*20 January 2023*

**C. Hinsley**

Every new student of mathematics hears that there is no general method of solving for the roots of quintic or higher-degree polynomials with rational coefficients and is inevitably led to the methods of Galois theory, which (among other things) allows one to determine the solvability of a particular polynomial. However, the jargon and machinery involved in the usual presentation of Galois theory is a barrier to newcomers to algebra. This is unnecessary; a drastically simplified presentation can be given for the desirable case of polynomials over $\mathbb{Q}$, after which the student can develop the full theory with context and a clear sense of direction. This is an attempt to deliver that simplified presentation.

Let $f$ be a polynomial of degree $n$ in one variable with rational coefficients; in other words, let $f \in \mathbb{Q}[x]$. Let $\overline{\mathbb{Q}}$ denote the algebraic closure of $\mathbb{Q}$. Since $\mathbb{C}$ is known to be algebraically closed and is familiar to most people, we can choose to embed $\overline{\mathbb{Q}}$ therein for concreteness:

$$ \overline{\mathbb{Q}} \cong \{z \in \mathbb{C} : \exists g \in \mathbb{Q}[x] : g(z) = 0\} \subsetneq \mathbb{C}. $$

The *splitting field* of $f$, $E$, is the smallest subset of $\overline{\mathbb{Q}}$ containing all of the roots of $f$. If we enumerate the roots of $f$ as $r_1, \dots, r_n \in \overline{\mathbb{Q}}$ so that $f(r_i) = 0, i = 1, \dots, n$, then $E$ may be described as

$$ E = \mathbb{Q}(r_1, \dots, r_n), $$

the smallest extension field of $\mathbb{Q}$ gained by adjoining elements $r_1$ through $r_n$ and taking the completion with respect to field operations. To see how the elements of $E$ look, we can inspect their general form. Take the polynomial ring $\mathbb{Q}[x_1, \dots, x_n]$; let $g$ be any polynomial in this ring and evaluate it at the roots of $f$ to get $g(r_1, \dots, r_n) \in E$. This yields a subring $R$ of $E$:

$$ R = \{g(r_1, \dots, r_n) : g \in \mathbb{Q}[x_1, \dots, x_n]\} \subseteq E. $$

- Because $\mathbb{Q}$ is a field, it turns out that $E = R$ precisely. But in general this does not hold (expand for details).

Now we have the splitting field $E$ of the polynomial $f \in \mathbb{Q}[x]$. The field extension $\mathbb{Q} \hookrightarrow E$ is usually denoted by $E/\mathbb{Q}$, though this is not really any kind of a quotient, properly speaking. Let $\sigma \in \text{Aut}(E)$ be an automorphism restricting to the identity on $\mathbb{Q}$: $\sigma|*\mathbb{Q} = id*\mathbb{Q}$. $\sigma$ is called a $\mathbb{Q}$-isomorphism in $E$. Define the group

$$ \text{Gal}(E/\mathbb{Q}) \overset{\text{def}}{=} \{\sigma : \sigma \text{ is a } \mathbb{Q} \text{-isomorphism in } E\}. $$

This group is referred to as the Galois group of the field extension $E/\mathbb{Q}$, or sometimes equivalently as the Galois group of the polynomial $f$, denoted $\text{Gal}(f)$ in the latter case.

**Remark.** The profinite groups, which include the finite groups, are precisely the Galois groups (of Galois extensions [3] of fields) [1][2].

We have obtained the Galois group $\text{Gal}(f)$ and hope to use it to determine whether $f$ is solvable by radicals; i.e., whether the roots $r_1, \dots, r_n$ of $f$ can be expressed as elements of the radical field extension $\mathbb{Q}(\alpha_1, \dots, \alpha_k)$ where each adjoined $\alpha_i$ satisfies $\alpha_i^m = q$ for some $m \in \mathbb{Z}_+, q \in \mathbb{Q}$. Galois proved that $f$ is solvable by radicals if and only if $\text{Gal}(f)$ is a *solvable* group: