Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The strangeness of $Aut(S_{6})$ suggests the following question:

Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be the zeros of a 6th degree polynomial $P$ over a field $F$.

Let $Q(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}) \in F(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})$ be a rational function whose stabilizer under the action of $S_{6}$ (by permutation of variables) is a transitive subgroup isomorphic to $S_{5}$.

Then the 6 images of $Q$ under the action of $S_{6}$ can be evaluated at $x_{i} = a_{i}$ to obtain a set of 6 elements of $K = F(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6})$ which form a union of orbits under the action of $Gal(K/F)$. Then the fundamental theorem on symmetric polynomials allows one to obtain an explicit formula for the coefficients of the polynomial $P^{*}$ with those 6 elements of $K$ as its roots, in terms of the coefficients of $P$.

My question is: Can $F$ and $Q$ be chosen so that this becomes an involution on 6th degree polynomials over $F$ (i.e., so that $P^{**} = P$)? (And, if so, how explicitly can $F$ and $Q$ be given?)

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.