Determine the class group for $ℚ(\sqrt{-77})$.

For $K=ℚ(\sqrt{-77})$ we have $Δ^2(K)=-308$, so that it suffices to look at $N(P) ⩽ 2 π^{-1} \sqrt{308}<\frac{4}{3} \sqrt{77}<36 / 3=12$. So, the class group is generated by $[P]$, for primes ideals $P ∣(p)$, for $p=2,3,5,7,11$. Since $[𝒪_K: ℤ[\sqrt{-77}]]=1$, Dedekind's Theorem applies to all rational primes $p$. The polynomial $x^2+77$ factors as: $\{x+1\}^2,\{x+1\}\{x-1\}, x^2+2, x^2, x^2$ modulo $2,3,5,7,11$, respectively, so that by Dedekind's Theorem, the factorisations into prime ideals are: $(2)=P_2^2$, where $P_2=(2, \sqrt{-77}+1)$ and where $N(P_2)=2^{\deg(x+1)}=2;(3)=P_3 P_3'$, where $P_3=(3, \sqrt{-77}+1), P_3'=(3, \sqrt{-77}-1)$ are distinct, and where $N(P_3)=3^{\deg(x+1)}=3, N(P_3')=3^{\deg(x-1)}=3 ;(5)=(5)$ is already a prime ideal; $(7)=P_7^2$, where $P_7=(7, \sqrt{-77})$ and where $N(P_7)=7^{\deg(x)}=7$; $(11)=P_{11}^2$, where $P_{11}=(11, \sqrt{-77})$ and where $N(P_{11})=11^{\deg(x)}=11$. Also, $[P_2]^2=[(2)]=[𝒪_K]$, Also, $[P_7]^2=[(7)]=[𝒪_K]$, Also, $[P_{11}]^2=[(11)]=[𝒪_K]$, and $[P_3][P_3']=[(3)]=[𝒪_K]$ so that $[P_3']=[P_3]^{-1}$. Hence the class group is generated by $[P_2],[P_3],[P_7],[P_{11}]$, where $[P_2],[P_7],[P_{11}]$ all have order at most 2.
We look for elements whose norms are divisible only by $2,3,7,11$. Now, $N((\sqrt{-77}))=|\operatorname{Norm}_{K / ℚ}(\sqrt{-77})|=77$, so $(\sqrt{-77}) ∣(77)=(7)(11)=P_7^2 P_{11}^2$, giving that $(\sqrt{-77})=P_7 P_{11}$, giving $[P_7][P_{11}]=[𝒪_K]$ and so $[P_{11}]=[P_7]^{-1}=[P_7]$, Hence the class group is generated by $[P_2],[P_3],[P_7]$. Also, $N((2+\sqrt{-77}))=|\operatorname{Norm}_{K / ℚ}(2+\sqrt{-77})|=81$, so $(2+\sqrt{-77}) ∣(81)=(3)^4=P_3^4 P_3^{\prime 4} ;$ but $2+\sqrt{-77} ∉P_3$ (since otherwise $1=\{2+\sqrt{-77}\}-\{1+\sqrt{-77}\} ∈ P_3$, contradicting that $P_3$ is a prime ideal), which means that $(2+\sqrt{-77})=P_3^{\prime 4}$; hence $[P_3']^4=[𝒪_K]$ and $[P_3]^4=[𝒪_K]$, so that $[P_3]$ has order dividing 4. Also, $N((7+\sqrt{-77}))=|\operatorname{Norm}_{K / ℚ}(7+\sqrt{-77})|=2 ⋅ 3^2 ⋅ 7$, so $(7+\sqrt{-77}) ∣(2)(3)^2(7)=P_2^2 P_3^2 P_3'^2 P_7^2$; hence, by norms, $(7+\sqrt{-77})$ is one of: $P_2 P_3^2 P_7, P_2 P_3 P_3' P_7$ or $P_2 P_3'^2 P_7$ and so $[P_7]$ is one of $[P_2]([P_3]^{-1})^2,[P_2][P_3]^2$ or $[P_2]$; in all cases, we may now say that the class group is generated just by $[P_2],[P_3]$.
In summary: the class group is generated by $[P_2],[P_3]$, where $[P_2]^2=[𝒪_K]$ and $[P_3]^4=[𝒪_K]$. Note that any $N((a+b \sqrt{-77}))=|\operatorname{Norm}_{K / ℚ}(a+b \sqrt{-77})|=a^2+77 b^2 ≠ 2,3,9,18$. This means that $P_2, P_3, P_3^2, P_2 P_3^2$ are all non-principal. Hence $[P_2] ≠[𝒪_K]$, giving that $[P_2]$ has order 2. Also, $[P_3] ≠[𝒪_K]$ and $[P_3]^2 ≠[𝒪_K]$ so that $[P_3]$ has order 4. Finally, $[P_2 P_3^2] ≠ 𝒪_K$, so that $[P_3]^2 ≠[P_2]^{-1}=[P_2]$. This forces the class group to be isomorphic to $C_2 × C_4$.

Let $p, q$ be odd primes such that $p q ≡ 5\pmod{12}$. Show that if the order of the class group of $ℚ(\sqrt{-p q})$ is not divisible by 3 then there are no integer solutions to $y^3=x^2+p q$.

Suppose that $y^3=x^2+p q$ for $x, y ∈ ℤ$. Then in $𝒪_K$, where $K=ℚ(\sqrt{-p q})$, we have $\{x+\sqrt{-p q}\}\{x-\sqrt{-p q}\}=y^3$, giving the equality of ideals: $(x+\sqrt{-p q})(x-\sqrt{-p q})=(y)^3$. We claim the ideals $(x+\sqrt{-p q})$ and $(x-\sqrt{-p q})$ in $𝒪_K$ are coprime. For imagine $P$ is a prime ideal which divides both (note that, since $P$ is prime, $N(P) ≠ 1)$. Then $x ± \sqrt{-p q} ∈ P$ so $2 \sqrt{-p q} ∈ P$. So $P ∣(2 \sqrt{-p q})$, and $1 ≠ N(P) ∣ N((2 \sqrt{-p q}))=2^2 ⋅ p q$, so that $2|N(P), p| N(P)$ or $q ∣ N(P)$. But also $P ∣(y)^3$, so that $N(P)|N((y)^3)=| \operatorname{Norm}_{K / ℚ}(y)|^3=y^6$, so that $2, p$ or $q$ divides $y^6$, and so $2, p$ or $q$ divides $y$. If $2 ∣ y$ then $2 ∤ x$ so $x^2+p q ≡ 1+1 ≡ 2\pmod4$ and $y^3 ≡ 0\pmod4$, a contradiction; if $p ∣ y$ then $p ∣ x$ and so $p^2 ∣ y^3-x^2=p q$, a contradiction. if $q ∣ y$ then $q ∣ x$ and so $q^2 ∣ y^3-x^2=p q$, a contradiction. Hence $(x+\sqrt{-p q})$ and $(x-\sqrt{-p q})$ are coprime, as claimed; by Unique Factorisation of Ideals they must each be a cube of ideals, say that $(x+\sqrt{-p q})=I^3$, for some ideal $I$. So $I^3$ is principal. Since 3 is coprime to the order 8 of the class group, $I$ is principal. Hence $I=(a+b \sqrt{-p q})$ for some $a, b ∈ ℤ$. It follows that $x+\sqrt{-p q}=u\{a+b \sqrt{-p q}\}^3$ for some unit $u ∈ 𝒪_K$. Only units are $± 1$, which are both cubes, so we may assume $u=1$. Hence \[ x+\sqrt{-p q}=\{a+b \sqrt{-p q}\}^3=\{a^3-3 ⋅ p q a b^2\}+\{3 a^2 b-p q b^3\} \sqrt{-p q} . \] So $1=b\{3 a^2-p q b^2\}$, and so $b= ± 1$. Hence $3 a^2=p q ± 1$. Since $p q ≡ 5\pmod{12}$, we can exclude $3 a^2=p q-1$, since $3 ∤ p q-1$. But if $3 a^2=p q+1 ≡ 6\pmod{12}$ then $a^2 ≡ 2\pmod4$, which is also impossible. Hence there are no solutions.

Show that the class group for $ℚ(\sqrt{-105})$ contains a subgroup isomorphic to $C_2 × C_2 × C_2$.

By Dedekind's Theorem, $x^2+105 ≡\{x+1\}^2, x^2, x^2, x^2 \bmod 2,3,5,7$, so $(2)=P_2^2,(3)=P_3^2,(5)=P_5^2$ and $(7)=P_7^2$, where $P_2=(2, \sqrt{-105}+1)$ and $P_i=(p_i, \sqrt{-105})$ for $i=3,5,7$. So, all of $[P_2]^2,[P_3]^2,[P_5]^2,[P_7]^2$ are equal to $[𝒪_K]$. Also, any principal ideal has norm $N((a+b \sqrt{-105}))=a^2+105 b^2 ≠ 2,3,5,6,10,15,30$, so that none of the following are principal $P_2, P_3, P_5, P_2 P_3, P_2 P_5, P_3 P_5, P_2 P_3 P_5$, and so none of their classes are equal to $[𝒪_K]$. It follows that the subgroup generated by $[P_2],[P_3],[P_5]$ is isomorphic to $C_2 × C_2 × C_2$

Show that for any $N ∈ ℕ$ there exists a quadratic number field for which the class number is greater than $N$.

Let $d=p_1 p_2 ⋯ p_r$, where $r ⩾ 3$ and $p_1, …, p_r$ are distinct odd primes such that $d ≡ 1\pmod4$, so that $-d ≡ 3\pmod4$ (which can be guaranteed by taking $r$ even). By Dedekind's Theorem, $x^2+d ≡ x^2\bmod$ each $p_i$, so each $(p_i)=P_i^2$, where $P_i=(p_i, \sqrt{-d})$, and so each $[P_i]^2=[𝒪_K]$. Any principal ideal has norm $N((a+b \sqrt{-d}))=a^2+d b^2 ≠ p_i$ and $≠ p_i p_j$ for any distinct $i, j$. Hence each $P_i$ and each $P_i P_j$ is non-principal, and so each $[P_i] ≠[𝒪_K]$ and each $[P_i][P_j] ≠[𝒪_K]$, giving $[P_i] ≠[P_j]^{-1}=[P_j]$. This means that all of $[P_1], …,[P_r]$ are distinct, and so the order of the class group can be made to be greater than any $N$.
[You may assume throughout this question that, for square-free $d ∈ ℤ$, the ring of integers of $ℚ(\sqrt{d})$ is $ℤ[\sqrt{d}]$ when $d ≡ 3\pmod4$.