Exercise 12.2.7

Proof.

(a)

Here the cubic roots are reals, so $$\sqrt[3]{1+2\sqrt{7}}\sqrt[3]{1-2\sqrt{7}}=\sqrt[3]{(1+2\sqrt{7})(1-2\sqrt{7})}=\sqrt[3]{1-28}=\sqrt[3]{-27}=-3.$$

Therefore $\sqrt[3]{1-2\sqrt{7}}=-3∕\beta \in K$.

(b)

Consider the following formula in $\mathbb{Q}(x,u,v)$ $$(x-u-v)(x-\mathit{\omega u}-{\omega }^{2}v)(x-{\omega }^{2}u-\mathit{\omega v})=x^3-3\mathit{uvx}-(u^3+v^3).$$

If we use the evaluation $u\mapsto \beta =\sqrt[3]{1+2\sqrt{7}},v\mapsto \gamma =\sqrt[3]{1-2\sqrt{7}}$, since

$$\mathit{uv}\mapsto -3,u^3+v^3\mapsto 2,$$

we obtain

$$x^3+9x-2=(x-(\beta +\gamma ))(x-(\mathit{\omega \beta }+{\omega }^2\gamma ))(x-({\omega }^2\beta +\mathit{\omega \gamma })),$$

so the root of $f$ are

$${\alpha }_1=\beta +\gamma ,{\alpha }_2=\mathit{\omega \beta }+{\omega }^2\gamma ,{\alpha }_3={\omega }^2\beta +\mathit{\omega \gamma }.$$

Since $\beta ,\gamma \in K$, ${\alpha }_1,{\alpha }_2,{\alpha }_3$ lie in $K^{\prime }=K(\omega )$:

$K(\omega )$ contains all roots of $f$.