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We know from Galois Theory that a polynomial is solvable by radicals if and only if its Galois group is solvable. On the other hand solvable by radicals for example means that the equation $X^n-1=0$ is always solvable by radicals (its Galois group is abelian), but this only means that we can find a solution by saying it is $1^{1/n}$, which is a radical. As for $n\le 6$ we can find its solution in the form $a+ib$ where $a,b$ are representable by real radicals. I guess this is not always possible for larger $n$. (I can see how it could be up to $n\le10$, but any higher?) Is there a theory concerning this kind of problem (whether a polynomial can be solved by "real radicals")?

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This might interest you... –  J. M. May 16 '11 at 13:42

2 Answers 2

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This article "Solution of Polynomials by Real Radicals" might interest you


It closes with the observation:

"We close with the observation that solvable polynomials with real roots but which are not solvable by real radicals seem to abound. For example, for any prime p, the polynomial $$f(x) = x^3 - 2px + p$$ has this property... It is amusing to solve this polynomial by Cardan's method to see where nonreal numbers come in."

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This paper indeed is very interesting, although I guess I was asking for a notion of polynomials, whose roots are $a+ib$ were $a,b$ are real radicals. With the results of the paper I might be able to work out that notion by myself. –  Peter Patzt May 17 '11 at 20:21
I guess the result in the paper is not exactly what I asked for. It basically shows something for $\alpha\in\mathbb R$ that is a real radical and whose minimal polynomial splits in $\mathbb R$. I think it be very interesting to see the result if you get rid off the restriction that its minimal polynomial needs to splitt in $\mathbb R$. Then it would be easy to generalize the result to complex numbers $a+ib$ where $a,b$ are real radicals. –  Peter Patzt May 18 '11 at 11:44

Since $1^{1/n} = 1$ there is no parameter to vary express the other roots of unity. This does not give the solutions to $x^n-1=\prod_{d|n}\Phi_d(x)=0$ it only solves one of the factors $(x-1)$.

One can express primitive roots of unity ($\Phi_n(x)=0$) in terms of smaller primitive roots, which may be varied, producing all $\varphi(n)$ primitive $n$th roots of unity.

This is in simultaneous induction with nonconstructive proof that solvable Galois group means solvable in radicals, or what it really says: Solvable Galois group means the number field is a sequence of roots radical extension and roots of unity extensions.

See a basic Galois theory textbook for Casus Irreducibilis regarding the fact that you need to have imaginary numbers as an intermediate in Cardano's formula. The theory on that is general so you can apply it to all cases.

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