# Let $K=\mathbb{Q}(\alpha \xi)$. Then choose the correct statements

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 $$K$$, $$\sigma (\alpha \xi)=\alpha \xi$$.

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.

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