6
$\begingroup$

Let $k$ be a field and $\mathrm{Br}(k)$ the Brauer group of $k$. Let $k \subset L$ be a field extension. Let $b \in \mathrm{Br}(k)$ and denote by $b \otimes L \in \mathrm{Br}(L)$ the base-change of $b$ to $L$.

If $b \otimes L = 0$, then does this exist a subextension $k \subset K \subset L$ such that $K/k$ has finite degree and such that $b \otimes K = 0$?

i.e. if $b$ is killed by some field extension $L$, then must $b$ be killed by some finite field extension of $k$ which is contained in $L$?

$\endgroup$

1 Answer 1

12
$\begingroup$

No: the conic $C:X^2+Y^2+1=0$ splits over the field $L=\mathbb{Q}(x)[y]/(x^2+y^2+1)$, since $(X,Y)=(x,y)$ is an $L$-point of $C$. However $L$ has no subfields algebraic over $\mathbb{Q}$ other than $\mathbb{Q}$ itself, since it is the function field of a geometrically irreducible variety.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.