= Solution
Let $\alpha=\sqrt[3]{3}$. The <polynomial> $X^3-3$ is <Eisenstein> at three, so the <number field> has <degree of a field extension> three. Use the permitted <ring of integers> $R=\mathbb Z[\alpha]$, whose <integral basis> is $1,\alpha,\alpha^2$. The <field traces> $\operatorname{Tr}(1)=3$, $\operatorname{Tr}(\alpha)=\operatorname{Tr}(\alpha^2)=0$, $\operatorname{Tr}(\alpha^3)=9$, and $\operatorname{Tr}(\alpha^4)=0$ give
$$
d_K=\det\begin{pmatrix}3&0&0\\0&0&9\\0&9&0\end{pmatrix}=-243.
$$
There is one real <field embedding> and one conjugate pair of nonreal <field embeddings>. The <Minkowski bound for ideal classes> is therefore
$$
\frac4\pi\frac{3!}{3^3}\sqrt{243}=\frac{8\sqrt3}{\pi}<5.
$$
Every <ideal class> has an <integral ideal> representative of <ideal norm> at most four, so it is enough to show that every <prime ideal> of <ideal norm> at most four is a <principal ideal>.
Modulo two,
$$
X^3-3\equiv(X+1)(X^2+X+1),
$$
where the quadratic factor is an <irreducible polynomial> over the <finite field> $\mathbb F_2$. The <Dedekind factorization theorem> gives
$$
(2)=P_2Q_2,\quad P_2=(2,\alpha-1),\quad N(P_2)=2,\quad
Q_2=(2,\alpha^2+\alpha+1),\quad N(Q_2)=4.
$$
Now
$$
N_{K/\mathbb Q}(\alpha-1)=2,
\qquad (\alpha-1)(\alpha^2+\alpha+1)=2.
$$
Thus $(\alpha-1)=P_2$, by containment and equality of <ideal norms>, and cancellation in the second identity gives $(\alpha^2+\alpha+1)=Q_2$. This explicitly handles the <prime ideal> of <ideal norm> four, as well as the one of <ideal norm> two.
Modulo three the <polynomial> is $X^3$, so
$$
(3)=P_3^3,\qquad P_3=(3,\alpha)=(\alpha),\qquad N(P_3)=3.
$$
These exhaust the <prime ideals> of <ideal norm> at most four. Every bounded representative factors into the listed <principal ideals>, and hence is a <principal ideal>. Therefore the <Class group of Q of cube root of three> is
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt[3]{3}))=1}.
$$
Back to article page
