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Cubic equations and Galois theory
By Anonymous on Tuesday, December 5, 2000 - 06:01 pm:

We all know the solution to the general quadratic ax2+bx+c=0. Does anyone know about the general solution to the cubic? I know that Galois theory proves that certain equations of higher degree have no solution. Does this apply to all higher degree polynomials? Could you explain the extreme BASICS of galois theory so that an A'level student could understand it? Maybe you know of how to reduce the difficulty of finding the solution to higher degree polynomials ( ie: by looking for symmetry in the coefficients).

Thanks for any help.

P.S: Can you solve this?:
x6+x2+x=0
Thanks!!

By Michael Doré (Md285) on Tuesday, December 5, 2000 - 06:21 pm:

For a derivation of the cubic formula see these threads in Asked NRICH:
Solving Cubic Equations
Quick factorisation of polynomials
Formula for solving cubic equations

I can't help with Galois theory I'm afraid... Perhaps someone else who knows a bit about that can comment.

And I'm afraid I can't solve your equation. Clearly one solution is x = 0, but then I can't find a neat way of analysing:

x5 + x + 1 = 0

By Dan Goodman (Dfmg2) on Tuesday, December 5, 2000 - 06:55 pm:

Well, I just tried this equation in Mathematica, and it has some nasty roots. You can simplify it by using -(-1)1/3 is a root, as is (-1)2/3. Finding the rest of the roots just involves solving a cubic, but it's algebraically messy, involving cube roots of 25-3 sqrt(69) and things like that.

I don't know if there is an easy introduction to Galois theory, having only just done the course in my third year at university. The basic idea is that you look at "field extensions" of the rationals. For instance Q[sqrt(2)] is the set of numbers a+b.sqrt(2) for rational numbers a and b. A "radical extension" of a field K (a field is basically something where you can add, subtract, multiply and divide elements, for instance the rationals, the reals or the complex numbers) is a field K[x1/n] where x is an element of K. For instance Q[sqrt(2)] and Q[fifthroot(-1)] are radical extensions of Q, the rationals. A more complicated one is Q[sqrt(2)][sqrt(1+sqrt(2))] which is a radical extension of Q[sqrt(2)] (let K=Q[sqrt(2)], x=1+sqrt(2) is in K and K[sqrt(x)] is Q[sqrt(2)][sqrt(1+sqrt(2))]).

You say that a polynomial equation p(x)=0 is soluble by radicals if there is a series of radical extensions of Q which has the roots of p(x)=0 in them. For instance x4+x2+1=0 is soluble by radicals, because if you let y=x2 then y2+y+1=0 which is soluble by radicals (just use the quadratic formula). But then you can just take the positive and negative square roots of y to get the solutions to the original equation.

To prove that there are quintics which aren't soluble by radicals, you need to introduce the idea of the "Galois group" of a polynomial, which I won't go into here. You then show that if an equation is soluble by radicals, then the Galois group of the polynomial is "soluble" (the definition of this I also won't go into here). However, you can show that S5 (the symmetry group on 5 elements) isn't a soluble group, and you can find a polynomial with Galois group S5, so this cannot be soluble by radicals.

By Vivien Easson (Vre20) on Friday, December 8, 2000 - 11:37 am:

I'll recommend Ian Stewart's Galois Theory again as a relatively unfrightening introduction. See also this discussion on Radicals for my comments on Galois Theory. Roughly it's all about symmetries of roots in the complex plane fixing the rationals.

For example, consider f(X) = X2 - 2. This polynomial has roots sqrt(2) and -sqrt(2), and splits into factors in Q[sqrt(2)]. The only symmetry of the roots which fixes Q - the rational numbers - is sqrt(2) <-> -sqrt(2), so the Galois group is cyclic of order 2 - it contains the identity symmetry (leave everything where it was) and the symmetry above, which is of order two since performing it twice puts everything back as it was.

Can you see what possible symmetries there are for the following?
f(X) = X2 - 3 f(X) = X4 - 2
f(X) = X2 - 5X + 6
f(X) = X5 - 1
f(X) = X5 - X - 1
Some of these are really hard...

Hope this helps.
Vivien

By Pooya Farshim (P2572) on Wednesday, December 13, 2000 - 07:02 pm:

x5+x+1=(x2+x+1)(x3-x2+1).