-1
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.