I have a question about an argument used in the proof of Kummer theorem in Kedlaya's notes (see page 8) www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
Let $K$ be a field which contains a n-th root $\zeta_n$. We consider $\alpha \in K* = K \backslash \{0\}$ with the property that $\alpha^{1/d} \not \in K$ for every proper divisor $d$ of $n$. The autor deduces that $x^n- \alpha$ is irreducible over $K$.
Why? The fact that $\alpha^{1/d} \not \in K$ states just that $x^n- \alpha$ doesn't splits in linear factors in $K$ but naively it could be split in non linear factors, right? Or where is the error in my reasonings?
Let $\zeta$ be a primitive $n$-th root of unity. Over the Kummer extension we have a factorisation $$X^n-\alpha=\prod_{k=0}^{n-1}(X-\zeta^k\beta).$$ If $f(X)$ is a non-trivial monic factor over $K$, then $f(X)=X^r+\cdots\pm\xi\beta^r$ where $\xi$ is some $n$-the root of unity. Therefore $\beta^r\in K$. As $\beta^n=\alpha\in K$, then $\beta^m\in K$ where $m=\gcd(n,r)$ and $\beta^m=\alpha^{1/d}$ for $d=n/m$.