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Math Help - A solved problem ..please check my argument.

  1. #1
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    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.
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  2. #2
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    Can any one check my argument please to ass if it's true or not ? Thank you very much !
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