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Abel's impossibility theorem states that the roots of a general polynomial (of degree 5 or higher) cannot be written using arithmetic operations and radicals. Radicals are solutions of a specific polynomial basis and it's natural to wonder if roots of an arbitrary polynomial can be written in terms of the roots of a different polynomial basis.

Let $p_1, p_2, \ldots$ be a basis for the space of polynomials, where $p_i$ is of degree $i$. A "generalized radical" is a solution to $p_i = \alpha$ for some $i$ and constant $\alpha$. For example, when $p_i = x^i$, we get the standard notion of radicals.

It is then natural to think of another polynomial basis, such as falling factorials or some polynomial basis consisting of orthogonal polynomials. Is there a polynomial basis under which the generalized notion of radicals is powerful enough to express the roots of arbitrary polynomials?

A related problem is whether there is a richer (but nontrivial) set of algebraic numbers than usual radicals that can capture the roots of arbitrary polynomials.

Is there anything in the literature related to these problems?

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    $\begingroup$ There is a paper of Abhyankar which I believe shows that the answer is "no". See my answer at mathoverflow.net/a/61558/297 . $\endgroup$
    – David E Speyer
    Commented Dec 13, 2019 at 3:03
  • $\begingroup$ What does "A 'generalized radical' is a solution to $p_i = \alpha$" mean? Does it mean that we are allowed to insert a symbol $\sigma_{i, \alpha}$ anywhere in an equation, and the equation is judged to be correct if some choice, for each such symbol, of a solution $\sigma$ of $p_i(\sigma) = \alpha$ makes it so? (I mean to say, is a generalised radical a specific element of the field, or a place-holder that can stand for any one of potentially many such elements?) $\endgroup$
    – LSpice
    Commented Dec 13, 2019 at 3:10
  • $\begingroup$ Yes, you can use any root of any equation p_i = alpha for any i and alpha. $\endgroup$
    – MCH
    Commented Dec 13, 2019 at 3:28
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    $\begingroup$ Quintics are solvable if you allow solutions of $x^5+x+a$ (but that's in addition to radicals, not in place of them). $\endgroup$
    – Gerry Myerson
    Commented Dec 13, 2019 at 5:00
  • $\begingroup$ Perhaps this Q&A on the Math.SE could give some insight on what happens when the degree $n$ is $>4$. $\endgroup$
    – Daniele Tampieri
    Commented Dec 13, 2019 at 8:26

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