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Put $K=\mathbb{Q}(\sqrt{-6})$. Its fundamental discriminant is $-24$. The Minkowski bound for ideal classes isso every ideal class contains an integral ideal of norm $1$, $2$ or $3$. The primes above $2$ and $3$ are ramified, their classes have order at most two, andshows that they represent the same class. This class is nontrivial because the norm form $a^2+6b^2$ does not represent $2$. Hence

Now let $H=K(\sqrt{2})=\mathbb{Q}(\sqrt{2},\sqrt{-3})$. The discriminant of a biquadratic field is the product of the discriminants of its three quadratic subfields, soThe relative discriminant ${\mathfrak{D}}_{H/K}$ therefore has norm one, proving that $H/K$ is unramified at every finite prime. Since $[H:K]=2=|\mathrm{Cl}(K)|$, the Hilbert class field of Q of square root minus six is

If $L$ has degree $n=r_1+2r_2$ and discriminant $d_L$, then every class in the ideal class group contains a nonzero integral ideal $\mathfrak{a}$ satisfying the Minkowski bound for ideal classes

Here $n=2$, $r_2=1$, and $d_K=-47$, so the Minkowski bound for ideal classes isEvery ideal class therefore has a representative of norm at most four. Such classes are generated by the prime ideals above $2$ and $3$. Putin additive notation. Since conjugate prime ideals represent inverse classes, the two principal-ideal factorizations from part (b) giveThus $y=-2x$ and $5x=0$, so the class group is generated by $x$ and has order dividing five.

It is not trivial. If ${\mathfrak{p}}_2$ were principal, there would be an element of norm two, butcannot equal two for integers $a,b$. Therefore $x\ne 0$. Since five is prime,

Suppose that $L/\mathbb{Q}$ were a Finite Galois extension with Galois group $S_3$ and unramified away from $7$. Its quadratic subfield is the fixed field of the alternating subgroup $A_3$. A quadratic field ramified only at $7$ must be $\mathbb{Q}(\sqrt{-7})$: the classification by fundamental discriminants shows that $-7$ is the only nontrivial quadratic discriminant supported at $7$. In particular, $7$ ramifies in this quadratic subfield.

Because $7$ does not divide $|S_3|=6$, all ramification at $7$ is tame ramification. Its inertia group is therefore cyclic. Its image in $S_3/A_3\simeq C_2$ is nontrivial because the quadratic subfield is ramified, so the inertia group is generated by a transposition and has order two; it cannot have order six because it is cyclic.

The discriminant exponent of a tame Galois extension is thereforeThere is no other finite ramification, so $|d_L|=7^3=343$.

On the other hand, the Minkowski bound for ideal classes implies the Minkowski lower bound for a number-field discriminantSince $0\le r_2\le 3$, the right side is smallest at $r_2=3$ and is greater than $980$. This contradicts $|d_L|=343$, so no such extension exists.

Two nonzero fractional ideals $I,J$ of ${\mathcal{O}}_K$ are in the same ideal class whenfor some $\alpha \in K^{\times }$, equivalently when $I{J}^{-1}$ is principal. Multiplication is well defined on classes:The identity is $[{\mathcal{O}}_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because ${\mathcal{O}}_K$ is a Dedekind domain. Thus these classes form the abelian ideal class group $\mathrm{Cl}(K)$.

For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne \alpha \in I$ withSince $(\alpha )\subseteq I$, the idealis integral, represents $[I{]}^{-1}$, and satisfiesHence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\mathrm{Cl}(K)$ is finite. This is the bounded-norm ideal representatives argument.

Now take $K=\mathbb{Q}(\sqrt{-33})$. Since $-33\equiv 3(\mathrm{mod}4)$,The imaginary-quadratic Minkowski bound for ideal classes isEvery class is therefore represented by an integral ideal of norm at most seven.

The ramified prime ideals above two and three arewithNeither is principal, since the norm equationhas no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thusare four distinct classes, all of order at most two.

The prime five is inert, so it contributes no ideal of norm five. The prime seven splits asSincecomparison of norms, all equal to $42$, confirms the factorization and givesThe conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[{\mathfrak{p}}_2{\mathfrak{p}}_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. ThereforeThis is the Ideal class group of Q of square root of minus thirty-three.

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