Julian Henry

polyglot / software engineer / author

Abel–RuffiniVIQuintic Symmetry

22 Aug 2026

Abel–Ruffini: I · II · III · IV · V · VI

The general polynomial has group $S_n$; $x^5-6x+3$ has group $S_5$; $x^5-2$ does not. Abel–Ruffini, and Galois’ sharper version, closed.

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 $\lbrace 1,\ldots,p\rbrace$. Adjacent transpositions generate $S_p$, as recorded at the end of Vol. III. 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 in Vol. V: $3$ 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: $\lbrace 1\rbrace\trianglelefteq C_5 \trianglelefteq C_5\rtimes C_4$. 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 $3$: $\lbrace 1\rbrace\trianglelefteq A_3\trianglelefteq S_3$. 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 $4$: $\lbrace 1\rbrace\trianglelefteq V_4\trianglelefteq A_4\trianglelefteq S_4$. 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

  • 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.

Compare the short proof of Vol. I. There, loops of coefficients stood in for automorphisms, and the tower of commutators stood in for the derived series. It kills the uniform formula with almost no machinery, but it says nothing about $x^5-6x+3$ on its own: a single polynomial has no coefficients to wiggle. That sharper statement is what the five volumes of field theory buy.

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.

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 series 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; Sheldon Axler, Linear Algebra Done Right, 4th ed., §4.12, for the $k$th-root walk; 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.

For the topological proof of Vol. I the original is V. I. Arnold’s 1963 lectures to Moscow schoolchildren, written up by V. B. Alekseev as Abel’s Theorem in Problems and Solutions (Kluwer, 2004); see also the Wikipedia summary of Arnold’s proof.

Videos.

Abel–Ruffini: Vol. I · Vol. II · Vol. III · Vol. IV · Vol. V · Vol. VI