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.

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

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