1) Let
be all the roots of some
, then for every
there is a permutation
of
such that
. (beacuse the set of all such posible expressions (
) are the roots of some polynomial
). But for every permutation
the set of polynomials
such that
is a linear subspace of
as a
vector space (of infinite dimension). But the a finite union of subspaces can´t be the whole space. So one of them must be all
, ie, there exist
such that
, in particular if
it gives that
.
2) There is another solution, maybe cleaner by noteing that any simple normal extension of
must be contained in
(since some conjugate of the generator is), in particular the splitting fields are such kind of extensions if
(primitive element theorem). So every irreducible polynomial
with coefficients in
must be purely inseparable, but in this case it trivially splits over
beacuse there is only one root.
I couldnt prove or disprove the following question ¿Is the problem still true if instead of a root in
we only ask that any polynomial in
have some nontrivial factorization in
?
¿Any ideas?
)
![g \in K[x]](http://data.artofproblemsolving.com/images/latex/9/e/7/9e7f3275a929aa283948760140f8080ec67a3872.gif)
