Why does $\mbox{Irr}(\alpha,K)$ have distinct roots in $N$?

Question

My textbook assumes that

1. $N\supset K$ is normal and finite
2. $\alpha\in N$ is separable over $K$
3. $N\supset K(\alpha)$ is separable

and deduces (in order to prove that $N\supset K$ is separable) that $\newcommand{\Irr}{\mbox{Irr}}\Irr(\alpha,K)$ has all its roots and all distinct in $N$.

I see why $N$ must contain all roots, since $N\supset K$ is normal. Also I see why $\alpha$ is NOT repeated as it is separable over $K$.

But how can I see that none of the other roots can be repeated roots?

Answer 1

Let $m(x)$ be the minimal polynomial of $\alpha$ over $K$. If it has a repeated root ($\alpha$ or another root - makes no difference), then that multiple root $\beta$ gives rise to a common factor $(x-\beta)$ of $m(x)$ and $m'(x)$.

But by Euclid's algorithm $r(x):=\gcd(m(x),m'(x))\in K[x]$. This is thus a non-trivial factor of $m(x)$ contradicting the assumption that $m(x)$ is irreducible in $K[x]$ as a minimal polynomial. Unless $r(x)=m(x)$, which in turn would imply $m'(x)=0$, which in turn implies that all the zeros of $m(x)$ have multiplicity $>1$.