3 added 283 characters in body

This is very far from a solution but just some ideas that could be possibly be used as a start.

First a simple observation on an Artin-Schreier type extension $X\rightarrow Y$, where $X:=\mathrm{Spec}\mathcal O_Y[t]/(t^q-t-f)$. Here $q=p^m$, $p$ a prime with $p\mathcal O_Y=0$ and $f\in\mathcal O_Y(Y)$. This a finite étale covering with Galois group $\mathbb F_q$. For any surjective group homomorphism $\varphi\colon\mathbb F_q\rightarrow\mathbb Z/p$ we get an induced $\mathbb Z/p$-covering $X_\varphi\rightarrow Y$. It can be described in the following way: There is an $\alpha\in\mathbb F_q^\ast$ such that $\varphi(\lambda)=\mathrm{Tr}(\alpha\lambda)$ and then $X_\alpha=\mathrm{Spec}\mathcal O_Y[t]/(t^p-t-\alpha^{-1}f)$. This is particularly relevant when we want to consider the cohomology of $X$. We get an action of $\mathbb F_q$ on $H^*(X,K)$, where $K$ is a finite extension of $\mathbb Q_\ell$ containing $p$'th roots of unity. Hence, the action on any isotypical component factors through some $\varphi$ and hence the kernel of $\varphi$ acts trivially on it. We can then use the fact that the invariants on cohomology is the cohomology of the quotient to conclude that $H^\ast(X,K)$ is the direct sum of the cohomology of $Y$ and the parts of the $H^\ast(X_\alpha,K)$ where $\mathbb Z/p$ acts non-trivially.

Assume now that we have a (smooth) affine algebraic group $G$ defined over $\mathbb F_q$ and a central subgroup $\mathbb G_a\subseteq G$ also defined over $\mathbb F_q$. We then put $H:=G/\mathbb G_a$ which is also affine which means that $G\rightarrow H$ has a section, in any case we choose a particular such section such that $G$ is the product $H\times\mathbb G_a$. The multiplication of $G$ is then given by a cocycle $\sigma\colon H\times H\rightarrow\mathbb G_a$. Using it we can describe the Lang torsor $Fg\cdot g^{-1}\colon G\rightarrow G$, where $F$ is the Frobenius map (with respect to $\mathbb F_q$), in terms of $\sigma$: Given $g=(h,z)$ an element of $H\times\mathbb G_a$ we have $Fg=(Fh,Fz)$ and $g^{-1}=(h^{-1},-z-\sigma(h,h^{-1}))$ and hence $Fg\cdot g^{-1}=(Fh\cdot h^{-1},Fz-z+\sigma(Fh,h^{-1})-\sigma(h,h^{-1}))$. In particular this means that the inverse image of $H\times{0}$ under the Lang torsor is defined by ${(h,z)|Fz-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})}$. As $Fz=z^q$ we get exactly an Artin-Schreier type covering of $H$ given by $z^q-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$ and hence if we want the cohomology of this covering we need only consider the cohomology of $H$ itself and ordinary Artin-Schreier coverings $t^p-t=\alpha^{-1}(\sigma(h,h^{-1})-\sigma(Fh,h^{-1}))$ for $\alpha\in\mathbb F_q^\ast$.

This is about as far as I've come with only a few comments on the group at hand. We have in that case that $m=ns$ and $H=\mathbb G_a^{n-1}$ with $\sigma(a,b)=\sum_ia_ib_{n-i}^{r^i}$, where $r=p^s$ (I have changed notation a little so that my $r$ is the OP's $q$ and my $q$ is the OP's $q^n$). Note that somewhat curiously $\sigma$ is biadditive with respect to the addition on $\mathbb G_a^{n-1}$. Perhaps more relevantly $\sigma$ is quasi-homogeneous; we can extend $G$ to elements $x_0^{-1}+x_1\tau+\cdots+x_n\tau^n$, with $x_0$ invertible and then $G$ is a normal subgroup so that we get a $\mathbb G_m$-action on $G$. Under this action $x_i$ is homogeneous of degree $r^i-1$ and with these weights $\sigma$ is quasi-homogeneous of degree $q-1$. Unfortunately $\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$ is not homogeneous as even though $\sigma(h,h^{-1})$ is homogeneous (of degree $q-1$), $\sigma(Fh,h^{-1})$ is not homogeneous. However, $\mu_{q-1}\subseteq\mathbb G_m$ acts trivially on both $\sigma(h,h^{-1})$ and $\sigma(Fh,h^{-1})$ so we get some symmetry.

Addendum: I forgot to mention that the cohomology of an Artin-Schreier extension $t^p-t=f$ is presumably best studied by computing $Rf_!$ on $\mathbb A^1$ and then see the cohomology of the Artin-Schreier extension as the fibre of its Fourier transform over the point $1$.

2 Typos

This is very far from a solution but just some ideas that could be possibly be used as a start.

First a simple observation on an Artin-Schreier type extension $X\rightarrow Y$, where $X:=\mathrm{Spec}\mathcal O_Y[t]/(t^q-t-f)$. Here $q=p^m$, $p$ a prime with $p\mathcal O_Y=0$ and $f\in\mathcal O_Y(Y)$. This a finite étale covering with Galois group $\mathbb F_q$. For any surjective group homomorphism $\varphi\colon\mathbb F_q\rightarrow\mathbb Z/p$ we get an induced $\mathbb Z/p$-covering $X_\varphi\rightarrow Y$. It can be described in the following way: There is an $\alpha\in\mathbb F_q^\ast$ such that $\varphi(\lambda)=\mathrm{Tr}(\alpha\lambda)$ and then $X_\alpha=\mathrm{Spec}\mathcal O_Y[t]/(t^p-t-\alpha^{-1}f)$. This is particularly relevant when we want to consider the cohomology of $X$. We get an action of $\mathbb F_q$ on $H^*(X,K)$, where $K$ is a finite extension of $\mathbb Q_\ell$ containing $p$'th roots of unity. Hence, the action on any isotypical component factors through some $\varphi$ and hence the kernel of $\varphi$ acts trivially on it. We can then use the fact that the invariants on cohomology is the cohomology of the quotient to conclude that $H^(X,K)$ H^\ast(X,K)$is the direct sum of the cohomology of$Y$and the parts of the$H^(X_\alpha,K)$H^\ast(X_\alpha,K)$ where $\Z/p$ \mathbb Z/p$acts non-trivially. Assume now that we have a (smooth) affine algebraic group$G$defined over$\mathbb F_q$and a central subgroup$\mathbb G_a\subseteq G$also defined over$\mathbb F_q$. We then put$H:=G/\mathbb G_a$which is also affine which means that$G\rightarrow H$has a section, in any case we choose a particular such section such that$G$is the product$H\times\mathbb G_a$. The multiplication of$G$is then given by a cocycle$\sigma\colon H\times H\rightarrow\mathbb G_a$. Using it we can describe the Lang torsor$Fg\cdot g^{-1}\colon G\rightarrow G$, where$F$is the Frobenius map (with respect to$\mathbb F_q$), in terms of$\sigma$: Given$g=(h,z)$an element of$H\times\mathbb G_a$we have$Fg=(Fh,Fz)$and$g^{-1}=(h^{-1},-z-\sigma(h,h^{-1}))$and hence$Fg\cdot g^{-1}=(Fh\cdot h^{-1},Fz-z+\sigma(Fh,h^{-1})-\sigma(h,h^{-1}))$. In particular this means that the inverse image of$H\times{0}$under the Lang torsor is defined by${(h,z)|Fz-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})}$. As$Fz=z^q$we get exactly an Artin-Schreier type covering of$H$given by$z^q-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$and hence if we want the cohomology of this covering we need only consider the cohomology of$H$itself and ordinary Artin-Schreier coverings$t^p-t=\alpha^{-1}(\sigma(h,h^{-1})-\sigma(Fh,h^{-1}))$for$\alpha\in\mathbb F_q^\ast$. This is about as far as I've come with only a few comments on the group at hand. We have in that case that$m=ns$and$H=\mathbb G_a^{n-1}$with$\sigma(a,b)=\sum_ia_ib_{n-i}^{r^i}$, where$r=p^s$(I have changed notation a little so that my$r$is the OP's$q$and my$q$is the OP's$q^n$). Note that somewhat curiously$\sigma$is biadditive with respect to the addition on$\mathbb G_a^{n-1}$. Perhaps more relevantly$\sigma$is quasi-homogeneous; we can extend$G$to elements$x_0^{-1}+x_1\tau+\cdots+x_n\tau^n$, with$x_0$invertible and then$G$is a normal subgroup so that we get a$\mathbb G_m$-action on$G$. Under this action$x_i$is homogeneous of degree$r^i-1$and with these weights$\sigma$is quasi-homogeneous of degree$q-1$. Unfortunately$\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$is not homogeneous as even though$\sigma(h,h^{-1})$is homogeneous (of degree$q-1$),$\sigma(Fh,h^{-1})$is not homogeneous. However,$\mu_{q-1}\subseteq\mathbb G_m$acts trivially on both$\sigma(h,h^{-1})$and$\sigma(Fh,h^{-1})$so we get some symmetry. 1 This is very far from a solution but just some ideas that could be possibly be used as a start. First a simple observation on an Artin-Schreier type extension$X\rightarrow Y$, where$X:=\mathrm{Spec}\mathcal O_Y[t]/(t^q-t-f)$. Here$q=p^m$,$p$a prime with$p\mathcal O_Y=0$and$f\in\mathcal O_Y(Y)$. This a finite étale covering with Galois group$\mathbb F_q$. For any surjective group homomorphism$\varphi\colon\mathbb F_q\rightarrow\mathbb Z/p$we get an induced$\mathbb Z/p$-covering$X_\varphi\rightarrow Y$. It can be described in the following way: There is an$\alpha\in\mathbb F_q^\ast$such that$\varphi(\lambda)=\mathrm{Tr}(\alpha\lambda)$and then$X_\alpha=\mathrm{Spec}\mathcal O_Y[t]/(t^p-t-\alpha^{-1}f)$. This is particularly relevant when we want to consider the cohomology of$X$. We get an action of$\mathbb F_q$on$H^*(X,K)$, where$K$is a finite extension of$\mathbb Q_\ell$containing$p$'th roots of unity. Hence, the action on any isotypical component factors through some$\varphi$and hence the kernel of$\varphi$acts trivially on it. We can then use the fact that the invariants on cohomology is the cohomology of the quotient to conclude that$H^(X,K)$is the direct sum of the cohomology of$Y$and the parts of the$H^(X_\alpha,K)$where$\Z/p$acts non-trivially. Assume now that we have a (smooth) affine algebraic group$G$defined over$\mathbb F_q$and a central subgroup$\mathbb G_a\subseteq G$also defined over$\mathbb F_q$. We then put$H:=G/\mathbb G_a$which is also affine which means that$G\rightarrow H$has a section, in any case we choose a particular such section such that$G$is the product$H\times\mathbb G_a$. The multiplication of$G$is then given by a cocycle$\sigma\colon H\times H\rightarrow\mathbb G_a$. Using it we can describe the Lang torsor$Fg\cdot g^{-1}\colon G\rightarrow G$, where$F$is the Frobenius map (with respect to$\mathbb F_q$), in terms of$\sigma$: Given$g=(h,z)$an element of$H\times\mathbb G_a$we have$Fg=(Fh,Fz)$and$g^{-1}=(h^{-1},-z-\sigma(h,h^{-1}))$and hence$Fg\cdot g^{-1}=(Fh\cdot h^{-1},Fz-z+\sigma(Fh,h^{-1})-\sigma(h,h^{-1}))$. In particular this means that the inverse image of$H\times{0}$under the Lang torsor is defined by${(h,z)|Fz-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})}$. As$Fz=z^q$we get exactly an Artin-Schreier type covering of$H$given by$z^q-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$and hence if we want the cohomology of this covering we need only consider the cohomology of$H$itself and ordinary Artin-Schreier coverings$t^p-t=\alpha^{-1}(\sigma(h,h^{-1})-\sigma(Fh,h^{-1}))$for$\alpha\in\mathbb F_q^\ast$. This is about as far as I've come with only a few comments on the group at hand. We have in that case that$m=ns$and$H=\mathbb G_a^{n-1}$with$\sigma(a,b)=\sum_ia_ib_{n-i}^{r^i}$, where$r=p^s$(I have changed notation a little so that my$r$is the OP's$q$and my$q$is the OP's$q^n$). Note that somewhat curiously$\sigma$is biadditive with respect to the addition on$\mathbb G_a^{n-1}$. Perhaps more relevantly$\sigma$is quasi-homogeneous; we can extend$G$to elements$x_0^{-1}+x_1\tau+\cdots+x_n\tau^n$, with$x_0$invertible and then$G$is a normal subgroup so that we get a$\mathbb G_m$-action on$G$. Under this action$x_i$is homogeneous of degree$r^i-1$and with these weights$\sigma$is quasi-homogeneous of degree$q-1$. Unfortunately$\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$is not homogeneous as even though$\sigma(h,h^{-1})$is homogeneous (of degree$q-1$),$\sigma(Fh,h^{-1})$is not homogeneous. However,$\mu_{q-1}\subseteq\mathbb G_m$acts trivially on both$\sigma(h,h^{-1})$and$\sigma(Fh,h^{-1})\$ so we get some symmetry.