Quintic Formula

Ideas from Youtube vedio: https://youtu.be/BSHv9Elk1MU?si=jpgd7e8tOTUTmqLX

Also see the visual Youtube vedio: https://youtu.be/9HIy5dJE-zQ?si=Wu3Qy9d0OF5dXm2j

Why there is ‘no’ quintic formula? (A proof without Galois theory)

Fundamental theorem of algebra

Fundamental theorem of algebra:
A degree $n$ polynomial equation has $n$ solutions in the complex plane. (counting repeated roots)

A visual proof:
Considering a degree $n$ polynomial $p(z)=az^n+bz^{n-1}+\cdots+c$. While $z$ is large enough, the polynomial is mainly depend on the largest term $p(z)\sim az^n$. Then when $\arg(z)$ varies from $0$ to $2\pi$, the polynomial will rotate the origin $n$ times in a orbit near circle. See the figure using $f(z)=z^2-z-2$ as an example below.

Then we decrease the amplitude of $z$, and the trajectory of the $f(z)$ will finally approach the constant term in the polynomial $c$.

And through the whole process, the trajectory will cross the origin at $n$ times (here $n=2$), which mean there are $n$ solutions in the complex plane for a degree $n$ polynomial equation.

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python script

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation

# ================== parameters ==================
theta_vals = np.linspace(0, 2 * np.pi, 200)
z_size = 0.1
z_vals = z_size * np.exp(1j * theta_vals)
f_vals = z_vals**2 - z_vals - 2

z_x = z_vals.real
z_y = z_vals.imag
f_x = f_vals.real
f_y = f_vals.imag

# set the limits for the plot
lim = 12
# all_x = np.concatenate([z_x, f_x])
# all_y = np.concatenate([z_y, f_y])
# margin = max(np.abs(all_x).max(), np.abs(all_y).max()) * 0.1
# lim = max(np.abs(all_x).max(), np.abs(all_y).max()) + margin

# ================== create figure ==================
fig, ax = plt.subplots(figsize=(8, 8))
ax.set_title(r"$z = $" + str(z_size) + r"$e^{i\theta}$ and $f(z)=z^2 - z - 2$", fontsize=12)
ax.set_xlim(-lim, lim)
ax.set_ylim(-lim, lim)
ax.set_aspect('equal')
ax.grid(alpha=0.3)
ax.axhline(0, color='k', lw=0.5)
ax.axvline(0, color='k', lw=0.5)

circle_line, = ax.plot(z_x, z_y, 'b-', lw=1, alpha=0.7, label=r"$|z|=$" + str(z_size))

point_z, = ax.plot([], [], 'ro', markersize=6, label=r"$z(\theta)$")

traj_line, = ax.plot(f_x, f_y, 'g-', lw=1.5, alpha=0.8, label=r"$f(z)$")

point_f, = ax.plot([], [], 'mD', markersize=5, label=r"$f(z(\theta))$")

ax.legend()

# ================== init function ==================
def init():
    point_z.set_data([], [])
    point_f.set_data([], [])
    return point_z, point_f

# ================== update function ==================
def update(frame):
    point_z.set_data([z_x[frame]], [z_y[frame]])
    
    point_f.set_data([f_x[frame]], [f_y[frame]])
    
    return point_z, point_f, traj_line

# ================== generate animation ==================
ani = FuncAnimation(fig, update, frames=len(theta_vals),
                    init_func=init, blit=True, interval=20, repeat=True)

ani.save("z=" + str(z_size) + ".gif", writer="pillow", fps=30)

plt.show()

The symmetry of roots

• Quintic formula:

  $$z^5+bz^4+cz^3+dz^2+ez+f=(z-z_1)(z-z_2)(z-z_3)(z-z_4)(z-z_5)$$

  $$(b,c,d,e,f)\rightarrow(z_1,z_2,z_3,z_4,z_5)$$

• For $z^2=w$:
  If we change the $w$ continuously, for example $w=4e^{i\theta}$, and you will find that the solution $(z_1,z_2)$ can be both $(2,-2)$ and $(-2,2)$, when $w=4$.
  
  In other word, the solution $z=z(w)$ is multivalued function, which mean there is no quadratic solution without square root ($\pm\sqrt{w}$ is multivalued).

  The visual symmetry of square root help us to prove the theorem

  Theorem: The quadratic equation cannot be solved with a rational function of $a,b,c$.

  Proof: Imagine $z_1=\dfrac{-b+\cdots}{2a}+\cdots$ could be written without roots.
  Then it is a continuous single-valued function.
  Now consider $(z-2e^{i\theta})(z+2e^{i\theta})=0$. Smoothly changing $\theta=0$ to $\theta=\pi$, we guarantee $z_1$ must change (by continuity). (See the figure before)
  But the equation and hence $a,b,c$ are unaltered, so $z_1$ has not changed (by single-valuedness).

So our idea to prove there is ‘no’ quintic formula can be:

Theorem: The quintic equation cannot be solved using a single root of $b,c,d,e,f$.

Proof: Imagine $z_1=\dfrac{\cdots+\sqrt[n]{\cdots}}{\cdots}+\cdots$ could be written with just one roots.
Then it is a continuous tamely multi-valued function.
Now consider $(z-z_1)(z-z_2)(z-z_3)(z-z_4)(z-z_5)=0$. Smoothly and cleverly swapping these, we guarantee $z_1$ must change (by continuity).
But hopefully $b,c,d,e,f,\sqrt[n]{\cdots}$ are unaltered, so $z_1$ has not changed (by tameness?).

So our goal turn to: Can we swap the solutions $z_i$ without changing $n$th roots of coefficients $b,c,d,e,f$?

Roots aren’t enough for cubic/quartic/…

Consider a $n$th root $\sqrt[n]{r(b,c,d,e,f)}$, and the five solutions set

values	$\alpha$	$\beta$	$\chi$	$\delta$	$\epsilon$
solutions	$z_1$	$z_2$	$z_3$	$z_4$	$z_5$

If we take a continuously change of $r$, which add $+2\pi m$ onto the phase of $r$ and swap $\alpha$ and $\beta$, i.e.,

values	$\alpha$	$\beta$	$\chi$	$\delta$	$\epsilon$
solutions	$z_2$	$z_1$	$z_3$	$z_4$	$z_5$

and then add $+2\pi n$ onto the phase of $r$ which swap $\beta$ and $\chi$, we have

values	$\alpha$	$\beta$	$\chi$	$\delta$	$\epsilon$
solutions	$z_2$	$z_3$	$z_1$	$z_4$	$z_5$

and then undo the first step, i.e., $-2\pi m$ onto the phase of $r$ and swap $\alpha$ and $\beta$ back,

values	$\alpha$	$\beta$	$\chi$	$\delta$	$\epsilon$
solutions	$z_3$	$z_2$	$z_1$	$z_4$	$z_5$

and finally undo the second step, i.e., $-2\pi n$ onto the phase of $r$ and swap \bata and $\chi$ back,

values	$\alpha$	$\beta$	$\chi$	$\delta$	$\epsilon$
solutions	$z_3$	$z_1$	$z_2$	$z_4$	$z_5$

So the $r$ keep unchanged while the five solutions set rearranged.
This process is a commutator $\sigma\tau\sigma^{-1}\tau^{-1}$.

Therefore we have prove that: There is no cubic/quartic/… formula with only one $n$th root.

Nested roots for cubic and quartic

Nested roots, e.g., $\sqrt[3]{\dfrac{2b^3-9abc+27a^2d+\sqrt{(2b^3-9abc+27a^2d)^2-4(b^2-3ac)^3}}{2}}$

One comutator keep the inner root unchanged, but the outer root is not guaranteed unchanged.

We can take commutator of commutators, i.e., (comm.1)(comm.2)(comm.1)$^{-1}$(comm.2)$^{-1}$, and this one will guarantee the outer root unchanged.

• Prove all commutators of commutators permuting three points are trivial.

  $$(\sigma\tau\sigma^{-1}\tau^{-1})(\pi\rho\pi^{-1}\rho^{-1})(\tau\sigma\tau^{-1}\sigma^{-1})(\rho\pi\rho^{-1}\pi^{-1})$$

  single operator $\sigma,\tau,\pi,\rho$	commutator $(\sigma\tau\sigma^{-1}\tau^{-1})$	commutator of commutators
  1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1	1 2 3 2 3 1 3 1 2	1 2 3
  all 3 point permutations	cyclic permutations	nothing

  (all 3 point permutations -> cyclic permutations -> nothing) is called derived series in Galois theory.

• The quartic formula has roots of roots of roots:

  It can be proved that the commutators-of-commutators-of-commutators are trivial.

Onto the quintic

We want to prove that: The solutions to the quintic cannot be expressed using any finite number of nested roots, i.e., there are commutators of commutators of commutators of commutators of commutators … which mix up five points!

Proof:
consider the operation below

The cyclic permutation of 3 points can be the commutator of two other such permutations of 3 points

$$\rho_{123}=\sigma\tau\sigma^{-1}\tau^{-1}$$

$$\sigma=\alpha\beta\alpha^{-1}\beta^{-1},\tau=\gamma\delta\gamma^{-1}\delta^{-1},\cdots$$

and this can generate a infinite series commutator of commutators that mix up the five points.
QED.