I am confused about something on page 6 in Springer's Linear Algebraic Groups.
- $k$ is an algebraically closed field
- $X$ is a closed set in $k^n$.
- $I(X) = \langle f\in k[T_1, \dots, T_n] : f(v) = 0\; \forall v\in X \rangle$
- $A = k[X] = k[T_1 ,\dots , T_n] / I(X)$
- $F$ is a subfield of $k$.
I understand that an $F$-structure on $X$ is a $F$-subalgebra $A_0 = F[X]$ of $k[X]$ such that $A_0$ is of finite type and $k\otimes_F A_0 \simeq k[X]$.
Question: What are the $F$-rational points of $X$? Springer says that it is all $F$-algebra homomorphisms from $F[X] \to F$. I thought that the $F$-points of a variety are just the points you get from restricting the solutions of the underlying equations to $F$.