# Prove that $F \subset \sigma L$ is also radical.

Suppose that we have finite extensions $F \subset L \subset M$ and $\sigma \in Gal(M/F)$ and assume that $F \subset L$ is radical. Prove that $F \subset \sigma L$ is also radical.

Since the extension $F \subset L$ is radical, there exist $\alpha_1, \dots , \alpha_n \in \overline{F}$ such that

$i) \ L =F( \alpha_1, \dots , \alpha_n)$

$ii) \ \alpha_1$is a root of the polynomial $t^{l_1} - a_1 \in F[t]$

$iii)$ For $2 \leq i \leq n, \alpha_i$ is a root of $t^{l_i}-a_i \in F(\alpha_1, \dots , \alpha_{i-1})[t]$

Now to prove that $F\subset \sigma L$ is radical, I started by taking $\sigma(\alpha_1),\dots ,\sigma(\alpha_n)$ and showing that they satisfy $i), ii)$ and $iii)$. For $i)$, since $F \subset L$ is radical,

$$\sigma L = \sigma F(\alpha_1, \, \dots, \alpha_n) = F(\sigma(\alpha_1,\dots,\alpha_n))$$

because $\sigma$ leaves $F$ fixed. Now to prove $ii)$, I need to show that $\sigma(\alpha_1)$ is a root of $t^{l_1}-a_1$. But since $\sigma(a_i)=a_i$ because $a_i \in F$, then

$$(\sigma(\alpha_1))^{l_1}-a_1= \sigma(\alpha^{l_1})-\sigma(a_1)=\sigma(\alpha^{l_1}-a_1)=\sigma(0)=0$$

But I'm stuck on proving $iii)$ (considering that the arguments above are tru, which I'm not particularly convinced of). So if anyone could give me any hints or show me the correct way to prove this, I'd be grateful. Thanks in advance!

Your work is correct. To prove $iii)$, note that if $\alpha_k$ is radical over $L_{k-1} = F(\alpha_1, \alpha_2, \ldots, \alpha_{k-1})$, then $\sigma(\alpha_k)$ is radical over $\sigma L_{k-1}$ where $\sigma \in \textrm{Gal}(M/F)$. This is because there exists a polynomial $X^q - a_k \in L_{k-1}[X]$ which has $\alpha_k$ as a root. But on the other hand, if $a_k \in L_{k-1}$ then $\sigma(a_k) \in \sigma L_{k-1}$, and we have $0 = \sigma(0) = \sigma(\alpha_k^q - a_k) = \sigma(\alpha_k)^q - \sigma(a_k)$ so that $\sigma(\alpha_k)$ is a root of $X^q - \sigma(a_k) \in \sigma L_{k-1}[X]$.