Brauer Group for a Global Field with $l$-roots of unity $l\neq \text{char}(F)$ Let $F$ be global field that contains the $l$-roots of unity with $l$ a prime number different with the characteristic of $F$ and $\text{Br}F$ the Brauer Group of $F$. 
How can i proof that all element $\alpha$ in the $l$-torsion subgruoup $\text{Br}_lF$ of the Brauer group have a cyclic splitting field of degree $l$?
For the paper that i'm reading that is a very known fact, but i can't find bibliography...
The known exact sequence
$$
0 \rightarrow \text{Br}F\rightarrow \bigoplus_v \text{Br}F_v\rightarrow \mathbb{Q/Z} \rightarrow 0
$$
can help?
Thanks.
 A: 1). This is not so straightforward. As stated, your hypotheses hint at Kummer theory, and at a cohomological description of Brauer groups which you can find e.g. in Gille & Szamuely book "Central Simple Algebras and Galois Cohomology", chap. 4 (Cambridge Studies, 2017). Let us first recall the following general results : for any field $F$, denote by $F_{sep}$ a separable closure, and $G_F=Gal(F_{sep}/F)$. Then $Br(F)$ is canonically isomorphic to the Galois cohomology group $H^2(G_F, {F_{sep}}^*)$ . It is pointed out in G-S's book that this  is much more tractable than the usual description $Br(F)\cong H^1(G_F,PGL_{\infty})$. For instance, given a Galois extension of degree $n$ and group $G$, Hilbert's thm. 90 (= nullity of $H^1(G, E^*))$ gives rise to an exact inflation-restriction sequence $0\to H^2(G, E^*) \to Br(F) \to Br(E)$, which means that $H^2(G, E^*)$ is canonically isomorphic to the relative Brauer group $Br(E/F)$ := the subgroup of classes of $Br(F)$ which split in $E$. It follows also easily Hilbert's thm. 90 that $Br(E)[n]\cong H^2(G_E, \mu_n)$, where $\mu_n$ is the group of $n$-th roots of unity.  
2). It is classically known from cohomology theory that the order $n$ of $G$ kills $H^2(G, E^*)$, i.e. $Br(E/F)\subset Br(E)[n]$ (@). We would like to have a stronger hold on property (@), at least when $G$ is cyclic, in which case $H^2(G,M)\cong M/N(M)$ for any $G$-module $M$, where $N$ denotes the norm map of $E/F$. Under your hypotheses, since $G$ acts trivially on $\mu_n$, Kummer theory yields an isomorphism $\chi: G\cong Hom(G_F, \mu_n)=H^1(G_F,\mu_n)$, and property (@) reads: $\mu_n \subset F^*/N(E^*)$. Using basic properties of the cohomological cup-product, it's not hard to check that the isomorphism $F^*/N(E^*)\cong H^2(G,E^*)$ sends the class $a$ mod $N(E^*)$ to the class of the cyclic algebra $(\chi, a)$ (which is killed by the order $n$ of $G$). In particular, the class of $(\chi, a)$ splits in $E$ iff $b$ is a norm in the extension $E/F$. Note that Kummer theory allows to write $E=F(\sqrt [n] b)$, and the definition of $\chi$ then shows that  $(\chi, a)$ splits iff $a$ is a norm from $F(\sqrt [n] b)$ (this statement is symmetrical w.r.t. $a$ and $b$).
3). Under the additional assumption that $n$ is a prime $p$, we have that  $Br(E/F)= Br(E)[n]$ is a cyclic group of order $p$, and a non trivial class $(\chi, a)$ splits iff $a$ is a norm from $F(\sqrt [n] b)$. Suppose moreover that $F$ is a global field. Then CFT can be applied, and the so-called Hasse principle for cyclic extensions asserts that this global normic property holds iff all the local corresponding properties hold, i.e. the local Hilbert symbols $(a,b)_v$ are trivial for all the valuations $v$ of $F$ (actually this needs to be checked only for a finite number of $v$'s). If $p\neq 2$, there is no problem, one can take $a=b$ because the Hilbert symbol is anti-symmetric. If $p=2$, extra care is needed because the archimedean local Brauer groups are isomorphic to $\mathbf Z/2$. But they can be avoided since there is only a finite number of them. I skip this  ./.
