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In Higher composition laws IV: The parametrization of quintic rings M. Bhargava gave an explicit parametrization of quintic rings by quadruples of $5\times5$ skew-symmetric matrices. His proof hinges on establishing a previously unknown fundamental resolvent map between the triple product of the resolvent ring $S$ and the dual of the quintic ring $R$. The mere existence of this map is astonishing in and of itself.

My question is, what can we say about sextic rings which are sextic resolvent rings of some quintic ring?

We can see that not all sextic rings can be sextic resolvent rings of quintics, since the discriminant relation

$$\displaystyle \operatorname{Disc}(S) = (16 \cdot \operatorname{Disc}(R))^3$$

holds. In particular, every sextic resolvent ring has discriminant equal to a cube.

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  • $\begingroup$ I had wanted to ask this question, only to be reminded that I asked it 3 years ago with no answers. The original question contained an obvious oversight which has now been corrected. $\endgroup$
    – Stanley Yao Xiao
    Commented Jan 5, 2021 at 12:10
  • $\begingroup$ In case of quintic fields, shouldn't the Galois group of the sextic resolvent be a subgroup of $S_5$ embedded in $S_6$ ? $\endgroup$
    – Henri Cohen
    Commented Jan 5, 2021 at 21:22
  • $\begingroup$ @HenriCohen Yes; the Galois action on the resolvent ring should be induced by the Galois action on the quintic ring it comes from, so the Galois group must indeed be a subgroup of $S_5$. $\endgroup$
    – Stanley Yao Xiao
    Commented Jan 5, 2021 at 21:49

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