Lang’s Theorem

We give the standard proof of Lang’s theorem. This post will use some algebraic geometry.

Definition 1. Fix a field \(k\). Then a group scheme over \(k\) is a group object in the category of \(k\)-schemes.

Thus, we have two ways of thinking about a group scheme.

• We could say that a group scheme as being equipped with morphisms \(m\colon G\times G\to G\) and \(i\colon G\to G\) and \(e\colon\operatorname{Spec}k\to G\) satisfying some compatibilities. For example, multiplication should associate, and multiplying by \(e\) should do nothing.
• We could say that the functor of points \(G\colon\mathrm{Sch}_k\to\mathrm{Set}\) factors through the category \(\mathrm{Grp}\).

The two points give the same definition by the Yoneda lemma.

Theorem 2 (Lang). Let \(G\) be a smooth connected group variety over a finite field \(k\). Then \(\mathrm H^1(k;G)=0\).

Let’s take a moment to recall what this cohomology group means.

Definition 3. Fix a finite group \(\Gamma\) acting on a finite group \(G\). Then we define \(\mathrm H^1(\Gamma;G)\) to be the pointed set of \(1\)-cocycles up to equivalence. Here, a \(1\)-cocycle is a map \(c\colon\operatorname{Gal}(L/k)\to G(L)\) such that \[c(\sigma\tau)=\sigma c(\tau)\cdot c(\sigma)\] for all \(\sigma,\tau\in\Gamma\). Two cocycles \(c\) and \(c’\) are equivalent if and only if there is some \(g\in G(L)\) for which \[\sigma(g)\cdot c(\sigma)=c’(\sigma)\cdot g\] for all \(\sigma\in\operatorname{Gal}(L/k)\). In particular, \(c\) is in the trivial class if and only if there is \(g\in G(L)\) for which \(c(\sigma)=\sigma(g)g^{-1}\).

Definition 4. Fix a group scheme \(G\) over \(k\). For a finite Galois field extension \(L/k\), we define \[\mathrm H^1(L/k;G)=\mathrm H^1(\operatorname{Gal}(L/k);G(L)).\] For any field \(k\), we define \[\mathrm H^1(k;G)=\operatorname{colim}\mathrm H^1(L/k;G),\] where the colimit is taken over finite Galois extensions \(L/k\).

Proof of Theorem 2. By definition of \(\mathrm H^1(k;G)\), it is enough to show that \(\mathrm H^1(L/k;G)\) for each finite Galois extension \(L/k\). Thus, we choose an arbitrary \(1\)-cocycle \(c\colon\operatorname{Gal}(L/k)\to G(L)\), which we would like to show lives in the trivial class. Thus, we need to find some \(g\) for which \(c(\sigma)=\sigma(g)g^{-1}\) for all \(\sigma\).

1. The main claim is that the map \(f\colon G\to G\) defined by \[f(g)=\mathrm{Frob}_q(g)g^{-1}\] is surjective, where \(q\) is the order of \(k\). Let’s explain why this claim completes the proof. The point is that \(\operatorname{Gal}(L/k)\) is cyclic generated by the Frobenius. Thus, we note that the claim provides us with some \(g\) for which \(c(\mathrm{Frob}_q)=\mathrm{Frob}_q(g)g^{-1}\), and then it is enough to check that \(c(\mathrm{Frob}_q^i)=\mathrm{Frob}_q^{i}(g)g^{-1}\) for all \(i\), which we may do by induction: given the statement at \(i\), we calculate \[c\left(\mathrm{Frob}_q^{i+1}\right)=\operatorname{Frob}_qc\left(\mathrm{Frob}_q^{i}\right)\cdot c(\mathrm{Frob}_q)=\mathrm{Frob}_q^{i+1}(g)\cdot g^{-1}\] by the induction. (Note \(\mathrm{Frob}_q\) is a group homomorphism because \(m\) is defined over \(k\).)

2. It remains to show the main claim. We will do this in a roundabout way. Let’s start by making some reductions.

   Consider the action of \(G\) on \(G\) by “twisted conjugation,” writing \(g\cdot h=\mathrm{Frob}_q(g)hg^{-1}\). Note that this is in fact a group action, and it is even algebraic: the only worry about algebraicity is \(\mathrm{Frob}_q\), but it is given locally (on affine space) by taking \(q\)th powers, which is algebraic. The main claim amounts to showing that the orbit of the identity \(e\) under this action is the entire group (for geometric points). In other words, we want to show that there is only a single orbit.

   It is actually enough to show that the orbits of this action are Zariski open. Indeed, once the orbits are Zariski open, we know that they partition \(G\), so considering the complement shows that the orbits are also Zariski closed, so the number of orbits is at most the number of connected components of \(G\). However, \(G\) is connected! Thus, there can only be one orbit.

3. Given \(h\in G\), we would like to show that the orbit of \(h\) is open. By varying \(h\), it is enough to show that the orbit has an open neighborhood around \(h\). This is equivalent to showing that the image of the map \(f_h\colon G\to G\) given by \(f_h(g)=\mathrm{Frob}_q(g)hg^{-1}\) is open in a neighborhood of \(h\). In fact, we will show that \(f_h\) is étale at \(e\), which then implies that \(f_h\) is flat and hence open in a neighborhood of \(e\).

   To show that \(f_h\) is étale, considering the cotangent complex tells us that it is enough to show that \(f_h\) is an isomorphism on Kähler differentials. (Note that \(G\) is already smooth over \(k\).) Equivalently, we need to provide an isomorphism on the Zariski tangent spaces. Computing on differentials is only mildly tricky. Here are the points we need.

   • Left translation \(\ell_a\colon g\mapsto ag\) by some \(a\) is an isomorphism, so the differential map \(d\ell_a\colon T_eG\to T_aG\) is an isomorphism of Zariski tangent spaces.
   • The multiplication map \(m\colon G\times G\to G\) is addition on the tangent spaces because the two composites \(G\to G\times G\to G\) are both the identity.
   • Once we know multiplication, it follows that inversion is negation because it must give the identity when composed with multiplication.
   • Because we are in characteristic dividing \(q\), one can calculate that \(d\mathrm{Frob}_q\) vanishes on affines by the power rule. This then glues to show that it vanishes for any variety over \(k\).

   Now, \(f_h\) is the composite \[G\xrightarrow{(\mathrm{Frob}_q,i)} G\times G\xrightarrow{(\mathrm{id},\ell_h)} G\times G\stackrel m\to G.\] On tangent spaces, we thus see that the composite takes some differential \(x\in T_eG\) to \((0,-x)\) to \((0,-d\ell_hx)\) to \(-d\ell_hx\). As discussed above, this is in fact an isomorphism of tangent spaces, so we are done. \(\blacksquare\)

Here is an amusing application, but we will be brief. In particular, we won’t bother to define anything.

Corollary 5. Let \(f\colon A\to B\) be an isogeny of abelian varieties over a finite field \(k\). Then \(\#A(k)=\#B(k)\).

Proof. Let the kernel of \(f\) be \(F\), so we have an exact sequence \[1\to F\to A\to B\to1\] of group schemes. Then there is a long exact sequence in cohomology, which by Lang’s theorem reduces to \[1\to F(k)\to A(k)\to B(k)\to\mathrm H^1(k;F)\to1.\] The corollary will follow as soon as we know that \(F(k)\) and \(\mathrm H^1(k;F)\) have the same size. Well, one can use the cohomology of cyclic groups to show that \[\mathrm H^1(k;F)\cong\frac{F(\overline k)}{(\mathrm{Frob}_q-1)F(\overline k)},\] so we are done after noting that \(F(k)\) and \(\mathrm H^1(k;F)\) are the kernel and cokernel (respectively) of \((\mathrm{Frob}_q-1)\colon F(\overline k)\to F(\overline k)\). \(\blacksquare\)

Remark 6. It turns out that \(A\) and \(B\) are isogenous over the finite field \(k\) if and only if \(\#A(L)=\#B(L)\) for any finite extension \(L\) of \(k\). This is a rather difficult theorem of Tate.