# Automorphism group of genus 1 curve

Suppose $E$ is a regular curve of genus 1 over a field $k$ that is not necessarily algebraically closed. The automorphism groups of $E$ forms a one parameter family. If $k=\mathbb{C}$, then $E$ could be constructed as $\mathbb{C}/\Lambda$, where $\Lambda$ is a lattice in $\mathbb{C}$. Then $z \rightarrow z+a$ gives a one parameter family of automorphism group! Does any one know how to visualise or construct the one parameter family automorphism group for general $k$? If $p$ and $q$ are two $k$ valued points, how could we find a automorphism that maps $p$ to $q$, which is very straightforward when $k=\mathbb{C}$ and in fact this gives us the one parameter family.

Also how could we see the 3 parameter family of automorphism group of regular curves of genus $0$ that is not isomorphic to $\mathbb{P}^1$.

• What do you mean by one-parameter family in this context? If $E$ is actually an elliptic curve, then $\text{Aut}(E)=E(k)\rtimes \text{Aut}_{\text{e.c}}(E)$ and, of course, $\text{Aut}_{\text{e.c}}(E)$ is finite. What sort of description are you looking for? If it doesn't have a point, I would move to a finite Galois extension where it does and then argue in terms of descent. – Alex Youcis Mar 4 '16 at 11:58
• @AlexYoucis I mean the dimension of the automorphism group is 1, provided we could make sense of the dimension. How to prove $\text{Aut}(E)$ is the semidirect product of $E(k)$ and $\text{Aut}_{e.c}(E)$, and the finiteness of $\text{Aut}_{e.c}$? I think this is the description I am look for! – Wenzhe Mar 4 '16 at 12:05
• Answered below. – Alex Youcis Mar 4 '16 at 12:12

## 1 Answer

You know that if $\varphi:E\to E$ preserves $e\in E(k)$, the identity point, then by rigidity it's automatically a group morphism. In general, if you have a morphism $\varphi:E\to E$ then you know that $t_{-\varphi(e)}\circ \varphi$ (here $t_x$ is the translation by an element of $E(k)$ which makes sense, unlike the translation by an arbitrary point of $E$) takes $e$ to $e$ and so is a group map. Thus, we see that every element of $\text{Aut}(E)$ is the composition of the translation by an element of $E(k)$ and a group map.

Moreover, note that $E(k)$ is normal in $\text{Aut}(E)$—evidently $E(k)$ self-normalizes (:P) and

$$\varphi\circ t_c\circ\varphi^{-1}=t_{\varphi^{-1}(c)}$$

where normality then follows from the fact that, as we've noted, $E(k)\text{Aut}_{e.c}(E)=\text{Aut}(E)$.

Thus, finally, it suffices to note that $\text{Aut}_{\text{e.c}}(E)\cap E(k)=\{\text{id}\}$. Indeed, the only translation $t_c$ taking $e$ to $e$ is $t_e$, but $t_e$ is the identity.

NB: The above decomposition depends on choosing a base point (equiv. group structure) on $E$ so is, in some sense, canonical.

As to the finietness of $\text{Aut}_{e.c.}(E)$, this is classical. Namely, it suffices to assume that $k=\overline{k}$. If $\text{Char}(k)=0$ then one gets that it's bounded in size by $6$ (think about the complex case, and do a push-pull to $\mathbb{C}$ form general $k$). In positive characteristic, you can get as big as order $24$.

EDIT: See section III.10 of Silverman's book for details about the finiteness of automorphism groups.

• Thank you. If $\phi$ is an automorphism group of $E$, then $\phi(e)$ is in $E(k)$. To show that for every $q \in E(k)$, there is an automorphism $\phi$ such that $\phi(e)=q$, I want to use the morphism $E \rightarrow E \times q \hookrightarrow E \times E \rightarrow E$, where the last map is the multiplication that gives $E$ group scheme structure, then $e$ will be mapped to $e+q=q$. Is this proof OK? I guess this morphism is the translation $t_q$ you used in your arguments. – Wenzhe Mar 5 '16 at 17:51
• @WZ_Bosons That's correct. – Alex Youcis Mar 5 '16 at 22:05