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

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?

The easiest case is to assume $$p=n$$ is an odd prime, $$d = p^g-1$$ and there is $$r^2 \equiv -d \bmod p$$.

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.

Then $$(p)^g = (1+\sqrt{-d})(1-\sqrt{-d})$$ means $$(1+\sqrt{-d}) = P^a \overline{P}^b$$.

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

As $$N(1+\sqrt{-d}) =1+d= p^g$$ and $$N(P^a) = N(P)^a = p^a$$ it means $$a = g$$.

Finally let $$o$$ be the order of $$P$$ in the classgroup, so $$P^o = (u+v\sqrt{-d})$$. If $$v=0$$ then $$p^o=N(P^o)=N((u)^o)= u^{2o}$$ leads to a contradiction. The same holds if $$u = 0$$. Thus $$u,v \ne 0$$ and $$N(P^o) = u^2+v^2d \ge 1+d=N(1+\sqrt{-d})=N(P^g)$$. Thus $$o \ge g$$ and hence $$o=g$$ as wanted.

• Thanks. And if not? – DesmondMiles Apr 19 at 13:10
• As you see the main step is that $I^g =(1+\sqrt{-d})$ where $1+\sqrt{-d}$ is the element with least norm and $u,v \ne 0$ – reuns Apr 20 at 4:45