I am studying the resolution of the quintic equations, which involves the so-called Tschirnhausen transform.

The idea is to cancel the fourth and third degree coefficients by a change of variable of the form

$$y=x^2+\alpha x+\beta$$

which, by a suitable choice of the coefficients, will turn the quintic equation

$$x^5+px^4+qx^3+rx^2+sx+t=0$$ into a "principal reduced form"


The method to obtain the coefficients of the reduced equation is based on the sums of powers of the roots of the two polynomials. It is also said in some references that we eliminate $x$ from the two first equations.

My question: given the degree of the transformation, I would expect the reduced form to map to a polynomial in $x$ of the tenth degree. What am I missing ?


In my own words, the explanation is this: the transformation holds at the five roots only, not at all $x$ and it is clearer to write

$$y_k=x_k^2+\alpha x_k+\beta,k=1,\cdots5.$$ Then obviously the polynomial with transformed roots is also quintic.

To obtain this, one can expand the polynomial $\displaystyle\prod_{k=1}^5(y-x_k^2-\alpha x_k-\beta)$ and eliminate the $x_k$. The coefficients of the expansion are obtained by Vieta, and can be reworked to be expressed in terms of the original coefficients, again by Vieta. The sums of the roots up to the tenth powers (Newton's sums) will be needed.

  • $\begingroup$ I had always wondered the same, though I had forgotten to ask about this after all this time... $\endgroup$ – Simply Beautiful Art Aug 24 '17 at 23:05
  • $\begingroup$ @SimplyBeautifulArt: In this post, I gave a simple explanation how to reduce the general quintic to the form $x^5+x+\beta=0$ using Tschirnhausen transformations. $\endgroup$ – Tito Piezas III Jan 8 '18 at 14:00
  • $\begingroup$ @TitoPiezasIII 'this post', requesting link? $\endgroup$ – Simply Beautiful Art Jan 9 '18 at 2:05
  • $\begingroup$ @SimplyBeautifulArt: Oops, I mean this post. $\endgroup$ – Tito Piezas III Jan 9 '18 at 3:12

This transformation is not just a change of variables to $y = x^2+\alpha x + \beta$.

Instead, it re-factors factors the polynomial as $$ (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)=(y-(x_1^2-\alpha x_1-\beta))(y-(x_2^2-\alpha x_2-\beta))(y-(x_3^2-\alpha x_3-\beta))(y-(x_4^2-\alpha x_4-\beta))(y-(x_5^2-\alpha x_5-\beta)) $$ choosing $\alpha$ and $\beta$ in such a way as to make the desired coefficients disappear.

In the end, of course, you would have to recover each root of the original equation by solving the quadratic defining each $y_k$.

Look here https://en.wikipedia.org/wiki/Bring_radical#Principal_quintic_form for a bit more footing.

  • $\begingroup$ Héhé, this is one of the sources that caused my questioning. $\endgroup$ – Yves Daoust Aug 25 '17 at 6:39
  • $\begingroup$ O.O So that's how they do it... $\endgroup$ – Simply Beautiful Art Aug 27 '17 at 21:25

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