Table of Contents

  1. Prerequisites
  2. Solving Polynomials upto degree 4
  3. Lagrange Resolvent
  4. Fields
  5. Root Permutations
  6. Groups
  7. Factorization of the Resolvent Polynomial
  8. Solvability

Part 2: Solving Polynomials upto degree 4

Quadratic Polynomial

Consider the polynomial $f(x) = x^2 + ax + b$. It can be solved by completing the square:

\[f(x) = x^2 + ax + b = \big ( x+\frac{a}{2} \big )^2 - \frac{a^2}{4} + b=0\]

Substituting, $u = x+a/2$, the polynomial becomes

\[u^2 - \big ( \frac{a^2-4b}{4} \big ) = 0\]

This can be solved using a square root.

What this means: Instead of solving for a polynomial whose roots are $r_1, r_2$, it is easier to first solve a polynomial whose roots are $r_1 + a/2 = (r_1 - r_2)/2$ and $r_2 + a/2 = (r_2 - r_1)/2$. The coefficients of such a polynomial can always be found using known values (the coefficients of the original polynomial) and the polynomial is solvable using a square root alone.

Cubic Polynomial: Cardano’s Method

Consider the polynomial

\[f(x) = x^3 + ax^2 + bx + c\]

Substituting, $y = x+a/3$ completes the cube, removing the second degree term. This turns the polynomial into the depressed form:

\[y^3 + py + q=0\]

Substitute $y = u+v$. This turns the polynomial into:

\[u^3 +v^3 + (3uv+p)(u+v) + q=0\]

Let $uv=-p/3$. This turns the polynomial into:

\[\begin{align*} u^3 + v^3 &= -q\\ uv &= -p/3 \end{align*}\]

We know the sum and product of $u^3$ and $v^3$. This can be solved using a quadratic equation. This gives two solutions, one corresponding to $(u^3, v^3)$ and the other corresponding to $(v^3, u^3)$. We can choose one of these without loss of generality.

Once we know $u^3$, we can get the three cube roots of $u^3$: $u, u\omega, u\omega^2$ where $\omega$ is the primitive cube root of unity. From $uv=-p/3$, we can get the corresponding value of $v$ for each case: $(u,v)$, $(u\omega, v\omega^2)$, $(u\omega^2, v\omega)$. This gives $3$ roots of the depressed cubic: $y=u + v$, $u\omega + v\omega^2$, $u\omega^2 + v\omega$.

What this means: Instead of solving for the cubic with roots $r_1$, $r_2$, $r_3$, we can instead solve the depressed cubic with roots $r_1+a/3$, $r_2+a/3$, $r_3+a/3$ with $a = -(r_1+r_2+r_3)$. Further, instead of solving for this depressed cubic, we can instead solve for a quadratic whose roots are $u, v$ which satisfy:

\[\begin{align*} r_1+a/3 &= u + v\\ r_2+a/3 &= u\omega + v\omega^2\\ r_3+a/3 &= u\omega^2 + v\omega \end{align*}\]

which can also be expressed as:

\[3u = r_1 + r_2\omega^2 + r_3\omega\] \[3v = r_1 + r_2\omega + r_3\omega^2\]

Hence, it is always possible to create a quadratic with roots $r_1 + r_2\omega^2 + r_3\omega$ and $r_1 + r_2\omega + r_3\omega^2$ whose coefficients are known values (can be expressed in terms of the coefficients of the original polynomial). Solving this, we can obtain the individual roots from the system of equations for $u, v$ and $r_1+r_2+r_3 = -a$.

Quartic Polynomial: Ferrari’s Method

Consider the polynomial

\[f(x) = x^4 + ax^3 + bx^2 + cx + d\]

Substituting, $y = x+a/4$ removes the third degree term. This turns the polynomial into the depressed form:

\[y^4 + py^2 + qy + r=0\]

We can rewrite this as:

\[y^4 = -py^2 - qy - r\]

Adding $m^2 + 2my^2$ on both sides, we can try making both sides perfect squares:

\[(y^2 + m)^2= (2m-p)y^2 - qy + (m^2- r)\]

The left side is a perfect square. The right side is a quadratic in $y$. For the right side to be a perfect square, we need the discriminant to be zero:

\[q^2 = 4(2m-p)(m^2- r)\]

This is a cubic in $m$ that can be solved. Using any one of these $m$’s, we can write the equation as equality of $2$ squares:

\[(y^2 + m)^2= (\sqrt{2m-p}y - \frac{q}{2\sqrt{2m-p}})^2\]

This breaks down into two possible quadratics in $y$ that can be solved.

\[(y^2 + m)= \pm (\sqrt{2m-p}y - \frac{q}{2\sqrt{2m-p}})\]

The quadratics are:

\[y^2 + \sqrt{2m-p}y + m-\frac{q}{2\sqrt{2m-p}} = 0\] \[y^2 -\sqrt{2m-p}y + m + \frac{q}{2\sqrt{2m-p}} = 0\]

Solving these quadratics, we can solve the quartic.

What this means: Instead of solving for the quartic with roots $r_1$, $r_2$, $r_3$, $r_4$ we can instead solve the depressed quartic with roots $t_1 = r_1+a/4$, $t_2=r_2+a/4$, $t_3=r_3+a/4$, $t_4=r_4+a/4$ with $a = -(r_1+r_2+r_3+r_4)$. Further, instead of solving for this depressed quartic, we can instead solve a cubic in $m$ that satisfies:

\[t_1 + t_2 = -\sqrt{2m-p}\] \[t_3 + t_4 = +\sqrt{2m-p}\]

Multiplying these we get:

\[t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 = p-2m\]

Since $p = \sum t_it_j$, this simplifies to:

\[2m = t_1t_2 + t_3t_4\]

Writing this in form of $r_i\text{s}$:

\[2m = r_1r_2 + r_3r_4 - a^2/8\]

Hence, it is equivalent to say that instead of solving the depressed quartic, we can solve a cubic whose roots are $r_1r_2 + r_3r_4$, $r_1r_3 + r_2r_4$, $r_1r_4 + r_2r_3$.

If we know $r_1r_2 + r_3r_4$, we can find $(r_1r_2, r_3r_4)$ by solving a quadratic, since we know both the sum and product ($r_1r_2r_3r_4 = d$). Also, if we know $r_1r_2 + r_3r_4$, we can also find $(r_1+r_2)(r_3 + r_4) = b - (r_1r_2 + r_3r_4)$. Once again, by the same logic, we can find $(r_1+r_2, r_3 + r_4)$ by solving a quadratic, since we know the sum ($r_1+r_2 + r_3 + r_4 = -a$) and product.

We know $(r_1+r_2, r_3 + r_4)$ and $(r_1r_2, r_3r_4)$. From these we can solve for $(r_1, r_2, r_3, r_4)$ using quadratics. However, to do this, we need to know which sum corresponds to which product. Both of these are unordered pairs of numbers. For $r_1+r_2$, we need to know which of the two numbers $(r_1r_2, r_3r_4)$ is actually $r_1r_2$. We can do this by constructing the two possible values for:

\[\frac{r_1 + r_2}{r_1r_2} + \frac{r_3 + r_4}{r_3r_4} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4}\]

Since the right side is symmetric in $r_1$, $r_2$, $r_3$, $r_4$, this is a known quantity. Only one of the two possible values that we can come up by matching $(r_1+r_2, r_3 + r_4)$ with $(r_1r_2, r_3r_4)$ is the correct value. This can help us identify the correct way to match the sums and products, which can then help us solve for the roots using quadratics.

Resolvent Polynomial

Here is a summary of the above methods:

  1. Quadratic: We first solve a polynomial with roots $t = r_1 - r_2$ under all possible permutations of $r_1$, $r_2$. This polynomial can be expressed in terms of known quantities and is a quadratic that lacks the linear term. It can be solved using only a square root.

  2. Cubic: We first solve a polynomial with roots $t =r_1 + \omega r_2 + \omega^2 r_3$ under all possible permutations of $r_1$, $r_2$, $r_3$. This polynomial can be expressed in terms of known quantities and is a quadratic polynomial that we already know how to solve. Once we know the two possible values of $t$, we can solve for the roots $r_i$ using a system of linear equations.

  3. Quartic: We first solve a polynomial with roots $t =r_1r_2 + r_3r_4$ under all possible permutations of $r_1$, $r_2$, $r_3$, $r_4$. This polynomial can be expressed in terms of known quantities and is a cubic polynomial that we already know how to solve. Once we know the three possible values of $r_1r_2 + r_3r_4$, we can find the values $r_1r_2$ and $r_1 + r_2$ by solving two quadratics. Further, we can solve for the individual roots $r_1$, $r_2$ using a quadratic. We can do this similarly for the other root pairs.

In all these approaches, the first step has been to identify another polynomial that is both easier to solve and can help with finding the roots of the original polynomial. We call this polynomial the “Resolvent Polynomial".

Continue Reading: Part 3: Lagrange Resolvent