Let . The polynomial is Eisenstein at three, so the number field has degree of a field extension three. Use the permitted ring of integers , whose integral basis is . The field traces , , , and give
There is one real field embedding and one conjugate pair of nonreal field embeddings. The Minkowski bound for ideal classes is therefore
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,
where the quadratic factor is an irreducible polynomial over the finite field . The Dedekind factorization theorem gives
Now
Thus , by containment and equality of ideal norms, and cancellation in the second identity gives . This explicitly handles the prime ideal of ideal norm four, as well as the one of ideal norm two.
Modulo three the polynomial is , so
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
Put . The given ring of integers has integral basis . The discriminant of the power basis, equivalently of , is
The field has signature : one real embedding and one conjugate pair of complex embeddings. The Minkowski bound for ideal classes is therefore
Every ideal class consequently has an integral representative of norm at most four.
It remains to inspect prime ideals above and . Modulo ,
The Dedekind factorization theorem gives
where
But
so the principal ideal is contained in and has the same norm; hence
Also
so
Both primes above are principal.
Finally,
Thus the unique prime above is the principal ideal . 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