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One can simultaneously eliminate three terms from the general quintic using a quartic Tschirnhausen transformation. The 4th parameter of the quartic allows it to be done in radicals and details are in this post. But I found out (back in 2006) that a rational cubic Tschirnhausen transformation can do the same.

As usual, we start with the principal quintic,

$$y^5+\color{blue}{5}ry^2+\color{blue}{5}sy+t = 0\tag{1}$$

(with the "5" added for simplicity later) but with the cubic transformation,

$$y^3+ay^2+by+c =z(y+d)\tag{2}$$

where the extra parameter allows the same Bring-Jerrard "trick". Let $c=3r$, and eliminating $y$ between (1) and (2), we get the new quintic,

$c_0z^5+c_1z^4+c_2z^3+c_3z^2+c_4z+c_5=0\tag{3}$

where,

$$c_1 = -6 d r^2 + 4 d^3 s + 3 r s - d^2 t - (3 d^2 r - 4 d s + t)(b - a d)\tag{4}$$

and so on for the other $c_i$. Solve for $a$, substitute into $c_1=c_2=0$ and, after factoring, one gets the quadratic that is only in the unknown $d$,

$(18 r^2 s^2 - 9 r^3 t - 2 s t^2) + (-27 r^3 s + 16 s^2 t - 3 r t^2) d + (27 r^4 - 32 s^3 + 12 r s t)d^2 = 0\tag{5}$

Solving for $d$, one can then set $c_3=0$ which is only a cubic in the last unknown $b$,

$$e_0 b^3+e_1b^2+e_2b+e_3=0\tag{6}$$

thus solving $c_1=c_2=c_3=0$ in radicals.

Notes:

  1. The discriminant of the quadratic in (5) in fact has the same discriminant (up to a numerical factor) as the principal quintic.
  2. If the process is applied to higher degrees, then the equation in $d$ in (5) also rises in degree. For the principal sextic, it is already a quartic. Thus, the transformation is in radicals only for deg $\leq 6$.

Question:

How do we explain why the transformation exists from basic principles (in the same manner Sen did in the post mentioned above)?

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    $\begingroup$ Well, it's still classified as a Tschirnhausen transformation, $$y=\frac{g(x)}{h(x)}$$ of the equation $f(x)=0$. The Brioschi quintic also employs a rational Tschirnhausen, $$y=\frac{ax+b}{c^{-1}x^2-3}$$ $\endgroup$ – Tito Piezas III Dec 23 '13 at 16:27
  • $\begingroup$ I have derived the process of transformation of principal quintic into a Bring-Jerrard quintic by your transformation from power-sum formulas. Would that be qualified as an answer to this question? $\endgroup$ – Balarka Sen Dec 24 '13 at 16:43
  • $\begingroup$ It would be interesting to look at. $\endgroup$ – Tito Piezas III Dec 24 '13 at 19:47

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