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I'm working through a final exam from 2 years ago. First task was to find the ideal class group of $\Bbb{Q}(\sqrt{-73})$. That is not the difficult work. I can give the 4 representants of the group by $\omega$, $\frac{\omega+1}{2}$, $\frac{\omega+2}{7}$ and $\frac{\omega-2}{7}$ with $\omega=\sqrt{-73}$. This group is isomorphic to $C_4$. A generator for the group is given by $C=\frac{\omega-2}{7}$ (I hope this is true :) ) But now the following task:

Compute all ideals $\mathfrak{a}$ from $\Bbb{Z}[\sqrt{-73}]$ such that $N(\mathfrak{a})<15$ with $\mathfrak{a}\in C$ (here is N the Norm of the ideal).

Can someone help me with this question?! Thanks :)

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I'm not sure what you mean with $\mathfrak a \in C$ since $C = \frac {\omega - 2 } 7$. I guess $C$ is the set of integral ideals. Using unique prime ideal factorization (the ring of integers of this field is $\mathbb Z[\sqrt{-73}]$), you can concentrate on prime ideals. Since they are the divisors of the principal ideals $(p)$ with $p \in \mathbb Z$ prime, I would deompose all the ideals $(p)$ up to a properly chosen bound. –  Hans Giebenrath Dec 16 '12 at 15:06
    
with $\mathfrak{a}\in C$ i denote the ideals $\mathfrak{a}$ which lies in the equivalenceclass of $C$ –  Trace Dec 16 '12 at 15:37
    
Now $C$ is an element of the field. Thus the class of $(C)$ (the ideal generated by $C$) is the trivial one. Therefore the class of $C$ is the set of all principal ideals. Do you want to compute all princial ideals with norm bounded by 15? Also your sentence about the ideal class group is confusing. All your representatives are principal ideals, thus they are equal in the ideal class group (trivial). –  Hans Giebenrath Dec 16 '12 at 15:46
    
Okay i thought over $\Bbb{Z}+\Bbb{Z\omega}$,$2\Bbb{Z}+\Bbb{Z\omega+1}$,$7\Bbb{Z}+\Bbb{Z\omega+2}$ and $7\Bbb{Z}+\Bbb{Z\omega-2}$ resp. –  Trace Dec 16 '12 at 16:06
    
And ...? Can you please tell us more. This is too little information. I don't know what you want to tell us. –  Hans Giebenrath Dec 16 '12 at 16:18

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