The following is a rewrite of the proof of Lemma G from Steinberg's On theorems of Lie-Kolchin, Borel, and Lang. Let $G$ be an algebraic group over a finite field $k=\mathbb{F}_q$ and let $A$ be the coordinate ring of $G$.
Background from algebraic groups: We write $\Delta$ for the map $A \to A \otimes_k A$ induced by pull back along the multiplication map $G \times G \to G$. Let $A^{\vee}$ be the dual vector space $\text{Hom}_k(A,k)$. The map $\Delta : A \to A \otimes A$ induces a dual map $\Delta^{\vee}: A^{\vee} \otimes A^{\vee} \to A^{\vee}$. Conceptually, $A^{\vee}$ can be thought of as measures on $G$ and $\Delta^{\vee}$ is convolution; in particular, if $\lambda_i$ is evaluation at a point $g_i \in G(k)$, then $\Delta^{\vee}(\lambda_1 \otimes \lambda_2)$ is evaluation at $g_1 g_2$.
For $\lambda \in A^{\vee}$ and $\theta \in A$, set $R_{\lambda}(\theta) = (\text{Id}\otimes \lambda) (\Delta(\theta)) \in A$. If $\lambda$ is evaluation at a point $g \in G(k)$, then $R_{\lambda}(\theta)$ is the right translation of $\theta$ by $g$. For $\theta \in A$, let $R(\theta) = \{ R_{\lambda}(\theta) : \lambda \in A^{\vee} \}$. If $G(k)$ were Zariski dense in $G$, then $R(\theta)$ would be the span of the right translates of $\theta$; in our setting, $G(k)$ is finite, but I still think of this as the motivation for $R(\theta)$.
Lemma 1 Let $\theta \in A$ and let $\phi \in R(\theta)$. Then $R(\phi) \subseteq R(\theta)$.
Proof We have $R_{\lambda_1}(R_{\lambda_2}(\theta)) = R_{\Delta^{\vee}(\lambda_2 \otimes \lambda_1)}(\theta)$. $\square$
Lemma 2 For any element $\theta$ of $A$, the vector space $R(\theta)$ is finite dimensional.
Proof Let $\Delta(\theta) = \sum \alpha_i \otimes \beta_i$, by the definition of the tensor product, the sum is finite. I claim that $R(\theta)$ is contained in the span of the $\alpha_i$. Indeed, $R_{\lambda}(\theta) = \sum \alpha_i \lambda(\beta_i)$. $\square$
Proof of Lang's theorem: Let $F$ be the Frobenius map $G \to G$, so $F^{\ast}(\theta) = \theta^q$ for $\theta \in A$. Let $L$ be the Lang map $g \mapsto g^{-1} F(g)$. We want to show that $A$ is finite over $L^{\ast} A$.
Let $\theta$ be an arbitrary element of $A$; we will show that $\theta$ is integral over $L^{\ast} A$. Since $A$ is finitely generated as a $k$-algebra, this will imply that $A$ is finite over $L^{\ast} A$.
Using Lemma 2, choose a basis $\theta_1$, $\theta_2$, ..., $\theta_r$ of $R(\theta)$. Using Lemma_1, we have
$$\Delta(\theta_i) = \sum_j \theta_j \otimes \beta_{ij}$$
for some $\beta_{ij}$ in $A$. In other words, for $x$ and $y \in G(S)$, for any $k$-algebra $S$, we have
$$\theta_i(xy) = \sum_j \theta_j(x) \beta_{ij}(y).$$
Now, put $x = g$ and $y = L(g)$. So we have
$$\theta_i(g)^q = \theta_i(F(g)) = \sum_j \theta_j(g) \beta_{ij}(L(g))$$
or, in other words,
$$\theta_i^q = \sum_j \theta_j L^{\ast}(\beta_{ij}).$$
We have now shown that $\theta_i^q$ is in the $L^{\ast}(A)$-module generated by $\theta_1$, $\theta_2$, ..., $\theta_n$. Thus, the ring $L^{\ast}(A) \langle \theta_1, \theta_2, \ldots, \theta_n \rangle$ is finitely generated as an $L^{\ast}(A)$ module and, in particular, $\theta$ is integral over $L^{\ast}(A)$. $\square$.
