I have a question about a step in proof of Kummer theorem in Kedlaya's notes (see page 8) www.math.mcgill.ca/darmon/courses/cft/refs/kedlaya.pdf
We start with a $\alpha \in K* = K \backslash \{0\}$ with the property that $\alpha^{1/d} \not \in K$ for every proper divisor $d$ of $n$. Then he concludes 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?
Indeed not. For instance take $K=\Bbb Q$, $n=4$ and $\alpha=-4$. Then $\alpha $ is not a square in $K$, but $$x^n-\alpha=x^4+4=(x^2+2x+2)(x^2-2x+2)$$ is reducible over $K$.