Exercise 6.16

Proof. Let $\gamma$ a repeated root of $f(x)$. Then $f(\gamma )=f^{\prime }(\gamma )=0$, so $x-\gamma$ is a common factor of $f$ and $f^{\prime }$. Thus $f\wedge f^{\prime }\ne 1$ ( $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f\wedge f^{\prime })\ge 1)$. Since $f\wedge f^{\prime }\mid f$ and $f$ is irreducible (with $f$ , $f\wedge f^{\prime }$ monic), we conclude $f\wedge f^{\prime }=f$, so $f\mid f^{\prime }$. In $ℂ$, this is impossible since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)\ge 1$, thus $f^{\prime }\ne 0$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f^{\prime })<\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$. $f(x)$ does not have repeated roots in $ℂ$. □