 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 $T^3-3$, isThe field has signature $(1,1)$: one real embedding and one conjugate pair of complex embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class consequently has an integral representative of norm at most four.

It remains to inspect prime ideals above $2$ and $3$. Modulo $2$,The Dedekind factorization theorem giveswhereButso the principal ideal $(\theta -1)$ is contained in ${\mathfrak{p}}_2$ and has the same norm; henceAlsosoBoth primes above $2$ are principal.

Finally,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