2. Deligne–Lusztig curves

Throughout this section, let ${\mathbb F}$ be an algebraic field extension of ${\mathbb F}_q$ . We consider the affine curve,

\begin{align}{\mathbf Y} = \textrm{Spec}\!\left(\frac{{\mathbb F}[x,y]}{xy^q - yx^q = 1}\right),\end{align}

and its projective closure,

\begin{align}{\mathbf Z}= \textrm{Proj} \!\left( \frac{{\mathbb F}[X,Y,Z]}{XY^q - YX^q = Z^{q+1}} \right).\end{align}

We also consider the projective curve,

\begin{align}{\mathbf W}= \textrm{Proj} \!\left( \frac{{\mathbb F}[U,V,W]}{UV^q + VU^q = W^{q+1}} \right).\end{align}

We would first like to show that $\text{Pic}({\mathbf Z})[p] = 0$ .

Proof. The polynomial $P(X,Y,Z) = Z^{q+1} - (XY^q - YX^q) \in {\mathbb F}[X,Y,Z]$ is prime, which follows from Eisenstein’s criterion for $P \in {\mathbb F}[X,Y][Z]$ , at the prime ideal (X). Therefore ${\mathbf Z}$ is integral. Furthermore, ${\mathbf Z}$ is smooth, because the system $\partial_X P = \partial_Y P = \partial_Z P = 0$ has no solutions over ${\mathbf Z}(\overline{{\mathbb F}})$ . For the isomorphism, let $\lambda \in {\mathbb F}_{q^2}$ with $\lambda^{q-1} = -1$ , and let $\mu \in \overline{{\mathbb F}}$ with $\mu^{q+1} = \lambda^q$ . The element $\mu$ lies in ${\mathbb F}_{q^4}$ , as,

\begin{align}\mu^{q^2} = (\lambda^{q})^{q-1}\mu = - \mu,\end{align}

so,

\begin{align}\mu^{q^4} = ({-}\mu)^{q^2} = -({-} \mu) = \mu.\end{align}

Then the claimed isomorphism is given by,

\begin{align}U = X, \quad V = \lambda Y, \quad W = \mu Z.\end{align}

Indeed,

\begin{align*}X (\lambda Y)^q + (\lambda Y) X^q &= \lambda^q \!\left(X Y^q - YX^q\right), \\&= \lambda^q Z^{q+1} = (\mu Z)^{q+1},\end{align*}

and similarly $U(\lambda^{-1}V)^q - (\lambda^{-1} V) U^q = (\mu^{-1} W)^{q+1}$ .

Proof. By Lemma 2·1, ${\mathbf Z}_{\overline{{\mathbb F}}} \cong {\mathbf W}_{\overline{{\mathbb F}}}$ , and thus the group $\text{Pic}\!\left({\mathbf Z}_{\overline{{\mathbb F}}}\right)[p] \cong \text{Pic}\!\left({\mathbf W}_{\overline{{\mathbb F}}}\right)[p] \cong J(\overline{{\mathbb F}})[p]$ , where J is the Jacobian of ${\mathbf W}$ . ${\mathbf W}$ is known as the Hermitian curve, defined by affine equation $w^{q+1} = v^q + v$ , and is maximal over ${\mathbb F}_{q^2}$ [ Reference Stichtenoth26 , lemma 6·4·4], hence $J(\overline{{\mathbb F}})[p] = 0$ by [ Reference Garcia and Tafazolian15 , corollary 2·5]. Then, because pullback induces an exact sequence $0 \rightarrow \text{Pic}({\mathbf Z}) \rightarrow \text{Pic}\!\left({\mathbf Z}_{\overline{{\mathbb F}}}\right)$ [ 25 , Tag 0CC5], and p-torsion is left exact, $\text{Pic}({\mathbf Z})[p] = 0$ .

Our next goal is to establish that $\text{Pic}({\mathbf Y})[p] = 0$ .

Lemma 2·3. ${\mathbf Z}(\overline{{\mathbb F}}) \setminus {\mathbf Y}(\overline{{\mathbb F}})$ consists of the $q+1$ points,

\begin{align}{\mathcal P} \,{:}\,{\raise-1.5pt{=}}\, \{(a\,:\,b\,:\,0) \mid (a\,:\,b) \in {\mathbb P}^1\left({\mathbb F}_q\right)\}.\end{align}

Furthermore, ${\mathcal P} = {\mathbf Z}\!\left({\mathbb F}_q\right)$ .

Proof. If $(a\,:\,b\,:\,c) \in {\mathbf Z}(\overline{{\mathbb F}})$ with $c =0$ , then $b^q a - a^q b = 0$ , so $b^q a = a^q b$ . If $a \neq 0$ , then $\left( {b}/{a}\right)^q = {b}/{a}$ , so ${b}/{a} \in {\mathbb F}_q$ , and $(a\,:\,b) \in {\mathbb P}^1\left({\mathbb F}_q\right)$ . Similarly, if $b \neq 0$ , $(a\,:\,b) \in {\mathbb P}^1\left({\mathbb F}_q\right)$ . Thus ${\mathbf Z}(\overline{{\mathbb F}}) \setminus {\mathbf Y}(\overline{{\mathbb F}})= {\mathcal P}$ . To see ${\mathbf Z}(\overline{{\mathbb F}}) \setminus {\mathbf Y}(\overline{{\mathbb F}}) = {\mathbf Z}\!\left({\mathbb F}_q\right)$ , there are no points $(a\,:\,b\,:\,c) \in {\mathbf Z}\!\left({\mathbb F}_q\right)$ with $c = 1$ , because if so then $1 = ab^q - ba^q = ab - ba = 0$ , as $a,b \in {\mathbb F}_q$ .

Therefore the closed points of ${\mathbf Z} \setminus {\mathbf Y}$ are ${\mathcal P}$ [ Reference Görtz and Wedhorn16 , proposition 5·4], which we enumerate by ${\mathcal P} = \{P_0,\, ... \, ,P_q\}$ . From [ Reference Weibel27 , exercise 5·12 (a)] we have an exact sequence,

\begin{align}{\mathbb Z}^{q+1} \longrightarrow \text{Cl}({\mathbf Z}) \longrightarrow \text{Cl}({\mathbf Y}) \longrightarrow 0,\end{align}

where the first map sends,

\begin{align}(m_0,\, ... \, ,m_q) \longmapsto \sum_{i = 0}^q m_i [P_i],\end{align}

and the second sends, for I a finite set of closed points of ${\mathbf Z}$ ,

\begin{align}\sum_{P \in I} n_P [P] \longmapsto \sum_{P \in I \setminus {\mathcal P}} n_P [P].\end{align}

Let $\Gamma = \langle [P_0] ,\, ... \, ,[P_{q}] \rangle \subset \text{Cl}({\mathbf Z})$ be the image of ${\mathbb Z}^{q+1}$ in $\text{Cl}({\mathbf Z})$ . The resulting exact sequence,

\begin{align}0 \longrightarrow \Gamma \longrightarrow \text{Cl}({\mathbf Z}) \longrightarrow \text{Cl}({\mathbf Y}) \longrightarrow 0,\end{align}

yields the long exact sequence,

\begin{align*} & 0 \longrightarrow \Gamma[p] \longrightarrow \text{Cl}({\mathbf Z})[p] \longrightarrow \text{Cl}({\mathbf Y})[p] \\&\quad \longrightarrow \Gamma / p\Gamma \longrightarrow \text{Cl}({\mathbf Z})/ p \text{Cl}({\mathbf Z}) \longrightarrow\text{Cl}({\mathbf Y})/ p\text{Cl}({\mathbf Y}) \longrightarrow 0,\end{align*}

from the right derived functors of $\textrm{Hom}_{{\mathbb Z}}({\mathbb Z} / p{\mathbb Z}, -)$ . Then from Proposition 2·2 and the above discussion we have the following.

Proposition 2·4. There is an exact sequence

\begin{align}0 \longrightarrow \text{Cl}({\mathbf Y})[p] \longrightarrow \Gamma / p\Gamma \longrightarrow \text{Cl}({\mathbf Z}) / p \text{Cl}({\mathbf Z}),\end{align}

where the map $\Gamma / p\Gamma \rightarrow \text{Cl}({\mathbf Z}) / p \text{Cl}({\mathbf Z})$ is that induced by the inclusion $\Gamma \hookrightarrow \text{Cl}({\mathbf Z})$ .

Remark. We note that if ${\mathbf Z} \setminus {\mathbf Y}$ contained exactly one degree 1 closed point Q, then we could establish that $\text{Pic}({\mathbf Y})[p] = 0$ almost immediately in the following way. In the exact sequence,

\begin{align}{\mathbb Z} \longrightarrow \text{Cl}({\mathbf Z}) \longrightarrow \text{Cl}({\mathbf Y}) \longrightarrow 0,\end{align}

the map ${\mathbb Z} \rightarrow \text{Cl}({\mathbf Z})$ is actually injective and split by the degree homomorphism, hence $\text{Cl}({\mathbf Z}) \cong {\mathbb Z} \times \text{Cl}({\mathbf Y})$ so,

\begin{align}0 = \text{Cl}({\mathbf Z})[p] \cong {\mathbb Z}[p] \times \text{Cl}({\mathbf Y})[p] = \text{Cl}({\mathbf Y})[p].\end{align}

In particular, this can be applied to show that the class groups of affine dehomogenisations of ${\mathbf Z}$ with respect to both X and Y both have no p-torsion.

We want to show that $\text{Cl}({\mathbf Y})[p] = 0$ , and so in light of Proposition 2·4, we want to show that,

\begin{align}\Gamma / p\Gamma \longrightarrow \text{Cl}({\mathbf Z}) / p \text{Cl}({\mathbf Z}),\end{align}

is injective. In order to do so, we now examine the structure of $\Gamma$ . First we compute the principal divisors of some rational functions on ${\mathbf Z}$ .

Lemma 2·6. Let $(a\,:\,b), (c\,:\,d) \in {\mathbb P}^1\left({\mathbb F}_q\right)$ with $(a\,:\,b) \neq (c\,:\,d)$ . Then the rational function,

\begin{align}f \,{:}\,{\raise-1.5pt{=}}\, \frac{bX - aY}{dX - cY},\end{align}

has associated principal divisor,

\begin{align}(f) = (q+1)\left[P_{(a\,:\,b)}\right] - (q+1)\left[P_{(c\,:\,d)}\right].\end{align}

Proof. Consider the morphism $\zeta \,:\, {\mathbf Z} \rightarrow {\mathbb P}^1$ corresponding to the extension of function fields ${\mathbb F}({\mathbb P}^1) \rightarrow {\mathbb F}({\mathbf Z})$ , which sends,

\begin{align}\frac{S}{T} \longmapsto \frac{bX - aY}{dX - cY},\end{align}

where ${\mathbb P}^1 = \textrm{Proj}({\mathbb F}[S,T])$ , and ${\mathbb F}({\mathbb P}^1) = {\mathbb F}(S/T)$ . On $\overline{{\mathbb F}}$ -points, $\zeta \,:\, {\mathbf Z} \rightarrow {\mathbb P}^1$ is given by,

\begin{align}\zeta(x\,:\,y\,:\,z) = (bx - ay \,:\, dx - cy).\end{align}

This extension ${\mathbb F}({\mathbb P}^1) \rightarrow {\mathbb F}({\mathbf Z})$ has degree $q+1$ because it differs by an automorphism of ${\mathbb P}^1$ from the extension ${\mathbb F}({\mathbb P}^1) \rightarrow {\mathbb F}({\mathbf Z})$ , defined by,

\begin{align}\frac{S}{T} \longmapsto \frac{X}{Y},\end{align}

which clearly has degree $q+1$ . Let $Q_0, Q_{\infty}$ be the closed points of ${\mathbb P}^1$ defined by $(0\,:\,1), (1\,:\,0) \in {\mathbb P}^1(\overline{{\mathbb F}})$ respectively. By [ Reference Liu23 , corollary 3·9], we have that,

\begin{align}(f) = \zeta^*(S/T) = \zeta^*([Q_0]) - \zeta^*([Q_{\infty}]),\end{align}

and $\deg(\zeta^*([Q_0])) = \deg(\zeta^*([Q_{\infty}])) = [{\mathbb F}({\mathbb P}^1)\,:\,{\mathbb F}({\mathbf Z})] = q+1$ . But $\zeta^*([Q_0])$ is some integer multiple of $\left[P_{(a\,:\,b)}\right]$ and $\zeta^*([Q_{\infty}])$ some integer multiple of $\left[P_{(c\,:\,d)}\right]$ , hence,

\begin{align}(f) = (q+1)\left[P_{(a\,:\,b)}\right] - (q+1)\left[P_{(c\,:\,d)}\right].\end{align}

Let $\Gamma^0 \subset \Gamma$ be the degree 0 subgroup of $\Gamma$ , and $\text{Cl}^0({\mathbf Z}) \subset \text{Cl}({\mathbf Z})$ the degree 0 subgroup of $\text{Cl}({\mathbf Z})$ .

Lemma 2·7. The function $\phi \,:\, {\mathbb Z} \times ({\mathbb Z} / (q+1) {\mathbb Z})^q \rightarrow \Gamma$ ,

\begin{align}\phi\,:\, \left(n_0,\, ...\,n ,n_q\right)\longmapsto n_0 [P_0] + n_1([P_1] - [P_0]) + ... + n_q([P_q] - [P_0])),\end{align}

is a surjective homomorphism. In particular, $\Gamma^0$ is a quotient of $({\mathbb Z} / (q+1) {\mathbb Z})^q$ .

Proof. For each $P_k \in {\mathcal P}$ , we can write $P_k = P_{(a_k\,:\,b_k)}$ for some $a_k, b_k \in {\mathbb F}_q$ . For each $0 \leq i \neq j \leq q$ , consider the rational function,

\begin{align}f = \frac{b_i X - a_iY}{b_j X - a_j Y}.\end{align}

Taking the divisor of f,

\begin{align}0 = (f) = (q+1)[P_i] - (q+1)[P_j],\end{align}

in $\Gamma$ , by Lemma 2·6. Therefore, $\phi$ is a well-defined homomorphism, which is surjective because $\{[P_0],\, ...\, ,[P_q]\}$ generate $\Gamma$ . Finally, as $\Gamma^0=\langle [P_1] - [P_0] ,\, ...\, ,[P_q] - [P_0] \rangle$ , then $\Gamma^0$ is a quotient of $({\mathbb Z} / (q+1) {\mathbb Z})^q$ .

We are finally in a position to prove the main result of this section.

Proof. We can split the degree homomorphism with $[P_0]$ , as $[P_0]$ has degree 1 and $\langle [P_0] \rangle$ is free [ Reference Weibel27 , exercise 5·12 (b)]. Then,

\begin{align*}\psi \,:\, \text{Cl}({\mathbf Z}) &\longrightarrow \text{Cl}^0({\mathbf Z}) \times {\mathbb Z}, \\Q &\longmapsto (Q - \deg(Q) [P_0], \deg(Q)),\end{align*}

is an isomorphism, which restricts to,

\begin{align}\Gamma \cong \Gamma^0 \times {\mathbb Z}.\end{align}

We then obtain the following commutative diagram,

Here, the vertical maps are induced from the inclusions of $\Gamma$ into $\text{Cl}({\mathbf Z})$ and of $\Gamma^0$ into $\text{Cl}^0({\mathbf Z})$ , the left horizontal maps are induced by $\psi$ , and the right horizontal maps are the standard identifications.

Now, by Lemma 2·7, $\Gamma^0$ is a quotient of $({\mathbb Z} / (q+1){\mathbb Z})^{q}$ , thus $\Gamma^0 / p \Gamma^0 = 0$ . Consequently, $\Gamma / p\Gamma \rightarrow \text{Cl}({\mathbf Z}) / p \text{Cl}({\mathbf Z})$ is an injection. Therefore, $\text{Cl}({\mathbf Y})[p] = 0$ , by the exact sequence of Proposition 2·4.

3. Rigid curves

Let F be a finite extension of ${\mathbb Q}_p$ with uniformiser $\pi$ and residue field ${\mathbb F}_q$ . Let K be a complete field extension of F with residue field ${\mathbb F}$ , such that ${\mathbb F}$ is an algebraic extension of ${\mathbb F}_q$ . Let R be the ring of integers of K and $\varpi \in K$ an element with $0 < |\varpi| < 1$ .

Let A be the affinoid algebra,

\begin{align}A = K \!\left\langle x , \frac{1}{x^q - x} \right\rangle,\end{align}

for which the associated rigid space $\text{Sp}(A)$ has admissible formal model $\text{Spf}(A_0)$ , where,

\begin{align}A_0 = R \!\left\langle x , \frac{1}{x^q - x} \right\rangle.\end{align}

Let $u \,{:}\,{\raise-1.5pt{=}}\, x^q - x \in A_0^\times \subset A^\times$ , and let B be the affinoid algebra,

\begin{align}B \,{:}\,{\raise-1.5pt{=}}\, A[z] /\!\left(z^{q+1} - u\right).\end{align}

Consider the ring extension,

\begin{align}B_0 \,{:}\,{\raise-1.5pt{=}}\, A_0[z] /\!\left(z^{q+1} - u\right).\end{align}

$B_0$ is $\varpi$ -torsion free, and the natural map,

\begin{align}B_0 = A_0[z] /\!\left(z^{q+1} - u\right) \longrightarrow \left. R \!\left\langle x , \frac{1}{x^q - x}, z \right\rangle \middle/\!\left( z^{q+1} - u \right) \right.,\end{align}

is an isomorphism (because $u \in A_0^\times$ is a unit), hence $B_0$ is an admissible R-algebra. The special fibre of $\text{Spf}(B_0)$ is,

\begin{align}\textrm{Spec}(B_0 \otimes_{R} {\mathbb F}) = \textrm{Spec} \!\left( {\mathbb F}\!\left[y, 1 / v, t \right] / \left(t^{q+1} - v\right) \right),\end{align}

where $v = y^q - y$ , and the generic fibre of $\text{Spf}(B_0)$ is $\text{Sp}(B_0 \otimes_R K) = \text{Sp}(B)$ .

Proof. First note that there is an isomorphism of ${\mathbb F}$ -algebras,

\begin{align}{\mathbb F}[r,s] / (rs^q - sr^q - 1) \xrightarrow{\sim} {\mathbb F}\!\left[y, 1/v, t \right] /\!\left(t^{q+1} - v\right),\end{align}

given by $r \mapsto 1/t$ , $s \mapsto y / t$ , with inverse $y \mapsto s/r$ , $t \mapsto 1/r$ . Thus $\textrm{Spec}(B_0 \otimes_R {\mathbb F}) \cong {\mathbf Y}$ , and $\text{Spf}(B_0)$ is a smooth admissible formal model of $\text{Sp}(B)$ . Therefore by [ Reference Heuer20 , lemma 3·6], the natural maps,

\begin{align}\text{Pic}(\text{Sp}(B)) \xleftarrow{\sim} \text{Pic}\!\left(\text{Spf}\!\left(B_0\right)\right) \xrightarrow{\sim} \text{Pic}\!\left(\textrm{Spec}\!\left(B_0 \otimes_R {\mathbb F}\right)\right),\end{align}

are isomorphisms and we’re done.

We can now state our main results. If K contains $\breve{F}$ the completion of the maximal unramified extension of F, then we can consider the rigid analytic space $\Sigma^1$ defined over any such K. For an overview of the construction and properties of $\Sigma^1$ see [ Reference Junger22 , section 2]. If $v \in {\mathcal T}$ is the central vertex of the Bruhat-Tits tree, then the open affinoid subset $\Sigma^1_{v} \,{:}\,{\raise-1.5pt{=}}\, r^{-1}(v) \subset \Sigma^1$ has coordinate ring isomorphic to,

\begin{align}\mathcal{O}\!\left(\Sigma^1_v\right) \cong A[z] /\!\left(z^{q^2 - 1} - \left(\pi u^{q-1}\right)\right),\end{align}

by [ Reference Junger22 , theorem 2·7].

Let $\omega$ be a primitive $(q^2 - 1)$ st root of $\pi$ in $\overline{F}$ . From now on we strengthen our assumption on the complete field extension K of F and assume that,

\begin{align}K \, contains \, \breve{F}(\omega) \, and \, {\mathbb F} \, is \, an \, algebraic \, extension \, of \, {\mathbb F}_q.\end{align}

We note that this forces ${\mathbb F}$ to be an algebraic closure of ${\mathbb F}_q$ , and that this assumption holds for any complete field extension K of $\breve{F}(\omega)$ which is contained in ${\mathbb C}_p$ .

Proof. Because K contains $\omega$ ,

\begin{align}\mathcal{O}\!\left(\Sigma^1_{v}\right) \cong B^{q-1},\end{align}

and therefore,

\begin{align}\text{Pic}\!\left(\Sigma^1_v\right) \cong \text{Pic} \!\left(\text{Sp}\!\left(B^{q-1}\right)\right) = \text{Pic}(\text{Sp}(B))^{q-1} \cong \text{Pic}({\mathbf Y})^{q-1},\end{align}

by Lemma 3·1. But then $\text{Pic}\!\left(\Sigma^1_v\right)[p] \cong \text{Pic}({\mathbf Y})[p]^{q-1}$ , which is zero by Theorem 2·8.

Recall that $\Sigma^1_v = r^{-1}(v)$ is the pre-image of v, the central vertex of the Bruhat–Tits tree. The vertex v is fixed by $\text{GL}_2(\mathcal{O}_F)$ , and because r is equivariant, $\text{GL}_2(\mathcal{O}_F)$ acts on $\Sigma^1_v$ .

Corollary 3·3. The natural map,

\begin{align}\mathcal{O}\!\left(\Sigma^1_v\right)^{\times} / \mathcal{O}\!\left(\Sigma^1_v\right)^{\times p^n} \longrightarrow H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, \mu_{p^n}\right),\end{align}

arising from the Kummer exact sequence is an isomorphism of $\text{GL}_2(\mathcal{O}_F)$ -modules.

Proof. Because K has characteristic 0, we can consider the Kummer exact sequence for rigid analytic spaces [ Reference de Jong and van der Put11 , section 3·2]. Then the result follows from Theorem 3·2 after taking the long exact sequence in étale cohomology, using that $\text{Pic}\!\left(\Sigma^1_v\right) \cong H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, {\mathbb G}_m\right)$ [ Reference de Jong and van der Put11 , proposition 3·2·4].

As a consequence, we may now compute $H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, {\mathbb Z}_p(1)\right)$ as the p-adic completion of $\mathcal{O}\!\left(\Sigma^1_v\right)^\times$ . This is completely explicit, as the group $\mathcal{O}\!\left(\Sigma^1_v\right)^{\times}$ has been computed by Junger [ Reference Junger22 , theorem 5·1].

Theorem 3·4. There is an isomorphism of ${\mathbb Z}_p$ -linear representations of $\text{GL}_2(\mathcal{O}_F)$ ,

\begin{align}H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, {\mathbb Z}_p(1)\right) \cong \varprojlim_{n \geq 1} \mathcal{O}\!\left(\Sigma^1_v\right)^{\times} / \mathcal{O}\!\left(\Sigma^1_v\right)^{\times p^n}.\end{align}

Proof. For all $n \geq 1$ the diagram,

commutes. Then by the definition of $H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, {\mathbb Z}_p(1)\right)$ and Corollary 3·3,

\begin{align}H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, {\mathbb Z}_p(1)\right) = \varprojlim_{n \geq 1} H^1_{\acute{\text{e}}\text{t}}\!\left(\Sigma^1_v, \mu_{p^n}\right) \xleftarrow{\sim} \varprojlim_{n \geq 1} \mathcal{O}\!\left(\Sigma^1_v\right)^{\times} / \mathcal{O}\!\left(\Sigma^1_v\right)^{\times p^n}. \end{align}