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Étale cohomology is sheaf cohomology on the site whose objects are étale morphisms $Y\to X$ and whose coverings are jointly surjective families of such morphisms. It provides cohomological invariants for algebraic varieties over fields of positive characteristic. In Deligne-Lusztig theory, compactly supported groups with $\ell$-adic coefficients, for $\ell$ different from the field characteristic, give finite-dimensional representations of the finite groups acting on the varieties.

A finite étale morphism is both a finite morphism and an étale morphism. Affine-locally it is represented by a finite locally free algebra with zero module of Kähler differentials. Its fibres are finite products of finite separable field extensions. For an elliptic curve with good reduction at a prime not dividing $m$, multiplication by $m$ on its smooth proper group model is finite étale of degree $m^2$: its differential is multiplication by the unit $m$. A division torsor of a section consequently gives an unramified local extension, the good-prime input to the Weak Mordell-Weil theorem.

Choose a rational Borel subgroup $B=TU$ with rational maximal algebraic torus $T$, and use the Lang map $\mathcal{L}(g)=g^{-1}F(g)$. Conjugating the pair $({}^{g}B,F({}^{g}B))$ by $g^{-1}$ gives $(B,{}^{\mathcal{L}(g)}B)$. Part (a) therefore givesMoreover $\mathcal{L}(gb)=b^{-1}\mathcal{L}(g)F(b)$, and $F(B)=B$, so this inverse image is stable under right multiplication by $B$. The flag variety identification consequently restricts to the required isomorphism

The second printed isomorphism is false as written: it omits a unipotent quotient. Take $G={\mathrm{SL}}_2({\overline{\mathbb{F}}}_q)$ with standard Frobenius endomorphism of an algebraic group and $w=1$. Then $X(1)$ consists of the $q+1$ rational lines in ${\mathbb{F}}_q^2$, so has dimension zero. Every Borel subgroup $B^{\prime }$ has one-dimensional unipotent radical $U^{\prime }$. The Lang map is a finite surjective étale morphism, so ${\mathcal{L}}^{-1}(U^{\prime })$ has dimension one. Quotienting by the finite group $T^{\prime F}$ preserves that dimension. Thus no choices of $T^{\prime }$ and $B^{\prime }$ can make the asserted two varieties isomorphic in this example.

Here is the corrected construction, including the missing quotient. Choose $x$ by the Lang theorem for algebraic groups so that $x^{-1}F(x)=\dot{w}$. DefineThen $F(T^{\prime })=x\dot{w}T{\dot{w}}^{-1}{x}^{-1}=T^{\prime }$, so $T^{\prime }$ is a rational maximal torus contained in $B^{\prime }$. Put $J=U_x\cap U^{\prime }=F^{-1}(U^{\prime })\cap U^{\prime }$. The correct unipotent quotient in a Deligne-Lusztig variety isThe semidirect product denotes the subgroup $J{T}^{\prime F}$ acting on the right. Both $U_x$ and $U^{\prime }$ are normalized by $T^{\prime }$, so the action is well defined. Alsofor $h\in {\mathcal{L}}^{-1}(U^{\prime })$, showing directly that the map lands in the Deligne-Lusztig variety.

To prove surjectivity and identify its fibres, use the intermediate varietyThe right $U$-action follows from the twisted-translation identity for the Lang map. Right multiplication by $T^{F_w}$ also acts, and $Y(\dot{w})/T^{F_w}\cong X(w)$. Indeed if $\mathcal{L}(g)\in B\dot{w}B$, write it as $u_{1}t\dot{w}{u}_2$. Replacing $g$ by $ga$, with $a\in T$, changes its torus factor to $a^{-1}t{F}_w(a)$. Surjectivity of the twisted Lang map on the connected algebraic torus makes this factor one. Once the factor is one, two representatives of the same flag differ by $ua$ with $u\in U$ and $a\in T^{F_w}$. This proves both surjectivity and the asserted finite covering, including its group of fibres.

Now map $Z={\mathcal{L}}^{-1}(U^{\prime })$ to $Y(\dot{w})$ by $h\mapsto hxU$. Given $\mathcal{L}(g)=u_1\dot{w}{u}_2$, replace $g$ by $g{u}_1$. ThenThus $h=g{u}_1{x}^{-1}$ belongs to $Z$ and maps to the original point $gU$. This proves surjectivity. Two preimages have the form $h$ and $hk$, where $k\in U_x$. Since $F(k)\in U^{\prime }$ and $\mathcal{L}(h)\in U^{\prime }$, the identityshows that $hk\in Z$ precisely when $k\in U^{\prime }$, hence precisely when $k\in J$. Therefore the fibres are exactly the right $J$-orbits.

These are isomorphisms of geometric quotients, not just bijections on points. The groups $U_x,U^{\prime }$ contain root subgroups for the same maximal algebraic torus, so their intersection $J$ is smooth. The map $Z\to Y(\dot{w})$ is a smooth surjection with fibre $J$: the differential of the Lang map is invertible, and the kernel of the differential of $h\mapsto hxU$ on $Z$ is the translate of $\mathrm{Lie}(U_x)\cap \mathrm{Lie}(U^{\prime })=\mathrm{Lie}(J)$. This gives $Z/J\cong Y(\dot{w})$. Finally $x{T}^{F_w}{x}^{-1}=T^{\prime F}$, so quotienting the remaining finite torus action gives the displayed corrected formula.

For $w=1$, $J=U_x$ is exactly the affine-line factor in the counterexample. More generally $Z/T^{\prime F}\to X(w)$ retains unipotent fibres of dimension $\dim J$; it becomes the printed isomorphism only when this intersection is trivial. The missing factor is also recorded in the quotient construction in Tiep's lectures, proof of Theorem 5.22.

We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.