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.