# $F$-rational points of variety/alg. group in Springer

I am confused about something on page 6 in Springer's Linear Algebraic Groups.

Setup:

• $k$ 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 $F$.

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I think you can just look at the images of the $T_i$ under the map $F[X] \to F$ to get coordinates. It's not obvious that they're in there, though. –  Dylan Moreland May 25 '12 at 17:14

In general, if $F$ is a field, and $I=(f_1,\ldots,f_m)$ is an ideal of $F[x_1,\ldots,x_n]$, then the set $\mathrm{Hom}_{F-alg}(F[X_1,\ldots,X_n]/I,F)$ is in canonical bijection with the $n$-tuples $(\alpha_1,\ldots,\alpha_n)$ such that $f_j(\alpha_1,\ldots,\alpha_n)=0$ for all $j$. The map is $\varphi\mapsto(\varphi(X_1),\ldots,\varphi(X_n))$. This is true when you replace the target of the homomorphisms by an arbitrary $F$-algebra $A$. Then you get the points $(a_1,\ldots,a_n)\in A^n$ on which the $f_j$ all vanish.
I can't help but add that this awkwardness about the notion of $F$-rational points goes away completely when you do everything with schemes.
Thanks for the answer. In my notation $F[X]$ is just some $F$-subalgebra of $k[X]$. Do I know that $F[X] = F[T_1, \dots, T_n] / I$ for some ideal $I$? –  Thomas May 25 '12 at 17:21
Your $F[X]$ is assumed to be of finite type over $F$, so it can be written as $F[T_1,\ldots,T_r]/I$ for some ideal $I$ of $F[T_1,\ldots,T_r]$. This is what it means to be of finite type over $F$. Note that the $r$ may be different from the $n$ in your original setup. The simplest setting in which you have an "$F$-structure" is when $I$ can be generated by polynomials with coefficients in $F$. Then you can take $A_0$ to be the quotient of $F[X_1,\ldots,X_n]$ by $I\cap F[X_1,\ldots,X_n]$. –  Keenan Kidwell May 25 '12 at 17:23
Ok, so you are saying that $F[X]$ is of finite type, so $F[X] = F[f_1, \dots ,f_r]$ and so with $I$ being the kernel of $k[T_1, \dots , T_r]$ ($T_i \mapsto f_i)$ we get what you write? –  Thomas May 25 '12 at 17:28