# Abel–Ruffini theorem

In algebra, the Abel–Ruffini theorem (also known as Abel's impossibility theorem) states that there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients. The theorem is named after Paolo Ruffini, who made an incomplete proof in 1799, and Niels Henrik Abel, who provided a proof in 1824.

## Interpretation

The theorem does not assert that some higher-degree polynomial equations have no solution. In fact, the opposite is true: every non-constant polynomial equation in one unknown, with real or complex coefficients, has at least one complex number as a solution (and thus, by polynomial division, as many complex roots as its degree, counting repeated roots); this is the fundamental theorem of algebra. These solutions can be computed to any desired degree of accuracy using numerical methods such as the Newton–Raphson method or the Laguerre method, and in this way they are no different from solutions to polynomial equations of the second, third, or fourth degrees. It also does not assert that no higher-degree polynomial equations can be solved in radicals: the equation $x^{n}-1=0$  can be solved in radicals for every positive integer $n$ , for example. The theorem only shows that there is no general solution in radicals that applies to all equations of a given degree greater than 4.

The solution of any second-degree polynomial equation can be expressed in terms of its coefficients, using only addition, subtraction, multiplication, division, and square roots, in the familiar quadratic formula: the roots of the polynomial $ax^{2}+bx+c$  (with $a\neq 0$ ) are

${\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.$

Analogous formulas for third-degree equations and fourth-degree equations (using square roots and cube roots) have been known since the 16th century. What the Abel–Ruffini theorem says is that there is no similar formula for general equations of fifth degree or higher. In principle, it could be that the equations of the fifth degree could be split in several types and, for each one of these types, there could be some algebraic solution valid within that type. Or, as Ian Stewart wrote, "for all that Abel's methods could prove, every particular quintic equation might be soluble, with a special formula for each equation." However, this is not so, but this impossibility is a strictly stronger result than the Abel–Ruffini theorem and is derived with Galois theory.

## Proof

The following proof is based on Galois theory and it is valid for any field of characteristic 0. Historically, Ruffini and Abel's proofs precede Galois theory. For a modern presentation of Abel's proof see the article of Rosen or the books of Tignol or Pesic.

One of the fundamental theorems of Galois theory states that a polynomial $P(x)\in F\left[x\right]$  is solvable by radicals over $F$  if and only if its splitting field $K$  over $F$  has a solvable Galois group, so the proof of the Abel–Ruffini theorem comes down to computing the Galois group of the general polynomial of the fifth degree, and showing that it is not solvable.

Consider five indeterminates $y_{1},y_{2},y_{3},y_{4}$ , and $y_{5}$ , let $E=\mathbf {Q} \left(y_{1},y_{2},y_{3},y_{4},y_{5}\right)$ , and let

$P(x)=\left(x-y_{1}\right)\left(x-y_{2}\right)\left(x-y_{3}\right)\left(x-y_{4}\right)\left(x-y_{5}\right)\in E[x]$ .

Expanding $P(x)$  out yields the elementary symmetric functions of the $y_{i}$ :

$s_{1}=y_{1}+y_{2}+y_{3}+y_{4}+y_{5}$ ,
$s_{2}=y_{1}y_{2}+y_{1}y_{3}+y_{1}y_{4}+y_{1}y_{5}+y_{2}y_{3}+y_{2}y_{4}+y_{2}y_{5}+y_{3}y_{4}+y_{3}y_{5}+y_{4}y_{5}$ ,
$s_{3}=y_{1}y_{2}y_{3}+y_{1}y_{2}y_{4}+y_{1}y_{2}y_{5}+y_{1}y_{3}y_{4}+y_{1}y_{3}y_{5}+y_{1}y_{4}y_{5}+y_{2}y_{3}y_{4}+y_{2}y_{3}y_{5}+y_{2}y_{4}y_{5}+y_{3}y_{4}y_{5}$ ,
$s_{4}=y_{1}y_{2}y_{3}y_{4}+y_{1}y_{2}y_{3}y_{5}+y_{1}y_{2}y_{4}y_{5}+y_{1}y_{3}y_{4}y_{5}+y_{2}y_{3}y_{4}y_{5}$ ,
$s_{5}=y_{1}y_{2}y_{3}y_{4}y_{5}$ .

The coefficient of $x^{n}$  in $P(x)$  is thus $(-1)^{5-n}s_{5-n}$ . Let $F=\mathbf {Q} (s_{1},s_{2},s_{3},s_{4},s_{5})$  be the field obtained by adjoining the symmetric functions to the rationals. Then $P(x)\in F\left[x\right]$ . Because the $y_{i}$ 's are indeterminates, every permutation $\sigma$  in the symmetric group on 5 letters $S_{5}$  induces a distinct automorphism $\sigma '$  on $E$  that leaves $\mathbf {Q}$  fixed and permutes the elements $y_{i}$ . Since an arbitrary rearrangement of the roots of the product form still produces the same polynomial, for example,

$\left(x-y_{3}\right)\left(x-y_{1}\right)\left(x-y_{2}\right)\left(x-y_{5}\right)\left(x-y_{4}\right)$

is the same polynomial as

$\left(x-y_{1}\right)\left(x-y_{2}\right)\left(x-y_{3}\right)\left(x-y_{4}\right)\left(x-y_{5}\right)$ ,

the automorphisms $\sigma '$  also leave $F$  fixed, so they are elements of the Galois group $\operatorname {Gal} (E/F)$ . Therefore, we have shown that $S_{5}\subseteq \operatorname {Gal} (E/F)$ ; however there could possibly be automorphisms there that are not in $S_{5}$ . But, since the Galois group of the splitting field of a quintic polynomial has at most $5!$  elements, and since $E$  is a splitting field of $P(x)$ , it follows that $\operatorname {Gal} (E/F)$  is isomorphic to $S_{5}$ . Generalizing this argument shows that the Galois group of every general polynomial of degree $n$  is isomorphic to $S_{n}$ .

The only composition series of $S_{5}$  is $S_{5}\geq A_{5}\geq \lbrace e\rbrace$  (where $A_{5}$  is the alternating group on five letters, also known as the icosahedral group). However, the quotient group $A_{5}/\lbrace e\rbrace$  (isomorphic to $A_{5}$  itself) is not abelian, and so $S_{5}$  is not solvable, so it must be that the general polynomial of the fifth degree has no solution in radicals. Since the first nontrivial normal subgroup of the symmetric group on $n$  letters is always the alternating group on $n$  letters, and since the alternating groups on $n$  letters for $n\geq 5$  are always simple and non-abelian, and hence not solvable, it also says that the general polynomials of all degrees higher than the fifth also have no solution in radicals. Q.E.D.

The above construction of the Galois group for a fifth degree polynomial only applies to the general polynomial; specific polynomials of the fifth degree may have different Galois groups with quite different properties, for example, $x^{5}-1$  has a splitting field generated by a primitive 5th root of unity, and hence its Galois group is abelian and the equation itself solvable by radicals; moreover, the argument does not provide any rational-valued quintic that has $S_{5}$  or $A_{5}$  as its Galois group. However, since the result is on the general polynomial, it does say that a general "quintic formula" for the roots of a quintic using only a finite combination of the arithmetic operations and radicals in terms of the coefficients is impossible.

The proof is not valid if applied to polynomials whose degree is less than 5. Indeed:

• the group $A_{4}$  is not simple, because the subgroup $\{e,(12)(34),(13)(24),(14)(23)\}$ , isomorphic to the Klein four-group, is a normal subgroup;
• the groups $A_{2}$  and $A_{3}$  are simple, but since they are abelian too ($A_{2}$  is the trivial group and $A_{3}$  is the cyclic group of order 3), that is not a problem.

The proof remains valid if, instead of working with five indeterminates, one works with five concrete algebraically independent complex numbers, because, by the same argument, $\operatorname {Gal} (E/F)=S_{5}$ .

## History

Around 1770, Joseph Louis Lagrange began the groundwork that unified the many different tricks that had been used up to that point to solve equations, relating them to the theory of groups of permutations, in the form of Lagrange resolvents. This innovative work by Lagrange was a precursor to Galois theory, and its failure to develop solutions for equations of fifth and higher degrees hinted that such solutions might be impossible, but it did not provide conclusive proof. The first person who conjectured that the problem of solving quintics by radicals might be impossible to solve was Carl Friedrich Gauss, who wrote in 1798 in section 359 of his book Disquisitiones Arithmeticae (which would be published only in 1801) that "there is little doubt that this problem does not so much defy modern methods of analysis as that it proposes the impossible". The next year, in his thesis, he wrote "After the labors of many geometers left little hope of ever arriving at the resolution of the general equation algebraically, it appears more and more likely that this resolution is impossible and contradictory." And he added "Perhaps it will not be so difficult to prove, with all rigor, the impossibility for the fifth degree. I shall set forth my investigations of this at greater length in another place." Actually, Gauss published nothing else on this subject.

The theorem was first nearly proved by Paolo Ruffini in 1799. He sent his proof to several mathematicians to get it acknowledged, amongst them Lagrange (who did not reply) and Augustin-Louis Cauchy, who sent him a letter saying: "Your memoir on the general solution of equations is a work which I have always believed should be kept in mind by mathematicians and which, in my opinion, proves conclusively the algebraic unsolvability of general equations of higher than fourth degree." However, in general, Ruffini's proof was not considered convincing. Abel wrote: "The first and, if I am not mistaken, the only one who, before me, has sought to prove the impossibility of the algebraic solution of general equations is the mathematician Ruffini. But his memoir is so complicated that it is very difficult to determine the validity of his argument. It seems to me that his argument is not completely satisfying."

The proof also, as it was discovered later, was incomplete. Ruffini assumed that all radicals that he was dealing with could be expressed from the roots of the polynomial using field operations alone; in modern terms, he assumed that the radicals belonged to the splitting field of the polynomial. To see why this is really an extra assumption, consider, for instance, the polynomial $P(x)=x^{3}-15x-20$ . According to Cardano's formula, one of its roots (all of them, actually) can be expressed as the sum of a cube root of $10+5i$  with a cube root of $10-5i$ . On the other hand, since $P(-3)<0$ , $P(-2)>0$ , $P(-1)<0$ , and $P(5)>0$ , the roots $r_{1}$ , $r_{2}$ , and $r_{3}$  of $P(x)$  are all real and therefore the field $\mathbf {Q} (r_{1},r_{2},r_{3})$  is a subfield of $\mathbf {R}$ . But then the numbers $10\pm 5i$  cannot belong to $\mathbf {Q} (r_{1},r_{2},r_{3})$ . While Cauchy either did not notice Ruffini's assumption or felt that it was a minor one, most historians believe that the proof was not complete until Abel proved the theorem on natural irrationalities, which asserts that the assumption holds in the case of general polynomials. The Abel–Ruffini theorem is thus generally credited to Abel, who published a proof in just six pages in 1824. However, this short number of pages was obtained at the cost of writing in a very terse style. This was due to the fact that he had the proof printed at his own expenses and he needed to save paper and money. A more elaborated version of the proof would be published in 1826.

Proving that the general quintic (and higher) equations were unsolvable by radicals did not completely settle the matter, because the Abel–Ruffini theorem does not provide necessary and sufficient conditions for saying precisely which quintic (and higher) equations are unsolvable by radicals. Abel was working on a complete characterization when he died in 1829.