$\sigma \in \mathrm{Gal}(K/k), \sigma \alpha \ne \alpha$, but why is $\alpha \in k$?

Question

Suppose that $k$ contains $\zeta$, a primitive $p$-th root of unity where $p$ is prime, and that $K$ is Galois over $k$ with $[K : k]=p$; and write $G=\operatorname{Gal}(K / k) \approx C_p$. Show that $K=k\left(\sqrt[p]{\alpha}\right)$ for some $\alpha \in k$.

Hint: Let $\sigma$ be a generator of $G$. Take $$\alpha=\sum_0^{p-1} \zeta^{\nu} \cdot \sigma^{\nu} \beta$$ for $\beta \in K$ and show that one can choose $\beta$ so that $\alpha \neq 0$.

But with a simple try, I find that $$ \sigma \alpha=\sum_{0}^{p-1} \zeta^{\nu} \cdot \sigma^{\nu+1} \beta\neq\alpha, $$ so why is $\alpha$ in k?

Answer 1

This is just conflicting notation. The $\alpha$ in the hint is not the $\alpha$ in the assertion.

Show that you can find $\beta\in K$ such that $\gamma:= \sum_{\nu=0}^{p-1}\zeta^\nu\cdot \sigma^\nu(\beta)\neq 0$. Observe that $\sigma(\gamma) = \zeta^{-1}\gamma$ and conclude $\gamma\in K\setminus k$ and $\alpha:=\gamma^p \in k$.