Note:For a correct comprehension of the question see the "important edit" at the end.

Consider a projective variety over $\mathbb C$, $X=\textrm{Proj}\frac{\mathbb C[T_0,\ldots,T_n]}{f_1,\ldots,f_m}$ and a field automorphism $\sigma\in Aut(\mathbb C)$. Now with $X^\sigma$ we indicate the base change of $X$ along $\textrm{Spec}\,\left(\sigma^{-1}\right):\textrm{Spec}\mathbb C\longrightarrow \textrm{Spec}\mathbb C$. $X^\sigma$ as scheme is equal to $X$, but the structural morphism that gives the structure of $\mathbb C$-scheme is $(\textrm{Spec}\sigma)\circ p$.

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I have proved that $X^\sigma\cong\textrm{Proj}\frac{\mathbb C[T_0,\ldots,T_n]}{f^{\sigma^{-1}}_1,\ldots,f^{\sigma^{-1}}_m}$ ($f^{\sigma^{-1}}_1$ means that change the polynomial coefficients of $f_i$ by applying $\sigma^{-1}$). Therefore if we think in terms of projective algebraic sets: $X$ is $Z(f_1,\ldots,f_m)\subseteq\mathbb P^n(\mathbb C)$ and $X^\sigma $ is $Z(f^{\sigma^{-1}}_i,\ldots,f^{\sigma^{-1}}_m)\subseteq\mathbb P^n(\mathbb C)$.

Now I don't understand why the unique scheme isomorphism $X^\sigma\longrightarrow X$ induced by the fidebered product corresponds to the following map:

$$Z(f^{\sigma^{-1}}_1,\ldots,f^{\sigma^{-1}}_m)\longrightarrow Z(f_1,\ldots,f_m)\quad\quad(\ast)$$


At level of schemes this map is simply the identity of the scheme $X$ so I don't understand the role of $\sigma$.

Important Edit: As Felipe Voloch says in the comments, I have merged two different definitions of "twist" and in the above diagram, the presence of the map "$\text{id}$" is wrong. I'd like more informations about the difference of the following two ways to obtain of $X^\sigma$:

  • One $X^\sigma$ is obtained by changing only the structural morphism of $X$ (see the article B.Koeck - Belyi Theorem revisited at notation 1.1 for this construction).
  • Another $X^\sigma$ comes from the base change induced by $\text{Spec}(\sigma)$ (I'm interested to this construction).

However my original question remains, I reformulate it for clarity: consider $X^\sigma=X\times_{\text{Spec}(\mathbb C)}\text{Spec}(\mathbb C)$, I need a formal proof of the fact that the canonical map $X^\sigma\longrightarrow X$ corresponds to $(\ast)$ through the functor that relates algebraic sets and $\mathbb C$-schemes (reduced, separated...)

  • 3
    $\begingroup$ If $f=f_1=x_0+ix_1$, then $Z(f)=(1:i)\subset\mathbb{P}^1$. If $\sigma$ is complex conjugation then $Z(f^{\sigma})=(1:-i)$ and the schemes are isomorphic but not identical so the map between them is not the identity. "An algebraist is someone who doesn't know the difference between i and -i" :-) $\endgroup$ Aug 2, 2014 at 8:49
  • $\begingroup$ You example is clear, but my trouble comes from the diagram depicted in my question. The map $\textrm{id}$ makes such a diagram commutative. $\endgroup$
    – Dubious
    Aug 2, 2014 at 8:52
  • 1
    $\begingroup$ @Galoisfan — Well, the structure morphism of $X^{\sigma}$ is not what you say that it is. If $\phi_{\sigma}$ denotes the canonical map $X^{\sigma} \to X$, then the structure morphism of $X^{\sigma}$ is $\mathrm{Spec}(\sigma) \circ p \circ \phi_{\sigma}$. Everything I say here is a tautology. The point is that you seemed to make a circular argument. You have to explain how you arrive at your version of the structure morphism of $X^{\sigma}$… $\endgroup$
    – jmc
    Aug 2, 2014 at 9:09
  • 1
    $\begingroup$ There is an alternative way of defining the twist by keeping the underlying set the same but changing the structure sheaf. But, if you use that, your description of the twist as a projective variety is wrong. $\endgroup$ Aug 2, 2014 at 9:16
  • $\begingroup$ @jmc you are right, I've simply completed diagram with the most simple, to me, maps. At this point I'd like a formal proof that the canonical map $\phi_\sigma$ is $P\mapsto\sigma(P)$... I don't see it $\endgroup$
    – Dubious
    Aug 2, 2014 at 9:19

1 Answer 1


let me share my thoughts on your question. For clarity, let us work first with affine $n$-space; a point is then just a Morphism $p: \mathbb{C}[T_1, \ldots, T_n] \to \mathbb{C}$ of $\mathbb{C}$-algebras with kernel $(T_1 - p_1, \ldots, T_n-p_n)$.

The base change map $X^\sigma \to X$ is then $\mathbb{C}[T_1, \ldots, T_n] \to \mathbb{C}[T_1, \ldots, T_n] \otimes_{\sigma^{-1}} \mathbb{C}, P \mapsto P \otimes 1$ - note, the important thing is the different $\mathbb{C}$-algebra-structure on the right, where $\lambda \in \mathbb{C}$ is sent to $1 \otimes \lambda$. With this in mind, the map $$ \mathbb{C}[T_1, \ldots, T_n] \otimes_{\sigma^{-1}} \mathbb{C} \to \mathbb{C}[T_1, \ldots, T_n], P\otimes \lambda \to \lambda (P)^{\sigma^{-1}}, $$ where $\sigma^{-1}$ acts on the coefficients, is an isomorphism of $\mathbb{C}$-algebras.

Rephrasing again: On the level of rings, $X^\sigma \to X$ is given by "$P \mapsto P^{\sigma^{-1}}$" for polynomials, if we identify both $X$ and $X^\sigma $ with $Spec(\mathbb{C}[T_1, \ldots, T_n])$. Now consider a point of $X^\sigma $, that ist, a map $p$ as above, its image under the geometric base change map corresponds to the pullback on the level of rings, that is, its image in $X$ is the composite map $\mathbb{C}[T_1, \ldots, T_n] \to \mathbb{C}, P \mapsto p(P^{\sigma^{-1}})$ and we see that its kernel is given by $(T_1 - \sigma(p_1), \ldots, T_n - \sigma(p_n))$

Now i only considered affine n-space instead of projective varieties, but this is really the key point - closed subschemes of it work analogously, the key algebra isomorphism just becomes $$ \mathbb{C}[T_1, \ldots, T_n]/(f_1, \ldots, f_m) \otimes_{\sigma^{-1}} \mathbb{C} \to \mathbb{C}[T_1, \ldots, T_n]/(f_1^{\sigma^{-1}}, \ldots, f_m^{\sigma^{-1}}) $$ Glueing the morphisms should now give you the desired projective identity.

I apologize if the answer is not right on point, but i wanted to share the affine considerations, i hope they help you. Best regards.


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