Compute the class group of $K=ℚ[\sqrt{-30}]$. [You may use standard theorems provided you state them clearly.]

Since $-30 ≡ 2 \bmod 4$, the ring of integers in $ℚ[\sqrt{-30}]$ is $ℤ[\sqrt{-30}]$ and the discriminant is $-120$. The field has one pair of complex embeddings. According to Minkowski's theorem, every ideal class has a representative $I$ such that \[ N(I) ≤ (\frac4π)^1\frac{2!}{2^2}\sqrt{120}<7 \] Thus, the prime factors of $I$ are divisors of $2,3,5$, or $7$. Since $x^2+30 ≡ x^2 \bmod 2, \bmod 3$ and $\bmod 5$, we see that \[ \text { (2) }=𝒫_2^2,(3)=𝒫_3^2,(5)=𝒫_5^2 \] for $𝒫_2=(2, \sqrt{-30}), 𝒫_3=(3, \sqrt{-30}), 𝒫_5=(5, \sqrt{-30})$ such that $N(𝒫_2)=2, N(𝒫_3)=3, N(𝒫_5)=5$. In particular, $[𝒫_2]=[𝒫_2]^{-1},[𝒫_3]=[𝒫_3]^{-1},[𝒫_5]=[𝒫_5]^{-1}$. Hence, the only possible prime divisors of $I$ as above are $𝒫_2, 𝒫_3$ and $𝒫_5$.
But note that $\sqrt{-30} ∈ 𝒫_2 𝒫_3 𝒫_5$ and $N(\sqrt{-30})=30=N(𝒫_2 𝒫_3 𝒫_5)$. So, $𝒫_2 𝒫_3 𝒫_5=(\sqrt{-30})$ is principal, and $[𝒫_5]=[𝒫_5]^{-1}=[𝒫_2][𝒫_3]$. Hence, a full set of representatives is \[ \{[𝒪_K],[𝒫_2],[𝒫_3],[𝒫_2 𝒫_3]\} \] Since there are no integral solutions to $a^2+30 b^2=2, a^2+30 b^2=3, a^2+30 b^2=6$, all these ideal classes are non-trivial. In particular, $[𝒫_3]=[𝒫_3]^{-1} ≠[𝒫_2]$, and similarly, $[𝒫_2 𝒫_3]$ is not equal to $[𝒫_2]$ or $[𝒫_3]$. Since all non-trivial elements have order 2 we get $Cl(K) ≃ ℤ / 2 × ℤ / 2$.

Find all integer solutions to \[ y^2=x^3-30 . \]

Assume $(x, y)$ is an integral solution so that $y^2=x^3-30$. If $2 ∣ y$, then $2 ∣ x$, but then $4 ∣(y^2-x^3)$, a contradiction. Hence, $y$ is odd, and therefore, $x$ is odd. Similarly, $3 ∤ x, 3 ∤ y, 5 ∤ x, 5 ∤ y$. Write the equation as \[ (y+\sqrt{-30})(y-\sqrt{-30})=x^3 . \] We claim that $y+\sqrt{-30}$ and $y-\sqrt{-30}$ are relatively prime. Otherwise, there would be a prime ideal $𝒫$ dividing both, and hence, dividing $2 y$ and $2 \sqrt{-30}$. We then get $N(𝒫) ∣ 4 y^2$ and $N(𝒫) ∣ 2^2 ⋅ 30$. Since 3 and 5 do not divide $4 y^2$, the only choice is to have $N(𝒫)=2$ or $N(𝒫)=4$, so $𝒫 ∣(2)$. But we have $(2)=𝒫_2^2$, where $𝒫_2=(2, \sqrt{-30})$. Therefore, $𝒫=𝒫_2$. Since $𝒫_2 ∣ \sqrt{-30}=𝒫_2𝒫_3𝒫_5$, this then implies $𝒫_2 ∣ y$. Taking norms, $4 ∣ y^2$, a contradiction.
If $(x)=\prod_i 𝒬_i$ for prime ideals $𝒬_i$, we get \[ (y+\sqrt{-30})(y-\sqrt{-30})=\prod_i 𝒬_i^3 \] But each factor $𝒬_i$ can only divide one of the factors on the left. Hence, we get \[ (y+\sqrt{-30})=I^3 \] for some product $I=\prod_k 𝒬_k$ of the $𝒬_k$'s. Now $I^3$ is principal, but $Cl(K)$ has order 4, $I^4$ is principal. This implies that $I$ itself is principal, generated by some $α$. So we get \[ y+\sqrt{-30}=u α^3 \] for some unit $u=c+d \sqrt{-30}$. We must have $c^2+30 d^2=N(u)=1$, which implies $d=0$ and $c= ± 1$. That is, $± 1$ are the only units in $𝒪_K$. Hence, we conclude that $y+\sqrt{-30}=α^3$ for some $α=a+b \sqrt{-30} ∈ 𝒪_K$. Write this as the equation \[ y+\sqrt{-30}=a^3-90 a b^2+(3 a^2 b-30 b^3) \sqrt{-30} . \] From $3 a^2 b-30 b^3=3(a^2-10 b^2) b=1$, we reach a contradiction. Therefore, the equation has no solutions.

Let $g>1$ be an integer. Suppose $n ⩾ 3$ is an odd integer and $d=n^g-1$ is squarefree. Show that the class group of $ℚ(\sqrt{-d})$ has an element of order $g$. [Hint: Consider the ideals $(1+\sqrt{-d})$ and $(1-\sqrt{-d})$.

Since $d$ is even and square-free, $d ≡ 2 \bmod 4$ and the ring of integers of $ℚ[\sqrt{-d}]$ is $ℤ[\sqrt{-d}]$. Also $d>0$. Consider the factorisation$$n^g=(1+\sqrt{-d})(1-\sqrt{-d})$$We claim that $(1+\sqrt{-d})$ and $(1-\sqrt{-d})$ are coprime. If some $𝒫$ were to divide both, it would also divide their sum, which is 2. But $𝒫 ∣ n^g$, from which we deduce $2 ∣ n$, contradicting the oddness of $n$. Therefore, we must have\begin{array}l(1+\sqrt{-d})=I^g\\(1-\sqrt{-d})=J^g\end{array}for coprime ideals $I$ and $J$. Thus, the ideal classes $[I],[J]$ have order dividing $g$. Note that $N(I)=N(J)=n$.
Suppose $I^m=(a+b \sqrt{-d})$ with $a, b ∈ ℤ$ and some divisor $m$ of $g$. If $b=0$, we would have $I^m=(a)$, and hence, $(a^{g / m})=(1+\sqrt{-d})$. This implies $a^{g / m}=u(1+\sqrt{-d})$ for some unit $u$ in $ℤ[\sqrt{-d}]$. Write $u=x+y \sqrt{-d}, x, y ∈ ℤ$. Then $N(u)=x^2+d y^2=1$. Since $d ≥ 3^2-1=8$, this implies $y=0$, and hence, $u= ± 1$. Therefore, we would have$$±a^{g / m}= 1+\sqrt{-d}$$a contradiction, since $a^{g / m} ∈ ℤ$. Therefore, we must have $b ≠ 0$.
Taking the norm of the equation $I^m=(a+b \sqrt{-d})$, we then get $n^m=a^2+b^2 d ≥ d=n^g-1$.
By definition $m|g$. If $m≤g-1$, then, $n^{g-1} ≥ n^g-1$, which implies $1 ≥ n^g-n^{g-1}=n^{g-1}(n-1)$. But $n ≥ 3$, giving a contradiction. Hence the order of $I$ is $m=g$.