Let $A$ be an affine $k$-algebra with associated ring spaced $(X, \mathcal O_X)$, $F$ a subfield of $k$, and $A_0$ an $F$-structure on $A$. From Springer, Linear Algebraic Groups:
A closed subset $Y$ of $X$ is $F$-closed if the ideal $\mathcal I(Y) = \{ f \in A: f(x) = 0, \forall x \in Y\}$ is defined over $F$ (i.e. taking the span of $\mathcal I(Y) \cap A_0$ over the field $k$ gets you back $\mathcal I(Y)$). A set is $F$-open if its complement is $F$-closed. An example of an $F$-open set is $D(f) = \{ x \in X : f(x) \neq 0\}$, where $f \in A_0$. These form a basis for the $F$-topology.
Is this really true? What if I do the following: $F$ has characteristic $p$, $a \in F$, but $b := \sqrt[p]{a} \not\in F$. Let $A = k[Y], A_0 = F[Y]$, then $A_0$ is an $F$-structure for $A$. The polynomial $f(Y) = Y^p - a$ lies in $A_0$, but I don't think the complement of $D(f) = \{ y \in k : f(y) \neq 0\}$ is $F$-closed.
This is because $\mathcal I \{ y \in k : f(y) = 0 \}$ is equal to the radical of $k[Y](Y^p - a) = k[Y](Y - b)^p$, which is $k[Y](Y - b)$. If you try to take the span over $k$ of the ideal $k[Y](Y- b) \cap F[Y] = F[Y](Y^p - a)$, you won't get back $k[Y](Y-b)$.
Is there something I'm not understanding here, or is Springer incorrect about these sets forming a basis for the $F$-topology? As far as I can see, they might not even be open.