Abel’s proof of insolvability of the quintic

For more than two centuries, the solution of polynomial equations had appeared to follow a compelling pattern. The quadratic equation had a formula. The cubic equation had yielded to the methods of the Italian algebraists, and the quartic had soon followed. Each success had the same general character: starting from the coefficients, one combined the ordinary arithmetic operations with successive extractions of square roots, cube roots, and other roots. It was therefore natural to expect that the equation of degree five would eventually submit to a still more elaborate formula of the same kind. The question was not whether particular quintic equations could be solved. Many can. For example, $X^5-a=0$ is solved by taking a fifth root of $a$ and multiplying it by the fifth roots of unity. Nor was the issue whether one could approximate quintic roots numerically; that can be done very effectively. The question was whether there existed one general formula: a finite algebraic recipe which, for every quintic, begins only with its coefficients and reaches its roots through rational operations and repeated extraction of roots.

Thus the object of the problem is the general equation

$\displaystyle f(X)=X^5-s_1X^4+s_2X^3-s_3X^2+s_4X-s_5.$

Can one construct each of its roots from the coefficients $s_1,\ldots,s_5$ by repeatedly using

$\displaystyle +,\quad -,\quad \times,\quad \div,\quad \sqrt[n]{\phantom{x}}\quad ?$

Abel’s theorem answers this question negatively. There is no formula of this kind for the general quintic.

The force of this conclusion is best appreciated by comparing it with the lower-degree cases. Abel did not discover merely that mathematicians had failed to find a sufficiently clever formula. He showed that no formula built from radicals could possibly exist. The obstruction is not computational complexity, nor a lack of ingenuity in manipulating symbols. It lies in the way the roots of a general quintic can be rearranged without changing the coefficients. A radical formula would have to select and distinguish an individual root from an unordered collection of five roots, but the allowable branch changes of radicals are too restricted to accomplish this.

For that reason, Abel begins not with the coefficients, but with the roots themselves. Let $x_0,x_1,x_2,x_3,x_4$ be five independent symbols, and write

$\displaystyle f(X)=\prod_{j=0}^{4}(X-x_j).$

The coefficients are the elementary symmetric functions of these five symbols. For instance,

$\displaystyle s_1=x_0+x_1+x_2+x_3+x_4, \\ s_2=x_0x_1+x_0x_2+\cdots + x_3x_4, \\ s_2=x_0x_1x_2+x_0x_2x_3+\cdots + x_2x_3x_4,$

and similarly for the remaining coefficients. A universal radical formula in the coefficients would remain valid in this symbolic situation, so it is enough to show that no one of the symbols $\displaystyle x_0,\ldots,x_4$ can be reached from the symmetric functions by radicals.

The central difficulty is already visible here. The coefficients do not remember the names of the roots. Every rearrangement of $x_0,x_1,x_2,x_3,x_4$ leaves $s_1,\ldots,s_5$ unchanged. Thus the coefficients know only the unordered five-element collection of roots, whereas a formula for one root would have to distinguish one particular member of that collection. Abel’s proof studies how rational functions of the roots change when the roots are rearranged, and compares those possible changes with the much more restricted branch changes introduced by radicals.

We may allow all roots of unity as known constants from the outset. This only makes radical formulas more powerful, so proving impossibility under this assumption is stronger. We may also reduce to prime radical steps. Indeed, if an expression requires an $\displaystyle n$ th root, one may introduce it through successive prime-factor root extractions. Thus a genuinely new radical step has the form

$\displaystyle u^p=R,$

where $\displaystyle p$ is prime and $\displaystyle R$ is built from quantities already available. Let $\displaystyle \zeta$ be a primitive $\displaystyle p$ th root of unity. The possible branches of $\displaystyle u$ are

$\displaystyle u, \zeta u, \zeta^2u,\ldots,\zeta^{p-1}u.$

The basic identity $1+\zeta+\zeta^2+\cdots+\zeta^{p-1}=0$ has a more precise version:

$\displaystyle \sum_{\nu=0}^{p-1}\zeta^{\nu m}= \begin{cases} p,&p\mid m,\\ 0,&p\nmid m.\end{cases}$

This is merely the finite geometric-series formula, but it becomes a powerful extraction device. Suppose that a quantity depending on $\displaystyle u$ has the form

$\displaystyle Y=a_0+a_1u+a_2u^2+\cdots+a_{p-1}u^{p-1}.$

Changing the branch of $\displaystyle u$ gives the $\displaystyle p$ values

$\displaystyle Y_\nu=a_0+a_1\zeta^\nu u+a_2\zeta^{2\nu}u^2+\cdots+a_{p-1}\zeta^{(p-1)\nu}u^{p-1}.$

Multiply the $\displaystyle \nu$ th expression by $\displaystyle \zeta^{-r\nu}$ and add over all branches. Every power of $\displaystyle u$ disappears except the $\displaystyle r$ th one:

$\displaystyle \sum_{\nu=0}^{p-1}\zeta^{-r\nu}Y_\nu = p a_r u^r.$

Thus

$\displaystyle a_ru^r= \frac1p\sum_{\nu=0}^{p-1}\zeta^{-r\nu}Y_\nu.$

This is the calculation that drives the proof. It says that if one knows all branch-values of an expression, one can recover each separate radical contribution. In modern language this is finite Fourier inversion, but nothing beyond the elementary root-of-unity identity is being used.

Before using this calculation, one must justify that arbitrary rational expressions involving $\displaystyle u$ can be reduced to the displayed polynomial form. Let

$\displaystyle V=\frac{F(u)}{G(u)}.$

Multiply the denominator by all its branch-conjugates:

$\displaystyle N_G=G(u)G(\zeta u)G(\zeta^2u)\cdots G(\zeta^{p-1}u).$

Replacing $\displaystyle u$ by $\displaystyle \zeta u$ merely permutes these factors, so $\displaystyle N_G$ is unchanged by branch replacement. Therefore only powers divisible by $\displaystyle p$ can occur in it. Since $\displaystyle u^p=R$ , the quantity $\displaystyle N_G$ belongs to the earlier stock of expressions. After multiplying numerator and denominator of $\displaystyle V$ by the remaining branch-conjugates of $\displaystyle G$ , and reducing powers of $\displaystyle u$ modulo the relation $\displaystyle u^p=R$ , every rational expression involving $\displaystyle u$ takes the form

$\displaystyle a_0+a_1u+a_2u^2+\cdots+a_{p-1}u^{p-1},$

with coefficients built from the quantities known before $\displaystyle u$ was introduced.

There is one further point. If $\displaystyle u$ is a genuinely new radical, then the powers

$\displaystyle 1,u,u^2,\ldots,u^{p-1}$

cannot satisfy a nontrivial relation with earlier coefficients. Suppose instead that

$\displaystyle b_0+b_1u+b_2u^2+\cdots+b_{p-1}u^{p-1}=0.$

Then the polynomial $\displaystyle T^p-R$ and the polynomial on the left have the common root $\displaystyle u$ . Their common factor has degree strictly between $\displaystyle 1$ and $\displaystyle p-1$ . Its roots are some proper nonempty collection of

$\displaystyle u,\zeta u,\zeta^2u,\ldots,\zeta^{p-1}u.$

The product of those roots is $\displaystyle \zeta^m u^d$ for some $\displaystyle 1\le d<p$ . Since the factor has earlier coefficients, this product is an earlier quantity; hence $\displaystyle u^d$ is already known. Because $\displaystyle p$ is prime, $\displaystyle d$ and $\displaystyle p$ are coprime, so integers $\displaystyle r,s$ exist with $\displaystyle rd+sp=1$ . Consequently,

$\displaystyle u=u^{rd+sp}=(u^d)^r(u^p)^s.$

Both factors on the right are old, making $\displaystyle u$ old as well. This contradiction proves the no-cancellation principle: a relation of degree less than $\displaystyle p$ in a genuinely new prime radical must vanish coefficient by coefficient.

Assume, in search of a contradiction, that one root of the generic quintic is obtained by radicals. Choose the final genuine prime radical occurring in such a formula and call it $\displaystyle u$ . After the normalization just discussed, the root can be written

$\displaystyle y=a_0+u+a_2u^2+\cdots+a_{p-1}u^{p-1},$

where $\displaystyle u^p=R$ , the coefficient of $\displaystyle u$ has been normalized to $\displaystyle 1$ , and all the other quantities have lower radical complexity. If the first nonzero radical coefficient originally occurred at a different power $\displaystyle u^j$ , one replaces $\displaystyle u$ by a suitable power of it; since $\displaystyle p$ is prime, every nonzero exponent modulo $\displaystyle p$ is invertible, so this causes no loss.

Substitute this expression into the original equation $f(y)=0.$ After expanding and reducing every high power of $\displaystyle u$ using $u^p=R$ , we obtain a relation

$\displaystyle c_0+c_1u+c_2u^2+\cdots+c_{p-1}u^{p-1}=0,$

where every $\displaystyle c_j$ has lower radical complexity. The no-cancellation principle forces

$\displaystyle c_0=c_1=\cdots=c_{p-1}=0.$

This is much stronger than saying that one chosen branch of $\displaystyle u$ happens to give a root. It says that the equation holds identically in the radical, so every branch produces a root:

$\displaystyle y_\nu= a_0+\zeta^\nu u+a_2\zeta^{2\nu}u^2+\cdots+a_{p-1}\zeta^{(p-1)\nu}u^{p-1}.$

These values are distinct. Otherwise, subtracting two of them would produce a forbidden nontrivial relation in $\displaystyle u$ , and the coefficient of $\displaystyle u$ would be nonzero. Thus the $\displaystyle p$ branch-values are distinct roots among $x_0,x_1,x_2,x_3,x_4.$ The root-of-unity calculation now recovers the radical itself from these roots:

$\displaystyle u= \frac1p\big( y_0+\zeta^{-1}y_1+\zeta^{-2}y_2+\cdots+ \zeta^{-(p-1)}y_{p-1}\big).$

Likewise, each quantity $a_ju^j$ is a rational function of the roots. Since $u$ is now a rational function of the roots, so are the coefficients $a_j$ and the radicand $R=u^p$ . Repeating the argument from the outside of the nested expression toward the inside shows that every genuinely necessary radical may be treated as a rational function of $x_0,x_1,x_2,x_3,x_4.$ This is the indispensable repair to the old Ruffini strategy. It is not enough simply to begin by assuming that radicals are rational functions of the roots. Abel’s branch calculation proves that one may do so.

We now need a result about rational functions of five independent root-symbols. Let $U=U(x_0,x_1,x_2,x_3,x_4).$ Suppose that, as the five roots are rearranged in every possible way, $\displaystyle U$ takes fewer than five values. Then it takes either one value or two values.

To see this, apply a five-cycle to the roots. Repeating that substitution gives an orbit of values whose size divides $\displaystyle 5$ ; hence its size is either one or five. But $\displaystyle U$ has fewer than five values in total, so every five-cycle must fix $\displaystyle U$ . Every three-cycle is a product of two five-cycles. For example, with the right-hand substitution performed first,

$\displaystyle (1,2,3)=(1,3,4,5,2)(1,5,4,3,2).$

Thus every three-cycle fixes $\displaystyle U$ . Every even rearrangement can be built from three-cycles, so every even rearrangement fixes $\displaystyle U$ . There are only two kinds of rearrangements left: even and odd. Hence $\displaystyle U$ can have at most one even value and one odd value.

The two-valued case is especially concrete. Define the alternating product

$\displaystyle D=\prod_{0\le i<j\le4}(x_i-x_j).$

Every even rearrangement leaves $\displaystyle D$ fixed, and every odd rearrangement changes its sign. Therefore $D^2$ is symmetric, hence rational in the coefficients; it is the discriminant of the generic quintic. Suppose that $\displaystyle U$ has the value $U_+$ under even rearrangements and $U_-$ under odd rearrangements. Then

$\displaystyle P=\frac{U_++U_-}{2}$

is symmetric, while

$\displaystyle Q=\frac{U_+-U_-}{2D}$

is also symmetric, because numerator and denominator both change sign under an odd rearrangement. Therefore every two-valued rational function has the form

$\displaystyle U=P+QD,$

where $\displaystyle P$ and $\displaystyle Q$ are rational functions of the coefficients.

Take the first genuine prime radical in the descended radical expression $r^p=R.$ Because it is the first radical, the radicand $\displaystyle R$ is a rational function of the coefficients and hence is symmetric in the roots. Since $\displaystyle r$ is now a rational function of the roots, every rearrangement of the roots sends it to another solution of $T^p-R=0.$ Thus a rearrangement sends $\displaystyle r$ to some branch $\displaystyle \zeta^\mu r$ .

Here there is a short decisive observation. Let $\displaystyle \tau$ be a transposition, so that applying $\displaystyle \tau$ twice does nothing. If $\tau(r)=\zeta^\mu r,$ then applying $\tau$ again gives

$\displaystyle r=\zeta^{2\mu}r.$

When $\displaystyle p$ is odd, this forces $\displaystyle \mu=0$ . Thus every transposition fixes $\displaystyle r$ . Since transpositions generate all rearrangements, $\displaystyle r$ would be symmetric, hence already rational in the coefficients. That contradicts the assumption that it was a genuinely new radical.

Therefore the first genuine radical must be quadratic. $r^2=R.$ Its possible values are $\displaystyle r$ and $\displaystyle -r$ . Since it is genuinely new, both values occur. The small-value result now applies: every even rearrangement fixes $\displaystyle r$ , and every odd rearrangement sends it to $\displaystyle -r$ . In the formula $r=P+QD,$ the average of the two values is zero, so $\displaystyle P=0$ . Hence $r=QD,$ with $Q$ symmetric. Thus adjoining the first genuine radical is, up to multiplication by a symmetric quantity, exactly the same as adjoining the alternating product $\displaystyle D$ , or equivalently the square root of the discriminant.

This is the first real obstruction. The first radical can distinguish even from odd rearrangements, but it cannot distinguish one root from another.

Once $D$ has been adjoined, every rational expression built from the coefficients and $\displaystyle D$ has the form $late A+BD, &fg=000000$ where $A$ and $B$ are symmetric. This is clear for addition and multiplication because $D^2$ is symmetric. It remains true for division, since

$\displaystyle \frac1{A+BD}= \frac{A-BD}{A^2-B^2D^2},$

and the denominator on the right is symmetric.

Suppose, toward a contradiction, that a further genuine prime radical appears: $w^q=A+BD,$ where $q$ is prime. Every even rearrangement leaves $D$ fixed and therefore leaves the radicand fixed. Hence an even rearrangement must send $w$ to one of the branches

$\displaystyle w,\eta w,\eta^2w,\ldots,\eta^{q-1}w,$

where $\displaystyle \eta$ is a primitive $q$ th root of unity.

First suppose that $q\ne3$ . Take any three-cycle. Applying it three times gives the identity. If it sends $w$ to $\eta^\mu w$ , applying it three times gives

$\displaystyle w=\eta^{3\mu}w.$

Since $q$ is prime and does not divide $3$ , this forces $\mu=0$ . Thus every three-cycle fixes $w$ , and therefore every even rearrangement fixes $w$ .

The only exceptional prime is $\displaystyle q=3$ . In this case use five-cycles instead. A five-cycle applied five times is the identity, so if it sends $\displaystyle w$ to $\eta^\mu w$ , then

$\displaystyle w=\eta^{5\mu}w.$

Because $\displaystyle 3$ does not divide $\displaystyle 5$ , we again obtain $\mu=0$ . Hence every five-cycle fixes $\displaystyle w$ . Since every three-cycle is a product of two five-cycles, every even rearrangement fixes $\displaystyle w$ in this case too.

Thus, for every prime $q$ , the supposed radical $w$ is fixed by all even rearrangements. It has at most two values under all rearrangements. By the two-valued formula,

$\displaystyle w=C+ED,$

with $C$ and $E$ symmetric. But this says that $w$ was already expressible from the coefficients and $\displaystyle D$ . It was not a new radical at all. This contradiction proves that no second genuine radical can occur.

We have proved that every expression obtainable from the coefficients of the generic quintic by radicals must already have the form $A+BD,$ with $A$ and $B$ symmetric. Every such expression is fixed by every even rearrangement of the roots.

But an individual root is not fixed by even rearrangements. For example, the three-cycle $(x_0,x_1,x_2)$ is even, yet it sends

$\displaystyle x_0\longmapsto x_1.$

The root-symbols are independent and distinct, so $x_0\ne x_1$ . Therefore $x_0$ cannot be an expression of the form $\displaystyle A+BD$ . The same applies to every individual root.

Hence no universal radical formula can recover the roots of the general quintic. The general equation of degree five is not solvable by radicals.

The force of the proof is not that the number five is somehow intrinsically too large. The decisive issue is the mismatch between two kinds of ambiguity. A radical changes branch by multiplication by roots of unity, so each radical step carries a cyclic ambiguity. The five roots of the generic quintic, however, can be rearranged in ways that survive even after the sign of a permutation has been detected. The discriminant square root records precisely that sign, separating even from odd reorderings. But once that square root has been adjoined, the remaining even rearrangements still move the individual roots, and no further prime radical can respond to those motions in a genuinely new way.

The root-of-unity averaging formula is the bridge that makes this argument work. It turns the several branch-values of a radical expression into explicit rational combinations of roots. That is why Abel can begin with a hypothetical nested radical formula and eventually study only rational functions of the five roots. The proof is therefore an analysis of substitutions and branch changes, carried out without the later vocabulary of Galois groups, but already containing the essential idea that made Galois theory possible.

The preceding proof was deliberately written in Abel’s language: roots regarded as independent quantities, rational expressions in those roots, the different values produced when the roots are rearranged, and the branch changes obtained by multiplying a radical by roots of unity. Modern algebra organizes the same ideas much more compactly. It does not replace Abel’s argument so much as reveal its underlying architecture.

Let $K=\mathbf{C}(s_1,s_2,s_3,s_4,s_5)$ be the field of rational functions in the coefficients of the generic quintic, and let $L=\mathbf{C}(x_0,x_1,x_2,x_3,x_4)$ be the field generated by all five roots. The coefficients are symmetric functions of the roots, so they lie in $K$ . Conversely, the fundamental theorem on symmetric rational functions says that every rational function of the roots unchanged by every rearrangement belongs to $K$ . Thus $K$ is exactly the field of completely symmetric rational functions in the roots. Every permutation of the five roots defines an automorphism of $L$ which fixes $K$ . Conversely, every automorphism fixing the coefficients must merely permute the roots, since it sends roots of the generic polynomial to roots of the same polynomial. Therefore the full symmetry group of the generic quintic is $\text{Gal}(L/K)=S_5.$ This is the modern meaning of Abel’s repeated study of “the values of a function when the roots are rearranged.” If $U$ is a rational function of the roots, its different values under permutations form its orbit under $S_5$ . The number of values of $U$ is the number of distinct images of $U$ under this group of automorphisms. A function with one value is fixed by all of $S_5$ , hence belongs to $K$ . A function with two values is fixed by all even permutations and changes only under odd ones; this is exactly the discriminant situation. The alternating product $D=\prod_{0\le i<j\le4}(x_i-x_j)$ changes sign under odd permutations and remains fixed under even permutations. Its square is symmetric: Thus adjoining $D$ gives the quadratic extension $K(D)/K.$ In group language, this corresponds to the sign map $S_5\longrightarrow{\pm1},$ whose kernel is the alternating group $A_5.$ This is the precise modern formulation of the earlier conclusion that the first possible radical can only distinguish even from odd substitutions. The discriminant square root does not identify an individual root. It only reduces the symmetry from all permutations to even permutations.

A radical formula gives a tower of extensions $K=K_0\subset K_1\subset\cdots\subset K_r,$ in which each new field is obtained by adjoining one radical: $K_{j+1}=K_j(u_j),\quad u_j^{p_j}\in K_j,$ where $p_j$ may be taken prime. After adjoining the appropriate roots of unity, changing the branch of $u_j$ amounts to multiplication by a power of a primitive $p_j$ th root of unity. The new ambiguity introduced at one stage is therefore cyclic and abelian. This is the modern content of Abel’s branch calculations with $u,\zeta u,\zeta^2u,\ldots,\zeta^{p-1}u.$ The root-of-unity averaging identity is the concrete ancestor of a standard fact about such extensions. When $u^p=R,$ the powers $1,u,u^2,\ldots,u^{p-1}$ form a basis over the preceding field, provided the radical is genuinely new. The weighted branch average isolates the coefficient of each basis element. In modern language, Abel’s calculation is the decomposition of an element into its eigenspaces for the cyclic branch action. The no-cancellation lemma says precisely that these basis elements are linearly independent over the earlier field.

The descent in the proof has an especially simple modern meaning. Abel showed that, if a radical formula for a root existed, then the essential radicals could be interpreted as rational functions of the roots themselves. Modern algebra phrases this by saying that one may compare the radical tower with the splitting field $L$ . The relevant question is not merely whether radicals can be introduced somewhere in a larger field, but whether the splitting field of the polynomial can be built by a sequence of extensions whose successive symmetry groups are cyclic and abelian. This leads to the modern criterion. A polynomial is solvable by radicals exactly when its Galois group is a solvable group. A finite group $G$ is called solvable when it can be reduced to the trivial group through successive normal subgroups whose quotients are abelian. Equivalently, define the successive commutator groups $G^{(0)}=G,\quad G^{(j+1)}=[G^{(j)},G^{(j)}].$ The group is solvable when this descending sequence eventually reaches the identity. The reason this matches radicals is that each radical extraction contributes only a cyclic, hence abelian, layer of symmetry. A tower made from radicals can dismantle only a group that can itself be dismantled through abelian quotients. For the generic quintic, the group is $S_5$ . Its first nontrivial commutator subgroup is $[S_5,S_5]=A_5.$ The group $A_5$ is simple and nonabelian: it has no proper nontrivial normal subgroup, and it is not commutative. In particular, it has no nontrivial abelian quotient. Once the discriminant has reduced the symmetry from $S_5$ to $A_5$ , there is no further abelian layer to peel away. This is the modern form of the earlier argument that, after the discriminant square root has been adjoined, no further genuine radical can separate the remaining even substitutions. Thus the final obstruction can be written in one line: $S_5\supset A_5\supset A_5\supset A_5\supset\cdots.$ The first quotient $S_5/A_5$ has order $2$ and corresponds to the discriminant square root. But the remaining group $A_5$ does not admit any nontrivial abelian quotient. A radical tower would require further cyclic quotients, and there are none. Therefore the generic quintic cannot be solved by radicals.

This modern conclusion is shorter than Abel’s original reasoning, but it should not obscure what Abel had already discovered. His roots-of-unity averages anticipate cyclic extensions; his functions with several values anticipate group orbits; his alternating product anticipates the discriminant and the subgroup $A_5$ ; and his argument that no further radical can genuinely appear anticipates the non-solvability of $A_5$ . Galois theory did not replace the substance of Abel’s insight. It supplied a language in which the structure Abel had uncovered could be stated all at once.

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