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")?
[Math] A question concerning polynomial solvable by radicals
abstract-algebragalois-theory
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Definition 1 Let $K$ be a field of characteristic $0$. Let $p$ be a prime number. Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$ and $X^p - \alpha^p$ is irreducible over $K$. Then we call $K(\alpha)/K$ a simple prime radical extension.
Definition 2 Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields. If $K_i/K_{i-1}$ is a simple prime radical extension for $i =1, \dots, n$, we call $L/K$ is a prime radical extension.
Definition 3 Let $L/K$ be a field extension. If there exists a prime radical extension $E/K$ such that $L$ is a subfield of $E$, we call $L/K$ a prime radically solvable extension.
Lemma 1 Let $K \subset M \subset L$ be a tower of fields. Suppose $M/K$ and $L/M$ are prime radically solvable extensions. Then $L/K$ is also a prime radically solvable extension.
Proof: This is proved in Lemma 10 of my answer to this question.
Lemma 2 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$ and $K$ has a $p$-th primitive root of unity. Then $L/K$ is a simple prime radical extension.
Proof: By this question, there is an element $\alpha$ of $L$ such that $L = K(\alpha)$ and $\alpha^p$ is an element of $K$. Since $p = [L\colon K]$, $X^p - \alpha^p$ is irreducible over $K$. Hence $L/K$ is a simple prime radical extension. QED
Lemma 3 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$. Then $L/K$ is a prime radically solvable extension.
Proof: Let $\Omega$ be an algebraic closure of $L$. Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$. Then $L(\zeta)/K(\zeta)$ is a Galois extension and $Gal(L(\zeta)/K(\zeta))$ is isomorphic to a subgroup of $Gal(L/K)$. Hence it is a cyclic group of order $p$ or $1$. By Lemma 2, $L(\zeta)/K(\zeta)$ is a prime radically solvable extension. On the other hand, by Lemma 6 of my answer to this question, $K(\zeta)/K$ is a prime radically solvable extension. Hence, by Lemma 1, $L(\zeta)/K$ is a prime radically solvable extension. Hence $L/K$ is a prime radically solvable extension. QED
Theorem Let $K$ be a field of characteristic $0$. Let $L/K$ be a finite Galois extension. Suppose $G = Gal(L/K)$ is sovable. Then $L/K$ is a prime radically solvable extension.
Proof: There exists a tower of subgroups of $G$: $G = G_n \supset G_{n-1} \supset \cdots \supset G_1 \supset G_0 = 1$, where each $G_i/G_{i-1}$ is a cyclic group of prime degree. Hence there exists a tower of fields: $K_0 = K \subset K_1 \subset \cdots \subset K_n = L$, where each $K_i/K_{i-1}$ is a cyclic Galois extension of prime degree. By Lemma 3, $K_i/K_{i-1}$ is a prime radically solvable extension. Hence, by Lemma 1, $L/K$ is a prime radically solvable extension. QED
As my Group theory professor told me, giving a root of $x^2-2=0$ the name $\sqrt 2$ doesn't magically "solve" the equation. Calling a number $\sqrt 2$ is just saying "it solves $x^2-2=0$". You haven't gained any new information. Similarly, if $p(x)$ is an irreducible polynomial, then simply writing down $p(x)=0$ could be considered to be "solving" the equation, in the sense that, in a certain technical sense, the fact that $x$ solves that equation contains all relevant information about $x$.
Solving by radicals means that we wish to express roots of a polynomial in terms of the roots of a particular family of polynomials $x^2-a$. One reason to do this is that it means that we can numerically approximate those roots using methods to numerically approximate roots of that simpler family of polynomials. This is a common strategy in mathematics. For example, in trigonometry we try to express all quantities using $\cos$ and $\sin$. In a sense, this doesn't mean we've "calculated" those quantities, since $\cos$ and $\sin$ are just names given to particular ratios of line segments, but it does mean that we can reduce the problem of approximating various quantities to just approximating $\cos$ and $\sin$.
The Bring radical represents the applicaiton of this strategy to quintics. We choose a particular one-dimensional (in the vector space sense) family of polynomials and invent a special symbol to refer to a particular root of a polynomial in that family. It can be shown that in this way we can express the roots of any quintic.
Best Answer
This article "Solution of Polynomials by Real Radicals" might interest you
http://www.jstor.org/stable/2323164
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."