Exercise 5.2.3

Proof.

(a)

As $\mathbb{Q}({\zeta }_n)$ contains ${\zeta }_n^k$ for all $k\in \mathbb{Z}$, $$\mathbb{Q}({\zeta }_n)=\mathbb{Q}(1,{\zeta }_n,{\zeta }_n^2,\cdots ,{\zeta }_n^{n-1}).$$

$\mathbb{Q}(1,{\zeta }_n,{\zeta }_n^2,\cdots ,{\zeta }_n^{n-1})=\mathbb{Q}({\zeta }_n)$ is the splitting field of $x^n-1$ over $\mathbb{Q}$.

Conclusion: $\mathbb{Q}\subset \mathbb{Q}({\zeta }_n)$ is a normal extension.

(b)

The minimal polynomial of $\sqrt[3]{2}\in \mathbb{Q}(\sqrt{2},\sqrt[3]{2})=L$ over $\mathbb{Q}$ is $f=x^3-2$. The roots of $f$ are $\sqrt[3]{2},\omega \sqrt[3]{2},{\omega }^2\sqrt[3]{2}$. But $\omega \sqrt[3]{2}\notin ℝ$, and $L\subset ℝ$, thus $\omega \sqrt[3]{2}\notin L$. So $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{2})$ is not a normal extension.

(c)

By Exercise 4.2.9, the polynomial $f=x^3-t$ is irreducible over ${𝔽}_3(t)$. Let $\alpha$ a root of $f$ in the spitting field $L$ of $x^3-t$ over $F$.

As the characteristic of $F$ is 3, $f=x^3-t={(x-\alpha )}^3$, where $\alpha \in L$. The splitting field of $f$ over $F$ is so $F(\alpha )$, thus $F\subset F(\alpha )$ is a normal extension.