Let $F(x,y) = a_6 x^6 + a_5 x^5 y + \cdots + a_0 y^6 \in \mathbb{Z}[x,y]$ be a binary sextic form with non-zero discriminant. By applying a $\operatorname{GL}_2(\mathbb{Z})$-translation if necessary, we can assume that $a_0 \ne 0$. Then $F$ is said to be a Klein form if its coefficients satisfy the quadratic equations

$$\displaystyle 10 a_0 a_4 - 5 a_1 a_3 + 2a_2^2 = 0,$$ $$\displaystyle 25 a_0 a_5 - 5 a_1 a_4 + a_2 a_3 = 0,$$ and $$\displaystyle 50 a_0 a_6 - 2 a_2 a_4 + a_3^2 = 0.$$

My question is, what is the Galois group of the splitting field of a generic binary sextic Klein form?

It is easy to show that the generic Galois group cannot be the full symmetric group $S_6$, by a theorem of Bhargava and Yang (see Theorem 5 in https://arxiv.org/abs/1312.7339), since Klein forms have non-trivial automorphisms in $\operatorname{PGL}_2(\overline{\mathbb{Q}})$. In fact, they have the largest possible $\operatorname{PGL}_2(\overline{\mathbb{Q}})$-automorphism group of any sextic form with non-zero discriminant; so one would expect that their Galois group is very small. In particular, its Galois group should at least be solvable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.