"$F$-structures can be described in algebraic terms"

Question

Let $(X, \mathcal O_X)$ be an affine variety (ringed space which is isomorphic to a closed subset of $k^n$). An $F$-structure on $(X, \mathcal O_X)$ is defined (Springer, Linear Algebraic Groups) to be an $F$-structure $F[X]$ on $k[X]$ (which gives us a subtopology on $X$ of $F$-open sets), along with a collection of $F$-subalgebras $\mathcal O_X^F(U)$ of $\mathcal O_X(U)$, for each $U$ $F$-open in $X$, such that the natural $k$-algebra homomorphism $$k \otimes_F \mathcal O_X^F(U) \rightarrow \mathcal O_X(U)$$ is an isomorphism.

Springer claims "The proof of (the proposition that says that the ring of regular functions $\mathcal O_X(X)$ is really just the ring $k[X]$ of polynomial functions $X \rightarrow k$) carries over" and gives us that $F[X] = O_X^F(X)$." That's fine, I'll verify what he claims later. My question is about what he says next "We conclude that affine $F$-varieties and their morphisms can be described in algebraic terms." What does he mean here?

Answer 1

As I had to deal with some similar questions lately, I thought I tell you about my understanding of Springer's words.

Note that Springer uses the phrase 'can be described in algebraic terms' not only in the context of affine $F$-varieties but also a little bit earlier when he talks about affine $k$-varieties and their morphisms (without any $F$-structure). I believe what he is talking about here is that if $X$ is an affine $k$-variety, then all the information about $X$ is contained in $k[X]$. As a topological space, it is the maximal spectrum of $k[X]$ with the Zariski topology. Moreover, the structure sheaf $\mathcal{O}_X$ is uniquely determined by $\mathcal{O}_X(X) \cong k[X]$, as we have $\mathcal{O}_X(D(f)) \cong k[X]_f$ for all $f \in k[X]$ and the $D(f)$ form a basis for the Zariski topology.

As Springer explains, we can also identify morphisms $X \to Y$ of affine $k$-varieties with morphisms $k[Y] \to k[X]$ of $k$-algebras in a canonical way. Putting things together, we see that there is no geometry left in our description of affine $k$-varieties. We could even say an affine $k$-variety is an affine $k$-algebra and morphisms between such varieties are just the morphisms of $k$-algebras in opposite order. In terms of category theory, we have an equivalence between the category of affine $k$-varieties and the category opposite to the category of affine $k$-algebras.

Furthermore, not much changes when we consider $F$-structure. If $X$ is an affine $k$-variety with an $F$-structure, then this $F$-structure is already uniquely determined by $F[X]$. In fact we have to have $\mathcal{O}_X(D(f))(F) \cong F[X]_f$ for all $f \in F[X]$ and the $D(f)$ again form a basis for the $F$-topology.

Just like before, the $F$-morphisms $X \to Y$ correspond to the $F$-algebra homomorphisms $F[Y] \to F[X]$. And again, all geometry needed to describe affine $F$-varieties vanishes. That is, all we need to know about an affine $F$-variety $X$ are the algebras $k[X]$ and $F[X]$ and $F$-morphisms can be identified with algebra homomorphisms in the opposite direction.

I believe what Springer wants to say is simply that we are able to interpret all geometric aspects of the situation in terms of algebras and their morphisms.

Edit: To see that the $F$-structure $F[X]$ of $k[X]$ determines the affine $F$-variety structure, we only have to show $\mathcal{O}_X(D(f))(F) \cong F[X]_f$ for all $f \in F[X]$. We have $F[X] = \mathcal{O}_X(X)(F)$, so we have restriction maps $F[X] \to \mathcal{O}_X(D(f))(F)$ for any $f \in F[X]$. We may thus consider $f \in \mathcal{O}_X(D(f))(F)$. Notice that multiplication with $f$ is a linear map $\mathcal{O}_X(D(f))(F) \to \mathcal{O}_X(D(f))(F)$ which becomes an isomorphism after tensoring with $k$. Hence it was an isomorphism to begin with. But then $f$ is invertible in $\mathcal{O}_X(D(f))(F)$ and so we have a map $F[X]_f \to \mathcal{O}_X(D(f))(F)$ which becomes an isomorphism after tensoring with $k$, and so again it is an isomorphism itself.

Answer 2

This is not a real answer to my own question, but what might be useful towards an answer. Namely the fact that if $A$ is an affine $k$-algebra (we can put a ringed space structure on the set of its maximal ideals, and treat that ringed space noncanonically as some Zariski closed set of $k^n$, for some $n$), and $A_0$ an $F$-subalgebra of finite type which is an $F$-structure on $A$, then we can find an $n \geq 1$, and a Zariski closed set $\mathscr X \subseteq k^n$, such that $I(\mathscr X)$ can be generated by polynomials in $F[X_1, ... , X_n]$, $A_0 \cong F[X_1, ... , X_n]/I(\mathscr X) \cap F[X_1, ... , X_n]$ as $F$-algebras, $A \cong k[X_1, ... , X_n]/I(\mathscr X)$, and the inclusion $A_0 \subseteq A$ can be identified with the $F$-algebra monomorphism $F[X_1, ... , X_n]/I(\mathscr X) \cap F[X_1, ... , X_n] \hookrightarrow k[X_1, ... , X_n]/I(\mathscr X)$.

This doesn't mean that $F$-structures on a given affine $k$-algebra are unique in any sense, or that the $F$-topologies on $Max(A)$ from different $F$-structures on $A$ are homeomorphic, but it does give us a "standard" way to look at any given $F$-structure.

Now Springer claims (and I think I know how to verify) that if $(X, \mathcal O_X, \mathcal O_X^F)$ is an affine $F$-variety, then $F[X]$ (that is, $F[X_1, ... , X_n]/I(\mathscr X) \cap F[X_1, ... , X_n]$, or $A_0$, or what have you) is the same thing as $\mathcal O_X^F(X)$. So we can describe $\mathcal O_X^F(X)$ in algebraic terms, sure. But what about $\mathcal O_X^F(U)$ for $U$ $F$-open? In particular for when $U$ isn't affine? How can we view that "algebraically?"