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Let $\alpha=\sqrt[5]{2} \in \mathbb{R}$ and $\xi=e^{\large \frac{2 \pi i}{5}}$. Let $K=\mathbb{Q}(\alpha \xi)$. Then choose the correct statements:

$(i)$ There exists a field automorphism $\sigma$ of $\mathbb{C}$ such that $\sigma (K)=K$ and $\sigma \neq identity$

$(ii)$ There exists a field automorphism $\sigma$ of $\mathbb{C}$ such that $\sigma(K) \neq K$

$(iii)$ There exists a finite extension $E$ of $\mathbb{Q}$ such that $K \subseteq E$ and $\sigma(K) \subseteq E$ for every field automorphism $\sigma$ of $E$

$(iv)$ For all field automorphism $\sigma$ of $K$, $\sigma (\alpha \xi)=\alpha \xi$.

Answer:

Here $K=\mathbb{Q}(\alpha \xi)$ is the finite extension obtained by adding $5$ roots of $2$, it is one kind of cyclotomic extension which is obtaibed by adding roots of unity to rational field $\mathbb{Q}$, here $\xi$ is the $5^{th}$ roots of unity.

Since for any finite extenstion $E$ of a field $F$, there exists an intermediate field $K$ such that $F \subseteq K \subseteq E$.

Thus $K \subseteq E$ and hence $\sigma(K) \subseteq E$.

So $(3)$ is true.

Also $(1)$ is true.

But how to judge option $(2)$ and $(4)$ ?

Please explain me because I want to learn deeply.

Help me.

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  • $\begingroup$ What is $(1)$? Do you mean $(i)$? And is $K=\Bbb Q(\alpha,\xi)$, the splitting field of $X^5-2$? $\endgroup$ – Dietrich Burde Apr 3 at 19:15
  • $\begingroup$ @DietrichBurde, yes sir. My typo it is. But $K=(\alpha \xi)$, the product $\endgroup$ – M. A. SARKAR Apr 4 at 4:37

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