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March 11th, 2014, 07:23 AM   #1
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does cyclic field implies Galois field

I am thinking about the following statement, and I wonder if this is true: Every cyclic field is Galois.
I only know a special case of that statement:
Prop. The splitting field of an irreducible polynomial is cyclic iff the discriminant is a rational square.

But every cubic polynomial has one real root. So if a cubic polynomial has the discriminant that is a rational square, then the splitting field (i.e Galois field) is cyclic and the real root generates the whole field. So all the remaining roots must also be real and belong to that field right? Do I get it right

Does the statement hold in general?

Many thanks
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March 11th, 2014, 03:27 PM   #2
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Re: does cyclic field implies Galois field

You will have to clarify what you mean by "cyclic field" since the definition I know requires the extension K to be Galois over F.

Compare here: http://www.math.ntu.edu.tw/~jkchen/F03AA/F03AAL17.pdf
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