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Let $$\displaystyle f(x) = a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 $$ be an irreducible sextic polynomial with integer coefficients. Write $\theta_1, \cdots, \theta_6$ for the roots of $f$, or rather the image of the roots under some embedding into the splitting field of $f$. In the course of my work, I have encountered the following quantities:

$$\displaystyle 3 \theta_1 \theta_3 \theta_5 - \theta_1^2 \theta_3 - \theta_1 \theta_5^2 - \theta_3^2 \theta_5,$$ $$\displaystyle 3 \theta_1 \theta_3 \theta_5 - \theta_1 \theta_3^2 - \theta_1^2 \theta_5 - \theta_3 \theta_5^2,$$ $$\displaystyle \theta_1^2 + \theta_3^2 + \theta_5^2 - \theta_1 \theta_3 - \theta_1 \theta_5 - \theta_3 \theta_5$$ and the same quantities with the indices $1,3,5$ replaced with $2,4,6$ respectively.

Edit: the last quantity I had before was not symmetric in the $\theta_i$'s, because it was reducible by a factor which depends only one two of the $\theta_i$'s. Once eliminated, one gets the symmetric version above.

I suspect that these quantities are coefficients of some covariant of the sextic. Unfortunately, the computational difficulty seems high and the pay-off likely low, so no one seems to have done this in the literature.

The case I am most interested is when the Galois group of $f(x)$ is equal to $S_3$, realized as the group generated by $(1 \text{ } 3 \text{ } 5)(2 \text{ } 4 \text{ } 6)$ and $(1 \text{ } 2)(3 \text{ } 6)(4 \text{ } 5)$ (the roots $\theta_i$ have been labelled with this in mind).

The motivation is the following. For quartic polynomials $g(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$, with $\theta_1, \cdots, \theta_4$ the roots of $g$, the quantities $$\displaystyle \theta_1 + \theta_2 - \theta_3 - \theta_4,$$ $$\displaystyle \theta_1 \theta_2 - \theta_3 \theta_4,$$ $$\displaystyle \theta_1 \theta_2(\theta_3 + \theta_4) - \theta_3 \theta_4(\theta_1 + \theta_2)$$ for example are simultaneously proportional to the coefficients of a quadratic covariant identified by Cremona in http://homepages.warwick.ac.uk/~masgaj/papers/r34jcm.pdf. Moreover, the covariant is guaranteed to be proportional to a rational polynomial precisely when the Galois group of $g$ fixes the quantity $\theta_1 \theta_2 + \theta_3 \theta_4$.

My question is, is there a natural covariant of a sextic form whose coefficients correspond to the quantities above?

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