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Let $O_K = \mathbb{Q}\left(\frac{1+\sqrt{65}}{2}\right)=\mathbb{Q}(\alpha)$ be the ring of algebraic integers of $K = \mathbb{Q}(\sqrt{65})$. I want to find the Ideal Class Group $G$ of $O_K$.

The hint I am given is to try to show that every ideal of norm $10$ is principal. By Dedekind's criterion we can factor

$$(2)=(2,\alpha)(2,\alpha+1)$$ $$(3) = (3)$$ $$(5) = (5,\alpha-3)(5,\alpha+2)$$ and the Minkowski Bound $\approx 5.1$.

Now I am having trouble showing that every ideal of norm $10$ is principal. Indeed, I am having trouble finding any ideal of norm $10$ as if $N(a+b\alpha) = a^2+ab-16b^2 = 10$ does not have any immediate solutions in mind. Even so, supposing we could show this, we would then that the class ideal group $G$ is generated by $a,b,c,d$ where $ac=ad=bc=bd=ab=bc=1$, i.e. cyclic generated by $a$. So we must determine the order of $a = (2,\alpha)$ in $G$. This does not seem straightforward either. Perhaps it just requires a lot of trials case by case, but in any case, I feel I am at an impasse.

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    $\begingroup$ How about $a=2$, $b=1$, giving $4+2-16=-10$? You don’t need $+10$, for $-10$ will do. $\endgroup$
    – Lubin
    May 7, 2018 at 22:05
  • $\begingroup$ I can't quite believe no one has noticed after 5 years of this question being here, but the hint you have been given is misleading, because you don't need to consider the ideals of norm 5. The Minkowski bound is $\frac{\sqrt{65}}{2}<5$ so you can skip all that work and go straight to proving $(2,\alpha)$ has order 2 $\endgroup$
    – S.L.
    Apr 20 at 14:16

2 Answers 2

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The order of $a=(2,\alpha)$ is two in the class group, let us compute $a^2$ modulo principal ideals: $$ \begin{aligned} a^2 &= (2,\alpha)\cdot (2,\alpha) \\ &= (2\cdot 2, \ 2\cdot \alpha, \ \alpha\cdot 2,\ \alpha\cdot\alpha) \\ &= \left(\ 4, \ 1+\sqrt{65}, \ \frac 14(1+65+2\sqrt{65})\ \right) \\ &= \left(\ 4, \ 1+\sqrt{65}, \ \frac 12(33+\sqrt{65})\ \right) \\ &= \left(\ 4, \ 1+\sqrt{65}, \ \frac 12(1+\sqrt{65})\ \right) \\ &= \left(\ 4, \ \frac 12(1+\sqrt{65})\ \right) \\ &= \left(\ 4, \ \frac 12(-7+\sqrt{65})\ \right) \\ &= \left(\ \frac 12(-7+\sqrt{65})\ \right) \ . \end{aligned} $$ We have used $33=32+1$, and $4=\frac 12(-7+\sqrt{65})\cdot \frac 12(7+\sqrt{65})$.


Note: There are computer algebra programs, that make it easy to work in the class group. sage for instance. The code is self-explanatory:

sage: K.<a> = QuadraticField( 65 )
sage: G = K.class_group()
sage: G
Class group of order 2 with structure C2 of Number Field in a with defining polynomial x^2 - 65
sage: G.order()
2
sage: G.gens()
(Fractional ideal class (2, 1/2*a - 1/2),)
sage: K.ideal( [2, (1+a)/2] )^2
Fractional ideal (-1/2*a + 7/2)
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In this ring, $\eta=8+\sqrt{65}$ is a unit. If $N(\alpha)=\pm10$, then $N(\pm\eta^r\alpha)=10$ so we can replace $\alpha$ by some $\pm\eta^r \alpha$ and so assume $1<\alpha<\eta$ say. Then if $\alpha '$ is the conjugate of $\alpha$ then $\alpha'=\pm10/\alpha$, so $10>|\alpha'|>10/\eta$. We can derive some bounds on $a=\alpha+\alpha'$ from this, and so get a finite number of possible $a$ in $\alpha=\frac12(a+b\sqrt{65})$ to check.

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  • $\begingroup$ Thanks. Any ideas on how to solve the second part of the question? $\endgroup$ May 7, 2018 at 13:14

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