$ is not algebraically closed then the kernel might still be trivial.

Explicit examples of non-identity $\psi$ with $\psi^* \omega_E = \omega_E$:

In characteristic 2, take $E: y^2 + y = x^3$, with $\omega_E = dx$. Let $\psi(x,y) = (x+1, y+x+\eta)$ where $\eta^2 + \eta = 1$. Note that $\psi^2$: $(x,y) \mapsto (x,y+1)$is multiplication by $-1$.

In characteristic 3, take $E: y^2 = x^3 - x$, with $\omega_E = dx/y = dy$. Let $\psi(x,y) = (x+1, y)$. However, for $y^2 = x^3 + x$ over the prime field ${\bf Z} / 3{\bf Z}$, the kernel is trivial because the only nontrivial automorphisms fixing both $0$ and $\omega_E$ are $(x,y) \mapsto (x \pm i,y)$ with $i^2 = -1$.

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  • $\begingroup$ thanks for answer! Do you mind if you could tell me what is the homomorphism $A\rightarrow{k^{*}}$ and why is it injective? Is it the map given by $\phi\mapsto{\lambda}$ where $\phi\in{A}$, $\lambda\in{\overline{k}(E)^{*}}$ and $\phi^{*}\omega_{E}=\lambda\omega_{E}$? how do you know that $\lambda\in{k^{*}}$? $\endgroup$ – antonioACR1 May 7 '16 at 4:52
  • $\begingroup$ oh ok nevermind, I think the injectivity is explained in the proof of corollary 5.6, chapter III in Silverman's book. $\endgroup$ – antonioACR1 May 7 '16 at 5:50
  • $\begingroup$ Since $A$ is finite, some power of $\phi$ is the identity, so $\lambda$ is invertible (and indeed a root of unity). But I messed up earlier; see the corrected answer. $\endgroup$ – Noam D. Elkies May 7 '16 at 21:55

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