# Abel–Ruffini theorem

(Redirected from Abel-Ruffini theorem)
Not to be confused with Abel's theorem.

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

A general solution to any quadratic equation can be given using the quadratic formula above. Similar formulas exist for polynomial equations of degree 3 and 4. But no such formula is possible for 5th degree polynomials; the real solution -1.1673... to the 5th degree equation below cannot be written using basic arithmetic operations and nth roots:

$x^5 - x + 1 = 0$

In algebra, the Abel–Ruffini theorem (also known as Abel's impossibility theorem) states that there is no general algebraic solution—that is, solution in radicals— to polynomial equations of degree five or higher.[1] The theorem is named after Paolo Ruffini, who made an incomplete proof in 1799, and Niels Henrik Abel, who provided a proof in 1823. Évariste Galois independently proved the theorem in a work that was posthumously published in 1846.[2]

## Interpretation

The theorem does not assert that 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; this is the fundamental theorem of algebra. Although the solutions cannot always be expressed exactly with radicals, they can be computed to any desired degree of accuracy using numerical methods such as the Newton–Raphson method or Laguerre method, and in this way they are no different from solutions to polynomial equations of the second, third, or fourth degrees.

The theorem only concerns the form that such a solution must take. The theorem says that not all solutions of higher-degree equations can be obtained by starting with the equation's coefficients and rational constants, and repeatedly forming sums, differences, products, quotients, and radicals (n-th roots, for some integer n) of previously obtained numbers. This clearly excludes the possibility of having any formula that expresses the solutions of an arbitrary equation of degree 5 or higher in terms of its coefficients, using only those operations, or even of having different formulas for different roots or for different classes of polynomials, in such a way as to cover all cases. (In principle one could imagine formulas using irrational numbers as constants, but even if a finite number of those were admitted at the start, not all roots of higher-degree equations could be obtained.) However some polynomial equations, of arbitrarily high degree, are solvable with such operations. Indeed, if the roots happen to be rational numbers, they can be expressed trivially as constants. The simplest nontrivial example is the equation xn = a, where a is a positive real number, which has n solutions, given by:

$x = \sqrt[n]{a}\cdot e^{i 2\pi k/n},\quad k=0,1,\dots,n-1.\$

Here the expression $e^{i 2\pi k/n}$, which appears to involve the use of the exponential function, in fact just gives the different possible values of $\sqrt[n]1$ (the n-th roots of unity), so it involves only extraction of radicals.

## Lower-degree polynomials

The solutions of any second-degree polynomial equation can be expressed in terms of addition, subtraction, multiplication, division, and square roots, using the familiar quadratic formula: The roots of the following equation are shown below:

$\textstyle{ax^2 + bx + c = 0, a \ne 0}$
$x=\frac{-b \pm \sqrt {b^2-4ac\ }}{2a}.$

Analogous formulas for third- and fourth-degree equations, using cube roots and fourth roots, had been known since the 16th century.

## Quintics and higher

The Abel–Ruffini theorem says that there are some fifth-degree equations whose solution cannot be so expressed. The equation $x^5 - x + 1 = 0$ is an example. (See Bring radical.) Some other fifth degree equations can be solved by radicals, for example $x^5- x^4 - x + 1 = 0$, which factors into $(x-1)(x-1)(x+1)(x+i)(x-i) = 0$. The precise criterion that distinguishes between those equations that can be solved by radicals and those that cannot was given by Évariste Galois and is now part of Galois theory: a polynomial equation can be solved by radicals if and only if its Galois group (over the rational numbers, or more generally over the base field of admitted constants) is a solvable group.

Today, in the modern algebraic context, we say that second, third and fourth degree polynomial equations can always be solved by radicals because the symmetric groups S2, S3 and S4 are solvable groups, whereas Sn is not solvable for n ≥ 5. This is so because for a polynomial of degree n with indeterminate coefficients (i.e., given by symbolic parameters), the Galois group is the full symmetric group Sn (this is what is called the "general equation of the n-th degree"). This remains true if the coefficients are concrete but algebraically independent values over the base field.

## Proof

The following proof is based on Galois theory (for a short explanation of Abel's proof that does not rely on prior knowledge in group theory see [3]). Historically, Ruffini and Abel's proofs precede Galois theory.

One of the fundamental theorems of Galois theory states that a polynomial f(x) ∈ F[x] is solvable by radicals over F if and only if its splitting field K over F has a solvable Galois group,[4] so the proof of the Abel–Ruffini theorem comes down to computing the Galois group of the general polynomial of the fifth degree.

Let $y_1$ be a real number transcendental over the field of rational numbers $Q$, and let $y_2$ be a real number transcendental over $Q(y_1)$, and so on to $y_5$ which is transcendental over $Q(y_1, y_2, y_3, y_4)$. These numbers are called independent transcendental elements over Q. Let $E = Q(y_1, y_2, y_3, y_4, y_5)$ and let

$f(x) = (x - y_1)(x - y_2)(x - y_3)(x - y_4)(x - y_5) \in E[x].$

Expanding $f(x)$ out yields the elementary symmetric functions of the $y_n$:

$s_1 = y_1 + y_2 + y_3 + y_4 + y_5$
$s_2 = y_1y_2 + y_1y_3 + y_1y_4 + y_1y_5 + y_2y_3 + y_2y_4 + y_2y_5 + y_3y_4 + y_3y_5 + y_4y_5$
$s_3 = y_1y_2y_3 + y_1y_2y_4 + y_1y_2y_5 + y_1y_3y_4 + y_1y_3y_5 + y_1y_4y_5 +y_2y_3y_4 + y_2y_3y_5 + y_2y_4y_5 + y_3y_4y_5$
$s_4 = y_1y_2y_3y_4 + y_1y_2y_3y_5 + y_1y_2y_4y_5 + y_1y_3y_4y_5 + y_2y_3y_4y_5$
$s_5 = y_1y_2y_3y_4y_5.$

The coefficient of $x^n$ in $f(x)$ is thus $(-1)^{5-n} s_{5-n}$. Let $F = Q(s_i)$ be the field obtained by adjoining the symmetric functions to the rationals (the $s_i$ are all transcendental, because the $y_i$ are independent). Because our independent transcendentals $y_n$ act as indeterminates over $Q$, every permutation $\sigma$ in the symmetric group on 5 letters $S_5$ induces a distinct automorphism $\sigma'$ on $E$ that leaves $Q$ fixed and permutes the elements $y_n$. Since an arbitrary rearrangement of the roots of the product form still produces the same polynomial, e.g.:

$(y - y_3)(y - y_1)(y - y_2)(y - y_5)(y - y_4)$

is still the same polynomial as

$(y - y_1)(y - y_2)(y - y_3)(y - y_4)(y - y_5)$

the automorphisms $\sigma'$ also leave $f$ fixed, so they are elements of the Galois group $G(E/F)$. So we have shown that $S_5 \subseteq G(E/F)$; however there could possibly be automorphisms there that are not in $S_5$. However, since the relative automorphism group for the splitting field of a quintic polynomial has at most 5! elements, it follows that $G(E/F)$ is isomorphic to $S_5$. Generalizing this argument shows that the Galois group of every general polynomial of degree $n$ is isomorphic to $S_n$.

And what of $S_5$? The only composition series of $S_5$ is $S_5 \ge A_5 \ge \{e\}$ (where $A_5$ is the alternating group on five letters, also known as the icosahedral group). However, the quotient group $A_5/\{e\}$ (isomorphic to $A_5$ itself) is not an abelian group, 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 $n \ge 5$ 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.

Note that 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, e.g. $x^5 - 1$ 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. Q.E.D.

## 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 theorem, however, was first nearly proved by Paolo Ruffini in 1799, but his proof was mostly ignored. He had several times tried to send it to different mathematicians to get it acknowledged, amongst them, French mathematician Augustin-Louis Cauchy, but it was never acknowledged, possibly because the proof was spanning 500 pages. The proof also, as was discovered later, contained an error. In modern terms, Ruffini failed to prove that the splitting field is one of the fields in the tower of radicals which corresponds to the hypothesized solution by radicals; this assumption fails, for example, for Cardano's solution of the cubic; it splits not only the original cubic but also the two others with the same discriminant. While Cauchy felt that the assumption was minor, most historians believe that the proof was not complete until Abel proved this assumption. The theorem is thus generally credited to Niels Henrik Abel, who published a proof that required just six pages in 1824.[5]

Insights into these issues were also gained using Galois theory pioneered by Évariste Galois. In 1885, John Stuart Glashan, George Paxton Young, and Carl Runge provided a proof using this theory.

In 1963, Vladimir Arnold discovered a topological proof of the Abel–Ruffini theorem,[6] which served as a starting point for topological Galois theory.[7]