Exercise 14.1.5

Let F be a subfield of the real numbers, and let f F [ x ] be irreducible of prime degree p > 2 . Assume that f is solvable by radicals. Prove that f has either a single real root or p real roots.

Answers

Proof. Since deg ( f ) = p is odd, f has at least a real root. Suppose that f has two distinct real roots α , β . By Theorem 14.1.1, since f is solvable by radicals, the splitting field of f over F is F ( α , β ) . In this case all roots of f are real, and these roots are distinct, since the characteristic of F is 0 , thus the irreducible polynomial f is separable.

We have proved that f has either a single real root or p real roots. □

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2022-07-19 00:00
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