If $L/K$ is a non-trivial radical extension then there exists a prime $p$ and an $\alpha \in L$ such that $\alpha \notin K$ but $\alpha^p \in K$

Question

I want to show that if $L/K$ is a non-trivial radical extension then there exists a prime $p$ and an $\alpha \in L$ such that $\alpha \notin K$ but $\alpha^p \in K$.

We know, from the definition of a radical extension, that there exists a chain of subfields $K = L_0 \subset L_1 \subset ... \subset L_n = L$ such that $L_i = L_{i-1} (\alpha_i)$ and $\alpha_i^{n_i} \in L_{i-1}$ for some $n_i > 0$.

Then since $L/K$ is non-trivial, we may without loss suppose that $L_0$ is strictly contained in $L_1$ (by removing any repetitions of the same field). Then $\alpha_1 \notin L_{0}=K$ but $\alpha_1^{n_1} \in L_{0}=K$.

The bit I'm struggling with is showing that there exists such an $n_1$ which is prime.

Answer 1

You can write a sequence $m_0,\dots,m_r$ where $m_0=1$, $m_{i+1}=p_i\cdot m_i$ where $p_i$ is prime, and $m_r=n_1$. Then take the maximal $i$ such that $\alpha^{m_i}\not\in K$ (so $i<r$), and you have $(\alpha^{m_i})^{p_i}=\alpha^{m_{i+1}}\in K$.