- #1

Bashyboy

- 1,421

- 5

## Homework Statement

Prove that if ##p(x) \in \Bbb{Q}[x]## is an irreducible polynomial, then ##p(x)## has no repeated roots in ##\Bbb{C}##.

## Homework Equations

I will appeal to the theorem I attempted to prove here: https://www.physicsforums.com/threads/repeated-roots-and-being-relatively-prime-w-derivative.927168/

Let ##k## be a subfield of the field ##K##. If ##f(x), g(x) \in k[x]##, then their gcd in ##k[x]## is equal to their gcd in ##K[x]##.

If ##k## is a field, then a polynomial ##p(x) \in k[x]## is irreducible if and only if ##\deg (p) = n \ge 1##and there is no factorization in ##k[x]## of the form ##p(x) = g(x) h(x)## in which both factors have a degree smaller than ##n##.

## The Attempt at a Solution

I will prove the contrapositive. Suppose that ##p(x)## has a repeated root in ##\Bbb{C}##. Then ##(p,p')_{\Bbb{Q}} = (p,p')_{\Bbb{C}} \neq 1##, and according to the theorem given in the link above, ##p## must have a repeated root in ##\Bbb{Q}##. This means that ##(x-a)^2 |p(x)##, where ##a \in \Bbb{Q}## is the repeated root of ##p##, and so ##p(x) = (x-a)^2 f(x)##. This means ##\deg(p(x)) = \deg((x-a)^2 f(x)) = 2 + \deg(f)## and hence ##\deg(f) < \deg (p)## and ##2 \le \deg (p)##. If ##2 = \deg(p)##, then ##\deg (f) = 0## and so ##f(x) = r \in \Bbb{Q}##, which gives us ##p(x) = r(x-a)^2 = r(x-a)(x-a)##, each factor having a degree smaller than ##p##'s. Thus ##p(x)## is not irreducible whether ##\deg(p) = 2## or ##\deg (p) < 2##.

How does this sound?