Irreducibility and Roots in ##\Bbb{C}##

[Bashyboy]

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?

 

[fresh_42]

Correct. The proof of the statement you used is the interesting part.