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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 $\text{Spec}(\sigma):\text{Spec}\, K\longrightarrow\text{Spec}\, K$.

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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.

  • 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

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.

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

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