Use of generic points to show that $k$-tori are split over $k_s$

Question

Let $k$ be a field of characteristic $p$, and let $T$ be a $k$-torus. To say that $k$-split is to say that every character $T \rightarrow \mathbf{G}_m$ is defined over $k$.

The following result is standard. It comes down to showing that if $T$ is split over a purely inseparable extension of $k$, then it is already split over $k$. I have seen a proof in Springer, Linear Algebraic Groups which uses some facts about derivations. But I wanted to understand this proof from Borel Tits, Groupes Reductifs:

I guess what they are doing in this proof is identifying $T$ with the group of invertible $n$ by $n$ diagonal matrices with coefficients in a universal field $\Omega$ containing $k$. Then if $t = (t_1, ... , t_n)$ is a point of $T$, I think they mean $k(t)$ to be the field $k(t_1, ... , t_n)$.

To say that $t$ is generic is to say that $t_1, ... ,t_n$ are algebraically independent over $k$. In the proof, they show that if $t$ is generic over $k$, then the given character $a$ maps $t^q = (t_1^q, .. , t_n^q)$ into $k(t^q)$. They seem to then conclude that $a(t^q) \subseteq k(t^q)$ for all points $t \in T$, which implies that the coefficients defining the morphism of varieties $a$ must lie in $k$, i.e. $a$ is defined over $k$. I don't understand how what is written in the proof shows anything: what if $t$ is not generic over $k$?

Answer 1

By $k(t)$, Borel-Tits mean the field of rational functions on $T$. It is, in general, not purely transcendental. The field $k(t^q)$ is the subfield of pull-backs of rational functions via the power map $\phi:T\to T:t\mapsto t^q$. Now they claim that $\overline k(t^q)\cap k(t)=k(t^q)$. Since $a^q(t)$ is an element of the left side it is of the form $b(t^q)$ with $b(t)\in k(t)$. But then the equality $a^q(t)=b(t^q)$ implies $a(t)=b(t)$. Thus $a(t)\in k(t)$. But a rational function which is a character is regular, i.e., $a(t)\in k[t]$.

By the way, the intersection $\overline k(t^q)\cap k(t)=k(t^q)$ works more generally: Let $\phi:X\to Y$ be a dominant morphism between two geometrically integral $k$-varieties (e.g., they is smooth with a $k$-rational point). Then $\overline k(Y)\cap k(X)=k(Y)$ inside $\overline k(X)$ . To see this, first use $(\overline k\otimes_kk(Y))\cap k(X)=k(Y)$ inside $\overline k\otimes_kk(X)$ and then $\overline k\otimes_kk(X)=\overline k(X)$ and $\overline k\otimes_kk(Y)=\overline k(Y)$ by geometric integrality.