I’d like to take a quick (one-post) break from simplicial methods. This summer, I will be studying étale cohomology and the proofs of the Weil conjectures through the HCRP program. I have currently been going through the basic computations in étale cohomology, and, to help myself understand one point better, would like to mention a very pretty and elementary argument I recently learned from Johan de Jong’s course notes on the subject (which are a chapter in the stacks project).

1. Motivation via étale cohomology

When doing the basic computations of the étale cohomology of curves, one of the important steps is the computation of the sheaf {\mathcal{O}_X^*} (that is, the multiplicative group of units), and in doing this one needs to know the cohomology of the generic point. That is, one needs to compute

\displaystyle H^*(X_{et}, \mathcal{O}_X^*)

where {X = \mathrm{Spec} K} for {K} a field of transcendence degree one over the algebraically closed ground field, and the “et” subscript means étale cohomology. Now, ultimately, whenever you have the étale cohomology of a field, it turns out to be the same as Galois cohomology. In other words, if {X = \mathrm{Spec} K}, then the small étale site of {X} is equivalent to the site of continuous {G = \mathrm{Gal}(K^{sep}/K)}-sets, and consequently the category of abelian sheaves on this site turns out to be equivalent to the category of continuous {G}-modules. Taking the étale cohomology of this sheaf then turns out to be the same as taking the group cohomology of the associated {G}-module. So, if you’re interested in étale cohomology, then you’re interested in Galois cohomology. In particular, you are interested in things like group cohomologies of the form

\displaystyle H^2(\mathrm{Gal}(K^{sep}/K), (K^{sep})^*).

2. Central simple algebras

Now it turns out that this {H^2} is the Brauer group. I might do a post on this someday, but it turns out that the Brauer group has an interpretation as the set (which has a natural group law) of finite-dimensional simple {K}-algebras with center {K}, modulo a suitable equivalence. In particular, the unit of the Brauer group is represented by the matrix ring {M_n(K)} (for any {n}). So let’s accept this fact, along with the other assertions I stated without proof. The upshot is that if {K} is a field of transcendence degree one over an algebraically closed field, or in other words, {K} is the function field of a curve, then we might very well be interested in classifying central simple {K}-algebras. In particular, we have the following result:

Theorem 1 If {K} is of transcendence degree one over an algebraically closed field, then any central simple {K}-algebra is isomorphic to {M_n(K)} for some {n}. In particular, the Brauer group of {K} is trivial.

Now, by the Artin-Wedderburn structure theory for simple algebras, we know that any simple {K}-algebra is of the form {M_n(D)} for {D} a division ring. Here if the algebra is central simple, then {D} must be central over {K}—that is, the center of {D} must be {K} itself. Thus, if we want to prove this fact, it suffices to show that if {K} is as above, the only finite-dimensional division ring over {K}, whose center is {K}, is just {K} itself.

3. The reduced norm

This result on central simple algebras is what we are going to prove, and the key point is that there will exist a multiplicative norm map

\displaystyle N: D \rightarrow K ,

as with the commutative case, except it’s not going to be quite the same.

Proposition 2 Let {A} be a central simple algebra of degree {n^2} over a field {k}. Then there is a multiplicative map

\displaystyle A \rightarrow k

which commutes with base change and will be the determinant if {k} is algebraically closed.

Now, indeed, we are going to define this map using descent. Namely, we know that locally in the fpqc topology, {A} looks like a matrix algebra {M_n(K')} where {K'} is a separable extension of {k}. That is, after we make an fpqc base change (e.g. to the algebraic closure), {A} becomes a matrix algebra. So we have the determinant map on the matrix algebra—since the determinant is an invariant of a matrix ring, being stable under automorphisms (because any automorphism of a matrix ring is given by conjugation, by the Skolem-Noether theorem)—it “glues.” In this way we can define the determinant by fpqc base change and descent. So we get this map

\displaystyle N: A \rightarrow k

called the reduced norm, and, if we choose a basis {e_1, \dots, e_{n^2}} for {A}, then

\displaystyle N(x_1 e_1 + \dots + x_{n^2} e_{n^2}) \in k

is a homogeneous polynomial of degree {n} in the variables {x_1, \dots, x_{n^2} \in k}. This is easy to see because it is true when {k} is replaced by its algebraic closure, so it holds true for {k} as well.

However, if {A} is a division algebra, then it is clear that the multiplicativity of the norm implies that it maps {A^*} to {k^*}. In particular, if {A \neq k}, then we get a homogeneous polynomial of degree {n} in {n^2} variables that has no root in {k}, from the above norm map. Motivated by this, we make:

Definition 3 A field {k} is quasi-algebraically closed or {C_1} if any homogeneous polynomial of degree {d} in {n> d} variables has a nontrivial root.

For instance, an algebraically closed field is (rather uninterestingly) {C_1}. It follows that:

Proposition 4 A {C_1} field has trivial Brauer group.

Indeed, we have seen that if a field has a nontrivial central simple extension of degree {n} (which without loss of generality can be assumed to be a division algebra), then the reduced norm gives an example of a polynomial of degree {n} in {n^2} variables with no nontrivial roots.

4. Function fields are {C_1}

So, we have now showed that being quasi-algebraically closed or {C_1} is sufficient for the vanishing of the Brauer group {H^2}, and we may state:

Theorem 5 (Tsen’s theorem) A field of transcendence degree {1} over an algebraically closed field is {C_1} and, in particular, its Brauer group is trivial.

We are actually going to prove a more general result. Say that a field is {C_r} if any homogeneous polynomial of degree {d} in {n > d^r} variables has a nontrivial root. This is obviously a weaker condition that being {C_1}.

Theorem 6 If {k} is algebraically closed and {K} an extension of transcendence degree {r}, then {K} is {C_r}.

To prove this, we can assume first that {K} is finitely generated, as any polynomial will have coefficients in a finitely generated subextension. This means that {K} is the function field of an irreducible variety over the ground {k}, as there is an equivalence of categories between finitely generated extensions of {k} and integral varieties up to birational equivalence. So there is an (irreducible) variety {X}, which by taking the projective closure of an open subset we may assume projective, and we have {k(X) = K} and {X} is of dimension {r}. We can assume {X } normal by taking the normalization. Given some homogeneous polynomial {F} of degree {d} in {n> d^r} variables, we want to find rational functions {f_1, \dots, f_n \in k(X)} such that

\displaystyle F(f_1, \dots, f_n) = 0.

The idea will be to stratify the space of rational functions by their poles. Let {H} be a very ample divisor on {X}; now sections of {\mathcal{O}(m H)} correspond to rational functions with poles at most {m} along {H}, for each {m \in \mathbb{N}}. Now {F} can be thought of as a map from

\displaystyle \Gamma(X, \mathcal{O}(mH))^n \rightarrow \Gamma(X, \mathcal{O}( d m H))

because {F} has degree {d}. So it is a polynomial map (of degree {d}) between two finite-dimensional {k}-vector spaces. If {k_m = \dim \Gamma(X, \mathcal{O}(m H))} and {l_m = \dim \Gamma(X, \mathcal{O}(m dH))}, we can think of the condition as saying that we have {l_m} homogeneous polynomials of degree {d} on a space of dimension {n k_m}, and we want to find a nontrivial common root.

So in other words, we are trying to see that the intersection of {l_m} hypersurfaces in affine space {\mathbb{A}^{n k_m}_k}, which we are given contains the origin, contains something more than the origin, at least for {m} large. By standard dimension theory, we can do this if

\displaystyle n k_m > l_m.

Indeed, cutting with a hypersurface can drop the dimension of an affine variety by at most one—at least if the intersection is nonempty. So we need to know something about {k_m} and {l_m}. However, we know that these are the Hilbert polynomials of the variety, and consequently for {p \gg 0},

\displaystyle \dim \Gamma(X, \mathcal{O}(p)) = \deg X \frac{p^r}{r!} + \mathrm{smaller \ terms}.

It follows that {\frac{l_m}{k_m} = d^r + o(1)}, and this will be less than {n} for {m \gg 0}, by assumption. It follows that for large {m}, there is always a nontrivial zero of the map { \Gamma(X, \mathcal{O}(mH))^n \rightarrow \Gamma(X, \mathcal{O}( d m H)) , } which shows that the field is {C_r}.

This computation is a key step in the cohomology of curves.