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How to characterize all polynomials $ p(x) $ such that every polynomial obtained from $ p(x) $ by permuting the coefficients of $ p(x) $ has a root in common with $ p(x) $ (in some field extension)?

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I don't understand your last question. If $p(x)$ is a polynomial and $\sigma$ is an automorphism of its splitting field, do we not always have $\sigma(p(x))=p(x)$, so that those two polynomials have all roots in common ? – Ewan Delanoy Jan 28 '11 at 8:58
forget that formulation of the question!! It is just to ask whether p(x) and q(x) have at least one root in common where q‌​(x) is the polynomial with coefficients equal to the coefficients of p(x)under th‌​e image of a permutation of coefficients of p(x)? – awllower Jan 28 '11 at 9:56
I have tried and, I think, failed to understand the question. This may well be my fault, but I'm voting to close, pending clarification. – Gerry Myerson Jan 29 '11 at 5:40
Please use the "edit" link below the question, then write a more coherent question. When you have done that, please flag for moderator attention so it can be reopened. – S. Carnahan Jan 29 '11 at 6:25
This question now has a meta thread:… – Ben Webster Jan 29 '11 at 19:19

If I understand the comment, we are to ignore the whole Galois group thing, and just talk about permuting the coefficients of $p(x)$, and whether the resulting polynomial $q(x)$ can have any common roots with $p(x)$. If $p(x)$ is irreducible then it can't have any common roots with $q(x)$ unless it equals $q(x)$. If $p(x)$ is not irreducible, well, $x^2+3x+2$ and $2x^2+3x+1$ share a root, it's not hard to construct other examples, it's not clear there's anything useful to be said about the situation. Permuting coefficients isn't "natural".

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Also, there's no relation between permuting coefficients and the solvability-by-roots thing. – Ewan Delanoy Jan 28 '11 at 12:10
I mainly agreed with Gerry Myerson except that permutting coefficients can be "natural" in some sense. Consider a polynomial, p(x), of the ring F[x] where F is a field and its splitting field E over F. Along with p(x) we take into account the polynomial with roots as the coefficients of p(x), denoted by q(x) and its splitting field K over F. Then for polynomials under our consideration we can express the roots of p(x) in terms of radicals of the coefficients of q(x). And hence we find a field extension of F such that E|F is a radical extension of K|F. And I have found less nontrivial examples. – awllower Jan 28 '11 at 14:26
True - of course, that's true for any polynomial with 1 as a root, since it just says the coefficients add up to zero, which is unaffected by permutations. – Gerry Myerson Jan 29 '11 at 11:28
If all the coefficients are the same, we have one kind of solution. Otherwise we have at least two which are different. Choose a permutation which has these two (say $a$ and $b$) as coefficients of $x$ and the constant term. Swap the two by a transposition. Any common root has to be a root of the difference, which is $(a-b)(x-1)$ so $a=b$ of $x=1$. Other transpositions (transpose two coefficients and take the difference) allow $x=0$ or roots of $x^m-1$, which would, of course, have to be roots of the original polynomial. But these roots would not survive all permutations. – Mark Bennet Jan 30 '11 at 15:40
Mark Bennet, may I suggest that you submit your comment as an answer so awllower can accept it and we can all move on? – Gerry Myerson Jan 30 '11 at 22:31

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