Thread: Galois 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.

Originally Posted by dori1123

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.

otherwise, there'd be one real root and two complex roots. then an element of the Galois group has to send the real root to itself and thus the Galois group would be of order at most 2.

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...

Originally Posted by dori1123

Should I prove this by contradiction then?

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).