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 $L$.



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> 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 $b \otimes K = 0$?
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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 $L$?



Brauer groups and field extensions
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.


