= Solution
Put $\theta=\sqrt[3]{3}$. The given ring of integers has integral basis $1,\theta,\theta^2$. The discriminant of the power basis, equivalently of $T^3-3$, is
$$
d_K=-27\cdot3^2=-243.
$$
The field has signature $(1,1)$: one real embedding and one conjugate pair of complex embeddings. The <Minkowski bound for ideal classes> is therefore
$$
\left(\frac4\pi\right)\frac{3!}{3^3}\sqrt{243}
=\frac{8\sqrt3}{\pi}<5.
$$
Every ideal class consequently has an integral representative of norm at most four.
It remains to inspect prime ideals above $2$ and $3$. Modulo $2$,
$$
T^3-3\equiv T^3+1
=(T+1)(T^2+T+1).
$$
The <Dedekind factorization theorem> gives
$$
(2)=\mathfrak p_2\mathfrak q_2,
$$
where
$$
\mathfrak p_2=(2,\theta-1),\quad N(\mathfrak p_2)=2,
\qquad
\mathfrak q_2=(2,\theta^2+\theta+1),\quad N(\mathfrak q_2)=4.
$$
But
$$
N_{K/\mathbb Q}(\theta-1)=3-1=2,
$$
so the principal ideal $(\theta-1)$ is contained in $\mathfrak p_2$ and has the same norm; hence
$$
\mathfrak p_2=(\theta-1).
$$
Also
$$
(\theta-1)(\theta^2+\theta+1)=\theta^3-1=2,
$$
so
$$
\mathfrak q_2=(\theta^2+\theta+1).
$$
Both primes above $2$ are principal.
Finally,
$$
(\theta)^3=(3),
\qquad
N((\theta))=|N_{K/\mathbb Q}(\theta)|=3.
$$
Thus the unique prime above $3$ is the principal ideal $(\theta)$. By <unique factorization of ideals in a number field>, every integral ideal of norm at most four is built from these principal prime ideals. Every ideal class is therefore trivial, and the <Class group of Q of cube root of three> is
$$
\boxed{\operatorname{Cl}\bigl(\mathbb Q(\sqrt[3]{3})\bigr)=1.}
$$
Back to article page
