2
$\begingroup$

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 $\mathbb{P}^1$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Alex Youcis Mar 4 '16 at 11:58
  • $\begingroup$ @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! $\endgroup$ – Wenzhe Mar 4 '16 at 12:05
  • $\begingroup$ Answered below. $\endgroup$ – Alex Youcis Mar 4 '16 at 12:12
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Wenzhe Mar 5 '16 at 17:51
  • $\begingroup$ @WZ_Bosons That's correct. $\endgroup$ – Alex Youcis Mar 5 '16 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.