Let $F$ be a global function field, for example $F={\mathbb F}\_q(t)$, the field of rational functions in one variable over a finite field ${\mathbb F}\_q\,$.



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> **Question.** What would be an example of a global function field $F$ and a finite Galois extension $E/F$ with non-cyclic Galois group $G=\{1,a,b,ab\}$ of order 4 such that all decomposition groups for $E/F$ are *cyclic*?
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**Motivation.** Let $E/F$ be as above. Consider the following three-dimensional $F$-tori:
\begin{align\*}
 &T\_1=R^1\_{E/F} {\Bbb G}\_m:=\ker\big[R\_{E/F} {\Bbb G}\_m \to {\Bbb G}\_{m,F}\big],\\
 &T\_2=(R\_{E/F} {\Bbb G}\_m)/{\Bbb G}\_{m,F}\,.
\end{align\*}
Then
$$Ш^1(F,T\_1)\cong{\Bbb Z}/2{\Bbb Z},\quad\ Ш^2(F,T\_2)\cong {\Bbb Z}/2{\Bbb Z}.$$
See Sansuc, J.-J., Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.
J. Reine Angew. Math. 327 (1981), 12–80, Example 5.6 and Remark 1.9.4.


Sansuc considers *number fields* $F={\Bbb Q}$ , $E={\Bbb Q}(\sqrt{13},\sqrt{17})$.
I am looking for a similar example with function fields.



Biquadratic extension of global function fields with cyclic decomposition groups
$E= \mathbb F\_q ( \sqrt{t}, \sqrt{t^2-1} ) $ over $F =\mathbb F\_q(t)$ does the trick if $q$ is congruent to $1$ mod $4$. It suffices to check that at each place where one of the extensions ramifies, the other is split, as this clearly gives a cyclic decomposition group, and unramified places are always cyclic.


The first extension ramifies at $0, \infty$. At $t=0$, the second extension $y^2=t^2-1 $ locally looks respectively like $y^2 = 0-1 = -1$ which is split since $q \equiv 1 \bmod 4$. At $t=\infty$, the second extension and $\left(\frac{y}{t}\right)^2= 1- \frac{1}{t^2}$ locally looks like $\left(\frac{y}{t}\right)^2 = 1- \frac{1}{\infty^2} = 1- 0 =1$ which is split.


The second extension ramifies at $1,-1$ where the first extension locally looks respectively like $y^2=1$ and $y^2=-1$, which again are split since $q\equiv 1\bmod 4$.


