Quintic Insolvability

Remark 1. I am aware that “insolvability” isn’t technically an English word. We’re just gonna use it—it has a ring to it.

Previously, we introduced the idea of the solvability of a group. It goes a bit like this: call a group solvable if it contains a sequence $H_0, H_1,\dots, H_n$ of subgroups such that

$1 = H_0 < H_1 < \cdots < H_n = G,$

and for each $i$ , $H_i \triangleleft H_{i+1}$ , $H_i \neq H_{i+1}$ , and $H_{i+1}/H_i$ is Abelian. This latter condition is responsible for classifying how “symmetric” a polynomial is, and therefore, its likelihood (or lack thereof) to have a formula involving radicals. But speaking of radicals, let’s properly formulate that idea.

Definition 1. Call a finite extension $\mathbb F(\alpha_1,\dots,\alpha_n) \supseteq \mathbb F$ radical if for each $i$ , there exists a positive integer $\beta_i$ such that $\alpha_i^{\beta_i} \in \mathbb F$ .

Intuitively, each $\alpha_i$ is the $\beta_i$ -th root of some element in $\mathbb F$ : given $\alpha_i^{\beta_i} = \gamma_i$ , we have $\alpha_i = \gamma_i^{1/\beta_i} \equiv \sqrt[\beta_i]{\gamma_i}$ . Of course, this expression requires some work in various fields to be well-defined, but it at least informs why solving degree- $n$ polynomials requires the use of the square root function $\sqrt[n]{\cdot}$ , at least for $n = 2,3,4$ .

Now let $G$ be any group. We shall first establish some basic ideas of solvable groups before connecting them to radical extensions.

Lemma 1. For any $g, h \in G$ , define $[g,h] := ghg^{-1}h^{-1}$ . Then $gh = hg$ if and only if $[g,h] = 1$ .

Proof. The right-hand side simplifies to $[g,h] = (gh)(hg)^{-1}$ .

Definition 2. Given subgroups $H, K \leq G$ , define

$[H,K] := \{[h,k]: h \in H, k \in K\}$

and the commutator subgroup of $G$ by $G' := [G, G]$ .

Example 1. $G' = \{1\}$ if and only if $G$ is Abelian.

Example 2. $\mathcal S_4' = \mathcal A_4$ .

Lemma 2. $G' \triangleleft G$ and $G/G'$ is Abelian. For any normal subgroup $N \triangleleft G$ , $G/N$ is Abelian if and only if $N$ contains $G'$ .

Proof. The subgroup $G' = [ G, G ]$ is generated by elements of the form $[g, h]$ . Furthermore, the subset $uGu^{-1}$ is generated by elements of the form $u[g,h]u^{-1}$ . A tricky computation yields

$\begin{aligned} u[g,h]u^{-1} &= [ugu^{-1}, uhu^{-1}] \in G'. \end{aligned}$

In particular, $uG'u^{-1} \subseteq G'$ , so that $G' \triangleleft G$ . Furthermore, it is not hard to check that $(gh)G' = (hg)G'$ , so that $G/G'$ is Abelian.

Now fix $N \triangleleft G$ . If $G/N$ is Abelian, then $(gh)N =(hg)N$ implies that $[g,h] \in N$ , so that $G' \leq N$ . On the other hand, if $G' \leq N$ , then by the third isomorphism theorem,

$G/N \cong (G/G')/(N/G')$

must be Abelian.

Lemma 3. Inductively define $G^{(0)} = G$ and $G^{(i+1)} := (G^{(i)})'$ . Then $G$ is solvable if and only if there exists some integer $n$ such that $G^{(n)} = 1$ .

Proof. $(\Rightarrow)$ Suppose there exists some integer $n$ such that $G^{(n)} = 1$ . By Lemma 2,

$1 = G^{(n)} \triangleleft G^{(n-1)} < \cdots < G^{(1)} = G,$

forms a subnormal series, so that $G$ is solvable.

$(\Leftarrow)$ Suppose $G$ is solvable. Then there exists a subnormal series

$1 = H_0 < H_1 < \cdots < H_n = G.$

We show the existence of $m \in \mathbb N$ such that $G^{(m)} = 1$ by induction on $n$ . If $n = 1$ , then set $m = 1$ . Suppose the result holds for general $k$ , and that $n = k+1$ :

$1 = H_0 < H_1 < \cdots < H_k < H_{k+1} = G.$

Since $G/H_k = H_{k+1}/H_k$ is Abelian, by Lemma 2, $G^{(1)} = G' \leq H_k.$ Define $H_i' := H_i \cap G^{(1)}$ . Then using the second isomorphism theorem, we can check that

$1 = H_0' < H_1' < \cdots < H_k' = G^{(1)}$

forms a subnormal series for $G^{(1)}$ , so that there exists an integer $m$ such that $G^{(m+1)} = (G^{(1)})^{(m)} = 1$ .

Lemma 4. Let $H_0,H_1,\dots, H_n$ denote a subnormal series of $G$ . If each $H_{i+1}/H_i$ is solvable, then so is $G$ . In particular, if $H \triangleleft G$ is solvable and $G/H$ is solvable, so is $G$ .

Proof Sketch. Chain the individual subnormal series together.

Theorem 1. Consider the field extensions $\mathbb Q \subseteq \mathbb L \subseteq \mathbb K \subseteq \bar{\mathbb Q}$ . If $\mathbb K \supseteq \mathbb Q$ is Galois, $\mathbb L \supseteq \mathbb Q$ is normal, and $\mathbb K \supseteq \mathbb Q$ is radical, then $\mathrm{Gal}(\mathbb L/\mathbb Q)$ is solvable.

Proof. Under the first two hypotheses, $\mathbb L \supseteq \mathbb Q$ is Galois, so that

$\displaystyle \mathrm{Gal}(\mathbb L / \mathbb Q) \cong \frac{\mathrm{Gal}(\mathbb K/\mathbb Q)}{\mathrm{Gal}(\mathbb K/\mathbb L)}.$

By Lemma 4, it suffices to show that $\mathrm{Gal}(\mathbb K/\mathbb Q)$ is solvable. Since $\mathbb K \supseteq \mathbb Q$ is radical, there exist $\alpha_i \in \mathbb K$ and $\beta_i \in \mathbb N^+$ such that $\alpha_i^{\beta_i} \in \mathbb Q(\alpha_1,\dots,\alpha_{i-1})$ and

$\mathbb K = \mathbb Q(\alpha_1,\dots,\alpha_n).$

Define $\beta := \mathrm{lcm}(\beta_1,\dots,\beta_n)$ and let $\zeta_\beta$ denote a primitive $\beta$ -th root of unity. Then

$\displaystyle \mathrm{Gal}(\mathbb K / \mathbb Q) \cong \frac{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q)}{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb K)}.$

Likewise, it suffices to check that $\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q)$ is solvable. Finally,

$\displaystyle \frac{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q)}{\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q(\zeta_\beta))} \cong \mathrm{Gal}(\mathbb Q(\zeta_\beta)/ \mathbb Q) \hookrightarrow \mathbb Z_\beta^*$

is cyclic, and thus Abelian, and thus, solvable.

It remains to check that $\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb F(\zeta_\beta))$ is solvable. Define

$\mathbb L_i := \mathbb Q(\alpha_1,\dots, \alpha_i, \zeta_\beta),$

so that $\mathbb L_0 = \mathbb Q(\zeta_\beta)$ and $\mathbb L_n = \mathbb K(\zeta_\beta)$ . Consider the series

$\mathbb Q(\zeta_\beta) = \mathbb L_0 < \mathbb L_1 < \cdots < \mathbb L_n = \mathbb K(\zeta_\beta).$

For each $i$ , $\mathbb L_i$ is a splitting field of the polynomial $x^{\beta_i} - \alpha_i^{\beta_i} \in \mathbb L_{i-1}[x]$ , and thus $\mathbb L_i \supseteq \mathbb L_{i-1}$ is Galois with $\mathrm{Gal}(\mathbb L_n/\mathbb L_i) \triangleleft \mathrm{Gal}(\mathbb L_n/\mathbb L_{i-1})$ . Therefore, the series

$1 = \mathrm{Gal}(\mathbb L_n/\mathbb L_n) < \cdots < \mathrm{Gal}(\mathbb L_n/ \mathbb L_1) < \mathrm{Gal}(\mathbb L_n/\mathbb L_0)$

is subnormal. Since

$\displaystyle \frac{\mathrm{Gal}(\mathbb L_j/\mathbb L_{i-1})}{\mathrm{Gal}(\mathbb L_j/\mathbb L_{i})} \cong \mathrm{Gal}(\mathbb L_i/\mathbb L_{i-1}) \cong \mathrm{Gal}(\mathbb L_{i-1}(\zeta_\beta)/\mathbb L_{i-1})$

is Abelian, $\mathrm{Gal}(\mathbb K(\zeta_\beta) / \mathbb Q(\zeta_\beta)) = \mathrm{Gal}(\mathbb L_n / \mathbb L_0)$ is solvable, as desired.

Now we have properly returned to radicals, and will define solvability by radicals in a relatively intuitive manner (relative to the content of Galois theory, that is).

Definition 3. We say that a polynomial $f \in \mathbb F[x]$ is solvable by radicals over $\mathbb F$ if there exists a radical extension $\mathbb K \supseteq \mathbb F$ such that $\mathbb F(\mathcal Z_f) \subseteq \mathbb K$ .

Lemma 5. If $\mathbb K \supseteq \mathbb Q$ is radical and $\mathbb L \supseteq \mathbb Q$ is the smallest normal extension contained in $\mathbb K$ , then $\mathbb L \supseteq \mathbb Q$ is radical.

Proof. Suppose $\bar{\mathbb F} \supseteq \mathbb K$ , so that

$\mathbb Q \subseteq \mathbb L \subseteq \mathbb K \subseteq \bar{\mathbb Q}.$

Since $\mathbb K \supseteq \mathbb Q$ is radical, write

$\mathbb K = \mathbb Q(\alpha_1,\dots,\alpha_n),\quad \alpha_i^{\beta_i} \in \mathbb Q(\alpha_1,\dots,\alpha_{i-1}).$

Let $\{\sigma_1,\dots,\sigma_m\}$ denote the set of $\mathbb Q$ -homomorphisms from $\mathbb K$ to $\bar{\mathbb Q}$ . We leave it as an exercise to check that

$\mathbb L = \sigma_1(\mathbb K) \cdot \cdots \cdot \sigma_m(\mathbb K),$

and that for each $j$ ,

$\sigma_j(\mathbb K) = \mathbb Q(\sigma_j(\alpha_1),\dots,\sigma_j(\alpha_n)),$

so that $\sigma_i(\mathbb K) \supseteq \mathbb Q$ is radical, so that $\mathbb L \supseteq \mathbb Q$ is radical too.

Finally, we can establish the insolvability of the quintic!

Theorem 2 (Abel-Ruffini Theorem). For any $n \geq 5$ , there exists a polynomial $f \in \mathbb Q[x]$ whose solution does not belong to any radical extension of $\mathbb Q$ .

Proof. Consider the irreducible (and thus, separable) polynomial

$f(x) = x^n - x - 1 \in \mathbb Q[x].$

Suppose for a contradiction that $f$ is solvable. By Definition 3, there exists a radical extension $\mathbb K \supseteq \mathbb Q$ such that

$\mathbb K \supseteq \mathbb Q (\mathcal Z_f) \supseteq \mathbb Q.$

By Lemma 5, $\mathbb Q (\mathcal Z_f) \supseteq \mathbb Q$ is radical. Since the extension $\mathbb Q (\mathcal Z_f) \supseteq \mathbb Q$ is normal and separable, it is Galois. By Theorem 1, $\mathrm{Gal}(\mathbb Q (\mathcal Z_f)/\mathbb Q)$ must be solvable. However, since $n \geq 5$ ,

$\mathrm{Gal}(\mathbb Q (\mathcal Z_f)/\mathbb Q) \cong \mathcal S_n,$

which is not solvable, a contradiction. Therefore, $f$ is not solvable by radicals.

In particular, there is no quintic formula; otherwise, $f$ would be solvable by radicals, a contradiction! Does this result mean that all quintics are not solvable by radicals? That would be too extreme of a claim. But more can be discussed about these queries and more.

But for now, we are done with (very introductory) Galois theory!

—Joel Kindiak, 25 Apr 26, 1822H

KindiakMath

Quintic Insolvability

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Quintic Insolvability

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