On the class group of $\Bbb Q(\sqrt{-d})$, $d = n^g - 1$

Question

Let $g > 1$ and $n \geq 3$ be integers such that $n$ is odd and $d = n^g - 1$ is squarefree. Prove that the class group of $\Bbb Q(\sqrt{-d})$ contains an element of order $g$.

Here is my attempt: The ring of integers is $\mathbb{Z}[\sqrt{-d}]$. Consider the ideals $\langle 1 \pm \sqrt{-d} \rangle$, whose product is $\langle n \rangle^g$. They are coprime (as any prime ideal factor has a norm which is either even or has a common prime factor with $d$ — in both cases we get a contradiction) so both must the $g$-th powers of ideals $I$ and $J$, say. We claim that the class $[I]$ of $I$ has order $g$ (not entirely sure this is correct). Now, $I^g$ is principal and if $I^k = \langle a + b\sqrt{-d} \rangle$, $k < g$ is, by $$a^2 + db^2 = \textrm{Norm}(I^k) = n^k < n^g - 1 = d$$ we obtain $b = 0$, i.e. $I^k = \langle a \rangle$ where $a \in \mathbb{Z}$. How to finish?

Answer 1

$I^k$ divides $I^g$, so $(1+\sqrt{-d})=I^g \subset I^k = <a>$. Hence, $a=\pm 1$, which implies that $I^k=\mathbb{Z}[\sqrt{-d}]$. Then $I^g=\mathbb{Z}[\sqrt{-d}]$ as well. However, $I^k=(1+\sqrt{-d})$ and we get a contradiction as required.

Answer 2

The easiest case is to assume $p=n$ is an odd prime, $d = p^g-1$.

If $(p)$ is prime in $\Bbb{Z}[\sqrt{-d}]$ then $N(1+\sqrt{-d})=1+d=p^g$ gives that $(1+\sqrt{-d})=(p^s)$ so that $(1+\sqrt{-d})=(1-\sqrt{-d})$ which is impossible because $(1+\sqrt{-d},1+\sqrt{-d})$ contains $(2,p^g)=(1)$.

Whence in $\Bbb{Z}[\sqrt{-d}]$ we have $(p) = P \overline{P}$ with $P= (p,\sqrt{-d}-r), \overline{P} = (p,\sqrt{-d}+r)$ two different maximal ideals of norm $p$, and $(1+\sqrt{-d}) = P^a \overline{P}^b$ with $a+b=g$.

Since $(1+\sqrt{-d},1-\sqrt{-d})$ contains $(2,p^g) = (1)$ we must have $a = 0$ or $b=0$, wlog it is $b=0$ so that $(1+\sqrt{-d}) = P^g$.

Finally let $o$ be the order of $P$ in the classgroup, so $P^o = (u+v\sqrt{-d})$. We can't have $v=0$ because it would give $P^o = (u)=(\overline{u})=\overline{P}^o$. Same for $u=0$.

Thus $u,v \ne 0$ and $N(P^o) = u^2+v^2d \ge 1+d=N(P^g)$. Thus $o \ge g$ and hence $o=g$ as wanted.