The quadratic formula gives one recipe that solves every $ax^2+bx+c$. Sixteenth-century algebraists found analogous — though much more complicated — recipes for cubics and quartics. Then something changes: there is no single recipe, using arithmetic operations and finitely nested roots, that solves every quintic.

That statement does not mean quintics have no roots. Every degree-five polynomial has five complex roots, counted with multiplicity. It also does not mean that no individual quintic can be solved by radicals: $x^{5}-2=0$ plainly can. The claim is about a universal method for arbitrary coefficients.

Two different claims. Abel–Ruffini says there is no one radical recipe for all quintics. Galois theory additionally tells us which individual quintics fail to have any radical solution. The common misreading — “no quintic can be solved by radicals” — is false.

The reason is symmetry. A polynomial’s coefficients do not care which root we call first. Some equations permit only a limited set of root rearrangements; others permit every rearrangement. Radical formulas can cope only with symmetries built in successive commutative layers. The general quintic has the full symmetry group $S_5$, and that group has an irreducible noncommutative core. That mismatch is Abel–Ruffini.

This note is one source for the walk. We take the seven stages below in order; each line is unpacked when we get there.

• The fundamental theorem of algebra
• Group theory: symmetries of a labeled family
• Field extensions: where the family lives
• Degree, and automorphisms of the home of the roots
• The tower theorem
• Quintic symmetry
• History, scope, references

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The fundamental theorem of algebra

A field is a set with addition and multiplication in which you can add, subtract, multiply, and divide by anything except zero, with the usual rules. $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are fields. A polynomial over a field $F$ is an expression $a_n x^n + \cdots + a_1 x + a_0$ with coefficients $a_i\in F$. Its degree is $n$ if $a_n\neq 0$. A root is a number $r$ (in some extension of $F$) with $p(r)=0$.

Theorem (fundamental theorem of algebra). Every non-constant polynomial $p\in\mathbb{C}[x]$ has at least one root in $\mathbb{C}$. Equivalently: $\mathbb{C}$ is algebraically closed. Equivalently: if $p$ has degree $n\ge 1$, then there exist $c,r_1,\ldots,r_n\in\mathbb{C}$ with

\[p(x) = c(x-r_1)\cdots(x-r_n).\]

The three statements are the same once you know polynomial division. Find a root $r_1$, divide by $x-r_1$, induct on degree.

This is an existence theorem. It does not produce a formula. It does not claim the roots lie in $\mathbb{Q}$, or even in $\mathbb{R}$. It does not claim they can be written by nesting radicals in the coefficients. For two centuries “the fundamental theorem of algebra” was, in practice, the search for such a formula. Gauss already suspected the search was the wrong problem. Abel and Ruffini proved it was.

What the rest of the argument needs from this theorem is only the family of roots. If you already accept that every non-constant polynomial has a complex root, you may skip the geometric proof below and go on to conjugate pairs.

What we will use

Two corollaries, and nothing more from the analytic side.

The family of roots. Given $p$ of degree $n$, we may speak of a complete family ${r_1,\ldots,r_n}$ in $\mathbb{C}$. Labels are arbitrary; later we will ask which rearrangements of the labels can be realized by maps that fix the coefficients.

Conjugate pairs. If $p$ has real coefficients and $p(z)=0$, then $p(\overline{z})=0$, because conjugation $z\mapsto\overline{z}$ is a field automorphism of $\mathbb{C}$ fixing $\mathbb{R}$. Non-real roots therefore come in pairs $\lbrace z,\overline{z}\rbrace$.

We will sit the splitting field $E=\mathbb{Q}(r_1,\ldots,r_n)$ inside $\mathbb{C}$. Conjugation permutes those roots, hence sends the generating set of $E$ to itself, hence restricts to a field automorphism of $E$ fixing $\mathbb{Q}$. (That restriction will later be an element of the Galois group.) As a permutation of the five labeled roots it fixes every real root and swaps each conjugate pair.

A real quintic cannot have zero real roots: $x^{5}$ dominates, so $p(x)\to+\infty$ as $x\to+\infty$ and $p(x)\to-\infty$ as $x\to-\infty$, and the intermediate-value theorem supplies a real root. Combined with conjugate pairing (even number of non-real roots) and degree five, there are exactly three possibilities. Write the permutation type of conjugation in each:

• Five real roots. Conjugation fixes all five labels. It is the identity in $S_5$, which is not a transposition.
• Three real roots and one conjugate pair. Conjugation swaps those two non-real roots and fixes the three real ones. Cycle type: a single $2$-cycle. That is a transposition.
• One real root and two conjugate pairs. Conjugation swaps two pairs and fixes the real root. Cycle type: a product of two disjoint transpositions, an even permutation of order $2$ — not a $2$-cycle.

Only the middle case will later feed an $S_5$ machine (a transposition plus a $5$-cycle generate $S_5$). The other two are silent: they do not put a transposition among the coefficient-fixing symmetries of the roots. That is why $x^{5}-2$, which has one real fifth root of $2$ and two conjugate pairs among $\sqrt[5]{2}\,\zeta_5^{k}$ for $k=1,2,3,4$, slips through as solvable — conjugation on its roots is of the third type, not the second. We will compute those symmetries later; the group has order $20$, not $120$.

A proof that does not hide the analysis (optional)

The theorem is not a theorem of algebra in the modern sense. Any proof uses some continuity. The argument below is geometric: identify $\mathbb{C}$ with the plane $\mathbb{R}^2$, view $\lvert p(z)\rvert$ as a height function over that plane, and find a closed disk on which the height is forced to dip to zero.

Write $p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_0$ with $a_n\neq 0$ and $n\ge 1$. For $\lvert z\rvert$ large, the leading term dominates. Precisely: the reverse triangle inequality gives $\lvert p(z)\rvert \ge \lvert a_n z^n\rvert - \lvert p(z)-a_n z^n\rvert$, and

\[\bigl\lvert p(z) - a_n z^n\bigr\rvert \le \bigl(\lvert a_{n-1}\rvert + \cdots + \lvert a_0\rvert\bigr)\,\lvert z\rvert^{n-1}.\]

If $\lvert z\rvert\ge 1$ the right-hand side is at most $M\lvert z\rvert^{n-1}$ with $M=\lvert a_{n-1}\rvert+\cdots+\lvert a_0\rvert$, so $\lvert p(z)\rvert \ge \bigl(\lvert a_n\rvert - M/\lvert z\rvert\bigr)\lvert z\rvert^n$. Choose $R\ge 1$ large enough that $M/R \le \lvert a_n\rvert/2$ and also $\tfrac12\lvert a_n\rvert R^n \gt \lvert p(0)\rvert$. Then $\lvert z\rvert \ge R$ implies

\[\lvert p(z)\rvert \ge \tfrac12 \lvert a_n\rvert\,\lvert z\rvert^n \gt \lvert p(0)\rvert.\]

(The closed condition $\lvert z\rvert \ge R$, not just $\lvert z\rvert \gt R$, is the one we want: height already exceeds $\lvert p(0)\rvert$ on and beyond the circle of radius $R$.)

The compact disk. Write

\[\overline{D}_R := \{\, z\in\mathbb{C} : \lvert z\rvert \le R \,\}\]

for the closed disk of radius $R$ centered at the origin — the filled circle, boundary included. As a subset of the plane $\mathbb{R}^2$ it is closed and bounded, hence compact by Heine–Borel. The modulus $\lvert p\rvert : \mathbb{C}\to[0,\infty)$ is continuous (a polynomial is continuous, and $\lvert\,\cdot\,\rvert$ is continuous). A continuous real-valued function on a compact set attains its minimum: there exists $z_0\in\overline{D}_R$ with $\lvert p(z_0)\rvert \le \lvert p(z)\rvert$ for every $z\in\overline{D}_R$.

That minimum cannot live on the boundary. The origin $0$ lies in $\overline{D}_R$, so $\lvert p(z_0)\rvert \le \lvert p(0)\rvert$. But on the circle $\lvert z\rvert = R$ we already have $\lvert p(z)\rvert \gt \lvert p(0)\rvert$. Therefore $\lvert z_0\rvert \lt R$: $z_0$ is an interior point of the disk. We claim $p(z_0)=0$.

The figure is this height function for the later quintic $p(z)=z^5-z-1$, with $R=1.6$. Five dark wells sit strictly inside the cyan circle; each well is a root. The real slice on the right is the same height restricted to $\overline{D}_R\cap\mathbb{R}=[-R,R]$.

Suppose not: $p(z_0)=a\neq 0$. Expand around $z_0$,

\[p(z_0+w) = a + c_k w^k + c_{k+1}w^{k+1} + \cdots + c_n w^n,\]

with $c_k\neq 0$ and $k\ge 1$. Choose a $k$th root of $-a/c_k$, call it $u$, so $c_k u^k = -a$. For small positive $t$, set $w=tu$. Then

\[p(z_0+tu) = a(1-t^k) + O(t^{k+1}).\]

For $t$ small enough, $\lvert p(z_0+tu)\rvert \lt \lvert a\rvert = \lvert p(z_0)\rvert$, contradicting minimality. Hence $p(z_0)=0$.

(If you prefer Liouville: if $p$ had no root then $1/p$ would be entire and bounded, hence constant, hence $p$ constant.)

The algebraic proofs in the literature assume two facts that themselves need the intermediate-value theorem: every real polynomial of odd degree has a real root, and every nonnegative real has a square root. From those one can show that $\mathbb{R}(i)$ is algebraically closed, which is the same theorem. We will not pretend this is a proof from the field axioms alone.

We now have a family of roots. The question is what we are allowed to do with them.

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Group theory: symmetries of a labeled family

Label the roots $r_1,\ldots,r_n$. A permutation of the labels is a bijection $\sigma$ of the set ${1,\ldots,n}$. The set of all such bijections is the symmetric group $S_n$. There are $n!$ of them. Composition of functions is the group law: $(\sigma\tau)(i)=\sigma(\tau(i))$, done right to left. The identity $\mathrm{id}$ does nothing; every $\sigma$ has an inverse.

This is the symmetry group of a labeled family of $n$ things. Whether a particular polynomial admits all $n!$ rearrangements as actual symmetries of its coefficients is a later question. First we need the group, and a few words for talking about groups.

A subgroup of $G$ is a subset that is itself a group under the same operation (contains the identity, inverses, and products). Two elements $a,b$ commute if $ab=ba$; a group is abelian if every pair commutes. The cyclic group $C_n$ is the group of $n$ rotations of a regular $n$-gon, equivalently ${\,1,g,g^2,\ldots,g^{n-1}\,}$ with $g^n=1$; it is abelian. A homomorphism $\varphi:G\to H$ is a map with $\varphi(ab)=\varphi(a)\varphi(b)$. Its kernel is ${\,g\in G:\varphi(g)=\mathrm{id}_H\,}$, always a subgroup. Two groups are isomorphic, written $G\cong H$, if there is a bijective homomorphism between them: the same group law, relabeled. A (left) coset of a subgroup $H\le G$ is a translate $gH={\,gh:h\in H\,}$; the distinct cosets partition $G$.

Small pictures

$S_2={\mathrm{id},(1\,2)}$. That is the quadratic: the two roots $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ are indistinguishable once you forget which sign you took. Swapping them is the only nontrivial symmetry.

$S_3$ has six elements: the identity, two $3$-cycles $(1\,2\,3)$ and $(1\,3\,2)$, and three transpositions $(1\,2)$, $(1\,3)$, $(2\,3)$. Already there is a distinction: some permutations reverse an ordering, some do not.

Parity

Fix the usual order $1 \lt 2 \lt \cdots \lt n$. An inversion of $\sigma$ is a pair $i \lt j$ with $\sigma(i) \gt \sigma(j)$. The permutation is even or odd according to the parity of the number of inversions.

Equivalently — and this is the definition we will actually use — consider the Vandermonde product

\[\Delta(x_1,\ldots,x_n) = \prod_{1\le i<j\le n}(x_i-x_j).\]

Permuting the variables either leaves $\Delta$ alone or multiplies it by $-1$, because each factor $x_i-x_j$ is sent to $\pm$ another factor. The square $\Delta^2=\prod_{i<j}(x_i-x_j)^2$ is the discriminant of the (monic) polynomial with those roots: it is unchanged by every permutation, hence is a polynomial in the coefficients. For $x^2+bx+c$ it is $b^2-4c$; the schoolbook $d=b^2-4ac$ of $ax^2+bx+c$ is $a^2$ times that quantity. We will need the name again when a prime is required not to divide the discriminant. Define

\[\mathrm{sgn}(\sigma) := \frac{\Delta(x_{\sigma(1)},\ldots,x_{\sigma(n)})}{\Delta(x_1,\ldots,x_n)}\in\{+1,-1\}.\]

This is a homomorphism $S_n\to{+1,-1}$: $\mathrm{sgn}(\sigma\tau)=\mathrm{sgn}(\sigma)\,\mathrm{sgn}(\tau)$. A transposition sends $\Delta$ to $-\Delta$, so $\mathrm{sgn}$ of a transposition is $-1$. (An adjacent transposition flips exactly one factor; a non-adjacent one, such as $(1\,3)$ in $S_4$, flips an odd number of factors. Either way the overall sign is $-1$.) Therefore $\mathrm{sgn}(\sigma)=(-1)^m$ whenever $\sigma$ is a product of $m$ transpositions. The number $m$ is not unique, but its parity is. See parity of a permutation for the inversion-count, adjacent-transposition, and cycle-index proofs that these notions coincide.

The even permutations form a subgroup, the alternating group $A_n=\ker\mathrm{sgn}$ — exactly the permutations sent to $+1$. It has index $2$ in $S_n$ (two cosets, even and odd), hence order $n!/2$ for $n\ge 2$. The odd permutations are the other coset $A_n\cdot(1\,2)$; they do not form a subgroup, because odd times odd is even.

A $k$-cycle $(a_1\,a_2\,\cdots\,a_k)$ is a product of $k-1$ transpositions, for instance

\[(a_1\,a_2\,\cdots\,a_k) = (a_1\,a_k)(a_1\,a_{k-1})\cdots(a_1\,a_2).\]

So a $k$-cycle is even if and only if $k$ is odd: $3$-cycles are even, transpositions are odd, $5$-cycles are even. In a disjoint-cycle decomposition (including $1$-cycles if you like), the permutation is odd if and only if the number of even-length cycles is odd.

Conjugating a cycle relabels its entries. If $\sigma=(a_1\,a_2\,\cdots\,a_k)$ and $\tau$ is any permutation, then

\[\tau\sigma\tau^{-1} = \bigl(\tau(a_1)\,\tau(a_2)\,\cdots\,\tau(a_k)\bigr).\]

The same rule applies factorwise to a product of disjoint cycles. This is the one computation used below to move $3$-cycles around $A_n$, to produce commutators in the simplicity argument, and to turn a $p$-cycle and a transposition into all adjacent transpositions.

Normal subgroups and quotients

A subgroup $N\le G$ is normal, written $N\trianglelefteq G$, if $gNg^{-1}=N$ for every $g\in G$. Equivalently, left and right cosets coincide, and the set of cosets $G/N$ inherits a group law. Normality is what lets a symmetry group be peeled into layers: the quotient $G/N$ is the residual symmetry after the layer $N$ has been accounted for. The kernel of any homomorphism is normal; $A_n\trianglelefteq S_n$ is the kernel of $\mathrm{sgn}$, and $S_n/A_n\cong C_2$, the cyclic group of order $2$.

For $n\ge 5$ (in fact $n\neq 6$), $A_n$ is the only nontrivial proper normal subgroup of $S_n$. We will need a piece of that: $A_n$ itself, for $n\ge 5$, has no nontrivial proper normal subgroups. Such a group is called simple. $A_5$ is the smallest non-abelian simple group.

Why simplicity matters

If a group is simple, it has no nontrivial proper normal subgroup, so it cannot be peeled into a smaller layer plus a quotient. Solvable groups are built from abelian pieces. A simple non-abelian group therefore cannot appear in a solvable series except as a dead end: you cannot start a stack of abelian quotients with it. That is why $A_5$ kills solvability of $S_5$: the only normal series of $S_5$ is ${1}\trianglelefteq A_5\trianglelefteq S_5$, and the factor $A_5$ is not abelian.

Contrast $S_4$, which can be peeled: ${1}\trianglelefteq V_4\trianglelefteq A_4\trianglelefteq S_4$, with abelian (in fact cyclic, after one refinement) quotients.

$A_5$ is simple

The calculation below establishes that $A_5$ has no nontrivial proper normal subgroups. Its role in the main argument is only the paragraph above: $A_5$ cannot be part of a solvable chain. Three facts.

(i) $A_n$ is generated by $3$-cycles, for $n\ge 3$. A $3$-cycle is even, so lies in $A_n$. Conversely every even permutation is a product of an even number of transpositions, and

\[(a\,b)(a\,c) = (a\,c\,b), \qquad (a\,b)(c\,d) = (a\,c\,b)(a\,c\,d)\]

when ${a,b}\cap{c,d}=\emptyset$. So products of two transpositions are products of $3$-cycles.

(ii) All $3$-cycles are conjugate in $A_n$ for $n\ge 5$. In $S_n$, any two $3$-cycles are conjugate: if $\sigma=(1\,2\,3)$ then $\tau\sigma\tau^{-1}=(\tau(1)\,\tau(2)\,\tau(3))$. For $n\ge 5$ there are two unused letters, say $4$ and $5$. If the conjugating $\tau$ is odd, replace it by $\tau’=\tau\cdot(4\,5)$. Then $\tau’$ is even, so lies in $A_n$, and $\tau’\sigma(\tau’)^{-1}=\tau\sigma\tau^{-1}$ because $(4\,5)$ does not meet ${1,2,3}$. Thus conjugacy of $3$-cycles still happens inside $A_n$.

(iii) Any nontrivial normal subgroup $N\trianglelefteq A_n$ ($n\ge 5$) contains a $3$-cycle. Take $\sigma\in N$, $\sigma\neq\mathrm{id}$. Because $N$ is normal, every $A_n$-conjugate of $\sigma$ is in $N$, and so is every commutator $\tau\sigma\tau^{-1}\sigma^{-1}$. The even cycle types that can occur in $A_5$ are $3$-cycles, products of two disjoint transpositions, and $5$-cycles. The first is already a $3$-cycle. The other two are handled by an explicit conjugation:

• A product of two disjoint transpositions, say $\sigma=(1\,2)(3\,4)$. (Even, so it can live in $A_n$.) For $n\ge 5$ there is a fifth letter. Conjugate by $\tau=(3\,4\,5)$:

  \[\tau\sigma\tau^{-1} = (1\,2)(4\,5).\]

  Then $\sigma\cdot(\tau\sigma\tau^{-1})=(1\,2)(3\,4)(1\,2)(4\,5)=(3\,4\,5)$, a $3$-cycle in $N$.

• A $5$-cycle, say $\sigma=(1\,2\,3\,4\,5)$. Let $\tau=(1\,2\,3)$. Conjugation by $\tau$ sends each letter $i$ in the cycle to $\tau(i)$, so $\tau\sigma\tau^{-1}=(2\,3\,1\,4\,5)$. The commutator, composing right to left, is the $3$-cycle $(1\,2\,4)$:

  \[\tau\sigma\tau^{-1}\sigma^{-1} = (1\,2\,3)(1\,2\,3\,4\,5)(1\,3\,2)(1\,5\,4\,3\,2) = (1\,2\,4).\]

  Track a letter if you want the arithmetic: $1\mapsto 2\mapsto 3\mapsto 1\mapsto 2$, then continue for $2,3,4,5$. (The same identity, unused letters fixed, works in $A_n$ for $n \gt 5$.)

Once $N$ contains one $3$-cycle, conjugacy (ii) puts every $3$-cycle in $N$, and generation (i) forces $N=A_n$. That is the whole of simplicity: a nontrivial normal subgroup cannot be proper.

Thus $A_5$ is simple. It is non-abelian: $(1\,2\,3)(3\,4\,5)\neq (3\,4\,5)(1\,2\,3)$. Therefore $A_5$ admits no chain of subgroups down to ${1}$ with abelian successive quotients, except the trivial two-step ${1}\trianglelefteq A_5$ whose quotient is not abelian. The same argument, with the extra room of unused letters, shows $A_n$ is simple for all $n\ge 5$.

Solvable groups

A finite group $G$ is solvable if there is a chain

\[\{1\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_k = G\]

in which each quotient $G_{i+1}/G_i$ is abelian. (For finite groups one may equivalently demand that the quotients be cyclic: a finite abelian group is a product of cyclics, and one can refine the chain.)

The picture is a stack of commutative layers. Each radical step $F(\sqrt[n]{a})/F$, once the $n$th roots of unity are present, has cyclic Galois group — one commutative layer. A nested-radical formula is a finite stack of such layers, so the Galois group of a polynomial solvable by radicals must be a group that can be built by stacking abelian pieces. That is the name. $S_3$ can (cyclic of order $3$, then $C_2$). $A_5$ cannot: it has no nontrivial abelian normal subgroup to start the stack.

$S_2$, $S_3$, $S_4$ are solvable.

• $S_2\cong C_2$, already abelian.
• ${1}\trianglelefteq A_3 \trianglelefteq S_3$ with quotients $C_3$ and $C_2$. Here $A_3=\langle(1\,2\,3)\rangle$.
• ${1}\trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$, where $V_4={\mathrm{id},(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)}$ is the Klein four-group, abelian of order $4$. Quotients: $C_2\times C_2$, $C_3$, $C_2$.

$V_4$ is normal in $S_4$ because conjugation preserves cycle type, and the three non-identity elements of $V_4$ are all the products of two disjoint transpositions in $S_4$. So $S_4$ permutes those three elements among themselves and leaves $V_4$ invariant. The quotient $A_4/V_4$ has order $3$, hence is cyclic. This is the group-theoretic shadow of Ferrari’s method: the resolvent cubic of a quartic is the quotient $S_4\to S_3\cong S_4/V_4$, and solving that cubic (solvable, because $S_3$ is) is the step that reduces a quartic to nested quadratics.

$S_n$ is not solvable for $n\ge 5$. Two facts, connected. First: for $n\neq 6$, $A_n$ is the unique nontrivial proper normal subgroup of $S_n$ (it is $\ker\mathrm{sgn}$, index $2$). So any normal series from ${1}$ up to $S_n$ has $A_n$ as its last proper term: ${1}\trianglelefteq\cdots\trianglelefteq A_n\trianglelefteq S_n$. Second: $A_n$ is simple and non-abelian, so the stretch ${1}\trianglelefteq\cdots\trianglelefteq A_n$ cannot be refined into abelian quotients — the only possibilities are to skip $A_n$ (impossible, by uniqueness) or to leave the non-abelian factor $A_n$ in the series. There is no analogue of $V_4$ sitting normally inside $A_5$. That is the group-theoretic half of Abel–Ruffini, stated before we have fields. The rest of the note is the identification: a radical formula produces a solvable group of symmetries of the root family, and the general quintic’s group is $S_5$. Then $S_5$ is too big.

One more computational fact, used twice below.

Lemma (adjacent transpositions generate $S_n$). The transpositions $(1\,2),(2\,3),\ldots,(n-1\,n)$ generate $S_n$. Any transposition $(i\,j)$ with $j \gt i+1$ is a conjugate of an adjacent one:

\[(i\,j) = (j-1\,j)\,(i\,j-1)\,(j-1\,j),\]

and inducting on $j-i$ writes $(i\,j)$ in the adjacent generators. Every permutation is a product of (not necessarily adjacent) transpositions, so the adjacent ones suffice. Equivalently: a $p$-cycle $(1\,2\,\cdots\,p)$ together with the transposition $(1\,2)$ generate $S_p$, because conjugating the transposition by powers of the cycle produces every adjacent transposition.

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Field extensions: where the family lives

The FTA placed the family ${r_1,\ldots,r_n}$ in $\mathbb{C}$. The coefficients may live in a much smaller field — classically $\mathbb{Q}$. The roots need not.

Take $x^2-2\in\mathbb{Q}[x]$. It is irreducible over $\mathbb{Q}$ (if $\sqrt{2}=p/q$ in lowest terms then $p^2=2q^2$, so $2$ divides $p$ and then $q$, contradiction). The family is ${\sqrt{2},-\sqrt{2}}$. Neither root is rational. Both live in the smallest field that contains $\mathbb{Q}$ and $\sqrt{2}$.

Adjoining one element

If $F$ is a field and $\alpha$ lives in some bigger field, write $F(\alpha)$ for the smallest field containing $F$ and $\alpha$. Concretely, if $\alpha$ is algebraic over $F$ — if it satisfies a polynomial with coefficients in $F$ — there is a unique monic polynomial of least degree with this property, the minimal polynomial $m_{\alpha,F}$. Then

\[F(\alpha) = \{ c_0 + c_1\alpha + \cdots + c_{d-1}\alpha^{d-1} : c_i\in F \},\]

where $d=\deg m_{\alpha,F}$, with multiplication reduced using $m_{\alpha,F}(\alpha)=0$. In particular $F(\alpha)$ is a $d$-dimensional vector space over $F$, with basis ${1,\alpha,\ldots,\alpha^{d-1}}$.

For $\alpha=\sqrt{2}$ over $\mathbb{Q}$, the minimal polynomial is $x^2-2$, and $\mathbb{Q}(\sqrt{2})={a+b\sqrt{2}:a,b\in\mathbb{Q}}$ with basis ${1,\sqrt{2}}$. The family of roots of $x^2-2$ already lives entirely in this field: $-\sqrt{2}$ is just $-1\cdot\sqrt{2}$.

Splitting fields: the home of the whole family

Adjoining one root is not always enough.

Existence: adjoin one root of an irreducible factor, repeat. Uniqueness up to an isomorphism fixing $F$: any two splitting fields are $F$-isomorphic. We work throughout with a splitting field sitting inside $\mathbb{C}$, which the FTA permits.

Example: $x^2-2$ over $\mathbb{Q}$. Splitting field $\mathbb{Q}(\sqrt{2})$. The family lives there.

Example: $x^3-2$ over $\mathbb{Q}$. The real cube root $\sqrt[3]{2}$ generates $\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{R}$. That field contains one root of $x^3-2$. The other two are $\sqrt[3]{2}\,\zeta_3$ and $\sqrt[3]{2}\,\zeta_3^2$, where $\zeta_3=e^{2\pi i/3}=-\frac12+i\frac{\sqrt{3}}{2}$ is a primitive cube root of unity. They are not real. So $\mathbb{Q}(\sqrt[3]{2})$ is not a splitting field. The splitting field is $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$. The family of three roots lives there and nowhere smaller.

This is the standing picture: coefficients in a base field $F$; family of roots in $\mathbb{C}$; splitting field $E=F(r_1,\ldots,r_n)$ the smallest house that holds them all.

Nested radicals are a special kind of house

The quadratic formula is already a radical tower, and it is worth writing it that way so the later obstruction has something to obstruct. For $ax^2+bx+c$ with $a\neq 0$, set $F=\mathbb{Q}(a,b,c)$ (or $\mathbb{Q}$, if $a,b,c\in\mathbb{Q}$). The discriminant $d=b^2-4ac$ lives in $F$. The tower is

\[F \subset F(\sqrt{d}),\]

a single square-root step of degree $1$ or $2$, and both roots $\frac{-b\pm\sqrt{d}}{2a}$ live in the top field. The family ${\sqrt{d},-\sqrt{d}}$ is swapped by the unique nontrivial automorphism, which is why the two choices of sign in the formula are not a defect: they are the Galois group.

A pure radical extension of $F$ is $F(\sqrt[n]{a})$ for some $a\in F$ and $n\ge 2$: adjoin a root of $x^n-a$. A radical tower over $F$ is a finite chain

\[F = F_0 \subset F_1 \subset \cdots \subset F_k\]

in which each $F_{i+1}=F_i(\alpha_i)$ with $\alpha_i^{n_i}\in F_i$ for some $n_i\ge 2$. A radical expression determines such a tower (each nested root is one step; a choice of branch is a choice of which root to adjoin). Conversely, the elements of a radical tower are the algebraic objects captured by nested radical constructions. Example: $\sqrt{2+\sqrt{2}}$ lives in $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2+\sqrt{2}})$. Cardano’s formula for a cubic is a messy tower of square roots and cube roots. Ferrari’s formula for a quartic is a longer such tower.

Definition. A polynomial $f\in F[x]$ is solvable by radicals over $F$ if there is a radical tower over $F$ whose top field contains a splitting field of $f$ — equivalently, contains the whole family of roots.

That is what “algebraic solution” meant to Abel. Not: the roots exist in $\mathbb{C}$. Not: they can be approximated. The roots can be written by starting from the coefficients and repeatedly extracting $n$th roots.

Two warnings, both classical.

First: adjoining one $n$th root is not the same as adjoining all of them. As with $x^3-2$, the real cube root of $2$ does not split $x^3-2$. A careful theory of radical towers inserts roots of unity when needed so that each step is a splitting field of a polynomial of the form $x^n-a$. We will come back to this; it is the gap in Ruffini’s argument.

Second: specific polynomials of every degree are solvable by radicals. $x^n-1$ is; the cyclotomic fields are radical (after Gauss) over $\mathbb{Q}$. Abel–Ruffini is a statement about the general polynomial, and, in the sharpened form, about those particular polynomials whose root-family is too symmetric.

Cardano’s formula, stripped of coefficients, has the shape

\[\sqrt[3]{\,u + \sqrt{u^2 + v^3}\,} + \sqrt[3]{\,u - \sqrt{u^2 + v^3}\,},\]

a square root nested inside two cube roots. That is a radical tower of length two (or three, if you count the two cube roots as separate steps, and four if you first adjoin $\zeta_3$ so the cube roots split). The family of three cubic roots lives in the top field; the Galois group is at most $S_3$, which we already know is solvable. Ferrari reduces a quartic to a cubic resolvent, then to quadratics — a longer tower, group at most $S_4$. There is no fifth-degree analogue of that reduction that stays inside radicals, and the reason will not be “we have not found it.” The reason is that $S_5$ has nothing like $V_4$ to quotient by.

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Degree, and automorphisms of the home of the roots

View an extension $E/F$ as a vector space over $F$. Its dimension, finite or infinite, is the degree $[E:F]$. (This is the standard definition.) For a simple algebraic extension, $[F(\alpha):F]=\deg m_{\alpha,F}$. So $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. Infinite degrees occur — $[\mathbb{Q}(x):\mathbb{Q}]=\infty$, because $1,x,x^2,\ldots$ are linearly independent over $\mathbb{Q}$ — but every splitting field of a polynomial is a finite extension, and we stay finite.

Automorphisms that fix the coefficients

An $F$-automorphism of $E$ is a field automorphism $\sigma:E\to E$ with $\sigma(c)=c$ for every $c\in F$. Write $\mathrm{Aut}(E/F)$ for the group of all of them.

Let $E$ be a splitting field of $f\in F[x]$, with family of roots ${r_1,\ldots,r_n}$. If $\sigma\in\mathrm{Aut}(E/F)$ and $f(r_i)=0$, then

\[f(\sigma(r_i))=\sigma(f(r_i))=0,\]

because $\sigma$ fixes the coefficients of $f$. So $\sigma$ permutes the family. Since $E=F(r_1,\ldots,r_n)$, the automorphism is determined by this permutation. We obtain an injective homomorphism

\[\mathrm{Aut}(E/F) \hookrightarrow S_n.\]

The image is the Galois group of $f$ over $F$, written $\mathrm{Gal}(f/F)$ or $\mathrm{Gal}(E/F)$. It is the group of realizable symmetries of the family: the rearrangements that can be carried out by a field automorphism fixing the coefficients. It is a subgroup of $S_n$. It need not be all of $S_n$. When it is all of $S_n$, the family is maximally symmetric, and that is the case that will kill radical formulae.

The splitting field of a separable polynomial is Galois

A polynomial is separable if its irreducible factors have distinct roots in a splitting field. We care because the Galois group injects into $S_n$ with $n=\deg f$ only when there are $n$ distinct letters to permute; a multiple root collapses two labels into one and the count $\lvert\mathrm{Aut}(E/F)\rvert=[E:F]$ fails. Over $\mathbb{Q}$ (characteristic $0$) an irreducible $f$ cannot share a root with its derivative $f’$ unless $f’$ is the zero polynomial, which it is not: $\deg f’\ge 0$ and the leading term of $f$ has not been killed by a characteristic. So every irreducible over $\mathbb{Q}$ is separable. (In characteristic $p$ one can have inseparables such as $x^p-a$.) We work in characteristic $0$ from here.

Theorem (standard; proof sketch below). Let $E$ be a splitting field of a separable polynomial $f\in F[x]$. Then:

1. $\lvert \mathrm{Aut}(E/F)\rvert = [E:F]$;
2. the fixed field ${\,x\in E : \sigma(x)=x\text{ for all }\sigma\in\mathrm{Aut}(E/F)\,}$ equals $F$;
3. $E/F$ is normal: every irreducible in $F[x]$ with one root in $E$ splits completely in $E$.

A finite extension with these properties is called Galois. Conversely, every finite Galois extension is the splitting field of a separable polynomial. This is the content of the discussion at Math.StackExchange 962898; the argument below is the standard one.

Why $\lvert G\rvert=[E:F]$. Always $\lvert \mathrm{Aut}(E/F)\rvert\le [E:F]$: an $F$-automorphism is determined by where it sends a primitive element (or, inductively, a sequence of adjoined roots), and each minimal polynomial has at most its degree many roots in $E$. For a splitting field of a separable $f$, each such choice extends — this is the isomorphism-extension theorem, which we use as a standard fact rather than prove: an $F$-isomorphism between two fields extends to an isomorphism of splitting fields of the same separable polynomial. Induct on the number of roots of $f$ outside $F$. If $f$ already splits in $F$, then $E=F$ and both sides are $1$. Otherwise let $\alpha$ be a root of an irreducible factor $p$ of $f$, of degree $d\ge 2$. There are $d$ distinct $F$-embeddings $F(\alpha)\to E$, one per root of $p$. Each extends to an automorphism of $E$ because $E$ is still a splitting field of $f$ over $F(\alpha)$. By induction $\lvert \mathrm{Aut}(E/F(\alpha))\rvert=[E:F(\alpha)]$. Counting:

\[\lvert \mathrm{Aut}(E/F)\rvert = d\cdot [E:F(\alpha)] = [F(\alpha):F]\,[E:F(\alpha)] = [E:F],\]

where the last equality is the tower law, proved in the next section; if you want the logic acyclic, postpone this count until after that proof — the two results are meant to be read as a pair.

Why the fixed field is $F$. Let $F’$ be the fixed field of $G=\mathrm{Aut}(E/F)$. Then $F\subset F’\subset E$, and $G=\mathrm{Aut}(E/F’)$. But $E$ is still a splitting field of the same separable $f$ over $F’$, so $\lvert \mathrm{Aut}(E/F’)\rvert=[E:F’]$. Combined with $\lvert \mathrm{Aut}(E/F)\rvert=[E:F]$ we get $[E:F’]=[E:F]$, hence $F’=F$.

Normality. If an irreducible $p\in F[x]$ has one root $\alpha\in E$, then any other root $\beta$ (in a splitting field) is the image of $\alpha$ under an $F$-embedding $F(\alpha)\to\overline{F}$, and such embeddings extend to automorphisms of $E$ when $E/F$ is a splitting field of a separable polynomial; thus $\beta\in E$.

Two examples, now with groups

$\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. Degree $2$, Galois group ${\mathrm{id},\,\sqrt{2}\mapsto-\sqrt{2}}\cong C_2\cong S_2$. The family of $x^2-2$ has two labels; both rearrangements are realized.

$\mathbb{Q}(\sqrt[3]{2},\zeta_3)/\mathbb{Q}$. Let $\alpha=\sqrt[3]{2}\in\mathbb{R}$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$. The polynomial $x^2+x+1$ is the minimal polynomial of $\zeta_3$ over $\mathbb{Q}(\alpha)$: it is irreducible over $\mathbb{R}$, hence over the real field $\mathbb{Q}(\alpha)$. So $[E:\mathbb{Q}(\alpha)]=2$ and $[E:\mathbb{Q}]=6$. (Symmetrically: $x^3-2$ stays irreducible over $\mathbb{Q}(\zeta_3)$, not because it is irreducible over $\mathbb{Q}$, but because a root in $\mathbb{Q}(\zeta_3)$ would embed a degree-$3$ field into a degree-$2$ extension, which the tower law forbids.) The Galois group has order $6$, hence is $S_3$. Explicitly: you may send $\alpha$ to $\alpha\zeta_3^k$ for $k=0,1,2$, and independently send $\zeta_3$ to $\zeta_3^{\pm 1}$.

Label the family $r_0=\alpha$, $r_1=\alpha\zeta_3$, $r_2=\alpha\zeta_3^2$. The automorphism $\sigma$ with $\sigma(\alpha)=\alpha\zeta_3$ and $\sigma(\zeta_3)=\zeta_3$ cycles the roots: $r_0\mapsto r_1\mapsto r_2\mapsto r_0$, a $3$-cycle. The automorphism $\tau$ with $\tau(\alpha)=\alpha$ and $\tau(\zeta_3)=\zeta_3^{-1}=\zeta_3^2$ swaps $r_1$ and $r_2$ and fixes $r_0$, a transposition. These generate $S_3$, and $\tau\sigma\tau^{-1}=\sigma^{-1}$, the usual presentation. Complex conjugation, on this labeling, is $\tau$: it fixes the real root and swaps the two non-real ones. Degree three, two non-real roots — the $p-2$ pattern already, except $S_3$ is solvable, so Cardano still works. The same pattern at $p=5$ will not.

The dictionary is now in place. Section II gave us groups of rearrangements. This section realized some of those rearrangements as automorphisms of the splitting field. The Galois group is the symmetry the coefficients can actually see. The lattice below is the same story for $x^3-2$: $\mathbb{Q}(\zeta_3)$ corresponds to a normal subgroup and is Galois over $\mathbb{Q}$; $\mathbb{Q}(\alpha)$ corresponds to a non-normal subgroup and is not.

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The tower theorem

Degrees multiply. That is the arithmetic of nested adjoining, and it is why a radical formula is a very particular kind of extension.

Theorem (tower law). If $K\subset L\subset M$ are fields, then

\[[M:K] = [M:L]\cdot[L:K].\]

The product is ordinary multiplication when both factors are finite, and a product of cardinals otherwise. In particular, if $M/K$ is finite then so are both steps; if either step is infinite, so is $M/K$. This is the multiplicativity formula and the tower law.

Proof, finite case

Let $d=[L:K]$ and $e=[M:L]$, both finite. Choose a basis ${u_1,\ldots,u_d}$ of $L$ as a $K$-vector space, and a basis ${w_1,\ldots,w_e}$ of $M$ as an $L$-vector space. The claim is that the $de$ products ${u_m w_n}$ form a basis of $M$ over $K$.

They span. Take $x\in M$. Write $x=\sum_{n=1}^e a_n w_n$ with $a_n\in L$. Write each $a_n=\sum_{m=1}^d b_{m,n} u_m$ with $b_{m,n}\in K$. Then

\[x = \sum_{n=1}^e\sum_{m=1}^d b_{m,n}\,(u_m w_n).\]

They are linearly independent. Suppose $\sum_{n,m} b_{m,n}(u_m w_n)=0$ with $b_{m,n}\in K$. Group as $\sum_n \bigl(\sum_m b_{m,n} u_m\bigr) w_n = 0$. Linear independence of the $w_n$ over $L$ forces $\sum_m b_{m,n} u_m = 0$ for each $n$. Linear independence of the $u_m$ over $K$ forces every $b_{m,n}=0$.

Hence $[M:K]=de$.

(The infinite case is the same argument with bases indexed by sets $A$ and $B$; the product basis is indexed by $A\times B$.)

What the formula forbids

If $[M:K]$ is prime, there is no field strictly between $K$ and $M$: the only factorizations of a prime are $1\cdot p$ and $p\cdot 1$. You cannot break the extension into smaller steps. That is why a prime-length cycle in a Galois group is powerful — under the correspondence it is an indivisible layer. Thus $\mathbb{C}/\mathbb{R}$, of degree $2$, has no intermediate field, and $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ has none either.

**Worked tower: $\mathbb{Q}(\sqrt{2},\sqrt{3})$. ** Let $L=\mathbb{Q}(\sqrt{2})$, so $[L:\mathbb{Q}]=2$. The polynomial $x^2-3$ remains irreducible over $L$ (if $\sqrt{3}=a+b\sqrt{2}$ then squaring and comparing rational and irrational parts yields a contradiction). So $[L(\sqrt{3}):L]=2$, hence $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4$. A basis is ${1,\sqrt{2},\sqrt{3},\sqrt{6}}$. This is a radical tower of two square-root steps, total degree $2\cdot 2=4$.

The same tower law reads the intermediate fields. Any proper subfield of a degree-$4$ extension has degree $2$ over $\mathbb{Q}$. There are three of them: $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$. Each is the fixed field of a subgroup of order $2$ in the Klein four-group $\mathrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})\cong C_2\times C_2$, generated by the independent sign-flips $\sqrt{2}\mapsto\pm\sqrt{2}$ and $\sqrt{3}\mapsto\pm\sqrt{3}$. Nested square roots, in this instance, produce an abelian Galois group of exponent $2$ — as solvable as a group can be.

A nested radical need not denest. $\sqrt{2+\sqrt{2}}$ generates a degree-$4$ extension of $\mathbb{Q}$ (its minimal polynomial is $x^4-4x^2+2$), sitting in the radical tower $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2+\sqrt{2}})$. Eisenstein’s criterion, used here and below: if a prime $q$ divides every coefficient of a monic $f\in\mathbb{Z}[x]$ except the leading $1$, and $q^2$ does not divide the constant term, then $f$ is irreducible over $\mathbb{Q}$. For $x^4-4x^2+2$ take $q=2$. Sometimes nested expressions collapse — $\sqrt{5+2\sqrt{6}}=\sqrt{2}+\sqrt{3}$ — and the tower law is how you prove the collapse: both sides generate the same degree-$4$ field. When they do not collapse, you are genuinely climbing. A prime-degree step has no room for an intermediate field, which is why a cyclic prime-degree radical extension cannot be “partially denested.”

Radical towers, measured

A radical formula builds numbers one controlled extraction at a time. Once the necessary roots of unity are available, the different choices for one extraction differ cyclically: for a square root, just a sign; for a cube root, multiplication by $1,\omega,\omega^2$. Thus radical constructions can only build root symmetries in successive commutative layers. “Solvable group” is the name for exactly that layered structure. The next two paragraphs confirm the claim rather than introduce it cold.

A pure radical step $F(\sqrt[n]{a})/F$ has degree dividing $n$: the minimal polynomial of $\sqrt[n]{a}$ over $F$ divides $x^n-a$. Once the $n$th roots of unity already live in $F$, the extension (if irreducible) is cyclic of degree $n$, with Galois group generated by $\sqrt[n]{a}\mapsto \zeta_n\sqrt[n]{a}$. That is Kummer theory: adjoining an $n$th root, in the presence of the $n$th roots of unity, is a cyclic Galois extension. Plainly: after adjoining those roots of unity, choosing an $n$th root differs only by multiplying by one of them, so that step has cyclic symmetry.

A nested-radical formula is therefore a tower whose successive Galois groups — after one inserts the missing roots of unity, so that each step splits $x^{n_i}-a_i$ completely — are cyclic. The composite Galois group, by the correspondence below, is assembled from cyclic quotients: it is solvable.

The insertion of roots of unity is not a minor bookkeeping point. A radical $\sqrt[n]{a}$ has $n$ conjugates, the other $n$th roots of $a$, and they differ by $n$th roots of unity. If those roots of unity are not already in the current field, the pure-radical step $F(\sqrt[n]{a})/F$ is typically not Galois (the real cube root of $2$, again). The refined tower that Galois theory wants looks like

\[F \subset F(\zeta_{n_1}) \subset F(\zeta_{n_1},\sqrt[n_1]{a_1}) \subset \cdots,\]

each step Galois, first abelian (cyclotomic), then cyclic (Kummer). Cyclotomic extensions are themselves solvable — $\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\times$ is abelian — so inserting them does not smuggle in non-solvable symmetry. It merely makes the correspondence apply at every rung.

The Galois correspondence, exhibited on a small example

Let $E/F$ be finite Galois, $G=\mathrm{Gal}(E/F)$. There is an inclusion-reversing bijection

\[\{\text{subgroups }H\le G\} \;\longleftrightarrow\; \{\text{fields }K\text{ with }F\subset K\subset E\}\]

sending $H$ to its fixed field $E^H$ and sending $K$ to $\mathrm{Gal}(E/K)$. Degrees match orders: $[E:E^H]=\lvert H\rvert$ and $[E^H:F]=[G:H]$. Moreover $K/F$ is Galois if and only if $\mathrm{Gal}(E/K)$ is normal in $G$, in which case $\mathrm{Gal}(K/F)\cong G/\mathrm{Gal}(E/K)$.

This is the dictionary between the group theory of §II and the towers of this section. For $\mathbb{Q}(\sqrt{2},\sqrt{3})$ it is the picture already drawn by the tower law: three intermediate fields of degree $2$, three subgroups of order $2$, lattices inverted.

A chain of fields $F=F_0\subset F_1\subset\cdots\subset F_k=E$ with each $F_{i+1}/F_i$ Galois cyclic corresponds to a chain of groups $G=\mathrm{Gal}(E/F_0)\trianglerighteq \mathrm{Gal}(E/F_1)\trianglerighteq\cdots\trianglerighteq{1}$ with cyclic quotients. That is a solvable series.

Theorem (Galois; used as a standard criterion, sketch only). A separable polynomial $f\in F[x]$ is solvable by radicals over $F$ if and only if $\mathrm{Gal}(f/F)$ is a solvable group.

This is the heart of Abel–Ruffini, and a complete proof is a chapter of a Galois-theory book. What follows is the shape of the argument, not a substitute for that chapter.

Sketch, radicals $\Rightarrow$ solvable group. A radical expression determines a radical tower. Insert roots of unity so that each step splits some $x^{n_i}-a_i$ (the cyclotomic extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ has abelian Galois group $(\mathbb{Z}/n\mathbb{Z})^\times$, hence is itself solvable, and inserts no extra obstruction). Each refined step is cyclic Galois by Kummer. The correspondence turns the field tower into a chain of cyclic quotients. The Galois group of the splitting field is therefore solvable.

Sketch, solvable group $\Rightarrow$ radicals. If $\mathrm{Gal}(E/F)$ is solvable, refine the series to cyclic quotients of prime order. After a cyclotomic base change, Kummer realizes each cyclic step as a pure radical extension. The roots then lie in a radical tower.

Ruffini assumed the radicals in a putative formula already lived inside the splitting field. That is extra. Cardano’s formula for $x^3-15x-20$ extracts cube roots of $10\pm 5i$, which do not lie in the splitting field over $\mathbb{Q}$ (the three roots are real). Abel’s theorem on natural irrationalities repairs the general-coefficient case; Galois’ criterion repairs all cases.

We now have the two halves. A radical formula $\Rightarrow$ solvable Galois group. $S_n$ for $n\ge 5$ is not solvable. It remains to see that the general quintic — and some perfectly concrete quintics over $\mathbb{Q}$ — actually have Galois group $S_5$. That is the symmetry of the quintic.

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Quintic symmetry

What a formula would be

Suppose there were a general radical formula for degree $n$: an expression built from the coefficients $a_1,\ldots,a_n$ by field operations and nested radicals, which, when the $a_i$ are specialized to the coefficients of any degree-$n$ polynomial, produces a root. Interpreting the $a_i$ as indeterminates, the expression would place a root in a radical tower over the field of rational functions $F=\mathbb{Q}(a_1,\ldots,a_n)$. The splitting field over $F$ would then sit in (a Galois closure of) a radical tower, and $\mathrm{Gal}(f/F)$ would be solvable.

So: if we can show that the Galois group of the general degree-$n$ polynomial is $S_n$, then for $n\ge 5$ there is no such formula.

The general polynomial has symmetry $S_n$

Why start from the roots? The coefficients of a polynomial are symmetric functions of the roots: they forget order. So the field generated by the coefficients is precisely the subfield of the root-field fixed by every permutation of the roots. If the roots are independent indeterminates, every permutation is available, and the Galois group is all of $S_n$.

Concretely for $n=3$. Let $t_1,t_2,t_3$ be indeterminates and

\[(x-t_1)(x-t_2)(x-t_3) = x^3 - s_1 x^2 + s_2 x - s_3,\]

with Vieta $s_1=t_1+t_2+t_3$, $s_2=t_1 t_2+t_1 t_3+t_2 t_3$, $s_3=t_1 t_2 t_3$. Any permutation of the $t_i$ leaves the $s_j$ unchanged. The map sending $t_i\mapsto t_{\sigma(i)}$ is therefore a field automorphism of $H=\mathbb{Q}(t_1,t_2,t_3)$ fixing $K=\mathbb{Q}(s_1,s_2,s_3)$. Distinct $\sigma$ move the $t_i$ differently, so give distinct automorphisms. Thus $\mathrm{Gal}(H/K)$ contains a copy of $S_3$, and since it embeds in $S_3$ it is $S_3$.

The same construction works in every degree. Let $x_1,\ldots,x_n$ be indeterminates and

\[P(x) = (x-x_1)\cdots(x-x_n) = x^n + b_1 x^{n-1} + \cdots + b_n,\]

so the $b_i$ are the elementary symmetric polynomials in the $x_j$ (Vieta: $b_1=-(x_1+\cdots+x_n)$, and so on). Let $H=\mathbb{Q}(x_1,\ldots,x_n)$ and $K=\mathbb{Q}(b_1,\ldots,b_n)$. Every permutation of the $x_i$ extends to a field automorphism of $H$ fixing $K$. Distinct permutations give distinct automorphisms, because they move the $x_i$ differently. Thus

\[\mathrm{Gal}(H/K) \cong S_n.\]

($H$ is the splitting field of the separable polynomial $P$ over $K$, so the extension is Galois.) The $b_i$ are algebraically independent — that is the fundamental theorem of symmetric polynomials, which we use as a named fact: every symmetric polynomial in the $x_j$ is a polynomial in the $b_i$, with no unexpected algebraic relations among the $b_i$. The map sending each indeterminate coefficient $a_i$ of the general polynomial $x^n+a_1 x^{n-1}+\cdots+a_n$ to the corresponding $b_i$ is therefore a field isomorphism $\mathbb{Q}(a_1,\ldots,a_n)\xrightarrow{\sim} K$. Carrying the Galois group along this isomorphism, the general polynomial of degree $n$ over $\mathbb{Q}(a_1,\ldots,a_n)$ has Galois group $S_n$.

If a universal radical formula existed, it would solve the polynomial with symbolic coefficients. But that symbolic — or general — quintic has Galois group $S_5$, and $S_5$ is not solvable. Therefore such a formula cannot exist.

Specialization is how the general fact infects particular polynomials. If a formula in the indeterminate coefficients existed, substituting rational numbers for the $a_i$ would give a radical expression for every quintic over $\mathbb{Q}$. So it is enough, to kill a uniform formula, that $S_5$ be the Galois group in the indeterminate case. To kill particular quintics, one still has to check that the Galois group remains non-solvable after specialization. Hilbert irreducibility (used as a named existence theorem, not proved here) says that infinitely many rational specializations of the general quintic still have Galois group $S_5$ or $A_5$. The $p-2$ real-roots theorem below is a hands-on way to catch some of them without that theorem.

Abel–Ruffini theorem. There is no formula, in the coefficients, by field operations and nested radicals, for the roots of the general polynomial of degree $n\ge 5$.

That is the original theorem: a statement about a formula in indeterminates. It does not, by itself, exhibit a polynomial with rational coefficients that cannot be solved by radicals. (In principle every particular quintic might have had its own private radical expression.) Galois’ criterion, plus the existence of polynomials over $\mathbb{Q}$ with Galois group $S_5$, closes that gap.

Concrete quintics with $S_5$ symmetry

We have shown that the general quintic (indeterminate coefficients) has group $S_5$. That kills a uniform formula. It does not, by itself, exhibit a polynomial with integer coefficients that is unsolvable: in principle every particular quintic might still have had its own private radical expression. Hilbert irreducibility says that “most” rational specializations of the general quintic keep Galois group $S_5$ or $A_5$. Here is how to check a specific case without that theorem.

The following machine is Math.StackExchange 3075225.

Theorem. Let $p\ge 5$ be prime, and let $f\in\mathbb{Q}[x]$ be irreducible of degree $p$, with exactly $p-2$ real roots (hence exactly one conjugate pair of non-real roots). Then $\mathrm{Gal}(f/\mathbb{Q})\cong S_p$.

Proof. Let $E\subset\mathbb{C}$ be a splitting field, $G=\mathrm{Gal}(E/\mathbb{Q})\le S_p$.

A group acts on a set if it permutes the set in a way compatible with the group law; the action is transitive if there is a single orbit — any point can be sent to any other. Irreducibility means all roots look algebraically interchangeable: the group $G$ acts transitively on them. The orbit has size $p$, so by the orbit-stabilizer theorem $p$ divides $\lvert G\rvert=[E:\mathbb{Q}]$ (equivalently: adjoining one root gives a subfield of degree $p$, and the tower law makes $p$ divide $[E:\mathbb{Q}]$). Cauchy’s theorem: a finite group whose order is divisible by a prime $p$ contains an element of order $p$. The only elements of order $p$ in $S_p$ are $p$-cycles (a product of disjoint cycles has order the lcm of the lengths, and $p$ is prime). So $G$ contains a $p$-cycle.

Complex conjugation is an automorphism of $\mathbb{C}$ fixing $\mathbb{Q}$. It preserves $E$ because it permutes the roots of $f$ (real coefficients). It fixes each of the $p-2$ real roots and swaps the two non-real ones. Restricted to $E$, it is therefore a transposition in $G$.

Lemma. A subgroup of $S_p$ ($p$ prime) that contains a $p$-cycle and a transposition is all of $S_p$.

Relabel so the transposition is $(1\,2)$. Write the $p$-cycle as $(1\,a_2\,\cdots\,a_p)$. Some power of it sends $1$ to $2$, because the cycle acts transitively; replacing the cycle by that power, we may assume it is $(1\,2\,3\,\cdots\,p)$. Conjugating the transposition by powers of the cycle produces the adjacent transpositions:

\[(1\,2\,3\,\cdots\,p)^k\,(1\,2)\,(1\,2\,3\,\cdots\,p)^{-k} = (k+1\,k+2),\]

indices modulo $p$ in ${1,\ldots,p}$. Adjacent transpositions generate $S_p$, as already recorded. Hence $G=S_p$.

The hypothesis that $p$ is prime was used twice: to guarantee that an element of order $p$ is a single $p$-cycle, and to guarantee transitivity of that cycle on all $p$ letters. For composite degree the same counting can leave you in a proper transitive subgroup (dihedral, Frobenius, $A_n$, \ldots). Prime degree plus a transposition is a sledgehammer.

Example: $f(x)=x^5-6x+3$. Eisenstein at $3$, as stated above: $3$ divides $6$ and $3$, $9$ does not divide $3$. Irreducible over $\mathbb{Q}$. Now count real roots. $f(-2)=-17$, $f(-1)=8$, $f(0)=3$, $f(1)=-2$, $f(2)=23$: three sign changes, so at least three real roots by the intermediate-value theorem. The derivative $f’(x)=5x^4-6$ has exactly two real zeros, $\pm(6/5)^{1/4}$. Rolle’s theorem: between any two real roots of $f$ lies a root of $f’$, so $f$ has at most three real roots. Exactly three real roots, two non-real. The theorem gives $\mathrm{Gal}(f/\mathbb{Q})\cong S_5$.

This polynomial is not solvable by radicals. Its five roots exist in $\mathbb{C}$, by the FTA. They do not lie in any radical tower over $\mathbb{Q}$.

Optional second method: $q(x)=x^5-x-1$. The $p-2$ argument is self-contained. A different factory for cycle types is the Dedekind–Frobenius theorem, which we state and use but do not prove: if a monic $q\in\mathbb{Z}[x]$ factors modulo a prime $\ell$ not dividing the discriminant $\Delta=\prod_{i<j}(r_i-r_j)^2$ (the square of the Vandermonde product from the parity section) as a product of distinct irreducibles of degrees $d_1,\ldots,d_k$, then $G$ contains an element of cycle type $d_1+\cdots+d_k$.

Modulo $2$, check by multiplying: $(x^2+x+1)(x^3+x^2+1)=x^5+x^4+x^2+x^4+x^3+x+x^3+x^2+1=x^5+x+1$ in $\mathbb{F}_2[x]$, and $x^5-x-1\equiv x^5+x+1\pmod{2}$. Two irreducibles of degrees $2$ and $3$, hence a permutation of type $2+3$ in $G$; cubing it yields a transposition. Modulo $3$, $q$ is irreducible, hence $G$ contains a $5$-cycle. A transposition and a $5$-cycle generate $S_5$. Same conclusion, different pair of hands.

Degree five is not the crime

The polynomial $x^5-2$ is solvable by radicals. Its splitting field is $\mathbb{Q}(\sqrt[5]{2},\zeta_5)$. The tower

\[\mathbb{Q} \subset \mathbb{Q}(\zeta_5) \subset \mathbb{Q}(\zeta_5,\sqrt[5]{2})\]

has degrees $\varphi(5)=4$ and $5$, product $20$ by the tower law. (You must adjoin $\zeta_5$ first, or at least somewhere: $\mathbb{Q}(\sqrt[5]{2})$ is a real field of degree $5$, hence not Galois over $\mathbb{Q}$, and does not contain the four non-real fifth roots of $2$. This is $x^3-2$ again, one degree up.) The Galois group has order $20$. It is the Frobenius group of affine transformations $x\mapsto ax+b$ over $\mathbb{F}_5$: an automorphism is determined by

\[\sqrt[5]{2}\mapsto \sqrt[5]{2}\,\zeta_5^{b}, \qquad \zeta_5\mapsto\zeta_5^{a},\]

with $a\in(\mathbb{Z}/5\mathbb{Z})^\times$ and $b\in\mathbb{Z}/5\mathbb{Z}$. The maps with $a=1$ form a normal cyclic subgroup $C_5$ (pure translation of the radical); the quotient is $C_4$ (the cyclotomic Galois group). The whole group is the semidirect product $C_5\rtimes C_4$: the set of pairs $(b,a)$ with the multiplication coming from $C_4$ acting on $C_5$ by $a\cdot b = a b$ in $\mathbb{F}_5$ — equivalently, the group of affine maps $x\mapsto ax+b$ already named. Solvable series: ${1}\trianglelefteq C_5 \trianglelefteq C_5\rtimes C_4$. And indeed $\sqrt[5]{2}$ is a radical, and the other roots are $\sqrt[5]{2}\,\zeta_5^k$.

Compare the two quintics as permutation groups acting on their five roots. For $x^5-2$, complex conjugation fixes the one real root and inverts $\zeta_5$, which (on a suitable labeling) is a product of two disjoint transpositions, an even permutation of order $2$ — not a transposition of two roots. The $p-2$ real-roots theorem never fires, because there are not three real roots to fix. For $x^5-6x+3$, conjugation is a transposition, the group is forced up to $S_5$, and solvability dies.

The same degree, two different symmetries. $S_5$ has order $120$; the Frobenius group has order $20$. One is solvable, one is not. Abel–Ruffini forbids a uniform formula because the general quintic has the larger symmetry. Galois forbids a radical expression for $x^5-6x+3$ because that quintic has the larger symmetry too.

Why degrees two, three, and four have formulae

$S_2$, $S_3$, $S_4$ are solvable. The Galois group of any polynomial of degree $\le 4$ embeds in one of these, hence is solvable, hence the polynomial is solvable by radicals. The quadratic formula, Cardano’s formula, and Ferrari’s formula are the radical towers those solvable series permit. One does not need to write the towers out to know they exist. (Writing them out is a different, older, and much messier story.)

The correspondence is visible in the groups we already wrote down.

• Degree $2$: $S_2\cong C_2$. One quadratic radical. The sign homomorphism is the Galois group.
• Degree $3$: ${1}\trianglelefteq A_3\trianglelefteq S_3$. The quadratic subextension is generated by $\sqrt{\Delta}$, where $\Delta=\prod_{i \lt j}(r_i-r_j)^2$ is the discriminant; adjoining it cuts $S_3$ down to $A_3\cong C_3$, which is then a pure cube-root step once $\zeta_3$ is present. Cardano’s nested square-then-cube is exactly this chain.
• Degree $4$: ${1}\trianglelefteq V_4\trianglelefteq A_4\trianglelefteq S_4$. The quotient $S_4/V_4\cong S_3$ is Cardano’s cubic resolvent. Solving it (by the previous bullet) lands you in a $V_4$-extension, which is a pair of quadratics. Ferrari is this chain written in coefficients.

There is no corresponding normal subgroup of $S_5$ with abelian — or even solvable — quotient except $A_5$, whose quotient is only $C_2$. Adjoining a square root of the discriminant cuts $S_5$ down to $A_5$, and $A_5$ does not budge. That is why “just adjoin $\sqrt{\Delta}$ and continue as in the cubic” dies on the quintic: you have spent your only normal subgroup, and what remains is simple.

For degree $\ge 5$ the embedding $\mathrm{Gal}(f/F)\hookrightarrow S_n$ is into a non-solvable group, which does not by itself prove unsolvability — a subgroup of a non-solvable group may be solvable, as $x^5-2$ shows. Unsolvability is the assertion that the Galois group equals (or contains) a non-solvable group, typically $A_n$ or $S_n$. The group $A_n$ for $n\ge 5$ is already enough: it is not solvable, so a polynomial with Galois group $A_5$ is equally unsolvable by radicals. Distinguishing $A_5$ from $S_5$ is the question of whether $\Delta$ is a square in the base field. Our examples were chosen to have a transposition, hence to meet $S_5$ rather than $A_5$; an odd permutation in the group is the whole distinction.

Precis (the four steps again)

• Existence. The fundamental theorem of algebra: a degree-$n$ polynomial has a family of $n$ roots in $\mathbb{C}$.
• Symmetry. The family may be rearranged. All rearrangements form $S_n$; the even ones form $A_n$. For $n\ge 5$, $A_n$ is simple and non-abelian, so $S_n$ is not solvable.
• Housing. The family lives in a splitting field over the coefficient field. Automorphisms of that field that fix the coefficients are the rearrangements the coefficients can see: the Galois group, a subgroup of $S_n$. The splitting field of a separable polynomial is Galois; the count of automorphisms equals the degree.
• Towers. Degrees multiply along a chain of fields. A nested-radical formula is a tower of pure radical steps. After inserting roots of unity, each step is cyclic Galois. The Galois group of a polynomial solvable by radicals is therefore a solvable group, and conversely.
• Mismatch. The general degree-$n$ polynomial has Galois group $S_n$. For $n\ge 5$ that group is not solvable. Hence there is no general radical formula. Particular polynomials over $\mathbb{Q}$, such as $x^5-6x+3$ and $x^5-x-1$, also have Galois group $S_5$, hence have no radical expression for their roots over $\mathbb{Q}$. Degree five is allowed to be solvable ($x^5-2$); $S_5$-sized symmetry is not.

That is the lack of an algebraic solution for quintic and higher equations, stated precisely: not that roots fail to exist, but that the symmetry of the root family cannot be assembled from the cyclic symmetries of nested radicals.

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History, scope, references

Paolo Ruffini nearly proved the general-coefficient case in 1799, in Teoria generale delle equazioni. Cauchy believed him; Abel, later, found the memoir too tangled to adjudicate. The gap is the one already named: Ruffini treated radicals as if they lived in the splitting field. Abel published a six-page proof in 1824 (he paid for the printing, and saved paper) and a fuller one in 1826. He was at work on the which quintics question when he died in 1829.

Évariste Galois, at eighteen, submitted the criterion — solvability by radicals if and only if the group of the equation is solvable — to the Paris Academy. It was rejected as too sketchy. He died in 1832. Liouville published the memoir in 1846. Wantzel, already aware of Galois, noted in 1845 that Abel’s argument hits the general polynomial, while Galois produces specific unsolvable ones.

What this note does not claim: that roots cannot be approximated (they can, to any precision); that no closed form of any kind exists (Bring’s radical and elliptic modular functions give expressions outside the radical language); that every quintic is unsolvable (only those whose Galois group is not a solvable group).

The sources woven here are the fundamental theorem of algebra; parity of a permutation; the degree of a field extension and the tower law; the splitting field of a separable polynomial is Galois; irreducible of prime degree with $p-2$ real roots has Galois group $S_p$; and the Abel–Ruffini theorem. For the $S_p$ lemma in cleaner notes, Keith Conrad’s Galois groups as permutation groups is the standard blurb. Dummit and Foote, Chapter 14, is the textbook companion if you want every lemma with a number.