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Exercise 14.1.5
Let be a subfield of the real numbers, and let be irreducible of prime degree . Assume that is solvable by radicals. Prove that has either a single real root or real roots.
Answers
Proof. Since is odd, has at least a real root. Suppose that has two distinct real roots . By Theorem 14.1.1, since is solvable by radicals, the splitting field of over is . In this case all roots of are real, and these roots are distinct, since the characteristic of is , thus the irreducible polynomial is separable.
We have proved that has either a single real root or real roots. □