[proofplan] The proof combines three ingredients. First, the [Galois Group of the General Polynomial](/theorems/1289) establishes that the Galois group of the general polynomial of degree $n$ over $K(u_1, \ldots, u_n)$ is the symmetric group $S_n$. Second, for $n \geq 5$, we show that $S_n$ is not a soluble group: the only normal subgroups of $S_n$ are $\{e\}$, $A_n$, and $S_n$ itself, so any composition series must pass through $A_n$, and the [Simplicity of $A_n$](/theorems/1286) guarantees that $A_n$ is simple and non-abelian, which obstructs solubility. Third, we apply the contrapositive of the [Radical Implies Soluble](/theorems/1287) theorem: since $\operatorname{Gal}(L/K(u_1, \ldots, u_n)) \cong S_n$ is not soluble, the general polynomial is not soluble by radicals. [/proofplan]

[step:Identify the Galois group of the general polynomial as $S_n$]Let $K$ be a field of characteristic $0$ and let $n \geq 5$ be an integer. Set $F := K(u_1, \ldots, u_n)$, where $u_1, \ldots, u_n$ are independent indeterminates over $K$. The general polynomial of degree $n$ over $K$ is \begin{align*} f(t) := t^n - u_1 t^{n-1} + u_2 t^{n-2} - \cdots + (-1)^n u_n \in F[t]. \end{align*} Let $L$ denote the splitting field of $f$ over $F$. By the [Galois Group of the General Polynomial](/theorems/1289), the extension $L/F$ is Galois with \begin{align*} \operatorname{Gal}(L/F) \cong S_n. \end{align*}[/step]

[guided]We begin by setting up the objects precisely. The general polynomial is constructed by taking the elementary symmetric polynomials as coefficients — the indeterminates $u_1, \ldots, u_n$ play the role of the elementary symmetric polynomials in the roots. The coefficient field is the rational function field $F = K(u_1, \ldots, u_n)$, which is the field of fractions of the polynomial ring $K[u_1, \ldots, u_n]$. The [Galois Group of the General Polynomial](/theorems/1289) applies here because $f$ is, by construction, the polynomial whose roots are abstract indeterminates $x_1, \ldots, x_n$ with $u_i = e_i(x_1, \ldots, x_n)$ (the $i$-th elementary symmetric polynomial). The splitting field $L$ is therefore $K(x_1, \ldots, x_n)$, the full rational function field in the roots. Every permutation $\sigma \in S_n$ acts on $L$ by permuting $x_1, \ldots, x_n$ and fixes $F$ (since the elementary symmetric polynomials are invariant under all permutations). The theorem confirms that this action gives an isomorphism $\operatorname{Gal}(L/F) \cong S_n$. The identification of the Galois group as the full symmetric group is the essential starting point: it reduces the question of solubility by radicals for the general polynomial to a purely group-theoretic question about $S_n$.[/guided]

[step:Show that $S_n$ is not soluble for $n \geq 5$]We prove that $S_n$ is not a soluble group when $n \geq 5$. [claim:$S_n$ is not soluble for $n \geq 5$] For $n \geq 5$, the symmetric group $S_n$ is not soluble. [/claim] [proof] Suppose, for contradiction, that $S_n$ is soluble. By the [Persistence of Solubility](/theorems/1285), every quotient of a soluble group is soluble. The alternating group $A_n$ is a normal subgroup of $S_n$ with index $[S_n : A_n] = 2$, and $A_n$ is also a subgroup of $S_n$. By part (1) of the [Persistence of Solubility](/theorems/1285), every subgroup of a soluble group is soluble, so $A_n$ would be soluble. A soluble group $G$ with $|G| > 1$ must possess a proper normal subgroup: the last nontrivial term $G_{r-1}$ in any composition series $\{e\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_0 = G$ satisfies $\{e\} \neq G_{r-1} \trianglelefteq G$ with $G_{r-1}/\{e\} \cong G_{r-1}$ abelian, and in particular $G_{r-1} \neq G$ (since if $G_{r-1} = G$ then $G$ itself is abelian, and an abelian simple group is cyclic of prime order; but $|A_n| = n!/2 \geq 60$ is not prime for $n \geq 5$). More precisely, solubility of $A_n$ requires the existence of a normal subgroup $N \trianglelefteq A_n$ with $\{e\} \subsetneq N \subsetneq A_n$ (since $A_n$ is non-abelian for $n \geq 5$, so the composition series cannot have length $1$). However, by [$A_n$ is Simple for $n \geq 5$](/theorems/1286), the alternating group $A_n$ has no normal subgroups other than $\{e\}$ and $A_n$ itself. Since $|A_n| = n!/2 \geq 60$ for $n \geq 5$, the group $A_n$ is non-abelian (it contains non-commuting $3$-cycles, e.g., $(1\,2\,3)$ and $(1\,2\,4)$ satisfy $(1\,2\,3)(1\,2\,4) = (1\,3)(2\,4) \neq (1\,4)(2\,3) = (1\,2\,4)(1\,2\,3)$). A non-abelian simple group cannot be soluble: solubility requires a composition series with abelian successive quotients, but the only composition series of a simple group $A_n$ is $\{e\} \trianglelefteq A_n$, and the single quotient $A_n/\{e\} \cong A_n$ is non-abelian. This contradicts the solubility of $A_n$. Therefore $S_n$ is not soluble. [/proof][/step]

[guided]The strategy is to derive a contradiction from the assumption that $S_n$ is soluble, by showing this would force the alternating group $A_n$ to be soluble — which is impossible because $A_n$ is simple and non-abelian. **Why does solubility of $S_n$ imply solubility of $A_n$?** The [Persistence of Solubility](/theorems/1285) states two closure properties: (1) every subgroup of a soluble group is soluble, and (2) every quotient of a soluble group is soluble. Since $A_n \leq S_n$ (the alternating group is a subgroup of index $2$ in the symmetric group), property (1) gives us solubility of $A_n$ for free, assuming $S_n$ is soluble. This is the key reduction. **Why can a simple non-abelian group not be soluble?** Recall that a group $G$ is soluble if and only if it has a composition series \begin{align*} \{e\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_0 = G \end{align*} in which every successive quotient $G_i / G_{i+1}$ is abelian. If $G$ is simple (no normal subgroups other than $\{e\}$ and $G$), then the only composition series is $\{e\} \trianglelefteq G$, and the sole quotient $G/\{e\} \cong G$ must be abelian for solubility. Therefore a simple group is soluble if and only if it is abelian (equivalently, cyclic of prime order). Since $A_n$ is non-abelian for $n \geq 5$ — for instance, the $3$-cycles $(1\,2\,3)$ and $(1\,2\,4)$ in $A_5$ do not commute: \begin{align*} (1\,2\,3)(1\,2\,4) &= (1\,3)(2\,4), \\ (1\,2\,4)(1\,2\,3) &= (1\,4)(2\,3), \end{align*} and $(1\,3)(2\,4) \neq (1\,4)(2\,3)$ — the group $A_n$ is simple but not abelian. Therefore $A_n$ is not soluble. **Why does the argument fail for $n \leq 4$?** The theorem [$A_n$ is Simple for $n \geq 5$](/theorems/1286) requires $n \geq 5$. For smaller values: $A_1$ and $A_2$ are trivial, $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ is cyclic (hence soluble), and $A_4$ has the normal subgroup $V_4 = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ (the Klein four-group), yielding the composition series $\{e\} \trianglelefteq V_4 \trianglelefteq A_4$ with abelian quotients. Correspondingly, $S_4$ is soluble, and the general quartic is indeed soluble by radicals (via Ferrari's method).[/guided]

[step:Conclude by contrapositive of the Radical Implies Soluble theorem]By the [Radical Implies Soluble](/theorems/1287) theorem: if a polynomial $f \in F[t]$ (where $F$ is a field of characteristic $0$) is soluble by radicals, then its Galois group $\operatorname{Gal}(L/F)$ is a soluble group. We apply the contrapositive. The general polynomial of degree $n$ has Galois group $\operatorname{Gal}(L/F) \cong S_n$ (established in the first step), and $S_n$ is not soluble for $n \geq 5$ (established in the second step). The hypotheses of the [Radical Implies Soluble](/theorems/1287) theorem are satisfied: $F = K(u_1, \ldots, u_n)$ has characteristic $0$ (since $K$ has characteristic $0$, so does every extension of $K$), the polynomial $f \in F[t]$ has splitting field $L$ over $F$, and $\operatorname{Gal}(L/F)$ is well-defined (since $L/F$ is Galois). Since the conclusion of the theorem — that $\operatorname{Gal}(L/F)$ is soluble — fails, the hypothesis must also fail: $f$ is not soluble by radicals. Therefore, for $n \geq 5$, the general polynomial of degree $n$ over a field $K$ of characteristic $0$ is not soluble by radicals.[/step]

[guided]The final step is purely logical: the [Radical Implies Soluble](/theorems/1287) theorem states \begin{align*} f \text{ soluble by radicals} \implies \operatorname{Gal}(L/F) \text{ soluble}. \end{align*} The contrapositive is \begin{align*} \operatorname{Gal}(L/F) \text{ not soluble} \implies f \text{ not soluble by radicals}. \end{align*} Before applying the contrapositive, we verify that all the conditions of the [Radical Implies Soluble](/theorems/1287) theorem are in scope: - The base field $F = K(u_1, \ldots, u_n)$ has characteristic $0$: the characteristic of a field extension coincides with that of the base field, and $\operatorname{char}(K) = 0$ by hypothesis. - The polynomial $f(t) = t^n - u_1 t^{n-1} + \cdots + (-1)^n u_n$ lies in $F[t]$. - The splitting field $L$ of $f$ over $F$ exists (by the existence of splitting fields for polynomials over any field). We have established that $\operatorname{Gal}(L/F) \cong S_n$ and that $S_n$ is not soluble for $n \geq 5$. By the contrapositive, $f$ is not soluble by radicals. It is worth noting the logical structure of the full Abel-Ruffini theorem. The result we have proved concerns the *general* polynomial — a single polynomial with indeterminate coefficients. This does not assert that every *specific* polynomial of degree $\geq 5$ is insoluble by radicals. Indeed, particular polynomials of degree $5$ (e.g., $t^5 - 2 \in \mathbb{Q}[t]$) may have soluble Galois groups and therefore be soluble by radicals. The theorem states that no *uniform* radical formula (analogous to the quadratic formula or Cardano's formula) can express the roots of an arbitrary polynomial of degree $n \geq 5$ in terms of its coefficients using only field operations and extraction of radicals.[/guided]