A few months ago, I read a nice elementary proof of Lang's theorem:


**Theorem:** Let $G$ be a connected linear algebraic group over $\overline{\mathbb{F}}\_p$ and let $F : G \to G$ be a Frobenius map. Then the Lang map $L : G \to G$ defined by $L(g) = g^{-1}F(g)$ is surjective.


The proof showed that $L$ was both finite and dominant, which I believe is quite a standard approach. The unique feature of this proof was that it showed finiteness by exhibiting a finite generating set for $\overline{\mathbb{F}}\_p[G]$ as an $\overline{\mathbb{F}}\_p[G]$-module (with the module structure induced by $L$). This was done by just playing around with the functions so it required no deep background. Unfortunately, I didn't realise at the time that I would need this later, so I didn't write down the reference and now I can't find it.


**Question:** Does anyone know where I can find this proof of Lang's theorem?


To clarify, I know that there are other proofs (e.g. in Digne-Michel) of Lang's theorem, but this elementary proof would fit perfectly into the context where I need it.



Reference request: Elementary proof of Lang's theorem
The following is a rewrite of the proof of Lemma G from Steinberg's [On theorems of Lie-Kolchin, Borel, and Lang](https://doi.org/10.1016/B978-0-12-080550-1.50032-9). 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 $L^{\ast}(A)$. $\square$.


