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 4: Fields

From now on, we consider all known quantities to come from a “Field”. A field is closed under the basic algebraic operations like addition, subtraction, multiplication, and division (of non-zero elements). However, root operations like square roots, cube roots, etc., are not supported in a field. We represent our field of known quantities as $K$. $K$ contains the coefficients of the given polynomial and other rational numbers. We sometimes also assume it to contain roots of unity.

Field Extension

Suppose $g(x)$ is an irreducible polynomial with coefficients from $K$. Let $t$ be a root of $g(x)$. Then, we can introduce a new element $t$ to the existing field $K$ using an extension. We represent this as $K(t)$. Since $K \in K(t)$ and $t \in K(t)$, any polynomial in $t$ with coefficients in $K$ must also be in $K(t)$. We can show that this is enough to form a new bigger field.

Consider the set of polynomials in some variable with coefficients from $K$. Two polynomials that have the same remainder on division by $g(x)$ is considered to be equal. Addition, subtraction and multiplication of two such polynomials would result in another polynomial.

For defining division, we need to first define a way to effectively invert any $a \in K(t)$ such that $a(x)b(x) = g(x)h(x) + 1$. We can always find this using Euclid’s method of finding GCD of two polynomials. Since $g(x)$ is irreducible, the GCD of $a, g$ is $1$. Hence all elements of $K(t)$ can be inverted and we can define division as multiplication by the inverse.

Splitting Field

The field extension $K(r_1, r_2, \ldots, r_n)$ which contains all the roots of the polynomial of the polynomial is called the Splitting Field. Solving a polynomial is equivalent to finding the Splitting Field. From Lagrange’s Theorem on Resolvents, we have shown that there is always a $t$ such that $K(t) = K(r_1, r_2, \ldots, r_n)$. The resolvent polynomial $g$ which has coefficients in $K$ and takes $t$ as a root is easy to find. We start with a polynomial whose roots are $t$ and all its values under the root permutations. This polynomial has coefficients in $K$. We factorize this polynomial into irreducible factors. The irreducible factor that contains $t$ as a root is our resolvent polynomial $g$.

So far we have found that:

  1. $t$ is a polynomial in the roots: $t = \psi(r_1, r_2, \ldots, r_n)$ and the roots $r_i$ can be expressed as a polynomial in $t$: $r_i = \phi_i(t)$. $g(x)$ is the polynomial whose roots are $t$ and its variants under all the root permutations of $r_i\text{s}$. If $g$ can be factorized, we take the irreducible factor containing the root $t$ and call it $g$.
  2. Lagrange’s Theorem in resolvents can be stated as: $K(t) = K(r_1, r_2, \ldots, r_n)$. Solving for the roots of the polynomial $f(x)$ is equivalent to solving for the resolvent polynomial $g(x)$.

Continue Reading: Part 5: Root Permutations