For two thousand years, mathematicians sought a formula for the roots of a polynomial in terms of its coefficients using only the four arithmetic operations and the extraction of roots — a formula analogous to the quadratic formula \begin{align*} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*} that resolves every degree-two equation over any field of characteristic not two. Cubic and quartic analogues were found in the sixteenth century by del Ferro, Tartaglia, Cardano, and Ferrari. Then for three centuries the problem for degree five stagnated. Abel proved in 1824 that no such formula exists for the general quintic. Galois, working a few years later, gave the complete answer: a polynomial equation is solvable by radicals if and only if its Galois group is solvable. This equivalence — connecting a statement about arithmetic operations on roots to a purely group-theoretic property of a permutation group — is one of the deepest correspondences in mathematics. The striking feature of Galois's theorem is that it answers a question about computation (can you write down the roots?) using algebra (what symmetries do the roots enjoy?). To understand why this works, we must first understand what "solvable by radicals" actually means as a precise mathematical concept, and then understand how it interacts with the Galois correspondence. [example: The Cubic Formula] To ground the abstraction, consider the depressed cubic $x^3 + px + q = 0$ over $\mathbb{Q}$. Cardano's formula gives the root \begin{align*} x = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}. \end{align*} To reach this expression starting from $\mathbb{Q}$, we adjoin $\Delta = \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}$ — a square root — to obtain an intermediate field $\mathbb{Q}(\Delta)$. We then adjoin the two cube roots of $-q/2 \pm \Delta$ to a further extension. The process is a tower: we build the root by passing through a chain of field extensions, each obtained from the previous by adjoining a single radical. This tower structure is exactly what "solvable by radicals" formalises. To see what can go wrong, consider the polynomial $f(x) = x^5 - 5x + 12$ over $\mathbb{Q}$. Its Galois group is $S_5$, which is not solvable. No tower of radical extensions, however long, contains all the roots of $f$. There is no Cardano-style formula for $f$. [/example] ## Definition The notion of solvability by radicals is defined field-theoretically. It captures the idea that the roots of a polynomial can be reached from the base field by iteratively extracting roots. The first ingredient is a radical extension — a single step in the tower. What does it mean to "extract a root"? Starting from a field $k$, we adjoin an element $\alpha$ satisfying $\alpha^n = a$ for some $a \in k$. This is the simplest possible extension one can build with a root extraction. We isolate this notion. [definition: Radical Extension] Let $K / k$ be a field extension. We say $K$ is a **radical extension** of $k$ if $K = k(\alpha)$ for some $\alpha \in K$ with $\alpha^n \in k$ for some $n \in \mathbb{N}$. [/definition] A radical extension adjoins a single $n$-th root. But to build complex expressions like the cubic formula, we need to stack such extensions on top of one another. [definition: Extension by Radicals] A field extension $K / k$ is an **extension by radicals** (or a **radical tower**) if there exists a tower of intermediate fields \begin{align*} k = k_0 \subset k_1 \subset k_2 \subset \cdots \subset k_r = K \end{align*} such that each step $k_{i+1} / k_i$ is a radical extension: $k_{i+1} = k_i(\alpha_i)$ with $\alpha_i^{n_i} \in k_i$ for some $n_i \in \mathbb{N}$. [/definition] The tower captures the iterated extraction of roots. The field $K$ contains all the intermediate radicals and the final roots, assembled by the tower. [definition: Solvable by Radicals] A polynomial $f \in k[x]$ is **solvable by radicals** over $k$ if there exists an extension by radicals $K / k$ such that $f$ splits completely in $K[x]$ — that is, every root of $f$ lies in $K$. [/definition] A polynomial is solvable by radicals if all its roots can be captured inside some radical tower over the base field. Notice: the definition does not require $K$ itself to be a splitting field; it only requires that the splitting field of $f$ sits inside some radical tower. [remark: The Tower Can Overshoot] The radical tower $K$ in the definition need not equal the splitting field of $f$. It may be larger. For example, the splitting field of $x^3 - 2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2}, \omega)$ where $\omega = e^{2\pi i/3}$. A radical tower containing this field might adjoin cube roots, sixth roots, and additional square roots beyond what the splitting field strictly requires. Solvability only asks that the roots can be found somewhere in a radical tower — it does not demand efficiency. [/remark] [example: The Quadratic Formula as a Radical Tower] For $f(x) = ax^2 + bx + c \in \mathbb{Q}[x]$ with $a \neq 0$, the tower \begin{align*} \mathbb{Q} \subset \mathbb{Q}(\sqrt{b^2 - 4ac}) \end{align*} is a single radical extension ($n = 2$, $\alpha = \sqrt{b^2 - 4ac}$). Both roots \begin{align*} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*} lie in $\mathbb{Q}(\sqrt{b^2 - 4ac})$. So every quadratic over $\mathbb{Q}$ is solvable by radicals with a tower of length one. [/example] ## Solvable Groups The group-theoretic condition that characterises solvability by radicals is captured by the notion of a solvable group. The connection is not obvious from the definitions — it requires the Galois correspondence — but the group-theoretic condition is cleanly stated on its own and is worth developing before we bring in the Galois theory. Solvable groups are those that can be "dismantled" step by step into abelian pieces. The precise requirement is on the successive quotients in a subnormal series. [definition: Subnormal Series] A **subnormal series** of a group $G$ is a chain of subgroups \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq \cdots \trianglelefteq G_r = G \end{align*} in which each $G_i$ is a normal subgroup of $G_{i+1}$. The quotients $G_{i+1} / G_i$ are called the **factors** of the series. [/definition] The normality condition is local: $G_i \trianglelefteq G_{i+1}$ does not require $G_i$ to be normal in all of $G$. This is a weaker requirement than a normal series (where each $G_i$ would be normal in $G$ itself), and the distinction matters: there are groups with subnormal series whose factors are all abelian, but no such normal series exists. [definition: Solvable Group] A group $G$ is **solvable** if it has a subnormal series \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_r = G \end{align*} whose factors $G_{i+1} / G_i$ are all abelian. [/definition] [remark: Equivalent Formulation via the Derived Series] The condition can also be phrased using the derived series. Define $G^{(0)} = G$ and $G^{(i+1)} = [G^{(i)}, G^{(i)}]$, the commutator subgroup. Then $G$ is solvable if and only if $G^{(r)} = \{e\}$ for some $r \in \mathbb{N}$. The derived series is the "fastest" descending series with abelian factors, and solvability asks that it eventually reaches the trivial group. [/remark] To understand which groups are solvable and which are not, it helps to build intuition through examples at both extremes. [example: Small Solvable Groups] Every abelian group $G$ is solvable: take $G_0 = \{e\}$ and $G_1 = G$; the factor $G / \{e\} \cong G$ is abelian by assumption. The symmetric group $S_3$ is solvable. It has the subnormal series \begin{align*} \{e\} \trianglelefteq A_3 \trianglelefteq S_3, \end{align*} where $A_3 = \{e, (123), (132)\} \cong \mathbb{Z}/3\mathbb{Z}$ and $S_3 / A_3 \cong \mathbb{Z}/2\mathbb{Z}$. Both factors are cyclic, hence abelian. The symmetric group $S_4$ is solvable via the series \begin{align*} \{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4, \end{align*} where $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ is the Klein four-group. The factors are $V_4 / \{e\} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, $A_4 / V_4 \cong \mathbb{Z}/3\mathbb{Z}$, and $S_4 / A_4 \cong \mathbb{Z}/2\mathbb{Z}$, all abelian. This is why the quartic is solvable by radicals. [/example] The pattern in these small cases — abelian factors like $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ stacking neatly — breaks down at degree five. The alternating group $A_5$ is the obstruction, and its non-solvability is the heart of the matter. [example: The Alternating Group $A_5$ Is Not Solvable] The alternating group $A_5$ is not solvable. To show this, we must show that no subnormal series with abelian factors exists. The key fact is that $A_5$ is simple: its only normal subgroups are $\{e\}$ and $A_5$ itself. To verify simplicity, let $N \trianglelefteq A_5$ be a nontrivial normal subgroup. The elements of $A_5$ are grouped by conjugacy class: - the identity: $1$ element, - $3$-cycles: $20$ elements (e.g., $(123)$), - $5$-cycles of type $(abcde)$: $24$ elements split into two classes of $12$. - double transpositions $(ab)(cd)$: $15$ elements. Since $N$ is a union of conjugacy classes containing the identity, $|N|$ must equal $1$ plus a sum of some of $\{20, 12, 12, 15\}$ and divide $|A_5| = 60$. The only such sum equal to a divisor of $60$ is $1$ itself (giving $N = \{e\}$) or $1 + 20 + 12 + 12 + 15 = 60$ (giving $N = A_5$). So $A_5$ is simple. Now suppose $A_5$ were solvable with series $\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_r = A_5$. Since all factors are abelian, the commutator $[A_5, A_5] \le G_{r-1}$. But $A_5$ is perfect: $[A_5, A_5] = A_5$ (since $A_5$ is simple and nonabelian, its commutator subgroup is all of $A_5$). So $G_{r-1} = A_5 = G_r$, and the series has not shortened. Repeating, no finite series can terminate at $\{e\}$. So $A_5$ is not solvable. [/example] The non-solvability of $A_5$ is the group-theoretic heart of Abel's theorem: the general quintic has Galois group $S_5$, which contains $A_5$ as a subgroup, and since $A_5$ is not solvable, neither is $S_5$. ## The Galois Criterion for Solvability The main theorem of this chapter connects solvability by radicals to the solvability of the Galois group. For a clean statement, we assume throughout that the base field $k$ has characteristic zero; this avoids the complications of inseparable extensions and primitive roots of unity in positive characteristic. Before stating the theorem, we need to understand how radical extensions interact with the Galois correspondence. The key structural fact is that adjoining an $n$-th root to a field produces a Galois extension — provided the field already contains the $n$-th roots of unity. When we adjoin an $n$-th root $\alpha$ of $a \in k$ to form $k(\alpha)$, the other roots of $x^n - a$ are $\zeta \alpha, \zeta^2 \alpha, \ldots, \zeta^{n-1} \alpha$, where $\zeta$ is a primitive $n$-th root of unity. If $\zeta$ already lives in $k$, then every automorphism of $k(\alpha)/k$ sends $\alpha$ to some $\zeta^j \alpha$, and the Galois group embeds into the cyclic group $\mathbb{Z}/n\mathbb{Z}$. Without $\zeta$ in the base field, the extension $k(\alpha)/k$ may fail to be Galois entirely — there may not be enough automorphisms to permute the roots. So the question "when is a radical extension Galois with abelian group?" reduces to "are the roots of unity already present?" This makes the following concept essential. [definition: Primitive Root of Unity] An element $\zeta \in K$ is a **primitive $n$-th root of unity** if $\zeta^n = 1$ and $\zeta^m \neq 1$ for all $1 \le m < n$. The field $\mathbb{Q}(\zeta_n)$ obtained by adjoining a primitive $n$-th root of unity $\zeta_n$ to $\mathbb{Q}$ is called the $n$-th **cyclotomic field**. [/definition] Over a field containing all $n$-th roots of unity, radical extensions are abelian Galois extensions. This is the content of the following theorem, which is the technical engine behind the Galois criterion. [quotetheorem:3337] The reason Galois theory requires roots of unity is that an automorphism $\sigma \in \operatorname{Gal}(K/k)$ must send $\alpha$ to another root of $x^n - a$, and these roots are exactly $\alpha, \zeta_n \alpha, \zeta_n^2 \alpha, \ldots, \zeta_n^{n-1} \alpha$. If $\zeta_n \in k$, then $\sigma$ is determined by which of these it sends $\alpha$ to, i.e., by a choice of $\zeta_n^j$. The map $\sigma \mapsto \zeta_n^j$ is a group homomorphism $\operatorname{Gal}(K/k) \to \langle \zeta_n \rangle \cong \mathbb{Z}/n\mathbb{Z}$, which accounts for the cyclic structure. We can now state the main theorem. [quotetheorem:1321] This theorem is the central result of classical Galois theory. Let us unpack both directions. **Forward direction: solvability by radicals implies solvable Galois group.** Suppose $f$ splits in a radical tower $k = k_0 \subset k_1 \subset \cdots \subset k_r = K$. We may assume $K$ is Galois over $k$ (if not, take the Galois closure; the Galois closure of a radical tower is still a radical tower over an extended base). After adjoining the appropriate roots of unity to make each step $k_i(\alpha_i) / k_i$ Galois, we obtain a refined tower in which each step is an abelian Galois extension. By the Galois correspondence, the corresponding chain of subgroups of $\operatorname{Gal}(K/k)$ is a subnormal series with abelian factors, so $\operatorname{Gal}(K/k)$ is solvable. Since $\operatorname{Gal}(L/k)$ is a quotient of $\operatorname{Gal}(K/k)$ (as $L \subset K$), and quotients of solvable groups are solvable, $\operatorname{Gal}(L/k)$ is solvable. **Backward direction: solvable Galois group implies solvability by radicals.** Suppose $G = \operatorname{Gal}(L/k)$ is solvable, with subnormal series $\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_r = G$ and abelian factors. Let $n = |G|$ and adjoin a primitive $n$-th root of unity $\zeta_n$ to $k$, forming $k' = k(\zeta_n)$. Over $k'$, the extension $L k' / k'$ has solvable Galois group (it is a quotient of $G$). The subnormal series of $\operatorname{Gal}(Lk'/k')$ produces — via the Galois correspondence — a tower of intermediate fields each obtained from the previous by a radical extension (using the Kummer theorem, since $k'$ contains all $n$-th roots of unity). Together with the cyclotomic extension $k'/k$ (which is itself an abelian, hence solvable, extension), we obtain a radical tower over $k$ containing all roots of $f$. [explanation: Why Characteristic Zero Is Required] The argument above uses two facts that require characteristic zero: first, that separable polynomials have splitting fields, and second, that the cyclotomic polynomial $x^n - 1$ has distinct roots (i.e., that $\gcd(n, \operatorname{char}(k)) = 1$ so the formal derivative $nx^{n-1}$ is nonzero). In characteristic $p > 0$, the polynomial $x^p - a$ factors as $(x - \alpha)^p$ in any extension (since $(x - \alpha)^p = x^p - \alpha^p = x^p - a$ by Frobenius), so the naive radical extension $k(\alpha)/k$ with $\alpha^p = a$ is inseparable — it is not Galois. The failure of Kummer theory in characteristic $p$ means the above argument breaks down, and the criterion for solvability by radicals in characteristic $p$ is more subtle, involving Artin-Schreier extensions. [/explanation] ## The Insolvability of the General Quintic The most celebrated application of the Galois criterion is the proof that the general polynomial of degree $n \ge 5$ is not solvable by radicals. To make this precise, we consider the general polynomial as a polynomial over the function field of its coefficients. What does "general polynomial of degree $n$" mean as a mathematical object? The classical quadratic formula applies to all quadratics $ax^2 + bx + c$ simultaneously — the coefficients $a, b, c$ are treated as indeterminates. The analogous setup for degree $n$ is to work over the field of rational functions $k = \mathbb{Q}(e_1, \ldots, e_n)$, where $e_1, \ldots, e_n$ are algebraically independent indeterminates playing the role of the elementary symmetric polynomials. [definition: General Polynomial of Degree $n$] The **general polynomial of degree $n$** over $\mathbb{Q}$ is the polynomial \begin{align*} f(x) = x^n - e_1 x^{n-1} + e_2 x^{n-2} - \cdots + (-1)^n e_n \in \mathbb{Q}(e_1, \ldots, e_n)[x], \end{align*} where $e_1, \ldots, e_n$ are algebraically independent over $\mathbb{Q}$. [/definition] The fundamental fact about the general polynomial is that its Galois group is as large as possible. [quotetheorem:1289] This theorem is the reason the general polynomial is the hardest possible case: its Galois group is the full symmetric group. For the Galois criterion to apply, we need to know when $S_n$ is solvable. [quotetheorem:3338] For $n \le 4$ the solvability of $S_n$ was verified above: $S_1$ and $S_2$ are abelian, $S_3$ has the series $\{e\} \trianglelefteq A_3 \trianglelefteq S_3$, and $S_4$ has the series $\{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$. For $n \ge 5$, the alternating group $A_n$ is simple (the argument generalises the $A_5$ case), so the only subnormal series of $S_n$ ending at $\{e\}$ must pass through $A_n$ and then jump to $\{e\}$. But $A_n$ is nonabelian, so the factor $A_n / \{e\} \cong A_n$ is not abelian. No subnormal series with abelian factors exists. Combining the three theorems: [quotetheorem:3312] This means: there is no algebraic expression in the coefficients $e_1, \ldots, e_n$, built from field operations and $m$-th root extractions for any $m$, that gives the roots of a general degree-$n$ polynomial when $n \ge 5$. Abel's theorem is a special case ($n = 5$); Galois's theorem gives the full picture. [remark: Specific Quintics Can Be Solvable] Insolvability of the general quintic does not mean no quintic is solvable by radicals. Polynomials whose Galois group is a proper solvable subgroup of $S_5$ — for example, any polynomial with abelian Galois group — are solvable by radicals. The polynomial $x^5 - 2$ has Galois group the Frobenius group of order 20, which is solvable, and its roots \begin{align*} \sqrt[5]{2}, \quad \sqrt[5]{2}\,\zeta_5, \quad \sqrt[5]{2}\,\zeta_5^2, \quad \sqrt[5]{2}\,\zeta_5^3, \quad \sqrt[5]{2}\,\zeta_5^4 \end{align*} are visibly expressible by radicals. The insolvability statement is about the general quintic — a polynomial with generic, algebraically independent coefficients — not about all quintics. [/remark] ## Explicit Criteria and the Lattice of Subgroups [illustration:s4-subgroup-lattice] In practice, determining whether a specific polynomial is solvable by radicals requires computing its Galois group and checking solvability. For polynomials of small degree, explicit criteria can be read off from discriminants and resolvent polynomials. This section develops these criteria for degrees three and four, and connects them to the group-theoretic analysis. The starting point for the cubic is the discriminant. It is an element of the base field built from the roots $x_1, x_2, x_3$ of a cubic $f(x) = x^3 + px + q \in \mathbb{Q}[x]$. [definition: Discriminant of a Cubic] For the cubic $f(x) = x^3 + px + q$ with roots $x_1, x_2, x_3$ in a splitting field, the **discriminant** is \begin{align*} \Delta(f) = \prod_{i < j} (x_i - x_j)^2 = -4p^3 - 27q^2. \end{align*} [/definition] The discriminant is a symmetric function of the roots and therefore lies in the coefficient field $\mathbb{Q}$. [quotetheorem:3339] Both $A_3$ and $S_3$ are solvable, confirming that every irreducible cubic is solvable by radicals, consistent with the cubic formula. [example: Galois Group of $x^3 - 3x + 1$] Let $f(x) = x^3 - 3x + 1$, so $p = -3$ and $q = 1$. The discriminant is \begin{align*} \Delta(f) = -4(-3)^3 - 27(1)^2 = -4(-27) - 27 = 108 - 27 = 81 = 9^2. \end{align*} Since $\Delta = 81$ is a perfect square in $\mathbb{Q}$, the Galois group is $A_3 \cong \mathbb{Z}/3\mathbb{Z}$. The splitting field has degree $3$ over $\mathbb{Q}$. The three roots \begin{align*} x_k = 2\cos\!\left(\frac{2\pi k}{9}\right), \quad k = 1, 2, 4 \end{align*} are all related by the field automorphism $x_1 \mapsto x_2 \mapsto x_4 \mapsto x_1$: since $f(x) = x^3 - 3x + 1$ has cyclic Galois group of order $3$, every root generates the splitting field over $\mathbb{Q}$, and the unique nontrivial automorphism of order $3$ cyclically permutes all three. No square root of the discriminant is needed — the splitting field is already totally real of degree $3$. [/example] For the quartic, the structure is more involved. The Galois group $G \le S_4$ can be any of five subgroups — $S_4, A_4, D_8, \mathbb{Z}/4\mathbb{Z}$, or $V_4$ — and distinguishing among them by inspection of the quartic's coefficients is not direct. The key insight is that $S_4$ acts on the three quantities $x_1 x_2 + x_3 x_4$, $x_1 x_3 + x_2 x_4$, $x_1 x_4 + x_2 x_3$ formed by pairing the four roots into two pairs. These three quantities are permuted among themselves by every element of $S_4$, so they are roots of a cubic polynomial with coefficients in the base field — the resolvent cubic. How many of those roots land in $\mathbb{Q}$ reveals how much of the pairing structure the Galois group fixes, which directly constrains which subgroup of $S_4$ it can be. [definition: Resolvent Cubic] Let $f(x) = x^4 + bx^2 + cx + d \in k[x]$ (a depressed quartic) with roots $x_1, x_2, x_3, x_4$. The **resolvent cubic** of $f$ is the cubic polynomial \begin{align*} g(y) = (y - x_1 x_2 - x_3 x_4)(y - x_1 x_3 - x_2 x_4)(y - x_1 x_4 - x_2 x_3). \end{align*} Its coefficients lie in $k$ (they are symmetric functions of the original roots), and $g \in k[y]$ can be computed explicitly: $g(y) = y^3 - by^2 - 4dy + (4bd - c^2)$. [/definition] A rational root of the resolvent cubic corresponds to a pairing of the four roots — say $\{x_1, x_2\}$ and $\{x_3, x_4\}$ — that is preserved by every automorphism in $G$. Preserving a pairing is a strong constraint: it forces $G$ to sit inside a subgroup of $S_4$ that stabilises that partition. Counting rational roots of the resolvent cubic therefore narrows the Galois group to a short list: [quotetheorem:3328] All of $S_4, A_4, D_8, \mathbb{Z}/4\mathbb{Z}$, and $V_4$ are solvable (each admits a subnormal series with abelian factors), confirming that all quartics are solvable by radicals. The jump from degree four to degree five is stark. For $n = 5$, there exist irreducible quintics $f \in \mathbb{Q}[x]$ with $\operatorname{Gal}(f) \cong S_5$. For such polynomials, no radical tower contains the roots. [example: An Irreducible Quintic with Galois Group $S_5$] Consider $f(x) = x^5 - 4x + 2 \in \mathbb{Q}[x]$. We claim $\operatorname{Gal}(f) \cong S_5$, and therefore $f$ is not solvable by radicals. **Irreducibility.** By the Eisenstein criterion at $p = 2$: the constant term $2$ and the coefficient $-4$ of $x$ are both divisible by $2$, the leading coefficient $1$ is not divisible by $2$, and the constant term $2$ is not divisible by $2^2 = 4$. Therefore $f$ is irreducible over $\mathbb{Q}$. **Counting real roots.** The derivative is $f'(x) = 5x^4 - 4$, which vanishes when $x^4 = 4/5$, giving two real critical points $x = \pm (4/5)^{1/4}$. Let $c = (4/5)^{1/4} \approx 0.9457$. Since $f''(x) = 20x^3$, we have $f''(c) > 0$ (local minimum at $x = c$) and $f''(-c) < 0$ (local maximum at $x = -c$). Computing the critical values: \begin{align*} f(c) &= c^5 - 4c + 2 = c \cdot c^4 - 4c + 2 = c \cdot \tfrac{4}{5} - 4c + 2 = -\tfrac{16}{5}c + 2 \approx -3.026 + 2 = -1.026 < 0, \\ f(-c) &= -c^5 + 4c + 2 = \tfrac{16}{5}c - 2 + 4 = \tfrac{16}{5}c + 2 \approx 3.026 + 2 > 0. \end{align*} Since $f(x) \to -\infty$ as $x \to -\infty$ and $f(-c) > 0$, there is exactly one root in $(-\infty, -c)$. Since $f(-c) > 0$ and $f(c) < 0$, there is exactly one root in $(-c, c)$. Since $f(c) < 0$ and $f(x) \to +\infty$ as $x \to +\infty$, there is exactly one root in $(c, +\infty)$. Thus $f$ has exactly three real roots and one pair of complex conjugate roots, giving exactly two nonreal roots. **The Galois group is $S_5$.** Since $f$ is irreducible of prime degree $5$, the Galois group $G = \operatorname{Gal}(f) \le S_5$ acts transitively on the five roots, so $5 \mid |G|$. By Cauchy's theorem, $G$ contains an element of order $5$, which must be a $5$-cycle in $S_5$. Complex conjugation is an automorphism of the splitting field that fixes the three real roots and swaps the two nonreal roots; it acts as a transposition in $S_5$. A subgroup of $S_5$ that contains both a $5$-cycle and a transposition generates all of $S_5$ (this is a standard result: conjugating the transposition by powers of the $5$-cycle produces enough transpositions to generate $S_5$). Therefore $G \cong S_5$. Since $S_5$ is not solvable, $f$ is not solvable by radicals over $\mathbb{Q}$. [/example] ## Abelian Extensions and the Kronecker-Weber Theorem The Galois criterion tells us that solvability by radicals is controlled by the solvability of the Galois group, and solvability allows arbitrarily many abelian factors in the subnormal series. A natural question arises: what happens when the Galois group itself is abelian — when a single abelian factor suffices? These are the simplest solvable extensions, and they turn out to have a remarkably rigid structure over $\mathbb{Q}$: every abelian extension of $\mathbb{Q}$ is carved out of a cyclotomic field. This is the Kronecker-Weber theorem, and it reveals that the roots of unity alone generate all abelian extensions of the rationals. A field extension $K / k$ is called **abelian** if it is Galois with abelian Galois group. Abelian extensions are automatically solvable, since an abelian group is solvable with the series $\{e\} \trianglelefteq G$. [quotetheorem:3340] This theorem says that any polynomial over $\mathbb{Q}$ with abelian Galois group has its splitting field inside $\mathbb{Q}(\zeta_n)$ for some $n$. Since $\zeta_n$ is itself a radical (it is an $n$-th root of unity), the splitting field is already inside the radical extension $\mathbb{Q}(\zeta_n) / \mathbb{Q}$. In particular: every abelian polynomial over $\mathbb{Q}$ is solvable by radicals via a single cyclotomic adjunction. [example: $\mathbb{Q}(\sqrt{d})$ Inside a Cyclotomic Field] For any squarefree $d \in \mathbb{Z}$, the quadratic field $\mathbb{Q}(\sqrt{d})$ is an abelian extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}/2\mathbb{Z}$. By Kronecker-Weber, $\mathbb{Q}(\sqrt{d}) \subset \mathbb{Q}(\zeta_n)$ for some $n$. Explicitly: the quadratic fields inside $\mathbb{Q}(\zeta_p)$ for an odd prime $p$ are exactly $\mathbb{Q}(\sqrt{p^*})$ where $p^* = (-1)^{(p-1)/2} p$. For example, $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\zeta_5)$ because $\zeta_5 + \zeta_5^{-1} = 2\cos(2\pi/5) = \frac{-1 + \sqrt{5}}{2}$, so $\sqrt{5} \in \mathbb{Q}(\zeta_5)$. To verify: $\cos(2\pi/5)$ satisfies the minimal polynomial $16x^4 - 12x^2 + 1$ via the double angle formulas, but $\cos(2\pi/5) + \cos(4\pi/5) = \frac{-1}{2}$ (sum of real parts of primitive 5th roots), so $\zeta_5 + \zeta_5^4 + \zeta_5^2 + \zeta_5^3 = -1$ and the Gauss sum $\tau = \sum_{a=1}^{4} \left(\frac{a}{5}\right) \zeta_5^a = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4$ satisfies $\tau^2 = 5$, giving $\sqrt{5} \in \mathbb{Q}(\zeta_5)$. [/example] The Kronecker-Weber theorem is the prototype of class field theory: it characterises abelian extensions of $\mathbb{Q}$ by cyclotomic fields. For abelian extensions of more general number fields, the role of cyclotomic fields is played by ray class fields — a much deeper story. ## Constructibility by Ruler and Compass The question of solvability by radicals has a classical geometric cousin: which lengths are constructible using ruler and compass? The two questions are related because constructibility corresponds to solvability by radicals of a very special kind — only square roots are allowed. A real number $\alpha$ is **constructible** if it can be obtained from $\mathbb{Q}$ by a finite sequence of additions, subtractions, multiplications, divisions, and square root extractions. Each step adds a degree-two extension, so a constructible number lies in a tower \begin{align*} \mathbb{Q} = k_0 \subset k_1 \subset \cdots \subset k_r \end{align*} with $[k_{i+1} : k_i] = 2$ for each $i$. [quotetheorem:3341] The proof uses the tower law: if $\alpha$ lies in a tower of quadratic extensions, then $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ divides the degree of the top field over $\mathbb{Q}$, which is a product of factors of $2$. Conversely, if $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2^m$, the Galois closure has a Galois group of order $2^m$, which is a $2$-group and hence solvable with a series of quotients $\mathbb{Z}/2\mathbb{Z}$, giving a tower of square root extensions. [example: The Regular Heptagon Is Not Constructible] The regular $7$-gon requires constructing $\cos(2\pi/7)$. This satisfies $[\mathbb{Q}(\cos(2\pi/7)) : \mathbb{Q}] = 3$ (since $\cos(2\pi/7)$ is a root of the cubic $8x^3 + 4x^2 - 4x - 1$). Since $3$ is not a power of $2$, $\cos(2\pi/7)$ is not constructible, and the regular heptagon cannot be constructed with ruler and compass. To verify the degree: the minimal polynomial of $\zeta_7 = e^{2\pi i/7}$ over $\mathbb{Q}$ is the 7th cyclotomic polynomial $\Phi_7(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, of degree $6$. Since $[\mathbb{Q}(\zeta_7) : \mathbb{Q}] = 6$ and $\mathbb{Q}(\cos(2\pi/7)) = \mathbb{Q}(\zeta_7 + \zeta_7^{-1})$ is the fixed field of complex conjugation in $\mathbb{Q}(\zeta_7)$, we get $[\mathbb{Q}(\cos(2\pi/7)) : \mathbb{Q}] = 3$. [/example] [example: The Regular 17-gon Is Constructible] Gauss proved in 1796 that the regular $17$-gon is constructible. The key is that $\varphi(17) = 16 = 2^4$, so $[\mathbb{Q}(\zeta_{17}) : \mathbb{Q}] = 16 = 2^4$ is a power of $2$. The Galois group $\operatorname{Gal}(\mathbb{Q}(\zeta_{17})/\mathbb{Q}) \cong (\mathbb{Z}/17\mathbb{Z})^\times \cong \mathbb{Z}/16\mathbb{Z}$ is cyclic of order $16 = 2^4$. It admits a subnormal series \begin{align*} \{e\} \trianglelefteq \mathbb{Z}/2\mathbb{Z} \trianglelefteq \mathbb{Z}/4\mathbb{Z} \trianglelefteq \mathbb{Z}/8\mathbb{Z} \trianglelefteq \mathbb{Z}/16\mathbb{Z} \end{align*} with all quotients equal to $\mathbb{Z}/2\mathbb{Z}$. By the Galois correspondence, the fixed fields form a tower of quadratic extensions from $\mathbb{Q}$ up to $\mathbb{Q}(\zeta_{17})$, and $\cos(2\pi/17) \in \mathbb{Q}(\zeta_{17})$ is reached by this tower. Hence $\cos(2\pi/17)$ is constructible. A regular $17$-gon is constructible by ruler and compass. More generally, the regular $n$-gon is constructible if and only if $\varphi(n)$ is a power of $2$, which holds exactly when $n = 2^a p_1 p_2 \cdots p_r$ where each $p_i$ is a distinct Fermat prime ($p_i = 2^{2^{k_i}} + 1$ for some $k_i \ge 0$). [/example] ## References - E. Artin, *Galois Theory* (1944). The definitive short treatment; Artin's formulation of the Galois correspondence is the standard. - I. Stewart, *Galois Theory* (4th ed., 2015). Accessible graduate textbook with full proofs and historical context. - J.-P. Serre, *Topics in Galois Theory* (2nd ed., 2008). Discusses inverse Galois problems and the structure of Galois groups over $\mathbb{Q}$. - D.J.H. Garling, *A Course in Galois Theory* (1986). Concise Cambridge treatment emphasising the solvability theorem. - N. Jacobson, *Basic Algebra I* (2nd ed., 1985). Chapter 4 contains a thorough development of Galois theory including the solvability criterion.