3
$\begingroup$

This question is closely related to this recent one.

Suppose that $s:X\longrightarrow\text{Spec}\, K$ is a variety over $K$ (i.e. a $K$ scheme, separated, proper and geometrically integral) and consider a field automorphism $\sigma\in\text{Aut}(K)$. Then we define the varierty $$X^\sigma:=X\times_{\text{Spec}\, K}\text{Spec}\,K$$ which is the base change of $X$ through the morphism $\text{Spec}(\sigma):\text{Spec}\, K\longrightarrow\text{Spec}\, K$.

enter image description here

The structural morphism of $X^\sigma$ is $p_2$. In general $X$ and $X^\sigma$ are not isomorphic as varieties, in fact the map $p_1$ is a morphism of schemes but not a morphism of $K$-schemes.

On the other hand should be true that $\mathbb P^n_K=\text{Proj} (K[x_0,\ldots,x_n])$ and $\left(\mathbb P^n_K\right)^\sigma$ are two isomorphic varieties, but I can't find this isomorphism. Do you have any idea?

I'm looking for a solution for long time, but without success.

$\endgroup$
  • 3
    $\begingroup$ It's just the action of $\sigma$ on coordinates of the projective space. Write this on the charts and you will see the isomorphism. $\endgroup$ – user40276 Dec 30 '14 at 22:16
4
$\begingroup$

The key is that $\mathbb{P}^n_K = \mathbb{P}^n_\mathbb{Z} \times_{\operatorname{Spec} \mathbb{Z}} \operatorname{Spec} K$. More generally, you can do this any choice of base schemes, and indeed, even in any category with pullbacks.

So consider a commutative diagram of the form below, $$\require{AMScd} \begin{CD} A_0 @>{\tilde{\sigma}}>> A_1 @>{q_1}>> B \\ @V{p_0}VV @V{p_1}VV @VV{p}V \\ C @>>{\sigma}> C @>>{q}> D \end{CD}$$ where the two squares are pullbacks and $\sigma : C \to C$ is an automorphism. It is straightforward to verify that $\tilde{\sigma} : A_0 \to A_1$ is an isomorphism. Moreover, if $q \circ \sigma = q$ (e.g. when $D$ is a terminal object) then we can arrange that $A_0 = A_1$ and $p_0 = p_1$, in which case it is immediate that $p_0 : A_0 \to C$ and $p_1 : A_1 \to C$ are isomorphic over $C$. Indeed, the point is that $$\begin{CD} A_0 @>{q_1 \circ \tilde{\sigma}}>> B \\ @V{p_0}VV @VV{p}V \\ C @>>{q \circ \sigma}> D \end{CD}$$ commutes and is a pullback square (by the pullback pasting lemma), so if $q \circ \sigma = q$, then there is a unique morphism $\tau : A_1 \to A_0$ making the following diagram commute, $$\begin{CD} C @<{p_1}<< A_1 @>{q_1}>> B \\ @| @V{\tau}VV @| \\ C @<<{p_0}< A_0 @>>{q_1 \circ \tilde{\sigma}}> B \end{CD}$$ and it is a straightforward exercise to verify that $\tau : A_1 \to A_0$ is an isomorphism.

$\endgroup$
  • $\begingroup$ Thanks! The point is that $q\circ\sigma=q$! $\endgroup$ – Dubious Dec 31 '14 at 11:45

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.