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Let $L/k$ be a field extension of algebraically closed fields of characteristic zero. Let $U$ be a smooth quasi-projective variety over $k$.

I am trying to understand why the base-change functor from $k$ to $L$ from

  1. the category of finite etale $k$-morphisms to $U$

to

  1. the category of finite etale $L$-morphisms to $U_L$

is faithful.

I have the feeling it is almost "trivial" and that there is a much more general statement concerning the base-change functor.

Does this appear anywhere, or could anybody provide some explanation?

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Just out of curiosity: in which context are you interested in a (nontrivial) extention of an algebraically closed field which is itself algebraically closed? –  Qfwfq Jun 17 '13 at 14:36
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An important example is $\mathbb{C}$ over $\overline{\mathbb{Q}}$ (links between number theory and complex geometry). –  Martin Brandenburg Jun 17 '13 at 14:40
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1 Answer

Let $L/k$ be a field extension. Then the base change functor $\mathsf{Sch}/k \to \mathsf{Sch}/L$ is faithful, i.e. if $f,g : X \to Y$ are morphisms of $k$-schemes with $f_L = g_L : X_L \to Y_L$, then $f=g$.

Proof: Choose an open affine covering of $Y$, pull it back to $X$ via $f$ and $g$ and take an affine refinement of the two affine coverings of $X$. This shows that we may assume that $X,Y$ are affine. But if $\alpha,\beta : A \to B$ are homomorphisms of $k$-algebras with $\alpha \otimes L = \beta \otimes L : A \otimes_k L \to B \otimes_k L$, then of course $\alpha=\beta $ since $B \to B \otimes_k L$ is injective. $\square$

Actually much more is true. See for example Grothendieck's exposé on descent theory in FGA or Vistoli's notes on descent theory.

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