Let $ \alpha=2^{1/5} $ and $\zeta=e^{2i\pi/5} $ and let $K=\mathbb{Q}(\alpha\zeta) $ then how many field automorphism of $K $ have?

Question

Let $\alpha=2^{1/5} $ and $\zeta =e^{2i\pi/5} $ .Let $K=\mathbb{Q}(\alpha\zeta) $. My first question does $K $ look like as below:

$K$ ={ $ b_0+b_1(\alpha\zeta)+b_2 (\alpha\zeta)^2+b_3 (\alpha\zeta)^3+b_4 (\alpha\zeta)^4 : b_i \in \mathbb{Q} $}

how many field automorphism of $K $ have ?

I think only one because any automorphism $g $ of $K $ will take $ \alpha\zeta $ to a root of the polynomial $x^5-2$ contained in $K $ but $K $ contains only one root of $x^5-2$ which is $ \alpha\zeta $ .

Am I right ? Any hint would be appreciated. Thanks in advance.

Edit: This question is different because I asked here how many automorphism of $K $ have? I am sure that I didn't ask this question anywhere else.

Answer 1

$K|_{\mathbb Q}$ is a degree $5$ extension with normal closure $\mathbb Q(2^{\frac{1}{5}}\zeta_5, 2^{\frac{1}{5}})$ which has degree $20$

First of all any automorphism fixes $\mathbb Q$ as it's a prime field. So $\alpha \zeta_5$ must go to another $5$th root of $2$. If any other $5$ th root of $2$ is there then $\mathbb Q(\alpha\zeta_5): \mathbb Q \geq 20$ , a contradiction.

So any automorphism fixes $\alpha \zeta_5$ as well and hence identify is the only such automorphism.