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Let $X$ be an algebraic variety (say, projective, irreducible and smooth), defined over a field $K$, and let $L$ be a Galois extension. I am interested in algebraic varieties $Y$, defined over $K$, such that there exists an isomorphism $\psi\colon X_L\to Y_L$ defined over $L$ (but not over $K$ in general).

The map $\psi$ induces a 1-co cycle from $Gal(L/K)$ to the group $Aut(X_L) $ of $L$-automorphisms, and it is easy to check that the image in $H^1 (Gal, Aut(X_L))$ only depends on the class of $Y$, up to $K$-isomorphisms. My question is the following: is every element of $H^1$ obtained? In other words, we have a map from the set of $L$-forms of $X$ to $H^1 (..)$, the map is injective but is it surjective?

If no, could it be true in some natural cases, if yes I would be happy to see a reference.

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I think this can be found in Serre, Galois Cohomology.… p. 121, Section Forms – Timo Keller Mar 1 '14 at 13:08
Indeed: this is exactly Chapter III, 1.3, Proposition 5. – abx Mar 1 '14 at 13:15
Thanks, this is exactly what I wanted. – Jérémy Blanc Mar 1 '14 at 13:36
More generally, such "forms" of $Y$ are classified by this cohomology group for any morphism which is of effective descent. I think Waterhouse's "An Introduction to Affine Group Schemes" is really helpful. – Jon Beardsley Mar 1 '14 at 15:55
Note, however, that without quasiprojectivity hypothesis, there may exist classes in $H^1$ which are not of this form. More precisely, given a (separated, of finite type) $L$-scheme $Y$ and a cocycle of $\mathop{\mathrm{Gal}}(L/K)$ with values in $\mathop{\mathrm{Aut}}(Y)$, there exists an algebraic space $X$ such that $X_L$ is isomorphic to $Y$, giving rise to the given cocycle. Mathieu Huruguen has given explicit examples of (non-quasiprojective) toric varieties $Y$ for which this phenomenon happens. – ACL Mar 1 '14 at 21:28
up vote 5 down vote accepted

CW answer, copied from the comments of Timo Keller and abx:

See J.P. Serre's “Galois Cohomology”, Chapter III, 1.3, Proposition 5.

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It's good practice to carry over specific attribution. – The Masked Avenger Mar 1 '14 at 20:51
@TheMaskedAvenger — You're absolutely right. Thanks for the edit. – jmc Mar 2 '14 at 18:58

In case you are interested in seeing a situation where the map is not surjective, consider the set of rational maps of degree $f:\mathbb{P}^1\to\mathbb{P}^1$, modulo the conjugation action of $\phi\in\text{PGL}_2$, i.e., $f^\phi=\phi\circ f\circ\phi^{-1}$. This is the natural action if one is interested in dynamics (iteration of the map $f$). The automorphism group of $f$ is $\text{Aut}(f)=\{\phi\in\text{PGL}_2:f^\phi=f\}$. Two maps $f_1$ and $f_2$ defined over $K$ are isomorphic over $\bar K$ if $f_1=f_2^\phi$ for some $\phi\in\text{PGL}_2(\bar K)$, and the set of twists of $f$ is the set of maps that are $\bar K$-isomorphic to $f$, modulo $K$-isomorphism. Just as in the case you're looking at, the set of twists injects into the cohomology group $H^1(G_{\bar K/K},\text{Aut}(f))$. But the image is not surjective. The image turns out to be exactly the elements of $H^1(G_{\bar K/K},\text{Aut}(f))$ that become trivial in $H^1(G_{\bar K/K},\text{PGL}_2(\bar K))$. You can read about this in the following places:

The Arithmetic of Dynamical Systems, Springer, Section 4.9, specifically Theorem 4.79.

The Field of Definition for Dynamical Systems, Compositio Math. 98 (1995), 269-304.

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This is a nice example! But not really of the type Jérémy asked for, since the quotient stack $\mathop{\rm Rat}_d/\mathop{\rm PGL_2}$ is not a scheme. (The GIT quotient exists and is affine, but does not represent the quotient.) – ACL Mar 2 '14 at 0:07
You're right, it's not of the type Jeremy asked about. But after one's seen those types of examples, it's good to see other cases where it's not true. And even better is the example with an explanation, as you've given in terms of quotient stacks. – Joe Silverman Mar 2 '14 at 0:31
Yes, thanks to both of you for your examples. I was indeed mainly interested in schemes (and even in projective ones), but it is nice to see what happens in general. – Jérémy Blanc Mar 2 '14 at 0:39
@JoeSilverman: You're right too! And since I heard you lecture about this GIT quotient, the fact that it is affine (hence quasiprojective) but that Weil's criterion did not apply still made me think how this could be so... (Remains to understand why!) – ACL Mar 2 '14 at 8:00

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