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Let $k|\mathbb{F}_q$ be a field extension. An $\mathbb{F}_q$-structure on a $k$-algebra $A$ is an $\mathbb{F}_q$-subalgebra $A _0$ of $A$ such that $A _0 \otimes _{\mathbb{F}_q} k \cong A$ via the canonical morphism $a \otimes \lambda \mapsto a \lambda$.

Now, my question is if this notion can be properly globalized to $k$-schemes? I saw a definition like: an $\mathbb{F}_q$-structure on a $k$-scheme $X$ is an $\mathbb{F}_q$-scheme $X _0$ such that $X \cong X _0 \times _{\mathrm{Spec}(\mathbb{F}_q)} \mathrm{Spec}(k)$ as $k$-schemes (see for example "Representations of finite groups of Lie type" by Digne and Michel, where $\cong$ is even replaced by $=$). But my problem is that here the particular choice of the canonical morphism as above does not appear so that on affines this definition is not the same as above. Is this a problem?

(The reason why I care about this is that I want to defined the (geometric) Frobenius on a $k$-Scheme with $\mathbb{F}_q$-structure as the "base change" of the canonical Frobenius (raising to the $q$-th power) on the $\mathbb{F}_q$-structure $X _0$.)

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I think that notion you cite from Digne and Michel is not a good one as you will not get a well-defined Frobenius. I suggest replacing $X_0$ by a pair $(X_0,p)$, where $p:X\to X_0$ is a morphism of $\mathbb{F}_q$ schemes such that $X$ is a product $X_0\times_{\mathbb{F}_q} \mathrm{Spec}(k)$ via the structure morphism $X\to \mathrm{Spec}(k)$ and via $p$. The Frobenius on $X$ will then be the unique map $F:X\to X$ of $k$-schemes such that $F\circ p= F_0\circ p$, where $F_0:X_0\to X_0$ denotes the canonical Frobenius.

To compare this to the definition for $k$-algebras, note that we could define two $\mathbb{F}_q$-structures $(X_0,p)$ and $(X_0',p')$ to be equivalent if there is an isomorphism $f:X_0\to X_0'$ such that $fp=p'$. Two equivalent $\mathbb{F}_q$-structures on $X$ will then give rise to the same(!) geometric Frobenius. In the case of a $k$-algebra $A$ every $\mathbb{F}_q$-structure will have a unique representative of the form $(\mathrm{Spec}(A_0),p)$, where $A_0\subset A$ and $p$ is induced by this inclusion.

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  • $\begingroup$ I assume that's why Digne and Michel used '$=$' in place of '$\simeq$'. This is a little sloppy, but I would read the equality sign as we are given an isomorphism between $X$ and $X_0\times Spec(k)$ that agrees with all structures', in this case, an isomorphism over $k$. We are given' means that the isomorphism is part of the data. $\endgroup$
    – t3suji
    Commented Jan 17, 2010 at 14:36
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    $\begingroup$ I agree with the answer: given a variety $X$ over a field $K$ and a subfield $k$ of $K$, then a model of $X$ over $k$ (or a $k$-structure on $X$) is a variety $X_0$ over $k$ together with an isomorphism $X_0\otimes_k K\to X. Digne and Michel are being sloppy. $\endgroup$
    – JS Milne
    Commented Jan 17, 2010 at 19:46
  • $\begingroup$ Okay, thanks! Is it correct that your definition is then more general in the case of affine schemes as the one I have given above (because I fixed a particular isomorphism)? I am not sure if this particular isomorphism is needed somewhere. As far as I can see, Digne and Michel just concentrate on affine and projective varieties... $\endgroup$
    – user717
    Commented Jan 17, 2010 at 20:59
  • $\begingroup$ In the second paragraph I tried to explain the relationship between the definition for schemes I have given and the one for algebras you have mentioned in the beginning. In both cases we fix one particular isomorphism (that's the difference to the one in Digne and Michel), but the one for schemes gives a slightly more general notion when applied to the affine case in that we do not require $A_0$ to be a subalgebra of $A$ but that we allow for an arbitrary injective morphism $A_0\to A$. That doesn't really change anything and if we pass to isomorphism classes as above we get the "same" notion $\endgroup$
    – Philipp Hartwig
    Commented Jan 17, 2010 at 21:10
  • $\begingroup$ Yes, that's right, I somehow missed this. $\endgroup$
    – user717
    Commented Jan 17, 2010 at 21:40

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