# a question on permutations of coefficients of polynomials

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)?

-
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: tea.mathoverflow.net/discussion/929/… – 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".
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