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 ${\mathbb F}_q\,$.

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?

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.


2 Answers 2


$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 like $y^2 = 0-1 = -1$ which is split since $q \equiv 1 \bmod 4$. We can write the second extension as $\left(\frac{y}{t}\right)^2= 1- \frac{1}{t^2}$, which at $t=\infty$ 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 $q\equiv 1\bmod 4$.

  • $\begingroup$ Many thanks for the promt and very helpful answer! $\endgroup$ Apr 21, 2023 at 14:35

For a different type of answer: over any field $K$, take two double coverings $C\rightarrow \mathbb{P}^1$, $D\rightarrow \mathbb{P}^1$ with disjoint branch loci; let $\sigma $ and $\tau $ be the corresponding involutions of $C$ and $D$. Now let $E:=C\times _{\mathbb{P}^1}D$. Then $\pi :E\rightarrow \mathbb{P}^1$ is a branched covering with Galois group $\mathbb{Z}/2\times \mathbb{Z}/2$. The involution $(\sigma ,\tau )$ of $E$ is fixed point free; this implies that all ramification points of $\pi $ have multiplicity 2, hence decomposition group $\mathbb{Z}/2$.

  • $\begingroup$ If $K$ has a nontrivial quadratic extension then I don't think this works, as the Galois group of the residue field is included in the decomposition group which can make it larger than $\mathbb Z/2$. $\endgroup$
    – Will Sawin
    Apr 21, 2023 at 23:15
  • $\begingroup$ @Will Sawin: Right. I was implicitely assuming that all branch points of the coverings are rational over $K$. $\endgroup$
    – abx
    Apr 22, 2023 at 6:17

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