Prove that if the Galois group of the splitting field of a cubic polynomial over is the cyclic group of order 3, then all the roots of the cubic polynomial are real.

How do I prove that the roots are real? Some hints please.

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- April 10th 2009, 11:51 AMdori1123Galois group of a cubic polynomial
Prove that if the Galois group of the splitting field of a cubic polynomial over is the cyclic group of order 3, then all the roots of the cubic polynomial are real.

How do I prove that the roots are real? Some hints please. - April 10th 2009, 01:05 PMNonCommAlg
- April 12th 2009, 05:52 PMdori1123
Should I prove this by contradiction then? If the Galois group is not the cyclic group of order 3, then some roots are complex?

Or assume that there is a root that's not real, then the Galois group is not of order 3... - April 13th 2009, 10:51 AMThePerfectHacker
Yes. If there is a complex root then there is another complex root that is its conjugate. If you let be complex conjugation then is an automorphism of order 2. Which contradicts that the Galois group is cyclic of order 3 (therefore all its automorphism must have order dividing 3).