Hmmm...it's not hard but it's not easy, either: if you know that the resultant of f(x) is in F and that the discriminant of f(x) is the resultant times some power of the leading coefficient of f(x) times (-1) to some power then we're done, otherwise we can try:

1) The discriminat is a symmetric function on the roots of f(x) (this is trivial)

2) If s_1,...,s_n are the symmetric pol's on the roots of f(x), thenanysymmetric pol. on these roots is in F[s_1,...,s_n] = the ring of pol's on s_1,...,s_n.

Of course, the problem is 2 and its proof isn't immediate (and I can't remember it right now, but some book must have it). I think the trick was about partial ordering the pol's in n variables in some way.

Try to work on this, and I'll try to make some time to consult some books.

Tonio