Insolvability of the quintic

Theorem

Insolvability of the quintic

T-094Home MU-303Threads structure

Statement

Let \(F\) be a field of characteristic \(0\) and let \(t_1,\dots,t_5\) be independent transcendentals over \(F\). Write \(s_1,\dots,s_5\) for the elementary symmetric polynomials in the \(t_i\), set \(K=F(s_1,\dots,s_5)\), and let \(f_t(X)=\prod_{i=1}^5(X-t_i)=X^5-s_1X^4+s_2X^3-s_3X^2+s_4X-s_5\in K[X]\) be the general quintic over \(K\). Then \(f_t\) is not solvable by radicals over \(K\): there is no tower \(K=K_0\subseteq K_1\subseteq\cdots\subseteq K_m\) with each \(K_{i+1}=K_i(\alpha_i)\), \(\alpha_i^{n_i}\in K_i\) for some \(n_i\ge 1\), such that the splitting field of \(f_t\) over \(K\) is contained in \(K_m\). Equivalently: there is no formula for the roots of a degree-five polynomial in terms of its coefficients built only from \(+,-,\times,\div\) and extraction of \(n\)-th roots, valid for all quintics simultaneously.

Why it matters

Every polynomial of degree \(\le 4\) can be solved "by formula": the quadratic formula is classical, and the cubic and quartic were cracked by del Ferro, Tartaglia, Cardano and Ferrari in the sixteenth century, each reducing the problem to nested radicals of the coefficients. It was natural to expect a degree-five analogue, and centuries of failed attempts culminated in Ruffini's flawed 1799 argument and Abel's rigorous 1824 proof that no such formula exists. Galois then explained why: solvability by radicals is controlled entirely by a finite group attached to the polynomial, and for the general quintic that group is too large and too rigid to be built out of abelian pieces.

This theorem is the historical origin of group theory as an independent subject — the notion of a "solvable group" is named for exactly this application — and it is the paradigm case of the Galois correspondence turning a question about algebraic formulas into a question about finite group structure.

Hypotheses

Characteristic 0 base field.Galois's radical-solvability criterion (Lemma, Statement 2 below) is proved using Kummer theory, which requires enough roots of unity to be adjoined without inseparability obstructions; in characteristic \(p\) one must also allow Artin–Schreier extensions \(\alpha^p-\alpha=a\) as "radical" steps, and the clean equivalence changes shape. Degree \(n=5\) (or more generally \(n\ge 5\)).For \(n\le 4\), \(S_n\) is solvable — e.g. \(S_4\supset A_4\supset V_4\supset\{e\}\) has abelian factors \(\mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_2\times\mathbb{Z}_2\) — so Cardano's and Ferrari's radical formulas exist and the theorem's conclusion fails. Genericity: \(s_1,\dots,s_5\) algebraically independent.Specialising to particular numbers can make the Galois group collapse to something solvable — \(x^5-1\) has abelian Galois group \(\mathbb{Z}_4\) over \(\mathbb{Q}\) and is solvable by radicals — so "the general quintic is unsolvable" does not mean "every quintic is unsolvable."

Only field operations and radicals are permitted in the formula.Allow transcendental tools beyond radicals — elliptic/theta functions or the Bring radical — and the quintic is uniformly solvable (Hermite 1858, Kronecker, Brioschi); the theorem is a statement about the radical vocabulary specifically, not about definability in general.

Proof

\text{A field extension } L/K \text{ is a } \textbf{radical extension} \text{ if there is a tower } K=K_0\subseteq K_1\subseteq\cdots\subseteq K_m=L \text{ with } K_{i+1}=K_i(\alpha_i),\ \alpha_i^{n_i}\in K_i.

Definition: \(f\in K[X]\) is solvable by radicals over \(K\) if its splitting field embeds in some radical extension of \(K\). A

f \text{ solvable by radicals over } K \iff \mathrm{Gal}(f/K) \text{ is a solvable group.}

Cited: Galois's radical-solvability criterion, established earlier in MU-303 via Kummer theory (radical steps correspond to cyclic Kummer extensions once enough roots of unity are present) combined with the Fundamental Theorem of Galois Theory. Valid since \(\operatorname{char}K=0\). B

\mathbb{Q}(t_1,\dots,t_5)^{S_5} = \mathbb{Q}(s_1,\dots,s_5) = K, \qquad \mathrm{Gal}\big(\mathbb{Q}(t_1,\dots,t_5)/K\big) \cong S_5.

\(S_5\) acts on \(\mathbb{Q}(t_1,\dots,t_5)\) by permuting the indeterminates \(t_i\); the Fundamental Theorem of Symmetric Polynomials identifies the fixed ring as generated by \(s_1,\dots,s_5\), and Artin's Fixed Field Theorem upgrades "fixed field of a finite group of automorphisms" to a Galois extension with that group as its Galois group and degree equal to \(|S_5|=120\). B

\mathbb{Q}(t_1,\dots,t_5) \text{ is the splitting field of } f_t(X)=\textstyle\prod_i(X-t_i) \text{ over } K, \quad \mathrm{Gal}(f_t/K)\cong S_5.

\(f_t\) splits completely in \(\mathbb{Q}(t_1,\dots,t_5)\) by construction (its roots are exactly \(t_1,\dots,t_5\)), and no smaller subfield containing \(K\) can split it since the \(t_i\) already generate that field; combine with Step 3. A

A_5 \text{ is simple: its conjugacy classes have sizes } 1,15,20,12,12 \text{ (summing to } 60\text{), and no proper subset containing } 1 \text{ sums to a divisor of } 60 \text{ other than } 60 \text{ itself.}

A normal subgroup is a union of conjugacy classes including \(\{e\}\), and its order divides \(|A_5|=60\) by Lagrange. Checking all subset sums of \(\{1,15,20,12,12\}\) that include the \(1\): the only ones dividing \(60\) are \(1\) and \(60\). Hence the only normal subgroups are \(\{e\}\) and \(A_5\). C

S_5 \text{ is not solvable.}

Suppose \(S_5\) were solvable. Subgroups of solvable groups are solvable (standard closure lemma), so \(A_5\le S_5\) would be solvable, meaning \(A_5\) admits a subnormal series with abelian factors. But by Step 5 the only such series is \(\{e\}\lhd A_5\), whose sole factor is \(A_5\) itself — nonabelian since e.g. \((1\,2\,3)\) and \((1\,2)(4\,5)\) do not commute. Contradiction. B

f_t \text{ is not solvable by radicals over } K.

Apply the criterion of Step 2 with \(\mathrm{Gal}(f_t/K)\cong S_5\) (Step 4), which is not solvable (Step 6). A

\text{No expression in } +,-,\times,\div \text{ and } n\text{-th roots of the coefficients } a_0,\dots,a_4 \text{ solves every quintic } X^5+a_4X^4+\cdots+a_0.

Suppose such a general formula existed. Substituting the algebraically independent coefficients \(-s_1,s_2,-s_3,s_4,-s_5\) for \(a_4,\dots,a_0\) is a legitimate specialisation (a ring homomorphism on the polynomial ring generated by the formula's intermediate radicands, since the \(s_i\) satisfy no algebraic relations to obstruct it), and the resulting expression would exhibit the roots of \(f_t\) as radical expressions in \(s_1,\dots,s_5\), i.e. would show \(f_t\) solvable by radicals over \(K\). This contradicts Step 7. Hence no such formula exists. \(\blacksquare\) B

Result

\mathrm{Gal}(f_t/K)\cong S_5 \text{ is not solvable} \;\Longrightarrow\; f_t \text{ is not solvable by radicals } \;\Longrightarrow\; \text{no general radical formula solves the quintic.}

Reading. The obstruction to a quintic formula is not a failure of cleverness but a structural fact about the symmetric group \(S_5\): it has no chain of normal subgroups with abelian quotients, because its only proper nontrivial normal subgroup \(A_5\) is simple and nonabelian. Since the Galois group of a "coefficients-as-independent-variables" quintic is the full \(S_5\), the tower of radical extensions that a formula would require simply cannot exist.

Scope. Applies to the general (generic-coefficient) polynomial of any degree \(n\ge 5\) over any characteristic-0 field, via the same argument with \(S_n\) in place of \(S_5\) (Step 6 generalises since \(A_n\) is simple for all \(n\ge5\)). It says nothing about any single numerical quintic in isolation — those must be checked one at a time via their own Galois group.

Corollaries & converses

The general polynomial of degree \(n\), for every \(n\ge5\), is unsolvable by radicals — same proof verbatim with \(S_n\), using that \(A_n\) is simple nonabelian for \(n\ge 5\).
A specific quintic \(f\in\mathbb{Q}[X]\) with \(\mathrm{Gal}(f/\mathbb{Q})\cong S_5\) (e.g. \(x^5-4x+2\), Worked Example 1) is individually unsolvable by radicals — the theorem's method transfers to any field once the Galois group is computed.
The converse of the underlying criterion (Step 2) genuinely holds both ways: it is stated and used here as an "\(\iff\)". So quintics with solvable Galois group (cyclic \(\mathbb{Z}_5\), dihedral \(D_5\), or Frobenius \(F_{20}\cong \mathbb{Z}_5\rtimes\mathbb{Z}_4\) — the three solvable transitive subgroups of \(S_5\) up to conjugacy, alongside the non-solvable \(A_5\) and \(S_5\)) are solvable by radicals, with an explicit radical tower recoverable from the group's composition series (Worked Example 2).
It does not follow that "most" quintics are unsolvable is somehow false for individual numbers — in fact a random monic integer quintic has Galois group \(S_5\) with density \(1\) (Hilbert irreducibility / van der Waerden), so unsolvability by radicals is the generic, not the exceptional, behaviour even numerically.

Fails without

Drop \(n\ge5\): for \(n=4\), \(S_4\) is solvable (\(S_4\supset A_4\supset V_4\supset\{e\}\), abelian factors \(\mathbb{Z}_2,\mathbb{Z}_3,V_4\)), and Ferrari's formula gives every quartic's roots by radicals — the theorem's conclusion reverses completely.
Drop "radicals only": the Bring–Jerrard normal form \(x^5+px+q\) (every quintic reduces to this via a solvable auxiliary Tschirnhaus transformation) is solved uniformly by Hermite (1858) using elliptic modular functions; allowing this wider toolkit restores a "general formula," so the restriction to radicals is load-bearing, not incidental.
Drop genericity, keep degree 5: \(x^5-1=(x-1)(x^4+x^3+x^2+x+1)\) has splitting field \(\mathbb{Q}(\zeta_5)\) with \(\mathrm{Gal}\cong(\mathbb{Z}/5\mathbb{Z})^\times\cong\mathbb{Z}_4\), abelian hence solvable; this specific quintic is solvable by radicals, showing the theorem is a statement about the general/generic polynomial, not a blanket ban on every degree-five equation.

Common errors

Concluding "the quintic has no roots" — false; every quintic has 5 roots in \(\mathbb{C}\) (or in its splitting field) by the Fundamental Theorem of Algebra. The theorem is about expressibility of those roots via radicals, not existence.
Claiming "no quintic is solvable by radicals" — false; \(x^5-2\), \(x^5-1\), and any reducible quintic are solvable. Only the general/generic quintic, and specific instances whose Galois group happens to be \(S_5\) or \(A_5\), are unsolvable.
Computing \(\mathrm{Gal}(f/\mathbb{Q})\) as a subgroup of \(S_5\) without first checking \(f\) is irreducible — if \(f\) factors, the group need not act transitively and the whole cycle-type argument (Worked Example 1) is unavailable.
Applying the "\(p\)-cycle + transposition generate \(S_p\)" lemma (Worked Example 1, Step 3) to a non-prime degree — it is false for composite \(n\): a \(4\)-cycle and a transposition in \(S_4\) need not generate \(S_4\) (e.g. \((1\,2\,3\,4)\) and \((1\,3)\) generate only the dihedral group of order 8).
Assuming the Galois group of "the general quintic" automatically tells you the Galois group of a specific numerical quintic without separate computation — specialisation can only decrease the group (it lands inside \(S_5\), but which conjugacy-restricted subgroup requires its own irreducibility/discriminant/mod-\(p\) analysis).

Discussion

Ruffini's 1799 memoir already argued essentially correctly that no radical formula could exist, but had a gap (he implicitly assumed any such formula's intermediate fields would be normal extensions, which needs proof). Abel closed the gap in 1824 with a fully rigorous, if group-theoretically implicit, argument — the Abel–Ruffini theorem. Galois, writing in the early 1830s and unpublished until 1846, reorganised the whole subject: he attached to every polynomial a permutation group of its roots and showed radical solvability is exactly solvability of that group, a statement that instantly explains both why degree \(\le4\) always works and why degree \(\ge5\) generically fails. The word "solvable" for groups is a direct legacy of this application.

It is worth separating two different theorems that are easy to conflate: Abel–Ruffini (no uniform formula for the general quintic) and the finer per-polynomial statement (a specific quintic is solvable by radicals iff its own Galois group, a particular transitive subgroup of \(S_5\), is solvable). The five transitive subgroups of \(S_5\) up to conjugacy — \(\mathbb{Z}_5\), \(D_5\) (order 10), \(F_{20}=\mathbb{Z}_5\rtimes\mathbb{Z}_4\) (order 20), \(A_5\) (order 60), \(S_5\) (order 120) — split exactly along the solvable/nonsolvable line, with only \(A_5\) and \(S_5\) failing.

The story does not end with "impossible." Hermite (1858), followed by Kronecker and Brioschi, showed that the Bring–Jerrard quintic \(x^5+px+q\) — to which every quintic can be reduced by a solvable radical Tschirnhaus transformation — can be solved using elliptic modular functions (theta constants), giving a genuine closed-form solution outside the radical vocabulary. This is the same phenomenon as the AGM/theta-function solutions to certain transcendental equations: enlarging the function class dissolves an obstruction that is specific to the narrower class.

A common misconception is treating this as a statement about numerical approximation or existence of roots — it is neither. Roots of any quintic can be found to arbitrary numerical precision (Newton's method, Sturm's theorem for isolating real roots) and exist exactly in \(\mathbb{C}\); the theorem is purely about which field-theoretic construction (radical towers) can reach them symbolically.

Worked examples

f(x) = x^5 - 4x + 2 \in \mathbb{Q}[X] \text{ is irreducible.}

Eisenstein's criterion at \(p=2\): \(2\) divides every non-leading coefficient (\(0,0,0,-4,2\)) and \(4\nmid 2\). A

f \text{ has exactly 3 real roots and one conjugate pair of non-real roots.}

\(f'(x)=5x^4-4\) vanishes at \(x=\pm a\), \(a=(4/5)^{1/4}\approx0.9457\). \(f(-a)\approx5.03\) (local max, positive), \(f(a)\approx-1.03\) (local min, negative); since \(f\to-\infty\) as \(x\to-\infty\) and \(f\to+\infty\) as \(x\to+\infty\), the Intermediate Value Theorem gives exactly one root in each of the three monotonic runs, and no more since \(f'\) has only two real zeros. B

\mathrm{Gal}(f/\mathbb{Q}) \le S_5 \text{ is transitive, contains a 5-cycle, and contains a transposition, hence equals } S_5.

Irreducibility (Step 1) makes \(\mathbb{Q}\)-conjugation transitive on the 5 roots, so \(5\mid|\mathrm{Gal}(f/\mathbb{Q})|\); by Cauchy's theorem the group contains an element of order 5, which in \(S_5\) must be a 5-cycle. Complex conjugation, restricted to the splitting field \(\subset\mathbb{C}\), is a field automorphism fixing \(\mathbb{Q}\); by Step 2 it swaps the two non-real roots and fixes the three real roots, so it acts as a single transposition. Lemma (proved here): if \(p\) is prime and \(H\le S_p\) contains a \(p\)-cycle \(\sigma\) and a transposition \(\tau=(a\,b)\), then \(H=S_p\). Relabelling so \(\sigma=(1\,2\,\cdots\,p)\), conjugation by powers of \(\sigma\) sends \(\tau\) to \((a{+}k,\,b{+}k)\) for every \(k\) (indices mod \(p\)); since \(\gcd(b-a,p)=1\) (as \(p\) is prime and \(a\ne b\)), the "difference-\((b-a)\)" graph on \(\{1,\dots,p\}\) with these edges is a single \(p\)-cycle graph, hence connected, and a set of transpositions whose graph is connected generates the full symmetric group on those points (standard union–find generation lemma). So \(H\) contains all of \(S_p\). C

S_5 \text{ is not solvable (Steps 5–6 of the main proof)}, \implies f \text{ is not solvable by radicals over } \mathbb{Q}.

Apply the Galois criterion (Step 2 of the main proof) to \(\mathrm{Gal}(f/\mathbb{Q})\cong S_5\). A

x^5-4x+2 \text{ has no expression by radicals over } \mathbb{Q}.

Reading. A concrete, perfectly ordinary integer quintic — not a generic or abstract one — is provably unsolvable by radicals, because its own Galois group happens to be all of \(S_5\).

Scope. The argument (Eisenstein + real-root count + prime-cycle lemma) is a general recipe for proving individual quintics unsolvable, whenever an Eisenstein prime and exactly 3 real roots can be found.

g(x) = x^5 - 2 \in \mathbb{Q}[X] \text{ is irreducible, with splitting field } K=\mathbb{Q}(\zeta_5,\,2^{1/5}).

Eisenstein at \(p=2\) gives irreducibility. The splitting field must contain the real root \(2^{1/5}\) and, dividing any two roots, a primitive 5th root of unity \(\zeta_5\); conversely adjoining both produces all five roots \(\zeta_5^k 2^{1/5}\). A

[K:\mathbb{Q}] = 20, \qquad \mathrm{Gal}(K/\mathbb{Q}) \cong F_{20} = \mathbb{Z}_5\rtimes\mathbb{Z}_4.

\([\mathbb{Q}(\zeta_5):\mathbb{Q}]=4\) (cyclotomic polynomial \(\Phi_5\) irreducible of degree \(\varphi(5)=4\)), \([\mathbb{Q}(2^{1/5}):\mathbb{Q}]=5\) by Step 1; the two degrees are coprime so \(\mathbb{Q}(\zeta_5)\cap\mathbb{Q}(2^{1/5})=\mathbb{Q}\) and \([K:\mathbb{Q}]=20\) by the tower/multiplicativity law. \(\mathbb{Q}(2^{1/5})/\mathbb{Q}\) is not normal (it misses the non-real roots), so \(\mathrm{Gal}(K/\mathbb{Q})\) is generated by \(\sigma:2^{1/5}\mapsto\zeta_5 2^{1/5}\) (order 5, normal since \(\langle\sigma\rangle\) fixes \(\mathbb{Q}(\zeta_5)\) pointwise... acting only on the radical) and \(\tau:\zeta_5\mapsto\zeta_5^2\) fixing \(2^{1/5}\) (order 4), with \(\tau\sigma\tau^{-1}=\sigma^2\), the standard presentation of the Frobenius group \(F_{20}\). B

F_{20} \text{ is solvable: } \{e\}\lhd \mathbb{Z}_5 \lhd F_{20}, \text{ with abelian factors } \mathbb{Z}_5 \text{ and } F_{20}/\mathbb{Z}_5\cong\mathbb{Z}_4.

\(\langle\sigma\rangle\cong\mathbb{Z}_5\) is normal (index 4, and normal by the relation \(\tau\sigma\tau^{-1}=\sigma^2\in\langle\sigma\rangle\)); the quotient has order 4 and is generated by the image of \(\tau\), which is cyclic. Both factors abelian, so \(F_{20}\) is solvable by definition. A

g \text{ is solvable by radicals over } \mathbb{Q} \text{ — and explicitly so: its roots are } \zeta_5^k\cdot 2^{1/5},\ k=0,\dots,4.

The Galois criterion (converse direction) guarantees a radical tower exists given Step 3; here it is visible directly, since \(\zeta_5=e^{2\pi i/5}\) is itself expressible by nested square roots (Gauss, via the quadratic Gauss sum construction for the constructible 17-gon's cousin, the constructible pentagon) and \(2^{1/5}\) is a bare radical. A

x^5-2 \text{ is solvable by radicals: } \mathrm{Gal} \cong F_{20} \text{ is solvable, consistent with (and not a counterexample to) the theorem.}

Reading. This example is the theorem's converse in action: a solvable Galois group guarantees, and here manifestly produces, a radical expression for the roots.

Scope. Illustrates that "insolvability of the quintic" is a statement about the general/generic case and about specific polynomials with \(\mathrm{Gal}\in\{A_5,S_5\}\); most textbook quintics with small coefficients (like this one) are in fact solvable.

Problems

Exhibit composition series with abelian factors for \(S_3\) and \(S_4\), confirming both are solvable.
Solution
\(S_3\supset A_3\supset\{e\}\) with factors \(S_3/A_3\cong\mathbb{Z}_2\) and \(A_3\cong\mathbb{Z}_3\), both abelian. \(S_4\supset A_4\supset V_4\supset\{e\}\) (\(V_4=\{e,(12)(34),(13)(24),(14)(23)\}\)) with factors \(S_4/A_4\cong\mathbb{Z}_2\), \(A_4/V_4\cong\mathbb{Z}_3\) (order \(12/4=3\)), and \(V_4\cong\mathbb{Z}_2\times\mathbb{Z}_2\), all abelian. Hence both groups are solvable.
Verify the conjugacy class sizes of \(A_5\) used in Step 5 of the main proof: \(1,15,20,12,12\).
Solution
\(|A_5|=60\). Identity: class size \(1\). Double transpositions \((a\,b)(c\,d)\): there are \(\binom{5}{2}\binom{3}{2}/2=15\) such elements in \(S_5\), and they stay a single class in \(A_5\) since the centralizer computation shows the \(S_5\)-class does not split (the centralizer in \(S_5\) already lies in \(A_5\)... concretely: 15 elements, orbit-stabilizer gives centralizer order \(60/15=4\), even order confirms it is one \(A_5\)-class). 3-cycles: \(5\cdot4\cdot3/3=20\) elements in \(S_5\); centralizer order \(60/20=3\), one class. 5-cycles: \(4!=24\) elements in \(S_5\), but a 5-cycle's \(S_5\)-centralizer has order \(120/24=5\), which is odd, so the class splits into two \(A_5\)-classes of size \(24/2=12\) each. Total: \(1+15+20+12+12=60\). ✓.
Show \(h(x)=x^5-6x+3\) is irreducible over \(\mathbb{Q}\), determine its number of real roots, and decide whether it is solvable by radicals.
Solution
Eisenstein at \(p=3\): \(3\mid\{0,0,0,-6,3\}\) and \(9\nmid3\), so \(h\) is irreducible. \(h'(x)=5x^4-6\) vanishes at \(x=\pm(6/5)^{1/4}\approx\pm1.0467\). \(h(1.0467)\approx1.0467^5-6(1.0467)+3\approx1.269-6.280+3=-2.011\lt0\) and \(h(-1.0467)\approx-1.269+6.280+3=8.011\gt0\); with \(h\to\mp\infty\) as \(x\to\mp\infty\), exactly 3 real roots (same sign pattern as Worked Example 1). By the prime-cycle+transposition lemma, \(\mathrm{Gal}(h/\mathbb{Q})=S_5\), which is not solvable, so \(h\) is not solvable by radicals.
Compute \(\mathrm{Gal}(x^5-3/\mathbb{Q})\) and confirm it is solvable, exhibiting the roots as radicals.
Solution
Identical structure to Worked Example 2 with \(2\) replaced by \(3\): \(x^5-3\) is irreducible (Eisenstein, \(p=3\)), splitting field \(\mathbb{Q}(\zeta_5,3^{1/5})\) of degree \(4\cdot5=20\) (degrees 4 and 5 coprime), Galois group \(F_{20}=\mathbb{Z}_5\rtimes\mathbb{Z}_4\), solvable via \(\{e\}\lhd\mathbb{Z}_5\lhd F_{20}\) with abelian factors \(\mathbb{Z}_5,\mathbb{Z}_4\). Roots explicitly: \(\zeta_5^k\cdot3^{1/5}\), \(k=0,\dots,4\), a bare radical expression.
Let \(k(x)=x^5-x-1\). Using the factorizations \(k(x)\equiv(x^2+x+1)(x^3+x^2+1)\pmod2\) and the fact that \(k(x)\) is irreducible mod \(3\), determine \(\mathrm{Gal}(k/\mathbb{Q})\) and decide solvability by radicals. (You may cite Dedekind's theorem: for \(p\nmid\mathrm{disc}(k)\), the cycle type of Frobenius at \(p\) in \(\mathrm{Gal}(k/\mathbb{Q})\le S_5\) matches the degree pattern of the factorization of \(k\) mod \(p\).)
Solution
Irreducibility mod 3 already forces \(k\) irreducible over \(\mathbb{Q}\), and by Dedekind's theorem gives an element of \(\mathrm{Gal}(k/\mathbb{Q})\) of cycle type \((5)\), i.e. a 5-cycle. The mod-2 factorization has degree pattern \((2,3)\) (both factors irreducible over \(\mathbb{F}_2\), checked by the absence of roots and the fact that neither divides the other), giving by Dedekind's theorem an element \(g\) of cycle type \((2,3)\); then \(g^3\) has order 2 and equals the 2-cycle part exactly (the 3-cycle part cubes to the identity), i.e. \(g^3\) is a genuine transposition. By the prime-cycle+transposition lemma (Worked Example 1, Step 3), \(\mathrm{Gal}(k/\mathbb{Q})=S_5\). Since \(S_5\) is not solvable, \(k=x^5-x-1\) is not solvable by radicals over \(\mathbb{Q}\).