F_q-structures on schemes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:21:35Z http://mathoverflow.net/feeds/question/12095 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12095/f-q-structures-on-schemes F_q-structures on schemes S1 2010-01-17T12:30:12Z 2010-01-20T18:27:46Z <p>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 <em>canonical</em> morphism $a \otimes \lambda \mapsto a \lambda$.</p> <p>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?</p> <p>(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 $X _0$.)</p> http://mathoverflow.net/questions/12095/f-q-structures-on-schemes/12101#12101 Answer by Philipp Hartwig for F_q-structures on schemes Philipp Hartwig 2010-01-17T14:03:08Z 2010-01-17T14:03:08Z <p>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. </p> <p>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.</p>