Exercise 7.1.9

Proof.

(a)

$f=x^3-2$ is irreducible over $\mathbb{Q}$, and has a root $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$, but $\omega \sqrt[3]{2}$ is a non real root of $f$, so is not in $\mathbb{Q}(\sqrt{2},\sqrt[3]{2})\subset ℝ$. Consequently, $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{2})$ is not a normal extension, so is not a Galois extension (Th. 7.1.1(c)).

(b)

Let $\alpha ,\beta ,\gamma$ the roots of $f$, where we suppose $\alpha \ne \beta$ (in fact the discriminant of $f$ is $-23$ : the three roots of $f$ are distinct). As $\alpha +\beta +\gamma =-1$, $\gamma =-1-\alpha -\beta \in \mathbb{Q}(\alpha ,\beta )$, thus $\mathbb{Q}(\alpha ,\beta )=\mathbb{Q}(\alpha ,\beta ,\gamma )$ is the splitting field of $f$, therefore $\mathbb{Q}\subset \mathbb{Q}(\alpha ,\beta )$ is a normal extension. Moreover the characteristic of $\mathbb{Q}$ is 0, thus this extension is separable (Prop. 5.3.7).

$\mathbb{Q}\subset \mathbb{Q}(\alpha ,\beta )$ is a normal and separable extension, so is a Galois extension (Th. 7.1.1(c)).

(c)

$t$ is a root of $f=x^p-t^p={(x-t)}^p\in {𝔽}_p(t^p)$. The only root of $f$ is $t$, and $t\notin {𝔽}_p(t^p)$, otherwise $t=u(t^p)∕v(t^p)$, where $u,v\in {𝔽}_p[t],u\wedge v=1$. Moreover $u{(t)}^p={({\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^d{a}_i{t}^i)}^p={\sum }_{i=0}^d{a}_i^p{t}^{\mathit{ip}}={\sum }_{i=0}^d{a}_i{t}^{\mathit{ip}}=u(t^p)$, and similarly for $v$.

Consequently, we would have $t=u{(t)}^p∕v{(t)}^p,u\wedge v=1$, which is impossible by Exercise 4.2.9.

The equation $f=x^p-t^p$ has so no root in $𝔽(t^p)$, where $p=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$ is prime. By Proposition 4.2.6, $f$ is irreducible over $𝔽(t^p)$: $f={(x-t)}^p$ is so the minimal polynomial of $t$ over $𝔽(t^p)$.

The minimal polynomial of $t\in 𝔽(t)$ is not separable, so ${𝔽}_p(t)∕{𝔽}_p(t^p)$ is not a Galois extension.

(d)

Let $f=x^2-\left(t+\frac{1}{t}\right)x+1\in ℂ(t+t^{-1})[x]$. Then $t$ and $t^{-1}$ are roots of $f$ in $ℂ(t)$. Moreover $t^{-1}\in ℂ(t)$, therefore $ℂ(t)=ℂ(t,t^{-1})$ is the splitting field of $f$ over $C(t+t^{-1})$. $ℂ(t+t^{-1})\subset ℂ(t)$ is so a normal extension, and is separable since the characteristic of $ℂ$, and of $ℂ(t+t^{-1})$, is zero.

$ℂ(t+t^{-1})\subset ℂ(t)$ is a Galois extension.

(e)

$t$ is a root of $x^n-t^n=(x-t)(x-\mathit{\zeta t})\cdots (x-{\zeta }^{n-1}t)\in ℂ(t^n)[x]$, where $\zeta =e^{2\mathit{i\pi }}∕n$.

As ${\zeta }^{k}t\in ℂ(t),0\le k\le n-1$, $ℂ(t)=ℂ(t,\mathit{\zeta t},\dots ,{\zeta }^{m-1}t)$ is the splitting field of the polynomial $x^n-t^n\in ℂ(t^n)[x]$, so $ℂ(t^n)\subset ℂ(t)$ is a normal extension. As the characteristic of $ℂ(t^n)$ is zero, this extension is also separable.

$ℂ(t^n)\subset ℂ(t)$ is a Galois extension.