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 $\mathbb{P}^1$.
 A: 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.
