• Dec 10th 2005, 09:56 PM
chester_l12
Let F have characteristic 0 and suppose the coefficients of the polynomial
f(x)=a_n x^n + a_(n-1) x^(n-1) + ... + a_1 t + a_{0}
be in F[x] and satisfy a_n=1, a_i = a_(n-i) for all i=0,...,n
(forexample f(t)=t^3-4t^2-4t+1

Show that if f(t) is irredicuble then n is even.
and if n=2k>4,then the galois group of this polynomial (over F) cannot be isomorphic S_n (the symmetric group)
• Dec 11th 2005, 10:18 AM
rgep
Try to write x^{-k}f as a polynomial g in y = x + 1/x. Then consider the field generated by the roots of g and its relation to the field generated by the roots of f.
• Dec 11th 2005, 03:40 PM
chester_l12
Galois theory
could you please explain a little more, i dont understand what you mean here
• Dec 11th 2005, 11:34 PM
rgep
A reciprocal polynomial f is one for which the coefficients read the same in either direction: if f is of degree n then f(x) = x^n f(1/x). Suppose f is reciprocal of degree n = 2k. Now s = x^k.(x+1/x)^k is also reciprocal of degree n, and hence so is f_1 = f - a_n.s. Now f_1/x is reciprocal of degree n-2 and so can be written as a power of x times a polynomial in (x+1/x). We conclude that any reciprocal polynomial of degree n = 2k is of the form x^k.g(x+1/x) where g is of degree k.

Now let K be the base field and consider the extension L of K by all the roots alpha_i of g: this is Galois with group a subgroup of S_k: the degree [L:K] <= k!. The extension F of L by the roots of all the x+1/x = alpha_i is a composite of k quadratic extensions so [F:L] <= 2^k. So [F:K] <= 2^k.k! and this cannot be n!. Hence Gal(F/K) cannot be s_n.