Exercise 9.9

Proof. Let $\pi$ be a prime in $D$, $\mathit{N\pi }\ne 3$, and $\alpha \in D,\pi \nmid \alpha$.

$\bar{\alpha }$ is a cube in ${(D∕\mathit{\pi D})}^{\text{∗}}$

$⟺x^3\equiv \alpha (\mathit{mod}\pi )$ has a solution in $D$

$⟺{\chi }_{\pi }(\alpha )=1$ (by Prop. 9.3.3(a))

$⟺{\alpha }^{\frac{\mathit{N\pi }-1}{3}}\equiv 1(\mathit{mod}\pi )$

$⟺{\bar{\alpha }}^{\frac{\mathit{N\pi }-1}{3}}=\bar{1}$.

The cubes in ${(D∕\mathit{\pi D})}^{\text{∗}}$ are then the roots of the polynomial $f(x)=x^{\frac{\mathit{N\pi }-1}{3}}-\bar{1}$ in $D∕\mathit{\pi D}$.

Let $q$ be the cardinal of the field $D∕\mathit{\pi D}$. Since $q=|D∕\mathit{\pi D}|=\mathit{N\pi }$, $\frac{\mathit{N\pi }-1}{3}\mid q-1$, $f(x)\mid x^{q-1}-1\mid x^q-x$. By Corollary 2 of Proposition 7.1.1, $f$ has $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=\frac{\mathit{N\pi }-1}{3}$ roots.

Conclusion: there are exactly $\frac{\mathit{N\pi }-1}{3}$ cubes in ${(D∕\mathit{\pi D})}^{\text{∗}}$. □

Proof. Let $\phi :{(D∕\mathit{\pi D})}^{\text{∗}}\to {(D∕\mathit{\pi D})}^{\text{∗}}$ be the group homomorphism defined by $\phi (x)=x^3$.

Then $\mathrm{im}(\phi )$ is the set of cubes in ${(D∕\mathit{\pi D})}^{\text{∗}}$.

The equation $x^3=\bar{1}$ has three distinct solutions $\bar{1},\bar{\omega },{\bar{\omega }}^2$ in $D∕\mathit{\pi D}$ if $\mathit{N\pi }\ne 3$ (see the demonstration of Proposition 9.3.1).

Thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{\bar{1},\bar{\omega },{\bar{\omega }}^2\}$ and $|\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )|=3$. Therefore $|\mathrm{im}(\phi )|=|{(D∕\mathit{\pi D})}^{\text{∗}}|∕|\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )|=(\mathit{N\pi }-1)∕3$. There exist exactly $\frac{\mathit{N\pi }-1}{3}$ cubes in ${(D∕\mathit{\pi D})}^{\text{∗}}$. □