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I’ve started studying the Tschirnhaus transformation and ran into a problem when the substitution degree is higher than two.

For example, for the cubic equation

$$ y^3 + p\,y + q = 0 $$

we can introduce the substitution

$$ z = y^2 + a\,y + b, $$

from which

$$ y^3 = z\,y - a\,y^2 - b\,y, \qquad y^2 = z - a\,y - b, $$

and substitute into the original equation to get a linear equation in $y$:

$$ (y^3 + p\,y + q = 0) \;\Longleftrightarrow\; (z\,y - a(z - a\,y - b) - b\,y + p\,y + q = 0) \;\Longleftrightarrow\; y = \frac{a\,z - a b - q}{z + a^2 + p - b}. $$

Plugging this back into $z = y^2 + a\,y + b$ and simplifying yields

$$ z^3 + (2p - 3b)\,z^2 + \bigl(a^2 p + p^2 + 3b^2 + 3a q - 4bp\bigr)\,z + \bigl(2b^2p + a^3q + apq - b^3 - q^2 - bp^2 - a^2bp -3abq\bigr) =0. $$

We can then eliminate the coefficients of $z^2$ and $z$ by imposing

$$ \begin{cases} 2p - 3b = 0,\\ a^2p + p^2 + 3b^2 + 3a q - 4b p = 0, \end{cases} $$

thereby reducing the equation to the form $z^3 - t = 0$.

However, when we use a cubic substitution

$$ z = y^3 + a\,y^2 + b\,y + c, $$

we end up needing to solve a quadratic in $y$, since the lowest power of $y$ we can express in terms of $z$ and lower powers of $y$ is $y^2$. For instance, for a quartic

$$ y^4 + p\,y^2 + q\,y + r = 0 $$

we write

$$ y^4 + p\,y^2 + q\,y + r = y\,(z - a\,y^2 - b\,y - c) + p\,y^2 + q\,y + r $$

$$ = z\,y - a(z - a\,y^2 - b\,y - c) - b\,y^2 - c\,y + p\,y^2 + q\,y + r \;\Longrightarrow\; (a^2 - b + p)\,y^2 + (z + ab - c + q)\,y + (ac - a\,z + r) = 0. $$

Solving this quadratic for $y$ gives

$$ y = \frac{-(z + a b - c + q) \pm \sqrt{(z + a b - c + q)^2 - 4\,(a^2 - b + p)\,(-a\,z + a\,c + r)}} {2\,(a^2 - b + p)}, $$

which substituted back into $z = y^3 + a\,y^2 + b\,y + c$ yields an equation of degree six.

Therefore, I’m not clear how the Tschirnhaus transformation works in the case of, say, reducing

$$ y^5 + u\,y^2 + v\,y + w = 0 $$

to

$$ z^5 + d_4\,z + d_5 = 0 $$

via the substitution

$$ z = y^4 + p\,y^3 + q\,y^2 + r\,y + s $$

as described, for example, in this answer: https://math.stackexchange.com/a/542228/1641603

Any clarification would be greatly appreciated. Thanks!

RobPratt
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Ivan_Rogers
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1 Answers1

2

Applying the transformation to equations

  • With Determinant of matrix

Applying a Tschirnhaus transformation to equations is equivalent to finding the Resultant of the 2 polynomials in $y$, treating $z$ as a parameter. The resulting equation in this case would've been $$\left|\begin{matrix} 1&0&0&0 & 1&0&0&0&0\\ 0&1&0&0 & p&1&0&0&0\\ 0&0&1&0 & q&p&1&0&0\\ u&0&0&1 & r&q&p&1&0\\ v&u&0&0 & s-z&r&q&p&1\\ w&v&u&0 & 0&s-z&r&q&p\\ 0&w&v&u & 0&0&s-z&r&q\\ 0&0&w&v & 0&0&0&s-z&r\\ 0&0&0&w & 0&0&0&0&s-z \end{matrix}\right|$$ Where the first 4(degree of second equation) contains the coefficient of the first equation, and the last 5(degree of first equation) contains the coefficient of the second equation, the number of row is total degree of the 2 equations. Expanding, we get a quintic in $z$. It being zero means the principal quintic shares roots ($y$) with the Tschirnhaus transformation.

  • With system of equation

When expanding $z^2,z^3,z^4,z^5$ we get equations higher than quintic, however using $$y^5=-uy^2-vy-w$$ To reduce high degree term into lower ones to get quartics, and then let $$z^5+Az^4+Bz^3+Cz^2+Dz+E=0$$ Substitute the expression of $y$, equate coefficient, then solve for $A,B,C,D,E$ in terms of $u,v,w,p,q,r,s$.

After that, just follow the process of elimination in the linked post in question.

Thinh Dinh
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  • Thank you for your answer.

    Is it correct that, by substituting the expression for $z$ in terms of $y$ into

    $$ z^5 + A z^4 + B z^3 + C z^2 + D z + E = 0 $$

    and then reducing all powers $\ge5$ via

    $$ y^5 = -u,y^2 - v,y - w, $$

    we arrive at a quartic

    $$ \lambda_1,y^4 + \lambda_2,y^3 + \lambda_3,y^2 + \lambda_4,y + \lambda_5 ;=;0, $$

    and that, in order to express the original coefficients $A,B,C,D,E$ in terms of $u,v,w,p,q,r,s$, one must solve the system

    $$ (\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5) = (0,0,0,0,0)? $$

     – Ivan_Rogers Jun 15 '25 at 08:33
  • @Ivan_Rogers Yes – Thinh Dinh Jun 15 '25 at 08:33
  • I would also like to know whether there exists a more compact method (apart from using the resultant) to obtain this expansion—perhaps one along the lines of what I described in my question (using successive expressions of powers of $y$ in terms of $z$) but applicable for arbitrary degree. – Ivan_Rogers Jun 15 '25 at 08:35
  • @Ivan_Rogers In that case you take the principal quintic (1) and reduce it via the quartic (2) to get a cubic (3), then reduce (2) via (3) to get a quadratic (4), then reduce (3) via (4) to get a linear. You would get $y$ as ratio of linears of $z$. Then just substitute into (1) to get a quintic in $z$. This process can be easily generalized – Thinh Dinh Jun 15 '25 at 08:40
  • Now everything is clear, thank you. – Ivan_Rogers Jun 15 '25 at 08:41
  • @Ivan_Rogers if you don't want substitution, you can continue reducing the quadratic via the linear to completely eliminate $y$ – Thinh Dinh Jun 15 '25 at 08:42
  • Let us continue this discussion in chat. – Ivan_Rogers Jun 15 '25 at 08:43