Why is it that the degree of a subextension of $K(a^{1/p})/K$ must have a degree dividing $p$?

Question

1. $p$ is a prime
2. $K$ is of characteristic not $p$
3. $a∈K$
4. $a^{1/p}∉K$
5. $a$ is not a root of unity

"Then if we pick an element $b∈K(a^{1/p})$, $b∉K$, then $K(b)/K$ is a non-trivial subextension, thus of degree dividing $p$."

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The only part I don't understand is that last sentence "thus of degree $p$". How did they reach that conclusion?

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I would really appreciate any thoughts/help.

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I apologize in advance if I ask too many questions about the notations used. I don't have a high level of knowledge in abstract algebra.

Answer 1

If $ K \subset L \subset M $ are fields, then we have generally that $ [M:K] = [M:L][L:K] $. Therefore, the question boils down to showing that $ [K(a^{1/p}) : K] = p $, which is equivalent to showing that $ X^p - a $ is irreducible in $ K[X] $ under the given conditions. The proof to this is simple: let $ \alpha $ be a root of $ X^p - a $ in a splitting field; then we know that the other roots of the polynomial are $ \zeta \alpha, \zeta^2 \alpha, \ldots, \zeta^{p-1} \alpha $ where $ \zeta $ is a primitive $ p $-th root of unity. Assume that there was a factorization

$$ X^p - a = f(X) g(X) $$

with $ f, g \in K[X] $ and nonconstant. Then, it is easy to see that we have $ \varepsilon \alpha^k \in K $ for some $ \varepsilon^p = 1 $ and $ 1 \leq k \leq p-1 $ (look at the constant terms of the polynomials). But then, $ k $ and $ p $ are coprime, so by Bezout we can find integers $ m, n $ such that $ mk + np = 1 $. The reader can then check that we have

$$ (\varepsilon \alpha^k)^m a^n = \varepsilon^m \alpha^{mk + np} = \varepsilon^m \alpha \in K $$

But we have that $ (\varepsilon^m \alpha)^p = \alpha^p = a $, and therefore $ a $ is a $ p $th power in $ K $: contradiction.