# Can we express the degree 10 and degree 15 Galois resolvents of sextic binary forms in terms of its basic invariants?

Let $V_6$ denote the 7 dimensional $\mathbb{C}$-vector space of binary sextic forms. For $T = \begin{pmatrix} t_1 & t_2 \\ t_3 & t_4 \end{pmatrix} \in \operatorname{GL}_2(\mathbb{C})$, $T$ acts on an element $F$ of $V_6$ by $F_T(x,y) = F(t_1 x + t_2 y, t_3 x + t_4 y)$. This induces an action of $\operatorname{GL}_2(\mathbb{C})$ on $V_6$. This action makes sense if we replace $\mathbb{C}$ with any subring $R$ of $\mathbb{C}$, or $\operatorname{GL}_2$ with any subgroup.

The action of $\operatorname{GL}_2(\mathbb{Z})$ on $V_6$ has four basic algebraically independent invariants of even degree, which is usually denoted by $I_2, I_4, I_6, I_{10}$ of degrees 2, 4, 6, 10 respectively, where $I_{10}$ can be taken to be the discriminant. Here $I_2, I_4$ also have special names. $I_2$ is the so-called apolar invariant which is defined for forms of even degree, and $I_4$ is the so-called Gundlefinger invariant. In particular, $I_4(F) = 0$ if and only if $F$ can be written as the sum of three sixth powers of linear forms over $\mathbb{C}$. There is also a degree 15 $\operatorname{SL}_2(\mathbb{Z})$ invariant, usually denoted by $I_{15}$, whose square is a polynomial in $I_2, I_4, I_6, I_{10}$.

In the degree 4 case, where we let $V_4$ denote the 5-dimensional vector space of binary quartic forms over $\mathbb{C}$, then there are just two basic invariants for the action of $\operatorname{GL}_2(\mathbb{Z})$ on $V_4$, usually denoted $I$ and $J$ of degrees 2 and 3 respectively. Here $I$ is the apolar invariant and $J$ is the Gundlefinger invariant for degree 4 binary forms. In this case, there is an interesting connection between these invariants and the Galois group of an element $F \in V_4(\mathbb{Z})$, where $V_4(\mathbb{Z})$ denotes the subset of $V_4$ consisting of all forms with integer coefficients. In particular, for an irreducible element $F$ of $V_4(\mathbb{Z})$, it is known that the Galois group $\operatorname{Gal}(F)$ of $F$ fails to have an element of order 3, i.e., it is a subgroup of the dihedral group $D_4$, if and only if a cubic resolvent factors over $\mathbb{Q}$. Let $\theta_1, \cdots, \theta_4$ be the roots of $F(x,1)$. Then $\operatorname{Gal}(F)$ fails to have an element of order 3 if and only if it does not act transitively on elements of the shape $\theta_1 \theta_{i_2} + \theta_{i_3} \theta_{i_4}$, where $\{i_2, i_3, i_4\} = \{2,3,4\}$. This happens only when one of these is rational, which we may take to be $\theta_1 \theta_2 + \theta_3 \theta_4$. Now let $p,q$ be any two non-zero rational numbers and put

$\beta_1(p,q) = p (\theta_1 \theta_2 + \theta_3 \theta_4) + q,$ $\beta_2(p,q) = p (\theta_1 \theta_3 + \theta_2 \theta_4) + q,$ $\beta_3(p,q) = p (\theta_1 \theta_4 + \theta_2 \theta_3) + q.$

Then a cubic resolvent for $F$ is simply the polynomial

$$q_{F,p,q}(x) = (x - \beta_1(p,q))(x - \beta_2(p,q))(x - \beta_3(p,q))$$

for some choice of $p,q$. It can be shown that for any binary quartic form $F \in V_4(\mathbb{Z})$, the polynomial $q_F(x)$ given by

$$q_F(x,y) = x^3 - 3I(F)x + J(F)$$

is a cubic resolvent. That is, $\operatorname{Gal}(F)$ is congruent to a subgroup of $D_4$ if and only if $q_F(x)$ has a rational root. In fact, by explicit computation, one shows that

$$\displaystyle q_F(x,y) = q_{F,3a_4, -a_2}(x)$$

where $a_4, a_2$ are respectively the $x^4$ and $x^2 y^2$ coefficients of $F(x,y)$.

For degree 6, there are two interesting resolvents $f_{10}(x)$ and $f_{15}(x)$ respectively of degree 10 and 15 (see the following paper by Thomas Hagedorn: http://www.sciencedirect.com/science/article/pii/S002186930098428X) which controls whether or not $\operatorname{Gal}(f)$ is solvable for $f \in V_6(\mathbb{Z})$. In particular, $S_6$ contains two maximal solvable subgroups, denoted by $G_{48}$ and $G_{72}$ respectively. Hagedorn showed that $\operatorname{Gal}(f) \subset G_{72}$ if and only if $f_{10}(x)$ has a rational root and $\operatorname{Gal}(f) \subset G_{48}$ if $f_{15}(x)$ has a root of multiplicity one (it turns out that this case is more complicated).

Note that the cubic resolvent $q_F(x)$ is `homogeneous': the $x^3, 3I(F)x, J(F)$ each has degree 3 (since $\deg I = 2, \deg J = 3$). Further note that the highest degree of the even invariants of binary sextics is 10 and the odd invariant has degree 15.

My question is: can the Galois resolvents $f_{10}(x)$ and $f_{15}(x)$ for sextic polynomials be expressed in the basic invariants of binary sextic forms? More precisely, can we write some translate $g_{10}(x), g_{15}(x)$ of $f_{10}(x)$ and $f_{15}(x)$ in the form

$\displaystyle g_{10}(x) = x^{10} + \mathcal{I}_1 x^9 + \mathcal{I}_2 x^8 + \cdots + \mathcal{I}_{10},$

$\displaystyle g_{15}(x) = x^{15} + \mathcal{J}_1 x^{14} + \cdots + \mathcal{J}_{15}$

where $\mathcal{I}_k, \mathcal{J}_l$ for $1 \leq k \leq 10$ and $1 \leq l \leq 15$ are invariants of $\operatorname{SL}_2(\mathbb{Z})$ on $V_6$?

This would generalize the construction for the cubic resolvent $q_F(x)$ for quartic forms.