# How to transform the general quintic to the Brioschi quintic form?

The quintic can be transformed, in radicals, to the one-parameter Brioschi form,

$$w^5-10\alpha w^3+45\alpha^2w-\alpha^2 = 0\tag1$$

Letting $w = 1/(x^2+20)$ and it becomes the rather nice,

$$(x^2+20)^2(x-5)+\alpha^{-1}=0\tag2$$

I was wondering if someone else would like to explain the transformation, or know an alternative method?

This would complete the summary in this post.

Since it has been several days and there is no answer, I guess it is ok to do it myself. To reduce the general quintic,

$$x^5+b_4x^4+b_3x^3+b_2x^2+b_1x+b_0=0\tag{1}$$

to the one-parameter Brioschi form,

$$w^5-10qw^3+45q^2w-q^2 = 0\tag{2}$$

which has a near-square discriminant,

$$D = 5^5q^8(-1+1728q)^2$$

is also done in two steps (like for the Bring-Jerrard form) .

Step 1: Transform the general quintic $$(1)$$ to the principal quintic (which is missing the $$x^4,x^3$$ terms) using a quadratic Tschirnhausen transformation $$y=x^2+mx+n$$ and is described in this post. One ends up with the form,

$$y^5+5ay^2+5by+c=0\tag{3}$$

Step 2: Find a second quadratic Tschirnhausen which establishes a relation between the roots of $$(2)$$ and $$(3)$$. A modern treatment can be found in Duke's and Tóth's "The Splitting of Primes in Division Fields of Elliptic Curves" (p.10). The transformation is,

$$y = \frac{pw+\lambda}{q^{-1}w^2-3}$$

where $$\lambda$$ is a quadratic purely in the coefficients of $$(3)$$, namely,

$$(a^4 + a b c - b^3)\lambda^2 - (11a^3\color{red}b - ac^2 + 2b^2c )\lambda+ (64a^2 b^2 - 27a^3c - b c^2) = 0\tag4$$

correcting a typo in the paper which is missing the variable in red. To find $$p,q$$, using any root $$\lambda$$, then,

$$p = \frac{ -8a \lambda^3 - 72b \lambda^2 - 72 c \lambda +j\, a^2}{a \lambda^2 + b \lambda + c}\tag5$$

$$q = \frac{1}{1728-j}\tag6$$

$$j = \frac{(a \lambda^2 - 3b \lambda - 3c)^3}{a^2(a c \lambda - b^2 \lambda - b c)}\tag7$$

provided $$j\neq 0, 1728$$. So it can be done using only quadratics. (Note: For the special case $$a=0$$, then a slightly different transformation can bypass division by zero.)

Example. Given,

$$y^5+5y^2+10y+2=0$$

so $$a,b,c = 1,2,2$$. To transform this to the form,

$$w^5-10qw^3+45q^2w-q^2 = 0$$

we let,

$$y = \frac{pw+\lambda}{q^{-1}w^2-3}$$

where,

$$\lambda = \tfrac{1}{3}\big(-17 + \sqrt{871}\big)$$

$$p = \tfrac{16}{27}\big(5071 - 179\sqrt{871}\big)$$

$$q = \tfrac{1}{43312^2}\big(615193 + 20894 \sqrt{871}\big)$$

using formulas $$(4)-(7)$$.

This completes the series of posts for the two reduced forms of the quintic. Now if someone knows how to reduce the general sextic into the Joubert sextic form...

• As per the footnote: Is there no straightforward way to remove the $x^5$ and $x^3$ terms from the general sextic? According to the footnotes on your Bring-Jerrard form post, one can even reduce it to $x^6+ax^2+bx+c$. – Simply Beautiful Art Jul 7 '18 at 20:56
• @SimplyBeautifulArt: Yes, there is a one-step way to remove the $x^5$ and $x^3$ from the general sextic $F(y)=0$. One uses a quadratic Tschirnhausen $x = y^2+ay+b$ where the two unknowns $a,b$ allow two terms of the sextic to be eliminated. The transformation's coefficients are determined by a cubic. If the $x^5$ and $x^2$ are to be removed, then the coefficients are now determined by a quartic. And so on. – Tito Piezas III Jul 8 '18 at 4:23