Theorem 59.67.8 (03RC)—The Stacks project

Processing math: 100%

Table of contents
Part 3: Topics in Scheme Theory
Chapter 59: Étale Cohomology
Section 59.67: Galois cohomology
Theorem 59.67.8 (cite)

numberstags

Theorem 59.67.8. Let $K$ be a $C_1$ field. Then $\text{Br}(K) = 0$ .

Proof. Let $D$ be a finite dimensional division algebra over $K$ with center $K$ . We have seen that

$D \otimes _ K K^{sep} \cong \text{Mat}_ d(K^{sep})$

uniquely up to inner isomorphism. Hence the determinant $\det : \text{Mat}_ d(K^{sep}) \to K^{sep}$ is Galois invariant and descends to a homogeneous degree $d$ map

$\det = N_\text {red} : D \longrightarrow K$

called the reduced norm. Since $K$ is $C_1$ , if $d > 1$ , then there exists a nonzero $x \in D$ with $N_\text {red}(x) = 0$ . This clearly implies that $x$ is not invertible, which is a contradiction. Hence $\text{Br}(K) = 0$ . $\square$

Comments (0)

There are also:

3 comment(s) on Section 59.67: Galois cohomology

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$ ). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Comments (0)

Post a comment