# Biquadratic extension of global function fields with cyclic decomposition groups

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.

$$E= \mathbb F_q ( \sqrt{t}, \sqrt{t^2-1} )$$ 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 $$q\equiv 1\bmod 4$$.

• Many thanks for the promt and very helpful answer! 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 $$\mathbb{Z}/2$$.

• 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 $\mathbb Z/2$. Apr 21, 2023 at 23:15
• @Will Sawin: Right. I was implicitely assuming that all branch points of the coverings are rational over $K$.
– abx
Apr 22, 2023 at 6:17