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These are parametrized by $H^1(Gal(\mathbb{Q}), Aut X)$, where X is some $\mathbb{Q}$-model of the curve.

It was established in Confusion about how the first cohomology classifies torsors that fiber bundles over $B$ with fiber $F$, structure group $G$ and transition maps with property $P$ are classified by $T$-torsors, where $T$ is the sheaf on $B$ of functions to $G$ with property $P$. $T$-torsors, in turn, are classified by $H^1(B, T)$.

Is there a way to interpret the aforementioned classification of $\mathbb{Q}$-models of a curve in these terms?

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Yes.

Roughly, the idea is that once your base object $X/B$ is fixed then for any object $Y$ you can consider the sheaf $Iso(X,Y)$ of isomorphisms from $X$ to $Y$. This has an action of the sheaf $Aut(X)$ and on any open cover $U$ where $Y|_U \cong X|_U$, making a choice of such an isomorphism gives an isomorphism of $Aut(X)(U)$-sets $Iso(X,Y)(U) \cong Aut(X)(U)$.

The difference from the standard argument in this case is that you need to use the etale topology instead of the Zariski topology. The etale covers of $\mathbb{Q}$ are finite field extensions $F$ and so the statement that $Y$ is "locally" equivalent to $X$ means that $Y_F \cong X_F$ for a large enough field extension $F$.

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So, to be precise... We are looking in the etale topology; B is spec(Q); G is the sheaf into Grps: Aut(X); and P is... what? –  Makhalan Duff Apr 29 '10 at 18:02
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I'd actually say that in the terminology you were originally using, G should be the group scheme of automorphisms of X and P should be simply algebraic maps. Then T(Spec(F)) is the set of algebraic maps Spec(F) -> Aut(X), or the Spec(F)-points of Aut(X), or the automorphism group of $X_F$ over Spec(F). –  Tyler Lawson Apr 29 '10 at 18:26
    
Huh.. I see. You say "group scheme of automorphisms". Is there any reason to believe it's representable? –  Makhalan Duff Apr 29 '10 at 18:40
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You can show representability of Aut(X) (or more generally the existence of Isom-schemes) by identifying an automorphism with its graph $\Gamma \subset X \times X$, and construct a representing object inside a certain Hilbert scheme. However, if you want to know more details about that you should ask another question so that a genuine algebraic geometer has the opportunity to answer. –  Tyler Lawson Apr 29 '10 at 18:51
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