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I know that there is a precise solution of the equations:

\begin{align} x + a & = 0 \\ x^2 + a\cdot x + b & = 0 \\ x^3 + a\cdot x^2 + b\cdot x + c & = 0 \\ x^4 + a\cdot x^3 + b\cdot x^2 + c\cdot x + d & = 0 \end{align}

Why there is no solutions for equation of degree $\ge 5$?

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1 Answer 1

up vote 7 down vote accepted

This is a very deep result which states that the polynomials of degree $5$ or higher do not have a general form of solution. Of course, some polynomials of degree $5$ do have solutions that we can find in explicit form, say $x^5$. The problem is we do not have a formula in terms of $\sqrt{}$, $\sqrt[3]{}, \dots, \sqrt[n]{}$, and field operations in terms of the coefficients.

The most beautiful proof of this fact lies beneath the symmetry group of the roots of a polynomial, and the idea is this : if we write a polynomial in split form say $(x-a_1)(x-a_2)\dots(x-a_5)$, there are some maps, called automorphisms, which fix the coefficients of the ground field (for instance the rational numbers), but act non-trivially on the roots. For instance, you could take the field of all numbers of the form $a+b \sqrt 2$ with $a,b \in \mathbb Q$, and produce the map $a+b \sqrt 2 \mapsto a - b \sqrt 2$. This map fixes the polynomial $x^2 - 2$ but permutes its roots together.

By studying those maps, one can see that composing them creates another of these maps, you can invert them, and one of them which does nothing also works (the identity map). Putting all those maps together gives us a group. When studying this abstract structure, one can make a correspondence between the fact that you can find a formula for the roots in terms of radicals (these are generating the so-called cyclic extensions of fields) and the fact that the underlying group has an interesting property called solvability. This correspondence is the big chunk of the "why" there is no formula to solve the quintic, because polynomials of degree $1$ to $4$ have corresponding groups that are subgroups of $S_1,S_2,S_3$ or $S_4$, and all those groups are solvable groups. The reason why there is no general formula is because a "generic polynomial" will have its corresponding group to be $S_n$ if the polynomial has degree $n$, but the group $S_n$ is not solvable for $n \ge 5$. Using the correspondence, we conclude that there are no formula involving radicals for the roots of all polynomials of degree $5$ or higher.

I didn't mean to give a proof ; you need a course in Galois Theory to understand the concepts behind this argument. I just meant to give you a feeling of how it works. If you're really interested, there's some deep algebra involved though. I don't suggest you get into it unless you have had a course in group theory and/or polynomials. If you want, I could give you a reference.

Hope that helps,

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