- Abel–Ruffini theorem
The

**Abel–Ruffini theorem**(also known as**Abel's impossibility theorem**) states that there is no general solution in radicals topolynomial equations of degree five or higher.**Misinterpretation**The content of this theorem is frequently misunderstood. It does "not" assert that higher-degree polynomial equations are unsolvable. In fact, the opposite is true: "every" polynomial with only real or complex coefficients can be solved with a complex number; this is the

fundamental theorem of algebra . Although these solutions cannot always be expressed exactly with radicals, they can be computed to any desired degree of accuracy using numerical methods such as theNewton-Raphson method orLaguerre 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" higher-degree equations have solutions which can be expressed by performing a finite number of operations of addition, subtraction, multiplication, division and root extraction on the equation's coefficients. Some polynomials of arbitrary degree are indeed solvable with a finite number of such operations. The simplest nontrivial example is the monomial equation $ax^n\; =\; b$, whose solution is $sqrt\; [n]\; \{b\; over\; a\}.$

**Lower-Degree Polynomials**The solutions of any second-degree polynomial equation can be expressed in terms of addition, subtraction, multiplication, division, and

square root s, using the familiarquadratic formula : The roots of the following equation are shown below: $extstyle\{ax^2\; +\; bx\; +\; c\; =\; 0\}$:$x=frac\{-b\; pm\; sqrt\; \{b^2-4ac\; \{2a\}.$

Analogous formulas for third- and fourth-degree equations, using

cube root s 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. (SeeBring radical .) Some other fifth degree equations "can" be solved by radicals, for example x^{5}- x^{4}- x + 1 = 0, which factorizes to (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 byÉvariste Galois and is now part ofGalois theory : a polynomial equation can be solved by radicals if and only if itsGalois group is asolvable 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 group s S_{"2"}, S_{"3"}and S_{"4"}are solvable groups, whereas S_{"n"}is not solvable for "n"≥5.**Proof**The following proof is based on

Galois theory . One of the fundamental theorems of Galois theory states that an equation is solvable in radicalsif and only if it has a solvableGalois group , 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 ofrational number s $Q$, 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:$f(x)\; =\; (x\; -\; y\_1)(x\; -\; y\_2)(x\; -\; y\_3)(x\; -\; y\_4)(x\; -\; y\_5)\; in\; E\; [x]\; .$

Multiplying $f(x)$ out yields the elementary

symmetric function s of the $y\_n$::$s\_1\; =\; y\_1\; +\; y\_2\; +\; y\_3\; +\; y\_4\; +\; y\_5$:$s\_2\; =\; y\_1y\_2\; +\; y\_1y\_3\; +\; cdots\; +\; y\_4y\_5$and so on up to :$s\_5\; =\; y\_1y\_2y\_3y\_4y\_5$The coefficient of $x^n$ in $f(x)$ is thus $s\_\{5-n\}$. Because our independent transcendentals $y\_n$ act as indeterminates over $Q$, every permutation $sigma$ in the symmetric group on 5 letters $S\_5$ induces an

automorphism $sigma\text{'}$ 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 $sigma\text{'}$ also leave $E$ fixed, so they are elements of the Galois group $G(E/F)$. Now, since $|S\_5|\; =\; 5!$ it must be that $|G(E/F)|\; ge\; 5!$, as there could possibly be automorphisms there that are not in $S\_5$. However, since the

splitting field of a quintic polynomial has at most 5! elements, $|G(E/F)|\; =\; 5!$, and so $G(E/F)$ must be 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 thealternating group on five letters, also known as theicosahedral group ). However, thequotient group $A\_5/\{e\}$ (isomorphic to $A\_5$ itself) is not anabelian 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 nontrivialnormal 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. 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.

**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. 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 byPaolo Ruffini in 1799, but his proof was mostly ignored and contained a gap. Ruffini assumed that a solution would necessarily be a function of the roots (in modern terms, he failed to prove that the splitting field is contained in the tower of radicals which corresponds to a solution expressed in radicals). While Cauchy felt that the assumption was minor, most historians believe that the proof was not complete until Abel proved this assumption. Ruffini's many proofs were quite innovative in usingpermutation group s. The theorem is generally credited toNiels Henrik Abel , who published a proof in 1824.Insights into these issues were also gained using Galois theory pioneered by Évariste Galois. In 1885,

John Stuart Glashan ,George Paxton Young , andCarl Runge provided a proof using this theory.**See also***

Theory of equations **References*** Edgar Dehn. "Algebraic Equations: An Introduction to the Theories of Lagrange and Galois". Columbia University Press, 1930. ISBN 0-486-43900-3.

* John B. Fraleigh. "A First Course in Abstract Algebra." Fifth Edition. Addison-Wesley, 1994. ISBN 0-201-59291-6.

* Ian Stewart. "Galois Theory". Chapman and Hall, 1973. ISBN 0-412-10800-3.

* [*http://www.everything2.net/e2node/Abel%2527s%2520Impossibility%2520Theorem Abel's Impossibility Theorem at Everything2*]

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