# Showing that if $E/F$ is a radical extension over $F$, then there is a radical tower where each extension is a simple radical extension of prime type.

So I want to show that there is a radical tower $F = B_0 \subset B_1 \subset \cdots B_t$ with each $B_{i+1}/B_i$ a simple radical extension of prime type, i.e. $B_{i+1} = B_i(\alpha)$ where $\alpha^p \in B_i$ for some prime $p.$ Could anybody check if the following proof is correct?

By the hypothesis that $E/F$ is a radical extension, there is a radical tower $$F = C_0 \subset C_1 \subset\cdots C_t$$ where $C_i/C_{i-1} = C_{i-1}(\alpha_i)$ with $\alpha^{m_i} \in C_{i-1}.$ If all the $m_i$ are prime, then the result follows immediately, so assume WLOG that $m := m_1$ is not prime (also define $\alpha := \alpha_1$ for simplicity). Then $C:=C_1 = F(\alpha)$ and $\alpha^m \in F.$ Suppose $m = pm'$ where $p$ is prime. Then there is a tower of fields $F \subset F(\alpha^{m'}) \subset F(\alpha).$ Now $F(\alpha^{m'})/F$ is a simple extension of type $p$, because $(\alpha^{m'})^p = \alpha^m \in F.$ Also, $F(\alpha)/F(\alpha^{m'})$ is a simple extension of type $m'$ because $F(\alpha) = F(\alpha^{m'})(\alpha)$ and $\alpha^{m'} \in F(\alpha^{m'}).$ Because $m$ has finitely many primes, this process must eventually end, and we are done.

Is this proof correct?

• It seems perfectly fine. – Starfall Sep 19 '16 at 10:08
• Great, thank you! – dhk628 Sep 19 '16 at 13:04