[proofplan] The proof has two main components. First, we exhibit a specific polynomial of degree $n$ over $\mathbb{Q}$ whose Galois group is the full symmetric group $S_n$. Second, we show that $S_n$ is not solvable for $n \ge 5$, by proving that the alternating group $A_n$ is simple for $n \ge 5$ and observing that the derived series of $S_n$ terminates at $A_n$ rather than at the trivial group. The Abel-Ruffini conclusion then follows from the fundamental theorem of Galois theory relating solvability by radicals to solvability of the Galois group. [/proofplan]

[step:Exhibit a polynomial $f \in \mathbb{Q}[x]$ of degree $n$ with Galois group $S_n$] We show that the generic polynomial of degree $n$ over $\mathbb{Q}$ has Galois group $S_n$. Consider the polynomial ring $\mathbb{Q}(t_1, \dots, t_n)[x]$ with indeterminate coefficients, and define \begin{align*} f(x) = \prod_{i=1}^{n}(x - t_i) = x^n - e_1 x^{n-1} + e_2 x^{n-2} - \cdots + (-1)^n e_n \in \mathbb{Q}(e_1, \dots, e_n)[x], \end{align*} where $e_1, \dots, e_n$ are the elementary symmetric polynomials in $t_1, \dots, t_n$. Let $k = \mathbb{Q}(e_1, \dots, e_n)$ and $K = \mathbb{Q}(t_1, \dots, t_n)$. Then $K$ is the splitting field of $f$ over $k$, since $f$ splits completely in $K$ as $\prod_{i=1}^n (x - t_i)$, and $K$ is generated over $k$ by the roots $t_1, \dots, t_n$. [claim:$\operatorname{Gal}(K/k) \cong S_n$] The Galois group of $K = \mathbb{Q}(t_1, \dots, t_n)$ over $k = \mathbb{Q}(e_1, \dots, e_n)$ is isomorphic to $S_n$. [/claim] [proof] Every $\sigma \in \operatorname{Gal}(K/k)$ is a $k$-automorphism of $K$ that fixes $k$ pointwise. Since $\sigma$ fixes $k = \mathbb{Q}(e_1, \dots, e_n)$, it fixes every elementary symmetric polynomial $e_j(t_1, \dots, t_n)$. The roots of $f$ in $K$ are $t_1, \dots, t_n$, and $\sigma$ must permute them (because $\sigma$ fixes the coefficients of $f$ and hence maps roots to roots). This defines an injective group homomorphism \begin{align*} \Phi: \operatorname{Gal}(K/k) &\to S_n \\ \sigma &\mapsto \text{the permutation } \pi \text{ such that } \sigma(t_i) = t_{\pi(i)}. \end{align*} The map $\Phi$ is injective because a $k$-automorphism of $K = k(t_1, \dots, t_n)$ is determined by its action on the generators $t_1, \dots, t_n$. For surjectivity, observe that every permutation $\pi \in S_n$ induces a $\mathbb{Q}$-automorphism $\sigma_\pi$ of $\mathbb{Q}(t_1, \dots, t_n)$ defined by $\sigma_\pi(t_i) = t_{\pi(i)}$. This is well-defined because $t_1, \dots, t_n$ are algebraically independent over $\mathbb{Q}$, so any permutation of them extends to a unique $\mathbb{Q}$-automorphism of the rational function field. Moreover, $\sigma_\pi$ fixes every symmetric polynomial in $t_1, \dots, t_n$, so $\sigma_\pi$ fixes $k = \mathbb{Q}(e_1, \dots, e_n)$ and hence $\sigma_\pi \in \operatorname{Gal}(K/k)$. Therefore $\Phi$ is an isomorphism and $\operatorname{Gal}(K/k) \cong S_n$. [/proof] To produce a concrete polynomial over $\mathbb{Q}$ (rather than over the function field $k$), one specialises the indeterminate coefficients. For any $n \ge 5$, there exist infinitely many specialisations $e_i \mapsto a_i \in \mathbb{Q}$ such that the resulting polynomial $f(x) = x^n - a_1 x^{n-1} + \cdots + (-1)^n a_n \in \mathbb{Q}[x]$ has Galois group $S_n$ over $\mathbb{Q}$. This follows from Hilbert's irreducibility theorem, which states that if $f(x) \in k(t_1, \dots, t_n)[x]$ is irreducible over $k = \mathbb{Q}(e_1, \dots, e_n)$, then for all but finitely many (in a density-zero set of) specialisations $e_i \mapsto a_i \in \mathbb{Q}$, the specialised polynomial remains irreducible with the same Galois group. The hypothesis of Hilbert's theorem is satisfied: $f(x) = \prod_{i=1}^n (x - t_i)$ is irreducible over $k$ because its splitting field $K/k$ has degree $[K:k] = |S_n| = n!$, so $f$ has no root in $k$ (any root would be a $t_i$, but $t_i \notin \mathbb{Q}(e_1, \dots, e_n)$ since the extension is non-trivial). Therefore infinitely many specialisations yield polynomials in $\mathbb{Q}[x]$ with Galois group $S_n$. Alternatively, for $n = 5$, the polynomial $g(x) = x^5 - 4x + 2 \in \mathbb{Q}[x]$ has Galois group $S_5$ over $\mathbb{Q}$. First, $g$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion with $p = 2$: the prime $2$ divides all non-leading coefficients ($0, 0, 0, -4, 2$), and $2^2 = 4$ does not divide the constant term $2$. Second, we verify that $\operatorname{Gal}(g/\mathbb{Q}) = S_5$ by reduction modulo primes: - **Modulo $3$:** $g(x) \equiv x^5 - x + 2 \pmod{3}$. This factors as $(x^2 + x + 2)(x^3 + 2x^2 + 2) \pmod{3}$, where both factors are irreducible over $\mathbb{F}_3$. The Frobenius element at $p = 3$ has cycle type $(2, 3)$, so $\operatorname{Gal}(g/\mathbb{Q})$ contains a permutation with cycle structure $(2,3)$. - Cubing the element of cycle type $(2, 3)$ yields a transposition in $\operatorname{Gal}(g/\mathbb{Q})$. Since $g$ is irreducible, $\operatorname{Gal}(g/\mathbb{Q})$ acts transitively on the five roots, so $5 \mid |\operatorname{Gal}(g/\mathbb{Q})|$, and by Cauchy's theorem the Galois group contains an element of order $5$, which must be a $5$-cycle. A transitive subgroup of $S_5$ containing both a transposition and a $5$-cycle equals $S_5$.[/step]

[guided]The strategy is to work first with the "generic" polynomial whose roots are formal indeterminates, rather than trying to exhibit a specific polynomial over $\mathbb{Q}$ directly. This is the cleanest way to see that $S_n$ arises as a Galois group, because the algebraic independence of the roots removes all the arithmetic complications. Let $t_1, \dots, t_n$ be algebraically independent indeterminates over $\mathbb{Q}$. Define the elementary symmetric polynomials: \begin{align*} e_1 &= t_1 + t_2 + \cdots + t_n, \\ e_2 &= \sum_{i < j} t_i t_j, \\ &\;\;\vdots \\ e_n &= t_1 t_2 \cdots t_n. \end{align*} Set $k = \mathbb{Q}(e_1, \dots, e_n) \subset K = \mathbb{Q}(t_1, \dots, t_n)$, and consider the polynomial \begin{align*} f(x) = \prod_{i=1}^n (x - t_i) = x^n - e_1 x^{n-1} + e_2 x^{n-2} - \cdots + (-1)^n e_n \in k[x]. \end{align*} Why does this polynomial have coefficients in $k$? Because expanding $\prod_{i=1}^n (x - t_i)$ by Vieta's formulas expresses every coefficient as $\pm e_j(t_1, \dots, t_n)$, and the $e_j$ are exactly the generators of $k$. The extension $K/k$ is Galois because $K$ is the splitting field of $f$ over $k$: the polynomial $f$ splits completely in $K$ as $\prod_{i=1}^n (x - t_i)$, and $K$ is generated over $k$ by the roots $t_1, \dots, t_n$. Now we compute the Galois group. Every $\sigma \in \operatorname{Gal}(K/k)$ fixes $k$ pointwise, hence fixes the coefficients of $f$, and therefore must permute the roots $\{t_1, \dots, t_n\}$. Since $K = k(t_1, \dots, t_n)$, the automorphism $\sigma$ is completely determined by its action on the roots, giving an injective homomorphism $\Phi: \operatorname{Gal}(K/k) \to S_n$. Conversely, any permutation $\pi \in S_n$ induces a $\mathbb{Q}$-automorphism $\sigma_\pi$ of $K = \mathbb{Q}(t_1, \dots, t_n)$ defined by $\sigma_\pi(t_i) = t_{\pi(i)}$. Why is this well-defined? Because $t_1, \dots, t_n$ are algebraically independent over $\mathbb{Q}$, so any permutation of them extends uniquely to a $\mathbb{Q}$-automorphism of the rational function field. Moreover, $\sigma_\pi$ fixes every symmetric polynomial $e_j(t_1, \dots, t_n)$ (by definition of symmetric polynomials), so $\sigma_\pi$ fixes $k$ and hence lies in $\operatorname{Gal}(K/k)$. This shows $\Phi$ is surjective, giving $\operatorname{Gal}(K/k) \cong S_n$. To produce a concrete polynomial over $\mathbb{Q}$ (rather than over the function field $k$), we specialise the indeterminate coefficients. Hilbert's irreducibility theorem guarantees that for "most" rational specialisations $e_i \mapsto a_i \in \mathbb{Q}$, the Galois group of the resulting polynomial $f(x) = x^n - a_1 x^{n-1} + \cdots + (-1)^n a_n \in \mathbb{Q}[x]$ over $\mathbb{Q}$ remains $S_n$. (The theorem requires $f$ to be irreducible over $k$, which holds: if $f$ factored over $k$, then some proper subset of $\{t_1, \dots, t_n\}$ would generate a proper intermediate field between $k$ and $K$, but the Galois group $S_n$ acts transitively on the roots, so no proper subset is preserved.) As a concrete example for $n = 5$: the polynomial $g(x) = x^5 - 4x + 2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion with $p = 2$ (the prime $2$ divides all non-leading coefficients $0, 0, 0, -4, 2$, and $4 = 2^2$ does not divide the constant term $2$). Its Galois group over $\mathbb{Q}$ is $S_5$: reducing modulo $3$ gives cycle type $(2, 3)$, and cubing this element yields a transposition; meanwhile $g$ is irreducible of degree $5$, so $5 \mid |\operatorname{Gal}(g/\mathbb{Q})|$, and by Cauchy's theorem the Galois group contains an element of order $5$, which must be a $5$-cycle. A transitive subgroup of $S_5$ containing a transposition and a $5$-cycle is all of $S_5$.[/guided]

[step:Show that $A_n$ is simple for $n \ge 5$] [claim:$A_n$ is simple for $n \ge 5$] For $n \ge 5$, the alternating group $A_n$ has no proper nontrivial normal subgroups. [/claim] [proof] Let $N \trianglelefteq A_n$ be a nontrivial normal subgroup with $N \ne \{e\}$. We show $N = A_n$ by establishing that $N$ contains every $3$-cycle, and the $3$-cycles generate $A_n$. [step:Show that $N$ contains a $3$-cycle] Let $\sigma \in N$, $\sigma \ne e$. Among all nontrivial elements of $N$, choose $\sigma$ to minimise the number of elements moved by $\sigma$ (i.e., the size of $\operatorname{supp}(\sigma) = \{i \in \{1, \dots, n\} : \sigma(i) \ne i\}$). **Case 1**: $\sigma$ is a $3$-cycle. Then $N$ contains a $3$-cycle and we are done. **Case 2**: $\sigma$ is a product of disjoint cycles, at least one of which has length $\ge 3$. Write $\sigma = (a_1 \; a_2 \; a_3 \; \cdots) \cdots$ where the displayed cycle has length $\ge 3$. Choose $b \in \{1, \dots, n\} \setminus \operatorname{supp}(\sigma)$ (possible since $|\operatorname{supp}(\sigma)| \le n$ and $n \ge 5$ provides at least one fixed point outside the first cycle). Let $\tau = (a_1 \; a_2 \; b) \in A_n$. Consider the commutator $[\sigma, \tau] = \sigma \tau \sigma^{-1} \tau^{-1}$. Since $\sigma \in N$ and $N \trianglelefteq A_n$, normality gives $\tau \sigma^{-1} \tau^{-1} \in N$. Therefore $[\sigma, \tau] = \sigma \cdot (\tau \sigma^{-1} \tau^{-1}) \in N$ as a product of two elements of $N$. The commutator $[\sigma, \tau]$ is nontrivial (since $\sigma$ moves $a_3$ but $\tau$ fixes $a_3$, the composition does not fix $a_3$) and moves strictly fewer elements than $\sigma$ (it fixes every element outside $\operatorname{supp}(\sigma) \cup \{b\}$, and $b \notin \operatorname{supp}(\sigma)$), contradicting the minimality of $|\operatorname{supp}(\sigma)|$. **Case 3**: $\sigma$ is a product of disjoint transpositions. Since $\sigma \in A_n$, the number of transpositions is even, so $\sigma$ moves at least $4$ elements. Write $\sigma = (a_1 \; a_2)(a_3 \; a_4) \cdots$. Since $n \ge 5$, choose $b \in \{1, \dots, n\} \setminus \{a_1, a_2, a_3, a_4\}$. Let $\tau = (a_1 \; a_2 \; b)$. The commutator $[\sigma, \tau] = \sigma \tau \sigma^{-1} \tau^{-1} \in N$ is nontrivial and moves fewer elements than $\sigma$, again contradicting minimality. In all cases we reach a contradiction unless $\sigma$ is a $3$-cycle. Hence $N$ contains a $3$-cycle. [/step]

[step:Show that $N$ contains all $3$-cycles] Suppose $N$ contains the $3$-cycle $(a \; b \; c)$. For any other $3$-cycle $(d \; e \; f)$ in $A_n$, there exists a permutation $\pi \in S_n$ with $\pi(a) = d$, $\pi(b) = e$, $\pi(c) = f$ (since $S_n$ acts transitively on ordered triples of distinct elements from $\{1, \dots, n\}$). If $\pi \in A_n$, then \begin{align*} \pi (a \; b \; c) \pi^{-1} = (d \; e \; f) \in N \end{align*} by normality of $N$ in $A_n$. If $\pi \notin A_n$, replace $\pi$ by $\pi \circ (g \; h)$ where $g, h \in \{1, \dots, n\} \setminus \{d, e, f\}$ (possible since $n \ge 5$); this does not affect the conjugation of $(a \; b \; c)$ to $(d \; e \; f)$ and produces an even permutation. Hence every $3$-cycle lies in $N$. [/step]

Since every element of $A_n$ is a product of $3$-cycles (every even permutation can be written as a product of $3$-cycles of the form $(1 \; 2 \; k)$), we conclude $N = A_n$. [/proof]

[guided] The simplicity of $A_n$ for $n \ge 5$ is the algebraic heart of the Abel-Ruffini theorem. We must show that if $N \trianglelefteq A_n$ is a nontrivial normal subgroup, then $N = A_n$. The proof strategy is a minimality argument: among all nontrivial elements $\sigma \in N$, pick one that moves the fewest elements (minimise $|\operatorname{supp}(\sigma)|$). We show this $\sigma$ must be a $3$-cycle by eliminating the other possibilities via commutator constructions. Let $\sigma \in N \setminus \{e\}$ minimise $|\operatorname{supp}(\sigma)|$. **Case 1**: $\sigma$ is a $3$-cycle. Then $N$ contains a $3$-cycle. **Case 2**: $\sigma$ has a cycle of length $\ge 3$. Write $\sigma = (a_1 \; a_2 \; a_3 \; \cdots) \cdots$. Choose $b \in \{1, \dots, n\} \setminus \operatorname{supp}(\sigma)$ (possible since $n \ge 5$ guarantees a fixed point). Let $\tau = (a_1 \; a_2 \; b) \in A_n$. The commutator $[\sigma, \tau] = \sigma \tau \sigma^{-1} \tau^{-1}$ lies in $N$: since $\sigma \in N$ and $N \trianglelefteq A_n$, we have $\tau \sigma^{-1} \tau^{-1} \in N$ by normality, so $[\sigma, \tau] = \sigma \cdot (\tau \sigma^{-1} \tau^{-1}) \in N$ as a product of elements of $N$. A direct computation shows $[\sigma, \tau] \ne e$ but $|\operatorname{supp}([\sigma, \tau])| < |\operatorname{supp}(\sigma)|$, contradicting minimality. **Case 3**: $\sigma$ is a product of disjoint transpositions. Since $\sigma \in A_n$, the number of transpositions is even, so $\sigma$ moves at least $4$ elements: $\sigma = (a_1 \; a_2)(a_3 \; a_4) \cdots$. Choose $b \in \{1, \dots, n\} \setminus \{a_1, a_2, a_3, a_4\}$ (possible since $n \ge 5$). Let $\tau = (a_1 \; a_2 \; b)$. Again $[\sigma, \tau] \in N$ by the same normality argument, and $[\sigma, \tau]$ is nontrivial with $|\operatorname{supp}([\sigma, \tau])| < |\operatorname{supp}(\sigma)|$, contradicting minimality. Thus $\sigma$ must be a $3$-cycle. Now we show $N$ contains *every* $3$-cycle. If $(a \; b \; c) \in N$, then for any other $3$-cycle $(d \; e \; f)$, there exists $\pi \in S_n$ with $\pi(a) = d$, $\pi(b) = e$, $\pi(c) = f$. If $\pi \in A_n$, then \begin{align*} \pi (a \; b \; c) \pi^{-1} = (d \; e \; f) \in N \end{align*} by normality. If $\pi \notin A_n$, replace $\pi$ by $\pi \circ (g \; h)$ where $g, h \in \{1, \dots, n\} \setminus \{d, e, f\}$ (possible since $n \ge 5$); this produces an even permutation that still conjugates $(a \; b \; c)$ to $(d \; e \; f)$. Hence every $3$-cycle lies in $N$. Since every even permutation is a product of $3$-cycles, $N = A_n$. The hypothesis $n \ge 5$ is used twice in an essential way: 1. In the commutator argument: we need "extra room" (a fifth element $b \notin \operatorname{supp}(\sigma)$) to construct a nontrivial commutator that moves fewer elements. 2. In the conjugation argument: we need two elements $g, h$ outside the target $3$-cycle to adjust the parity of the conjugating permutation. For $n = 4$, the group $A_4$ is not simple: the Klein four-group $V_4 = \{e, (1\;2)(3\;4), (1\;3)(2\;4), (1\;4)(2\;3)\}$ is a proper nontrivial normal subgroup. [/guided]

[/step]

[step:Prove that $S_n$ is not solvable for $n \ge 5$]Recall that a group $G$ is solvable if its derived series terminates at the trivial group. The derived series is defined by $G^{(0)} = G$ and $G^{(i+1)} = [G^{(i)}, G^{(i)}]$ (the commutator subgroup). Compute the first two terms of the derived series of $S_n$: \begin{align*} S_n^{(0)} &= S_n, \\ S_n^{(1)} &= [S_n, S_n] = A_n. \end{align*} The second equality holds because the commutator subgroup $[S_n, S_n]$ equals $A_n$: every commutator is an even permutation (each element appears twice in $\sigma \tau \sigma^{-1} \tau^{-1}$, so the total sign is $1$), so $[S_n, S_n] \le A_n$; conversely, every $3$-cycle $(a \; b \; c) = [(a \; b), (a \; c)]$ is a commutator, and $3$-cycles generate $A_n$, so $A_n \le [S_n, S_n]$. For the next term: \begin{align*} S_n^{(2)} = [A_n, A_n]. \end{align*} Since $A_n$ is simple and $[A_n, A_n] \trianglelefteq A_n$ is a normal subgroup, either $[A_n, A_n] = \{e\}$ or $[A_n, A_n] = A_n$. But $A_n$ is non-abelian for $n \ge 3$ (for instance, $(1\;2\;3)(1\;2\;4) \ne (1\;2\;4)(1\;2\;3)$), so $[A_n, A_n] \ne \{e\}$. Therefore $[A_n, A_n] = A_n$. The derived series of $S_n$ is thus $S_n \supsetneq A_n = A_n = A_n = \cdots$, which never reaches $\{e\}$. Hence $S_n$ is not solvable for $n \ge 5$.[/step]

[guided]A group $G$ is solvable if there exists a chain of subgroups $G = G_0 \triangleright G_1 \triangleright \cdots \triangleright G_m = \{e\}$ with each quotient $G_i / G_{i+1}$ abelian. Equivalently, the derived series $G \supset [G, G] \supset [[G, G], [G, G]] \supset \cdots$ eventually reaches $\{e\}$. For $S_n$: the commutator subgroup $[S_n, S_n]$ is $A_n$. This is because: - Every commutator $\sigma \tau \sigma^{-1} \tau^{-1}$ has sign $\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)\operatorname{sgn}(\sigma)^{-1}\operatorname{sgn}(\tau)^{-1} = 1$, so $[S_n, S_n] \le A_n$. - Conversely, every $3$-cycle $(a\;b\;c)$ can be written as the commutator $(a\;b)(a\;c)(a\;b)^{-1}(a\;c)^{-1} = (a\;b)(a\;c)(a\;b)(a\;c) = (a\;b\;c)$. Since $3$-cycles generate $A_n$, we get $A_n \le [S_n, S_n]$. Now the derived series is stuck at $A_n$ forever: $[A_n, A_n]$ is a normal subgroup of $A_n$, and since $A_n$ is simple (for $n \ge 5$, as we proved above) and non-abelian, the only possibility is $[A_n, A_n] = A_n$. The derived series never descends below $A_n$, so it never reaches $\{e\}$, and $S_n$ is not solvable. This is precisely where the simplicity of $A_n$ is decisive. For $n \le 4$, the group $A_n$ is not simple ($A_4$ has the Klein four-group as a normal subgroup, and $A_3 \cong \mathbb{Z}/3\mathbb{Z}$, $A_2 = A_1 = \{e\}$ are abelian), and indeed $S_n$ is solvable for $n \le 4$.[/guided]

[step:Conclude that the general quintic is not solvable by radicals]By the Galois-theoretic criterion for solvability by radicals, a polynomial $f \in \mathbb{Q}[x]$ is solvable by radicals if and only if its Galois group $\operatorname{Gal}(K_f / \mathbb{Q})$ is a solvable group, where $K_f$ is a splitting field of $f$ over $\mathbb{Q}$. We have exhibited polynomials of degree $n$ (for any $n \ge 5$) whose Galois group over $\mathbb{Q}$ is $S_n$, and we have proved that $S_n$ is not solvable for $n \ge 5$. Therefore these polynomials are not solvable by radicals. In particular, the general polynomial of degree $n \ge 5$ (with indeterminate coefficients) has Galois group $S_n$, which is not solvable. Hence there is no formula expressing the roots of the general degree-$n$ polynomial in terms of its coefficients using only the field operations and extraction of radicals.[/step]

[guided]The connection between solvability of the Galois group and solvability by radicals is the following: adjunction of a radical $\sqrt[m]{a}$ corresponds to a cyclic (hence abelian) extension at the Galois-theoretic level. A tower of radical extensions therefore corresponds to a tower of abelian extensions, which is exactly the definition of a solvable Galois group. Making this precise requires the theory of radical extensions and the primitive element theorem, which we take as established. With this criterion in hand, the conclusion is immediate: for $n \ge 5$, we have $\operatorname{Gal}(K/k) \cong S_n$, and $S_n$ is not solvable, so the polynomial $f$ is not solvable by radicals. This is the Abel-Ruffini theorem. Note that the theorem does not say that *no* polynomial of degree $\ge 5$ is solvable by radicals. For instance, $x^5 - 1$ is solvable by radicals (its roots are the fifth roots of unity, and its Galois group is $\mathbb{Z}/4\mathbb{Z}$, which is abelian hence solvable). The theorem says that there *exist* polynomials of degree $\ge 5$ that are not solvable by radicals, or equivalently, that there is no general formula for degree $\ge 5$.[/guided]