Re: A question about constructible roots

On 17 Sep., 02:14, Kent Holing <K...@xxxxxxxxxxx> wrote:
Can something similar be said about the irreducible quintic, sextic, etc.?
Kent Holing


My proof relied on 2-transitivity of the Galois group, while in
general you have only
simple transitivity.
I got 2-transitivity from the fact that most subgroups of S_4 could be
neglected
because they had a tower of subgroups of index 2 in each step.
This is not possible for bigger n.
Also, it was easy to show that no two sums xi+xj could be equal,
simply beacause
there are so few summands in -a = x1+x2+x3+x4; this will at least
become more difficult
for higher degrees.

hagman

.