A solved problem ..please check my argument.
I have the following question:
Let K=F(a) and K/F is finite and separable and [F(a):F]=p
1) Show that a has exactly p distinct Galois conjugates.
2) let a=a_1, a_2, ..., a_p be the Galois conjugates of a, if
a_2 is in K then show that K is Galois with a cyclic group.
for the first one, I answered as follows:
since K/F is separable, then a is a root of an irreducible and separable polynomial f(x)
of degree p and since it is separable then it has p roots, and those are conjugates
of a , hence a has at most p conjugates. On the other hand,
since [F(a):F]=p <= [E:F] = Aut(E/F) where E is the splitting field of f then
a has at least p conjugates and the result follows.
I have no idea about the second. If we can show that K is normal then it would be
Galois, but to show that K is normal we need to prove that all a_i are in K.
It follows from 1, that E is the splitting field of F hence it is normal and hence Galois.
Since the order of Gal(K/F) is prime then it is cyclic.
Please check my argument.