[*Bring's curve*](https://en.wikipedia.org/wiki/Bring%27s_curve) or *Bring's surface* with genus 4 and $5!=120$ automorphisms can be given by the homogeneous equations,


$$x\_1+x\_2+x\_3+x\_4+x\_5 = x\_1^2+x\_2^2+x\_3^2+x\_4^2+x\_5^2 = \\x\_1^3+x\_2^3+x\_3^3+x\_4^3+x\_5^3 = 0$$


This is also a property of the Bring quintic $x^5+ax+b = 0$ since its roots $x\_i$ obey,


$$\sum\_{i=1}^5 x\_i^k = 0,\quad k = 1,2,3$$


But there is also a surface with **six** coordinates such that,


$$y\_1 + y\_2 + y\_3 + y\_4 + y\_5 + y\_6 = y\_1^2 + y\_2^2 + y\_3^2 + y\_4^2 + y\_5^2 + y\_6^2 = \\ y\_1^4 + y\_2^4 + y\_3^4 + y\_4^4 + y\_5^4 + y\_6^4 = y\_1^7 + y\_2^7 + y\_3^7 + y\_4^7 + y\_5^7 + y\_6^7 = 0$$


This seems to be a property of the sextic $y^6+ay^3+by+c = 0$ since its roots $y\_i$ obey,


$$\quad\sum\_{i=1}^6 y\_i^k = 0,\quad k = 1,2,4,\color{red}7$$


**Questions:**


1. The general quintic and sextic can be transformed in radicals into $x^5+ax+b = 0$ and $y^6+ay^3+by+c = 0$, respectively. So does it follow if the first surface has $5!=120$ automorphisms, then the second has $6!=720$? (In [this table](https://en.wikipedia.org/wiki/Hurwitz%27s_automorphisms_theorem#Automorphism_groups_in_low_genus), no genus $< 12$ has more than 720.)
2. If the surface in $x$ has genus 4, what is the genus of the surface in $y$?
3. Any other special property, such as *why* is it valid for the high exponent $k=7$?



Bring's curve $\sum_{i=1}^5 x_i^k = 0$ for $k = 1,2,3$ and an analogue $\sum_{i=1}^6 y_i^k = 0$ for $k = 1,2,4,7$
The equations $\sum\_{i=1}^6 y\_i^k = 0$ for $k=1,2,4,7$ do not cut out a curve because the $k=1,2,4$ equations imply the one for $k=7$. (The 7th power sum is a polynomial in the power sums of degree 1 through 6, and there is no way to get 7 as a sum of numbers in $\{3,5,6\}$.) So you have some surface with 720 automorphisms.


If some family of sextics $y^6 + a y^3 + b y + c = 0$ is a sextic resolvent of $x^5 + Ax + B = 0$, then the corresponding curve -- call it $C$ -- is isomorphic to the Bring curve, and thus also has genus 4 and automorphisms by $S\_5$, acting by permutations of the $y\_i$ by a transitive subgroup of $S\_6$. If you find enough $S\_6$-invariant equations in $y\_1,\ldots,y\_6$ to cut out a 1-dimensional variety containing $C$, then that variety is the union of 6 curves each of which is isomorphic with the Bring curve and has automorphism group one of the six transitive subgroups of $S\_6$ that's isomorphic with $S\_5$.


