A field extension $M/K$ is called radical if there is a chain of subfields $$K=M_0 \subset M_1 \subset M_2 \subset \cdots \subset M_n=M$$ s.t. $M_i=M_{i-1}(\alpha_i)$ with $\alpha_i ^{n_i} \in M_{i-1}$ for some integer $n_i>0$.

Why does $M/K$ has to be finite?

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    $\begingroup$ Because each subextension is finite ( of degree at most $\;n_i\;$) and the total extension's degree is at most the product of all the subextensions'. $\endgroup$ – DonAntonio Dec 16 '16 at 13:26

Since $M_1 = M_0(\sqrt[n_0]{\alpha_0})$ for some $\alpha_0 \in M_0$, then $[M_1 : M_0] \leq n_0 < \infty$. Similarly $[M_2 : M_1] < \infty$, and so on.

Thus $[M_n : F] = [M_n : M_{n-1}][M_{n-1} : M_{n-2}]\cdots[M_1 : M_0] < \infty$.


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