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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.

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    $\begingroup$ For a textbook reference you could look at Theorem 4.1.12 of Geck's "An Introduction to Algebraic Geometry and Algebraic Groups". This is basically the proof given by David Speyer below but it is written up nicely. $\endgroup$
    – Jay Taylor
    Oct 20 at 10:59
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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 $L^{\ast}(A)$. $\square$.

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A one-paragraph proof that may not require deep background (the author advertises that it "can be used at an early stage in text books on algebraic groups") is given by Peter Müller in Algebraic groups over finite fields, a quick proof of Lang’s theorem.

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  • $\begingroup$ I had seen this one before, but thank you for the suggestion. Unfortunately it's assuming a bit more than I want (specifically the existence of a closed orbit and the condition about needing to show that $g^{-1}xF(g)=x$ has only finitely many solutions). I have a vague memory that what I am looking for might be in Steinberg's "On theorems of Lie-Kolchin, Borel, and Lang" which appears as reference [6] in the above. Unfortunately I can't find a copy that I have access to. $\endgroup$ Oct 19 at 12:51
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    $\begingroup$ you can read Steinberg's paper on Google Books --- my browser omits page 352, but shows the other pages of the paper. $\endgroup$ Oct 19 at 13:28
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    $\begingroup$ It is Lemma G in Steinberg's paper "On theorems of Lie-Kolchin, Borel, and Lang" on pages 351--352. If $A=k[G]$, using the morphism $\sigma^{\sharp}: A\rightarrow A$ (where $\sigma: G\rightarrow G$ is the Frobenius map), the gist of the proof is to show that a basis $\{f_j\}$ of left translations of an $f\in A$ implies that $(\sigma^{\sharp}-1)A[\{f_j\}]$ is f.g. as a module over $(\sigma^{\sharp}-1)A$ and contains $f$. Thus, $A$ is integral over $(\sigma^{\sharp}-1)A$ and hence $\sigma-1$ is finite. $\endgroup$
    – F Zaldivar
    Oct 19 at 13:55
  • $\begingroup$ @FZaldivar, at that point, isn't $\sigma$ any endomorphism with a finite fixed-point set, not necessarily Frobenius? $\endgroup$
    – LSpice
    Oct 19 at 17:57
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    $\begingroup$ @LSpice At least the way the proof is written in Steinberg's paper, no. He uses that $\sigma^{\#}(f) = f^q$ and that therefore, by proving $\sigma^{\#}(f_j)$ is in the $(\sigma^{\#}-1)A$-linear span of the $f_j$, one can deduce that the algebra $(\sigma^{\#}-1)A[f_1, f_2, \ldots, f_r]$ is finite over $(\sigma^{\#}-1)A$. I wouldn't be surprised if you could do something clever like what you say, though. $\endgroup$ Oct 19 at 19:21

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