2
$\begingroup$

Suppose $k$ is a field and $K$ is a Galois extension of $k$. If $X$ is a variety defined over $k$, let $\text{Aut}(X/k)$ be the group of $k$ automorphisms of $X$, i.e. isomorphisms from $X$ to $X$ in the category of $k$-varieties, $\require{AMScd}$ \begin{CD} X @>f>> X\\ @V V V @VVV\\ \text{Spec}\,k @>>\text{Id}> \text{Spec}\,k \end{CD} Then a twist of $X$ is a variety $X'$ over $k$, such that there is a $K$-isomorphism $\phi$ $\require{AMScd}$ \begin{CD} X' \times_{\text{Spec}\,k} \text{Spec}\,K @>\phi>> X\times_{\text{Spec}\,k} \text{Spec}\,K\\ @V V V @VVV\\ \text{Spec}\,K @>>\text{Id}> \text{Spec}\,K \end{CD} Then from literatures (e.g. Chapter X of The Arithmetic of Elliptic Curves by Silverman), there is a map associated to $\phi$ defined by \begin{equation} \xi: \text{Gal}(K/k) \rightarrow \text{Aut}(X/k),~\sigma \rightarrow \phi^\sigma \circ \phi^{-1} \end{equation} Question 1: What is $\phi^{\sigma}$? Is it given by the composition \begin{equation} X' \times_k K \xrightarrow{\sigma} X' \times_k K \xrightarrow{\phi} X \times_k K \end{equation} where the first morphism is given by $\sigma$ acting on the second factor of $X' \times_k K $? If so, how to see $\phi^\sigma \circ \phi^{-1}$ is a $k$-automorphism of $X$?

Question 2: In Silverman's book, it says that $\xi$ measures $\phi$'s failure to be defined over $k$, could anyone explain the intuitions behind this statement? any interesting examples?

$\endgroup$
  • $\begingroup$ $\phi^\sigma$ should be what you get by applying $\sigma$ to the coefficients of the formula(s) describing $\phi$. It is definitely not a composition. $\endgroup$ – Lubin Apr 22 '18 at 17:51
  • $\begingroup$ @Lubin Thank you. $\endgroup$ – Wenzhe Apr 22 '18 at 19:50
5
$\begingroup$

The morphism $\phi^\sigma$ is defined as follow : the automorphism $\sigma$ defines an automorphism still denoted $\sigma : \operatorname{Spec}K\to\operatorname{Spec}K$. By pull-back, it defines for any $X$ an automorphism $$\sigma:X\times_k \operatorname{Spec}K\to X\times_k\operatorname{Spec}K$$

This defines a right action of $\operatorname{Gal}(K/k)$ on any $X_K=X\times_k\operatorname{Spec}K$. Now we have $\phi^\sigma=\sigma\circ\phi\circ\sigma^{-1}$. Note that this morphism is a $K$ morphism. This is because the diagram : $$\require{AMScd} \begin{CD} X'\times_k\operatorname{Spec}K@>\sigma^{-1}>>X'\times_k\operatorname{Spec}K@>\phi>>X\times_k\operatorname{Spec}K@>\sigma>>X\times_k\operatorname{Spec}K\\ @VVV@VVV@VVV@VVV\\ \operatorname{Spec}K@>\sigma^{-1}>>\operatorname{Spec}K@=\operatorname{Spec}K@>\sigma>>\operatorname{Spec}K \end{CD}$$ commutes and that the bottom row is obviously the identity.

(Note : if $G$ acts on objects $A$ and $B$ (on the right), the formula $\phi^\sigma=\sigma\circ\phi\circ\sigma^{-1}$ is the usual (right) action on $Hom(A,B)$. Note also that if $X=\operatorname{Spec}k[x_1,...,x_n]/(I), X'=\operatorname{Spec}k[y_1,...,y_m]/(J)$ and $\phi:X'_K\to X_K$ is given by $x_i\mapsto P_i(y_1,...,y_m)$, then $\phi^\sigma$ is given by $x_i\mapsto \sigma P_i(y_1,...,y_m)$ as Lubin said.)

However, it seems that this is an error in Silverman's book or an abuse of notation : $\xi_\sigma:=\phi^\sigma\circ\phi^{-1}$ is definitively not a automorphism of $C/k$, only of $C_K/K$. In fact this morphism is not invariant by Galois and we even have the cocycle equality : $\xi_{\sigma\tau}=(\xi_\sigma)^\tau\xi_\tau$.


As for question 2), $\phi$ is defined over $k$ iff it is invariant by Galois (this is obvious with the local description), in other words iff $\phi^\sigma=\phi$ for every $\sigma$, this is in turn equivalent to the cocycle $(\xi_\sigma)$ being trivial.

Have you look at the examples in Silverman's book ?

$\endgroup$
  • $\begingroup$ Thank you. I will check the examples in Silverman's book. $\endgroup$ – Wenzhe Apr 23 '18 at 18:53

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.